Solution Manual for Applied Partial Differential Equations with Fourier Series and Boundary Value Pr

Chapter 1. Heat Equation Section 1.2 1.2.9 (d) Circular cross section means that P = 2πr, A = πr2 , and thus P/A = 2/r, where r is the radius.

Also γ = 0.

1.2.9 (e) u(x, t) = u(t) implies that

du 2h =− u. dt r The solution of this first-order linear differential equation with constant coefficients, which satisfies the initial condition u(0) = u0 , is · ¸ 2h u(t) = u0 exp − t . cρr cρ

Section 1.3 1.3.2 ∂u/∂x is continuous if K0 (x0 −) = K0 (x0 +), that is, if the conductivity is continuous.

Section 1.4 1.4.1 (a) Equilibrium satisfies (1.4.14), d2 u/dx2 = 0, whose general solution is (1.4.17), u = c1 + c2 x. The boundary condition u(0) = 0 implies c1 = 0 and u(L) = T implies c2 = T /L so that u = T x/L. 1.4.1 (d) Equilibrium satisfies (1.4.14), d2 u/dx2 = 0, whose general solution (1.4.17), u = c1 + c2 x. From the boundary conditions, u(0) = T yields T = c1 and du/dx(L) = α yields α = c2 . Thus u = T + αx.

1.4.1 (f) In equilibrium, (1.2.9) becomes d2 u/dx2 = −Q/K0 = −x2 , whose general solution (by integrating twice) is u = −x4 /12 + c1 + c2 x. The boundary condition u(0) = T yields c1 = T , while du/dx(L) = 0 yields c2 = L3 /3. Thus u = −x4 /12 + L3 x/3 + T .

1.4.1 (h) Equilibrium satisfies d2 u/dx2 = 0. One integration yields du/dx = c2 , the second integration yields the general solution u = c1 + c2 x. x=0: x=L:

c2 − (c1 − T ) = 0 c2 = α and thus c1 = T + α.

Therefore, u = (T + α) + αx = T + α(x + 1).

1.4.7 (a) For equilibrium: d2 u x2 du = −1 implies u = − + c1 x + c2 and = −x + c1 .

2 dx 2 dx du From the boundary conditions du dx (0) = 1 and dx (L) = β, c1 = 1 and −L + c1 = β which is consistent 2 only if β + L = 1. If β = 1 − L, there is an equilibrium solution (u = − x2 + x + c2 ). If β 6= 1 − L, there isn’t an equilibrium solution. The difficulty is caused by the heat flow being specified at both ends and a source specified inside. An equilibrium will exist only if these three are in balance. This balance can be mathematically verified from conservation of energy: Z Z L du du d L cρu dx = − (0) + (L) + Q0 dx = −1 + β + L. dt 0 dx dx 0

If β + L = 1, then the total thermal energy is constant and the initial energy = the final energy: ¶ Z L Z Lµ 2 x f (x) dx = − + x + c2 dx, which determines c2 .

2 0 0 If β + L 6= 1, then the total thermal energy is always changing in time and an equilibrium is never reached.

1

Section 1.5

¡ du ¢ d 1.5.9 (a) In equilibrium, (1.5.14) using (1.5.19) becomes dr r dr = 0. Integrating once yields rdu/dr = c1 and integrating a second time (after dividing by r) yields u = c1 ln r + c2 . An alternate general solution is u = c1 ln(r/r1 ) + c3 . The boundary condition u(r1 ) = T1 yields c3 = T1 , while u(r2 ) = T2 yields c1 = (T2 − T1 )/ ln(r2 /r1 ). Thus, u = ln(r21/r1 ) [(T2 − T1 ) ln r/r1 + T1 ln(r2 /r1 )].

1.5.11 For equilibrium, the radial flow at r = a, 2πaβ, must equal the radial flow at r = b, 2πb. Thus β = b/a. ¡ 2 du ¢ d 1.5.13 From exercise 1.5.12, in equilibrium dr r dr = 0. Integrating once yields r2 du/dr = c1 and integrat2 ing a second time (after dividing by r ) yields u = −c1 /r + c2 . The boundary conditions ¡ u(4) ¢ = 80 and u(1) = 0 yields 80 = −c1 /4 + c2 and 0 = −c1 + c2 . Thus c1 = c2 = 320/3 or u = 320 1 − 1r .

3

2

Chapter 2. Method of Separation of Variables Section 2.3

³ ´ ³ ´ dφ dφ kh d 1 dh 1 d = r .

Dividing by kφh yields = r = −λ or 2.3.1 (a) u(r, t) = φ(r)h(t) yields φ dh r dr dr kh dt rφ dr dr ³ ´ dt dφ dh 1 d = −λφ. dt = −λkh and r dr r dr 2

2

2

2

2.3.1 (c) u(x, y) = φ(x)h(y) yields h ddxφ2 + φ ddyh2 = 0. Dividing by φh yields φ1 ddxφ2 = − h1 ddyh2 = −λ or d2 φ d2 h dx2 = −λφ and dy 2 = λh.

4

4

d φ 1 dh 1 d φ 2.3.1 (e) u(x, t) = φ(x)h(t) yields φ(x) dh dt = kh(t) dx4 . Dividing by kφh, yields kh dt = φ dx4 = λ.

2

2

2

2

2.3.1 (f) u(x, t) = φ(x)h(t) yields φ(x) ddt2h = c2 h(t) ddxφ2 . Dividing by c2 φh, yields c21h ddt2h = φ1 ddxφ2 = −λ.

2.3.2 (b) λ = (nπ/L)2 with L = 1 so that λ = n2 π 2 , n = 1, 2, . . .

2.3.2 (d) √ √ (i) If λ > 0, φ = c1 cos λx + c2 sin λx. φ(0) = 0 implies c1 = 0, while dφ dx (L) = 0 implies √ √ √ c2 λ cos λL = 0. Thus λL = −π/2 + nπ(n = 1, 2, . . .). (ii) If λ = 0, φ = c1 + c2 x. φ(0) = 0 implies c1 = 0 and dφ/dx(L) = 0 implies c2 = 0. Therefore λ = 0 is not an eigenvalue. √ √ (iii) If λ < 0, let √ λ = −s √ and φ = c1 cosh sx + c2 sinh sx. φ(0) = 0 implies c1 = 0 and dφ/dx(L) = 0 implies c2 s cosh sL = 0. Thus c2 = 0 and hence there are no eigenvalues with λ < 0.

2.3.2 (f) The simpliest method is to let x0 = x − a. Then d2 φ/dx02 + λφ = 0 with φ(0) = 0 and φ(b − a) = 0.

2 Thus (from p. 46) L = b − a and λ = [nπ/(b − a)] , n = 1, 2, . . ..

P∞ −k(nπ/L)2 t . The initial condition yields 2.3.3 From (2.3.30), u(x, t) = n=1 Bn sin nπx L e R P∞ nπx 3πx 2 L nπx 2 cos L = n=1 Bn sin L . From (2.3.35), Bn = L 0 2 cos 3πx L sin L dx.

2.3.4 (a) Total heat energy = satisfies (2.3.35).

RL 0

cρuA dx = cρA

P∞

n=1 Bn e

2

−k( nπ L ) t 1−cos nπ nπ L

, using (2.3.30) where Bn

2.3.4 (b) heat flux to right = −K0 ∂u/∂x total heat flow to right = −K0 A∂u/∂x ¯ ¯ heat flow out at x = 0 = K0 A ∂u ∂x x=0 ¯ ∂u ¯ heat flow out (x = L) = −K0 A ∂x x=L ¯L RL ¯ ∂u ∂u d 2.3.4 (c) From conservation of thermal energy, dt u dx = k ∂u ∂x ¯ = k ∂x (L) − k ∂x (0). Integrating from 0 0 t = 0 yields ¸ Z L Z L Z t· ∂u ∂u u(x, t) dx − u(x, 0) dx = k (L) − (0) dx . ∂x ∂x |0 {z } |0 {z } | 0 {z | {z } } heat energy initial heat integral of integral of at t energy flow in at flow out at x=L x=L p p 2 2.3.8 (a) The general solution of k ddxu2 = αu (α > 0) is u(x) = a cosh αk x + b sinh αk x. The boundary condition u(0) = 0 yields a = 0, while u(L) = 0 yields b = 0. Thus u = 0.

3

2

d φ 2.3.8 (b) Separation of variables, u = φ(x)h(t) or φ dh dt + αφh = kh dx2 , yields two ordinary differential 2

1 dh α 1 d φ equations (divide by kφh): kh dt + k = φ dx2 = −λ. Applying the boundary conditions, yields the eigenvalues λ = (nπ/L)2 and corresponding eigenfunctions φ = sin nπx L . The time-dependent part are P ∞ nπx −k(nπ/L)2 t −λkt −αt −αt exponentials, h = e e . Thus by superposition, u(x, t) = e , where n=1 bn sin L e R P∞ 2 L nπx the initial conditions u(x, 0) = f (x) = n=1 bn sin nπx yields b = f (x) sin dx.

As t → ∞, n L L 0 L u → 0, the only equilibrium solution. q q −α 2.3.9 (a) If α < 0, the general equilibrium solution is u(x) = a cos −α x + b sin x. The boundary k q qk −α condition u(0) = 0 yields a = 0, while u(L) = 0 yields b sin −α k L = 0. Thus if k L 6= nπ, u = 0 is q nπx the only equilibrium solution. However, if −α k L = nπ, then u = A sin L is an equilibrium solution.

¡ π ¢2 2.3.9 (b) Solution obtained in 2.3.8 is correct. If − αk = L , u(x, t) → b1 sin πx L , the equilibrium solution. ¡ ¢ ¡ π ¢2 2 α π α If − k < L , then u → 0 as t → ∞. However, if − k > L , u → ∞ (if b1 6= 0). Note that b1 > 0 if f (x) ≥ 0. Other more unusual events can occur if b1 = 0. [Essentially, the other possible equilibrium solutions are unstable.]

Section 2.4 2.4.1 The solution is given by (2.4.19), where the coefficients satisfy (2.4.21) and hence (2.4.23-24). ¯ RL RL 2 L nπx ¯L 2 nπ (a) A0 = L1 L/2 1dx = 21 , An = L2 L/2 cos nπx dx = · sin L L nπ L ¯L/2 = − nπ sin 2 (b) by inspection A0 = 6, A3 = 4, others = 0. ¯L RL R πx 2 πx ¯ 2 −4 L πx nπx (c) A0 = −2 sin dx = cos L 0 L π L ¯ = π (1 − cos π) = 4/π, An = L 0 sin L cos L dx 0

(d) by inspection A8 = −3, others = 0.

2.4.3 Let x0 = x − π. Then the boundary value problem becomes d2 φ/dx02 = −λφ subject to φ(−π) = φ(π) and dφ/dx0 (−π) = dφ/dx0 (π). Thus, the eigenvalues are λ = (nπ/L)2 = n2 π 2 , since L = π, n = 0, 1, 2, ... with the corresponding eigenfunctions being both sin nπx0 /L = sin n(x−π) = (−1)n sin nx => sin nx and cos nπx0 /L = cos n(x − π) = (−1)n cos nx => cos nx.

Section 2.5 2

2

2.5.1 (a) Separation of variables, u(x, y) = h(x)φ(y), implies that h1 ddxh2 = − φ1 ddyφ2 = −λ. Thus d2 h/dx2 = −λh subject to h0 (0) = 0 and h0 (L) = 0. Thus as before, λ = (nπ/L)2 , n = 0, l, 2, . . . with h(x) = ¡ ¢2 2 cos nπx/L. Furthermore, ddyφ2 = λφ = nπ φ so that L n = 0 : φ = c1 + c2 y, where φ(0) = 0 yields c1 = 0 nπy n 6= 0 : φ = c1 cosh nπy L + c2 sinh L , where φ(0) = 0 yields c1 = 0.

The result of superposition is u(x, y) = A0 y +

∞ X

An cos

nπy nπx sinh .

L L

An sinh

nπx nπH cos , L L

n=1

The nonhomogeneous boundary condition yields f (x) = A0 H +

∞ X n=1

so that A0 H =

1 L

Z L f (x) dx and An sinh 0

4

nπH 2 = L L

Z L f (x) cos 0

nπx dx.

L

2

2

2.5.1 (c) Separation of variables, u = h(x)φ(y), yields h1 ddxh2 = − φ1 ddyφ2 = λ. The boundary conditions φ(0) = 0 and φ(H) = 0 yield an eigenvalue problem in y, whose solution is λ = (nπ/H)2 with φ = sin nπy/H, n = 1, 2, 3, . . . The solution of the x-dependent equation is h(x) = cosh nπx/H using dh/dx(0) = 0. By superposition: u(x, y) =

∞ X

An cosh

n=1

nπx nπy sin .

H H

The nonhomogeneous boundary condition at x = L yields g(y) = RH nπy 2 An is determined by An cosh nπL H = H 0 g(y) sin H dy.

P∞

n=1 An cosh

nπy nπL H sin H , so that

2.5.1 (e) Separation of variables, u = φ(x)h(y), yields the eigenvalues λ = (nπ/L)2 and corresponding ¡ ¢2 2 eigenfunctions φ = sin nπx/L, n = 1, 2, 3, ... The y-dependent differential equation, ddyh2 = nπ h, L nπy nπy dh satisfies h(0) − dy (0) = 0. The general solution h = c1 cosh L + c2 sinh L obeys h(0) = c1 , ¡ ¢ nπy nπy nπ nπ nπ obeys dh while dh dy = L c1 sinh L + c2 cosh L dy (0) = c2 L . Thus, c1 = c2 L and hence hn (y) = nπy L cosh nπy L + nπ sinh L . Superposition yields u(x, y) =

∞ X

An hn (y) sin nπx/L,

n=1

where An is determined from the boundary condition, f (x) = An hn (H) =

2 L

P∞

n=1 An hn (H) sin nπx/L, and hence

Z L f (x) sin nπx/L dx .

0

2.5.2 (a) From physical reasoning (or exercise 1.5.8), the total heat flow across the boundary must equal RL zero in equilibrium (without sources, i.e. Laplace’s equation). Thus 0 f (x) dx = 0 for a solution.

2.5.3 In order for u to be bounded as r → ∞, c1 = 0 in (2.5.43) and c̄2 = 0 in (2.5.44). Thus, u(r, θ) =

∞ X

An r−n cos nθ +

n=0

∞ X

Bn r−n sin nθ.

n=1

(a) The boundary condition yields A0 = ln 2, A3 a−3 = 4, other An = 0, Bn = 0. (b) The boundary conditions yield (2.5.46) with a−n replacing an . Thus, the coefficients are determined by (2.5.47) with an replaced by a−n 2.5.4 By substituting (2.5.47) into (2.5.45) and interchanging the orders of summation and integration " # Z ∞ ¢ 1 π 1 X ³ r ´n ¡ u(r, θ) = + cos nθ cos nθ̄ + sin nθ sin nθ̄ f (θ̄) dθ̄. π −π 2 n=1 a Noting the trigonometric addition formula and cos z = Re [eiz ], we obtain " # Z ∞ ³ ´n X 1 π 1 r in(θ−θ̄) u(r, θ) = f (θ̄) − + Re e dθ̄. π −π 2 a n=0 Summing the geometric series enables the bracketed term to be replaced by 2

1 1r 1 − ar cos(θ − θ̄) 1 1 1 2 − 2 a2 + = . − + Re = − 2 2 2 1 + ar22 − 2r 1 − ar ei(θ−θ̄) 1 + ar 2 − 2r a cos(θ − θ̄) a cos(θ − θ̄)

5

2 2.5.5 (a) The eigenvalue problem is d2 φ/dθ It can be √ = −λφ subject to dφ/dθ(0) = 0 and √ φ(π/2) = 0. √ shown that λ > 0 so that φ = cos λθ where φ(π/2) = 0 implies that cos λπ/2 = 0 or λπ/2 = −π/2 + nπ, n = 1, 2, 3, . . . The eigenvalues are λ = (2n − 1)2 . The radially dependent term satisfies (2.5.40), and hence the boundedness condition at r = 0 yields G(r) = r2n−1 . Superposition yields

u(r, θ) =

∞ X

An r2n−1 cos(2n − 1)θ.

n=1

The nonhomogeneous boundary condition becomes f (θ) =

∞ X

An cos(2n − 1)θ

or An =

n=1

4 π

Z π/2 f (θ) cos(2n − 1)θ dθ.

0

2.5.5 (c) The boundary conditions of (2.5.37) must be replaced by φ(0) = 0 and φ(π/2) = 0. Thus, L = π/2, so that λ = (nπ/L)2 = (2n)2 and φ = sin nπθ L = sin 2nθ. The radial part that remains bounded at √ r = 0 is G = r λ = r2n . By superposition, u(r, θ) =

∞ X

An r2n sin 2nθ .

n=1

To apply the nonhomogeneous boundary condition, we differentiate with respect to r: ∞ ∂u X = An (2n)r2n−1 sin 2nθ . ∂r n=1

The bc at r = 1, f (θ) =

P∞

4 n=1 2nAn sin 2nθ , determines An , 2nAn = π

R π/2 0

f (θ) sin 2nθ dθ.

2.5.6 (a) The boundary conditions of (2.5.37) must be replaced by φ(0) = 0 and φ(π) = 0. Thus L = π, so that the eigenvalues are λ = (nπ/L)2 = n2 and corresponding eigenfunctions φ = sin nπθ/L = √ λ n sin nθ, n = 1, 2, 3, ... The radial part which is bounded at r = 0 is G = r = r . Thus by superposition u(r, θ) =

∞ X

An rn sin nθ .

n=1

The bc at r = a, g(θ) =

P∞

n=1 An a

n

sin nθ, determines An , An an = π2

Rπ 0

g(θ) sin nθ dθ.

2.5.7 (b) The boundary conditions of (2.5.37) must be replaced by φ0 (0) = 0 and φ0 (π/3) = 0. This will 2 yield a cosine series with L = π/3, λ = (nπ/L)2 = (3n) and φ = cos nπθ/L = cos 3nθ, n = 0, 1, 2, . . .. √ λ The radial part which is bounded at r = 0 is G = r = r3n . Thus by superposition u(r, θ) =

∞ X

An r3n cos 3nθ .

n=0

R π/3 P∞ The boundary condition at r = a, g(θ) = n=0 An a3n cos 3nθ, determines An : A0 = π3 0 g(θ) dθ R π/3 and (n 6= 0)An a3n = π6 0 g(θ) cos 3nθ dθ.

2.5.8 (a) There is a full Fourier series in θ. It is easier (but equivalent) to choose radial solutions that satisfy the corresponding homogeneous boundary condition. Instead of rn and r−n (1 and ln r for n = 0), we choose φ1 (r) such that φ1 (a) = 0 and φ2 (r) such that φ2 (b) = 0 : ½ ½ ln(r/b) ln(r/a) ¡ r ¢n ¡ a ¢n n = 0 φ2 (r) = ¡ r ¢n ¡ b ¢n n = 0 . φ1 (r) = − n = 6 0 − r n 6= 0 a r b

6

Then by superposition u(r, θ) =

∞ X

cos nθ [An φ1 (r) + Bn φ2 (r)] +

n=0

∞ X

sin nθ [Cn φ1 (r) + Dn φ2 (r)] .

n=1

The boundary conditions at r = a and r = b, f (θ) =

∞ X

cos nθ [An φ1 (a) + Bn φ2 (a)] +

n=0

g(θ) =

∞ X

∞ X

sin nθ [Cn φ1 (a) + Dn φ2 (a)]

n=1

cos nθ [An φ1 (b) + Bn φ2 (b)] +

n=0

∞ X

sin nθ [Cn φ1 (b) + Dn φ2 (b)]

n=1

easily determine An , Bn , Cn , Dn since φ1 (a) = 0 and φ2 (b) = 0 : Dn φ2 (a) = π1

Rπ −π

f (θ) sin nθ dθ, etc.

2.5.9 (a) The boundary conditions of (2.5.37) must be replaced by φ(0) = 0 and φ(π/2) = 0. This is a sine series with L = π/2 so that λ = (nπ/L)2 = (2n)2 and the eigenfunctions are φ = sin nπθ/L = sin 2nθ, n = 1, 2, 3, . . .. The radial part which is zero at r = a is G = (r/a)2n − (a/r)2n . Thus by superposition, ·³ ´ ¸ ∞ X r 2n ³ a ´2n u(r, θ) = An − sin 2nθ. a r n=1 h¡ ¢2n ¡ ¢ i P∞ 2n The nonhomogeneous boundary condition, f (θ) = n=1 An ab sin 2nθ, determines An : − ab h¡ ¢2n ¡ ¢ i R 2n π/2 An ab − ab = π4 0 f (θ) sin 2nθ dθ.

2.5.9 (b) The two homogeneous boundary conditions are in r, and hence φ(r) must be an eigenvalue problem.

2 By separation of variables, u = φ(r)G(θ), d2 G/dθ2 = λG and r2 ddrφ2 +r dφ dr +λφ = 0 . The radial equation p is equidimensional (see p.78) and solutions are in the form φ = r .

Thus p2 = −λ (with λ > 0) so √ √ √ √ √ ±i λ ±i λ ln r that p = ±i λ. r =e . Thus real solutions are cos( λ ln√r) and sin( λ ln r).√ It is more convenient to use independent solutions which simplify at r = a, cos[ λ ln(r/a)] and sin[ λ ln(r/a)].

Thus the general solution is √ √ φ = c1 cos[ λ ln(r/a)] + c2 sin[ λ ln(r/a)]. √ The homogeneous condition φ(a) = 0 yields 0 = c1 , while φ(b) = 0 implies sin[ λ ln(r/a)] = 0 i. Thus h √ λ ln(b/a) = nπ, n = 1, 2, 3, ... and the corresponding eigenfunctions are φ = sin nπ ln(r/a) ln(b/a) . The √ nπθ solution of the θ -equation satisfying G(0) = 0 is G = sinh λθ = sinh ln(b/a) . Thus by superposition · ¸ nπθ ln(r/a) u= An sinh sin nπ . ln(b/a) ln(b/a) n=1 ∞ X

The nonhomogeneous boundary condition, f (r) =

∞ X n=1

An sinh

· ¸ nπ 2 ln(r/a) sin nπ , 2 ln(b/a) ln(b/a)

will determine An . One method (for another, see exercise 5.3.9) is to let z = ln(r/a)/ ln(b/a). Then a < r < b, lets 0 < z < 1. This is a sine series in z (with L = 1) and hence · ¸ Z 1 nπ 2 ln(r/a) An sinh =2 f (r) sin nπ dz.

2 ln(b/a) ln(b/a) 0 But dz = dr/r ln(b/a). Thus An sinh

nπ 2 2 = 2 ln(b/a) ln(b/a)

7

· ¸ ln(r/a) f (r) sin nπ dr/r. ln(b/a) 0

Z 1

Chapter 3. Fourier Series Section 3.2 P∞ P∞ 3.2.2 (a) x ∼ a0 + n=1 an cos nπx/L + n=1 bn sin nπx/L. From (3.2.2), a0 = 0 since f (x) is odd, RL 2L (−1)n+1 . (n 6= 0)an = 0 since f (x) is odd, and bn = L1 −L x sin nπx/L dx = nπ P∞ P∞ 3.2.2 (c) sin πx/L ∼ a0 + n=1 an cos nπx/L + n=1 bn sin nπx/L. By inspection, b1 = 1, all others = 0.

3.2.2 (f) From (3.2.2),

(n 6= 0)

a0 an

= =

bn

=

RL 1 2LR 0 dx = 1/2 1 L L 0 cos nπx/L dx = 0 ¯L R 1−cos nπ 1 L −1 nπx ¯ sin nπx/L dx = cos L 0 nπ L ¯ = nπ 0

Thus bn = 2/nπ, n odd, but bn = 0, n even.

3.3.2 (d) From (3.3.6), Bn = L2 ½

x2 ex

3.3.10 f (−x) =

R L/2 0

¯L/2 ¡ ¢ ¯ 2 = nπ sin nπx/L dx = −2 1 − cos nπ nπ cos nπx/L¯ 2 .

0

−x < 0 (or x > 0) −x > 0 (or x < 0).

Thus. ½

x2 + ex 2 −x x ½ 2+ e x x − e f0 (x) = 12 [f (x) − f (−x)] = 12 e−x − x2

x<0 x>0 x<0 x>0

fe (x) = 12 [f (x) + f (−x)] = 12

RL 3.3.13 bn = L2 0 f (x) sin nπx/L dx, given that f (x) is even around L/2. Note (perhaps by graphing) that sin nπx/L is odd around L/2 for n even. Thus f (x) sin nπx/L is odd around L/2 for n even, and hence bn = 0 for n even.

Section 3.4 3.4.1 (a)

Rb a

=

R c− a

Rb + c+ . Thus ¯c− ¯b ¯b ¯ ¯ ¯ c− ¯ ¯ Z b Z b Z b ¯ ¯ ¯ ¯ dv du du ¯ ¯ ¯ dx = uv ¯ + uv ¯ − dx = uv ¯ + uv ¯¯ − dx. u v v ¯ ¯ dx dx dx a a a ¯ ¯ ¯ a ¯ c+ a

3.4.3 (a) We want to determine the sine coefficients of df /dx: coefficients of f are given

df dx

∞ X

2 L

f=

an cos

n=0

nπx L

(n 6= 0) an =

c+

=

P∞

n=1 bn sin

Z L f cos 0

nπx L , where the cosine

nπx dx.

L

Here by integration by parts 2 bn = L

Z L 0

df nπx 2 sin dx = dx L L

"

# ¯L Z nπx ¯¯x0 − nπ L nπx nπx ¯¯ f sin − f cos dx . + f sin ¯ L 0 L ¯x0 + L 0 L

nπ 0 Thus bn = L2 sin nπx L (α − β) − L an .

8

P∞ ¡ nπ ¢2 nπx ∂2u bn sin nπx . Thus from n=1 L bn cos L since u(0) = 0 and u(L) = 0. ∂x2 ∼ − n=1 L ¡ ¢ ¡ ¢ RL P∞ dbn P 2 2 ∞ dbn nπx nπ nπx nπ 2 L p. 119, n=1 dt sin L ∼ −k n=1 L bn sin L + q. Thus dt + k L bn = L 0 q sin nπx L dx.

3.4.9 ∂u ∂x =

P∞ ¡ nπ ¢

3.4.12 The P∞ eigenfunctions of the related homogeneous problem are cos nπx/L, n = 0, 1, 2, . . .. Thus u ∼ nπx/L, which can be differentiated (if u is continuous) since it is a cosine series: n=0 An (t) Pcos ∞ ∂u/∂x ∼ n=0 An (−nπ/L) sin nπx L . This can be differentiated P∞ again (if ∂u/∂x is continuous) only because ∂u/∂x(0) = 0 and ∂u/∂x(L) = 0: ∂ 2 u/∂x2 ∼ − n=0 An (nπ/L)2 cos nπx/L. Thus from p.

119 ¸ ∞ · ³ nπ ´2 X dAn nπx 3πx +k An cos = e−t + e−2t cos . dt L L L n=0 The right hand side is a simple cosine series (with only two non-zero terms). Thus  −t n=0  e ³ ´ 2 dAn nπ e−2t n = 3 +k An =  dt L 0 otherwise .

RL RL The initial conditions are A0 (0) = L1 0 f (x) dx and (n 6= 0) An (0) = L2 0 f (x) cos nπx L dx. The solution of the differential equations are n 6= 0, 3

2

An (t) = A0 (t) =

An (0)e−k(nπ/L) t A0 (0) + 1 − e−t , obtained by integration

A3 (t) =

A3 (0)e−k(nπ/L) t + e

2

2

−2t

−e−k(nπ/L) t , 2 k( 3π L ) −2

obtained by using the method of undetermined coefficients (judicious guessing) for the particular solution. This works if e−2t is not a homogeneous solution, i.e., −2 6= −k(3π/L)2 .

Section 3.5 3.5.4 (a) Using 3.4.13 with f (x) cosh x(f (0) = 1, f (L) = cosh L),¤ P= ∞ £ 2 nπx n sinh x ∼ L1 (cosh L − 1) + n=1 nπ L bn + L ((−1) cosh L − 1) cos L . Since this is a cosine series, it may be differentiated ¸ ∞ ·³ X 2nπ nπx nπ ´2 n cosh x ∼ − bn + 2 ((−1) cosh L − 1) sin .

L L L n=1 Thus bn = −

¡ nπ ¢2 L

n

2nπ 1−(−1) cosh L n bn − 2nπ L2 [(−1) cosh L − 1] or bn = L2 1+(nπ/L)2 .

R L 1 L n=1 − nπ bn cos nπx/L, where A0 = L 0 sinh x dx = ¡ L ¢2 P∞ 1 bn sin nπx/L. Thus n=1 − nπ L (cosh L − 1). Integrating again yields cosh x − 1 = A0 x +

3.5.4 (b) Integrating yields sinh x = A0 + ∞ X

P∞

"

µ

bn 1 +

n=1

L nπ

¶2 # sin nπx/L = 1 + A0 x.

h ¡ L ¢2 i 2 2L Using (3.3.8) and (3.3.l2) bn 1 + nπ = nπ [1 − (−1)n ] + L1 (cosh L − 1) nπ (−1)n+1 or bn =

2 1 − (−1)n cosh L . ¡ ¢2 nπ 1+ L nπ

3.5.7 Evaluate (3.5.6) at x = L/2: µ ¶ L2 4L2 1 1 1 1 1 L2 /8 L2 = − 3 1 − 3 + 3 − . . . or 1 − 3 + 3 − 3 = = π 3 /32.

8 4 π 3 5 3 5 7 4L2 /π 3 9

Section 3.6 3.6.1 The complex Fourier coefficient is defined by (3.6.7): 1 cn = 2L Thus

Z L f (x)e −L

inπx/L

1 dx = 2L4

Z x0 +4

einπx/L dx .

x0

¯x +4 1 L inπx/L ¯¯ 0 1 cn = e = einπx0 /L (einπ4/L − 1) . ¯ 2L4 inπ 2inπ4 x0

Equivalently, cn =

1 einπ(x0 +4/2)/L (einπ4/2L − e−inπ4/2L ) or 2inπ4 cn =

1 inπ(x0 +4/2)/L e sin(nπ4/2L) . nπ4

10

Chapter 4. Vibrating Strings and Membranes Section 4.4

√ 4.4.1 (a) Natural frequencies are c λ, but λ = (nπ/L)2 . Thus frequencies are nπc/L, n = 1, 2, 3, ... √ √ 4.4.1 (b) Natural frequencies boundary condition φ(0) = 0 implies φ = c1 sin λx, while √ are c λ. The dφ/dx(H) = 0 yields λH = (m − 12 )π with m = 1, 2, 3. Thus the frequencies are (m − 12 )πc/H and the eigenfunctions are sin(m − 21 )πx/H. 2

2

4.4.2 (c) By separation of variables, u = φ(x)h(t), ddt2h = −λh and T0 ddxφ2 + (α + λρ0 )φ = 0. With φ(0) = 0 2 and φ(L) = 0, (α + λρ0 )/T √0 = (nπ/L)√, n = 1, 2, 3, . . . and φ = sin nπx/L. In general h(t) involves a linear combination of sin λt and cos λt, but the homogeneous initial condition u(x, 0) = 0 implies there are no cosines. Thus by superposition u(x, t) =

∞ X

An sin

p

λn t sin nπx/L,

n=1

q √ (nπ/L)2 T0 −α where the frequencies of vibration are λn = . The other initial condition, f (x) = ρ0 √ P∞ A λ sin nπx/L, determines A n n n=1 n Z p 2 L An λn = f (x) sin nπx/L dx.

L 0 00

0

00

4.4.3 (b) By separation of variables, u = φ(x)h(t), ρ0 hhT+βh = φφ = −λ. The boundary conditions φ(0) = 0 0 and φ(L) = 0 yield λ = (nπ/L)2 with φ = sin nπx/L, n = 1, 2, 3, . . .. The time-dependent equation has constant coefficients, ³ nπ ´2 ρ0 h00 + βh0 + T0 h = 0, L and hence can be solved by substitution h = ert . This yields the quadratic equation ³ nπ ´2 ρ0 r2 + βr + T0 = 0, L whose roots are r=

−β ±

p β 2 − 4ρ0 T0 (nπ/L)2 .

2ρ0

Since β 2 < 4ρ0 T0 (π/L)2 , the discriminant is < 0 for all n: s β T0 ³ nπ ´2 β2 + iwn , where wn = − 2. r=− 2ρ0 ρ0 L 4ρ0 Real solutions are h = e−βt/2ρ0 (sin wn t, cos wn t). Thus by superposition u=e

−βt/2ρ0

∞ X

(an cos wn t + bn sin wn t) sin

n=1

nπx .

L

RL ∂u The initial condition u(x, 0) = f (x) determines an , an = L2 0 f (x) sin nπx L dx, while ∂t (x, 0) = g(x) is ∞ X P∞ nπx β − a little more complicated, g(x) = n=1 bn wn sin nπx an sin , and thus L 2ρ0 L n=1 | {z } f (x)

bn wn =

2 βan + 2ρ0 L

11

Z L g(x) sin 0

nπx dx.

L

Chapter 5. Sturm-Liouville Eigenvalue Problems Section 5.3 5.3.1 By separation of variables, u = φ(x)h(t), · ¸ 1 d2 h 1 d2 φ = T + αφ = −λ.

0 h dt2 ρ0 φ(x) dx2 Thus,

d2 h d2 φ = −λh and T0 2 + αφ + λρ0 φ = 0.

2 dt dx √ √ For the time-dependent equation (if λ > 0), h = c1 cos λt + c2 sin λt. The spatial equation is in Sturm-Liouville form: ρ = T0 constant, q = α(x), σ = ρ0 (x). d d 5.3.3 Hd2 φ/dx2 = dx (Hdφ/dx)−dH/dx dφ/dx. Thus dx (Hdφ/dx)+dφ/dx(Hα−dH/dx)+(λβ +γ)Hφ = 0.

To be in standard Sturm-Liouville form, ·Z x ¸ dH = αH or H = c1 exp α(t) dt , let c1 = 1. dx

Then ρ(x) = H, q(x) = γH, and σ(x) = βH. 1 dh 1 h dt = φ

5.3.4 (b) By separation of variables, u = φ(x)h(t),

³ 2 ´ k ddxφ2 − v0 dφ = −λ. The boundary value dx

2

rx problem is k ddxφ2 − v0 dφ implies kr2 − v0 r + λ = 0 or dx + λφ = 0. φ = e p p v0 ± i 4λk − v02 v0 ± v02 − 4λk = . r= 2k 2k

To satisfy the boundary conditions φ(0) = 0 and φ(L) = 0: p v0 4λk − v02 nπ nπx = and φ(x) = e 2k x sin .

2k L L Thus, by superposition, u=

∞ X

v0

An e 2k x sin

n=1

∞ X v0 nπx −λn t nπx −λn t e = e 2k x An sin e .

L L n=1

The initial value problem can be solved v0

f (x) = e 2k x

∞ X

An sin

n=1

so that An = v02

Note that λ = 4k + k

¡ nπ ¢2 L

2 L

Z L

v0

f (x)e− 2k x sin

0

nπx L nπx dx.

L

.

√ 5.3.9 (c) Since it is equidimensional, φ = xr , which implies r(r−1)+r+λ =³0 or r2 = −λ. If λ³> 0, r =´±i λ, ´ √ √ √ √ where x±i λ = e±i λ ln x . Thus the general solution is φ = c1 cos λ ln x + c2 sin λ ln x . The ³√ ´ √ boundary condition φ(1) = 0 implies c1 = 0, and hence φ(b) = 0 yields sin λ ln b = 0 or λ ln b = d nπ, n = 1, 2, . . .. Thus λ = (nπ/ ln b)2 . If λ = 0 the differential equation becomes dx (xdφ/dx) = 0.

Thus, the general solution is φ = c1 + c2 ln x. The condition φ(1) = 0 yields c1 = 0, and then φ(b) = 0 implies c2 = 0. Thus λ = 0 is not an eigenvalue.

12

Section 5.4 5.4.2 The eigenfunctions satisfy (5.4.6) with the boundary conditions being dφ/dx(0) = 0 and dφ/dx(L) = 0.

From the Rayleigh quotient (5.4.16) λ ≥ 0. Also λ = 0 only if dφ/dx = 0 for all x. The boundary conditions imply λ = 0 is an eigenvalue with φ = constant = 1 the corresponding eigenfunction. Thus by superposition using (5.4.5) u=

∞ X

an φn (x)e−λn t (with λ1 = 0, φ1 = 1).

n=1

The initial conditions, f (x) =

∞ X

an φn (x) ,

n=1

yield

RL

f φn cρ dx an = R0 L , φ2n cρ dx 0 since the eigenfunctions are orthogonal with weight cρ. As t → ∞, RL f cρ dx u → a1 = R0 L , cρ dx 0 a weighted average of the initial temperature distribution. This can also be shown using conservation of total thermal energy. Since the ends are insulated, the final thermal energy (t → ∞) equals the initial thermal energy (t = 0).

5.4.3 From (5.2.11) the eigenfunctions, denoted φn (r), are orthogonal with weight r. The time-dependent part satisfies (5.2.10), and hence h(t) = e−λkt . By superposition u=

∞ X

an φn (r)e−λkt .

n=1

The initial condition, f (r) =

∞ X

an φn (r) ,

n=1

determines an ,

Ra f (r)φn (r)r dr . an = 0 R a 2 φ (r)r dr 0 n

5.4.6 By separation of variables, u = φ(x)h(t), 2

T0 ddxφ2 + λρ0 φ = 0 d2 h dt2 + λh = 0 .

The boundary conditions are of the Sturm-Liouville type, and therefore the eigenfunctions denoted φn (x) are orthogonal √ with weight √ ρ0 (x). We call the eigenvalues λn . Then the time-dependent problem has solutions cos λn t and sin λn t . Initially at rest means ∂u/∂t(x, 0) = 0, so that only cosines are needed. Thus by superposition ∞ X p u= An φn (x) cos λn t. n=1

The initial position yields f (x) =

∞ X n=1

13

An φn (x) ,

determining An ,

RL An =

0

f (x)φn (x)ρ0 (x) dx .

RL φ2n ρ0 dx 0

Section 5.5 5.5.1 (g) To be self-adjoint (if p is constant), u(L)

dV du dv du (L) − v(L) (L) = u(0) (0) − v((0) (0). dx dx dx dx

The derivatives at x = L may be eliminated from the boundary conditions. After some algebra, we obtain · ¸ dv du dv du (αδ − βγ) u(0) (0) − v(0) (0) = u(0) (0) − v(0) (0). dx dx dx dx Thus, it is self-adjoint only if αδ − βγ = 1.

5.5.9 Multiply the differential equation by φ and integrate: Z 1 4 Z 1 d φ φ 4 dx + λ ex φ2 dx = 0. dx 0 0 Note that d4 φ d φ 4 = dx dx Thus

¯1

3 ¯ φ ddxφ3 ¯

0

¯1 ¯

d2 φ ¯ − dφ dx dx2 ¯

0

µ 3 ¶ µ 3 ¶ µ ¶ µ 2 ¶2 d φ dφ d3 φ d d φ d dφ d2 φ d φ φ 3 − = φ 3 − + .

3 2 dx dx dx dx dx dx dx dx dx2

R1 R 1 ³ 2 ´2 dx + λ 0 ex φ2 dx = 0. + 0 ddxφ2

From the boundary conditions, the boundary contributions vanish. Thus R 1 ³ 2 ´2 − 0 ddxφ2 dx λ= , R1 ex φ2 dx 0 so that λ ≤ 0.

Section 5.6 5.6.1 (c) From (5.6.6) using p = 1, q = 0, σ = 1 and using the boundary conditions, R 1 ¡ T ¢2 u2T (1) + 0 du dx λ1 ≤ .

R 1 2 dx u dx 0 T Any trial function should satisfy the boundary conditions (and be continuous with no zeroes). Geometrically, a simple example is a parabola (uT (0) = 0, uT (1) = 1, duT /dx(1) = −1, from the boundary conditions. To obtain this parabola, we substitute uT = ax + bx2 into the condition at x = 1 : a + 2b + a + b = 0, a simple choice being a = 3, b = −2 so that uT = 3x − 2x2 and uT /dx = 3 − 4x. Thus, R1

¯1 3¯ 1 (3 − 4x) ¯ 1 − 12

1 1 + (3 − 4x) dx 1 − 12 (−1 − 27) 40/12 1 0 = =4 . λ1 ≤ R 1 0 = = R 1 4 2 2 2 3 4 4/5 6 3−3+ 5 (3x − 2x ) dx (9x − 12x + 4x ) dx 0 0 2

14

Section 5.7 5.7.1 c2 = T0 /ρ. In this example 1 ≤ c2 ≤ 1 + α2 . Thus from the bottom of page 197 with L = 1 p p ¡ ¢ π 2 ≤ λ1 ≤ π 2 1 + α2 or π ≤ λ1 ≤ π 1 + α2 . √ √ The circular frequency (cycles per 2π units of time) is λ, but the actual frequency is λ/2π (cycles per 1 unit of time): √ 1 λ 1p ≤ ≤ 1 + α2 .

2 2π 2

Section 5.8

√ √ √ 5.8.2 (c) By extension of Fig. 5.8.2 (a), 0 < λ1 L < π2 , π < λ2 L < 3π λ3 L < 5π In general 2 , 2π < 2 , etc. √ √ ¡ ¢ √ 1 (n − 1)π < λn L < n − 2 π. The eigenfunction sin λx is graphed as a function of λx. The √ endpoint x = L occurs at λL . For the first eigenfunction this occurs before π/2; the eigenfunction has no zeroes for 0 < x < L. For√the second eigenfunction, the end occurs before 3π/2 and after π; the eigenfunction has one zero at λ2 x = π. Etc.. In general, the nth eigenfunction has n − 1 zeroes.

√ √ 5.8.3 (b) If λ > 0, then √ φ = c1√ cos λx + c2√ sin λx. dφ/dx(0) √ = 0 implies √ c2 = 0. √The boundary condition at x = L yields − λ sin λL + h cos λL = 0, or tan λL = h/ λ = hL/ λL. Thus, √ the straight 0 0 line in Fig. 5.8.1 must be replaced by the hyperbola (z = c/x , where c = hL and x = λL). Finally, ¡ ¢ √ √ √ 1 0 < λ1 L < π2 , π < λ2 L < 3π , . . ..

In general, (n − 1)π < λ L < n − π, n = 1, 2, 3, . . .. n 2 2 √ Asymptotically for large n, λn L ∼ (n − 1)π.

15

5.8.7 (a)

Rπ³ 0

2

2

d v d u u dx 2 − v dx2

´

¯π £ du ¤ ¯ dv dv dv du dx = u dx − v du dx ¯ = 2 dx (0) dx (π) − dx (0) dx (π) .

0

√ √ √ √ 5.8.7 (b) If λ > 0, then φ(0) = 0 implies φ = sin λx. The other condition yields sin λπ = 2 λ = 2 λπ/π.

From a graphical solution, there is only one solution since the slope 2/π of the straight line is less than 1, the slope at the origin of the sinusoidal function. In fact, we note that λ = 1/4 (although this is not important). √ √ √ 5.8.7 (c) √ If λ = −s < 0, then φ(0) = 0 implies φ = sinh sx. The other condition yields sinh sπ = 2 s = 2 sπ/π. There are no solutions (by graphing the sinh function and the straight line) since the slope 2/π of the straight line is less than 1, the slope at the origin of the hyperbolic function.

5.8.7 (d) If λ = 0, then φ(0) = 0 implies φ = x. The other condition yields π = 2, and hence λ = 0 is not an eigenvalue.

5.8.7 (e) Since the boundary conditions for this eigenvalue problem are not of the form (5.3.2), this is not a regular Sturm-Liouville problem. Thus, there may be complex eigenvalues (with a non-zero imaginary part), not obtained in parts (b) - (d). √ √ √ √ √ 5.8.8 (c) If λ > 0, φ = c1 cos √ λx+c2 sin λx, so that dφdx = λ(−c1 sin λx+c2 cos √λx). The √ condition √ at x = 0, yields c1 = c2 λ, so that the eigenfunction is any multiple of φ = sin λx + λ cos λx. √ √ The boundary condition at x = 1 then yields tan λ = 2 √ λ/(λ − 1). This is graphed using √ Fig. 5.8.1 2x0 0 with the straight line replaced by z = (x02 −1) , where x = λ. Note the singularity at λ = 1 < π/2. √ Thus λn ∼ (n − 1)π.

16

√ √ √ 5.8.10 (a) h = 1, L = 1 and thus, from (5.8.15), tan λ = − λ with√π/2 < λ1 < π from (5.8.l9). Using a PC, tan(2.03) = −2.022419 and tan(2.02) = −2.074373. Thus λ1 ∼ 2.03 or λ1 ≈ 4.12. √ √ √ 5.8.10 (b) h = 1, L = 1 and thus, from √ (5.8.15), tan λ1 = − λ1 with π/2 < λ1 < π from (5.8.l9). Using Newton’s method (letting x = λ1 ) with f (x) = x + tan x , yields xn+l = xn −

f (xn ) xn + tan xn = xn − . f 0 (xn ) 1 + sec2 xn

√ In this manner, we obtain λ1 = limn→∞ xn = 2.028758 . . . or λ1 = 4.11586 . . . √ √ √ √ √ √ √ RL RL √cos λL . But tan λL = − λ/h.

5.8.13 0 sin2 λx dx = 12 0 (1−cos 2 λx) dx = L2 − sin42√λλL = L2 − sin λL 2 λ √ √ √ √ RL − λ 1 Therefore, sin λL = √ h 2 and cos λL = √ 1 2 . Thus 0 sin2 λx dx = L2 + 2h[1+(λ/h 2 )] .

1+(λ/h )

1+(λ/h )

Section 5.9

¡ ¢ Rx 5.9.1 (b) To satisfy φ(0) = 0, we use (5.9.9) and hence dφ/dx ≈ (σp)−1/4 (λσ/p)1/2 cos λ1/2 0 (σ/p)1/2 dx0 .

Thus µ ¶ Z L 1 1/2 1/2 λ (σ/p) dx0 ≈ n + π.

2 0 5.9.3 (a)

¡ 0 ¢ 1/2 1/2 σ exp[. . .] £ A00 + Aiλ ¡ ¢ ¤ A + iλ1/2 2A0 σ 1/2 + 21 σ −1/2 σ 0 A − λσA exp[. . .] ¡ ¢ By substituting into the differential equation, we obtain A00 = iλ1/2 2A0 σ 1/2 + 21 σ −1/2 σ 0 A + qA = 0. ¡ 1/4 ¢ P∞ d σ An+1 = 5.9.3 (e) Let A = n=0 An λ−n/2 . Thus A0n+1 σ 1/2 + 14 σ −1/2 σ 0 An+1 = 2i (A00n + qAn ) or dx R x i −1/4 (A00n + qAn ). By integrating, An+1 = 2i σ −1/4 0 σ −1/4 (A00n + qAn ) dx0 .

2σ φ0 φ00

= =

Section 5.10 5.10.2 (b) From (3.5.5) or (3.3.21-23), x = L2 − 4L π2 for a cosine series Z L

0

f 2 dx =

P∞

1 n=1 n2 cos nπx/L (odd only). From (5.10.14) with σ = 1

∞ X

αn2

n=0

Z L

cos2 nπx/L dx.

0

3 With f = x, α0 = L/2, (n odd) αn = n−4L 2 π 2 , (n even) αn = 0, we obtain (dividing by L ) P 4 ∞ 8 1 1 1 π 1 1 1 n=1 π 4 n4 (odd only) or 96 = 1 + 34 + 54 + 74 + . . .

3 = 4 +

17

Chapter 6. Numerical Methods Section 6.2 6.2.6 We use (6.2.13): ∂2u ∂ ∂u ∂ u(x + ∆x, y) − u(x − ∆x, y) = ( )= ∂x∂y ∂y ∂x ∂y 2∆x =

1 [u(x + ∆x, y + ∆y) − u(x − ∆x, y + ∆y) − u(x + ∆x, y − ∆y) + u(x − ∆x, y − ∆y)].

2∆y 2∆x

Section 6.3 (0)

6.3.4 (a) At t = 0, uj

= f (xj ) = fj . Thus, from (6.3.24) N −1 X

N −1 X nπxj nπj fj = βn sin = βn sin , L N n=1 n=1

since xj = j∆x and ∆x = L/N . We multiply by sin mπj N and sum from j = 1 to j = N − 1 (analogous to integrating over x): N −1 X

fj sin

j=1

N −1 N −1 X X mπj mπj nπj = βn sin sin .

N N N n=1 j=1

The discrete eigenfunctions are orthogonal (see exercise 6.3.3), and hence the inner summation is nonzero only for n = m. Thus N −1 X

fj sin

j=1

6.3.4 (b) Using the double angle formula N −1 X j=1

sin

N −1 X mπj mπj = βm sin2 .

N N j=1

PN −1

j=1 sin

1 2 mπj N = 2

PN −1

j=1 (1−sin

2mπj N ).

However

2mπ 2mπN N −1 X 2mπj 2mπj ei N − ei N = 0, = Im ei N = Im 2mπ N 1 − ei N j=1

since ei2mπ = 1, where the sum of a finite geometric series has been used. Thus, N −1 X j=1

sin2

mπj 1 = (N − 1).

N 2

6.3.6 (d) From the top of page 238, the scheme is stable if and only if s≤ 6.3.9

1 1−cos

(N −1)π N

1 . For n = 10, from a calculator or pc s ≤ 1−cos 9π = 0.5125....

10

1 1 (a) Using the centered difference, f1 = (∆x) 2 (u2 − 2u1 + u0 ) and f2 = (∆x)2 (u3 − 2u2 + u1 ), where ∆x = L3 . From the boundary conditions u0 = u3 = 0.

− → 1 → 6.3.9 (b) From part (a), A− u = f , where A = (∆x) 2

·

18

−2 1

1 −2

¸ .

(m)

6.3.10 (a) As on page 240, we let uj

be the probability that a particle is located at point j at time m∆t.

(m) (m−1) (m−1) (m−1) Thus uj = a uj−1 + a uj+1 + buj .

6.3.10 (b) Using the Taylor series of a function of two variables, (m−1)

uj

∂u + O(∆t)2 ∂t

− ∆t

∂u ∂u (∆x)2 ∂ 2 u − ∆x + + O[(∆t)2 , ∆x∆t, (∆x)3 ] ∂t ∂x 2 ∂x2

(m)

− ∆t

∂u (∆x)2 ∂ 2 u ∂u + ∆x + + O[(∆t)2 , ∆x∆t, (∆x)3 ] ∂t ∂x 2 ∂x2

= uj

(m−1)

= uj

uj+1

− ∆t

(m)

(m−1)

uj−1

(m)

= uj

Thus, (m)

uj

(m)

= (2a + b) uj

− (2a + b)∆t

∂u ∂2u + a(∆x)2 2 + O[(∆t)2 , ∆x∆t, (∆x)3 ]. ∂t ∂x

2

ka ∂ u From 2a + b = 1, we obtain ∂u ∂t = s ∂x2 since

lim∆x→0

and

∆t→0

(∆x)2 k ∆t = s .

6.3.14 (c) The eigenvalues are determined from (6.3.50): ¯ ¯ ¯ 1−λ 2 −3 ¯¯ ¯ ¯ 2 4−λ −6 ¯¯ = (1 − λ)[(4 − λ)(2 − λ) + 2] − 4(2 − λ) − 2 ¯ ¯ 0 1/3 2 − λ ¯ = −λ3 + λ2 (1 + 6) + λ(−6 − 10 + 4) = −λ(λ2 − 7λ + 12) = −λ(λ − 4)(λ − 3).

Thus, λ = 0, 3, 4. According to the Gershgorin circle theorem, every eigenvalue lies in the region composed of the interiors of 3 circles: ¯ ¯ ¯ ¯ ¯λ − 1¯ ≤ 5, ¯λ − 4¯ ≤ 8,

and

¯ ¯ ¯λ − 2¯ ≤ 1 , 3

from (6.3.62). In this case all three eigenvalues lie in the first two circles, but none lie in the third circle.

Section 6.4 6.4.1 Here we do not assume ∆x = ∆y. From (6.2.16) (m+1)

uj,l

(m)

(m)

− uj,l

∆t

= k[

(m)

(m)

uj+1,l − 2uj,l + uj−1,l (∆x)2

(m)

+

(m)

(m)

uj,l+1 − 2uj,l + uj,l−1 (∆y)2

].

We substitute (6.4.2) and obtain Q−1 eiα∆x − 2 + e−iα∆x eiβ∆y − 2 + e−iβ∆y = k[ + ]. ∆t (∆x)2 (∆y)2 −1 Thus, Q = 1 + 2k∆t[ cos(α∆x) + cos(β∆y)−1 ]. To be stable −1 < Q < 1. Since cos(α∆x)> −1 and (∆x)2 (∆y)2 1 1 cos(β∆y) > −1, we are guaranteed stability if 4k∆t[ (∆x) 2 + (∆y)2 ] ≤ 2, which yields the desired result.

19

Section 6.5 (m+1)

6.5.5 (a) The partial difference equation is

uj

(m)

− uj ∆t

(m)

+c

(m)

uj+1 − uj−1 = 0 using (6.2.13).

2∆x

6.5.5 (b) We substitute (6.5.6) into the results of part (a) and obtain

¯ ¯

iα∆x Q−1 −e−iα∆x ∆t + ce = 0. Thus Q = 1 − c ∆x isin(α∆x). Stability is determined by ¯Q¯ .

2∆x ¯ ∆t¯ √ ¯Q¯ = 1 + σ 2 > 1, where σ = c ∆t sin(α∆x), this scheme is unstable. ∆x (m+1)

(m−1)

− uj 2∆t

(m)

(m)

uj+1 − uj−1 = 0 . By substituting 2∆x 1 iα∆x −iα∆x Q− Q −e 1 (6.5.6), we obtain 2∆t + c e = 0. This is equivalent to Q − Q + 2iσ = 0, where σ = 2∆x ¯ ¯ √ √ ∆t 2 2 c ∆x sin(α∆x). Thus Q +2iσQ−1 = 0 or Q = −iσ± 1 − σ 2 . If σ < 1, then ¯Q¯ = σ 2 + 1 − σ 2 = 1.

2

6.5.6 Using (6.2.13), the partial difference equation is

uj

Since

+c

The numerical ¯scheme is ¯stable if σ < 1. This (stability of the numerical √ scheme) is guaranteed if ∆t 2 c ∆x σ 2 − 1). In this case, the < 1 since ¯sin(α∆x)¯¯ ≤ 1.

However, if σ > 1, then Q = i(−σ ± ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ scheme is unstable since Q > σ for one root and σ > 1 in this case.

20

Chapter 7. Higher Dimensional PDES Section 7.3 7.3.1 (a) The result of product solutions u(x, y, t) = φ(x, y)h(t) is dh/dt = −λkh and ∇2 φ = −λφ, which ¡ ¢2 2 is further separated φ(x, y) = f (x)g(y) : ddxf2 = −µf with f (0) = f (L) = 0 yielding µ = nπ ,f = L ¡ mπ ¢2 d2 g nπx sin L , n = 1, 2, ... and dy2 = −(λ − µ)g with g(0) = g(H) = 0 yielding λ − µ = H , g = ¡ nπ ¢2 ¡ mπ ¢2 mπy + H , φ = sin nπx sin mπy H , m = 1, 2, . . . Thus λnm = L L sin H , n = 1, 2, . . . , m = 1, 2, . . . By superposition ∞ X ∞ X nπx mπy −λnm kt u= anm sin sin e .

L H n=1 m=1 The initial condition implies f=

∞ X ∞ X

anm sin

n=1 m=1

mπy nπx sin , L H

yielding (by two-dimensional orthogonality) Z HZ L 4 nπx mπy f sin sin dx dy. anm = LH 0 L H 0 As t → ∞, u → 0 (the only equilibrium temperature if u = 0 along the entire boundary).

7.3.1 (c) The result of product solutions u(x, y, t) = φ(x, y)h(t) is dh/dt = −λkh and ∇2 φ = −λφ, which is further separated φ(x, y) = f (x)g(y): ¡ nπ ¢2 d2 f d2 g 0 0 , f = cos nπx dx2 = −µf with f (0) = f (L) = 0 yielding µ = L L , n = 0, 1, 2, ...and dy 2 = −(λ − ¡ ¢2 ¡ ¢2 , g = sin mπy , m = 1, 2, . . . Thus λnm = nπ + µ)g with g(0) = g(H) = 0 yielding λ − µ = mπ H H L ¡ mπ ¢2 mπy nπx , φ = cos L sin H , n = 0, 1, 2, . . . , m = 1, 2, . . . By superposition H u=

∞ X ∞ X

anm cos

n=0 m=1

The initial condition implies f=

∞ X ∞ X

mπy −λnm kt nπx sin e .

L H

anm cos

n=0 m=1

mπy nπx sin , L H

yielding (by two-dimensional orthogonality) RH RL anm =

mπy f cos nπx dx dy L sin H ½ 1 RH RL 2 mπy 2 LH cos2 nπx 1 L sin H dx dy = 0 0 4 LH 0

0

n=0 n 6= 0

.

As t → ∞, u → 0 (the only equilibrium temperature if u = 0 along the entire boundary).

7.3.2 (b) The result of product solutions u(x, y, z, t) = φ(x, y, z)h(t) is dh/dt = −λkh and ∇2 φ = −λφ, which is further separated φ(x, y) = f (x)g(y)Q(z): ¡ nπ ¢2 d2 f d2 g 0 0 0 , f = cos nπx dx2 = −µf with f (0) = f (L) = 0 yielding µ = L L , n = 0, 1, 2, . . . dy 2 = −νg with g (0) = ¡ ¢ 2 2 d Q 0 0 , g = cos mπy g 0 (L) = 0 yielding ν = mπ H , m = 0, 1, 2, . . . dz 2 = −(λ−µ−ν)Q with Q (0) = Q (W ) = ¡H`π ¢2 ¡ ¢ ¡ ¢ ¡ ¢ 2 2 nπ `π 2 0 yielding λ − µ − ν = W , Q = cos `πz ,φ = + mπ + W W , ` = 0, 1, 2, . . . Thus λnm` = L H mπy `πz cos nπx cos , cos , n = 0, 1, 2, . . . , m = 0, 1, 2, . . . , ` = 0, 1, 2, . . .

L H W By superposition u=

∞ X ∞ X ∞ X n=0 m=0 `=0

anm` cos

nπx mπy `πz −λnm` kt cos cos e .

L H W

21

The initial condition implies f=

∞ X ∞ X ∞ X

anm` cos

n=0 m=0 `=0

nπx mπy `πz cos cos , L H W

yielding (by three-dimensional orthogonality) RW RH RL mπy `πz f cos nπx L cos H cos W dx dy dz anm` = R W0 R H0 R L0 .

2 mπy cos2 `πz dx dy dz f cos2 nπx L cos H W 0 0 0 The normalization integrals in the denominator are easy, but one must distinguish between n = 0 and n 6= 0 (and similarly with m and `). As t → ∞ (λ000 = 0, all others > 0) Z WZ HZ L 1 u(x, y, z, t) → a000 = f dx dy dz, LHW 0 0 0 the average of the initial temperature distribution since all sides are insulated and thus all the initial energy must be in this mode.

7.3.4 (b) The eigenvalue problems both yield cosine series. Thus, by superposition u(x, y, t) =

∞ X ∞ X

Anm cos

m=0 n=0

mπy nπx cos hnm (t) .

L H

However, the solution of (7.3.l7) satisfying h(0) = 0 is ½ t n = 0, m = 0 hnm (t) = sin ωnm t otherwise, £ ¤ 2 where ωnm = c2 (nπ/L)2 + (mπ/H)2 . To satisfy the other initial condition, f=

∞ X ∞ X

Anm cos

m=0 n=0

mπy 0 nπx cos h (0), L H nm

and thus by orthogonality (2-dimensional) RH RL mπy f cos nπx 0 L cos H dx dy Anm hnm (0) = R 0H R 0L .

2 mπy dx dy cos2 nπx L cos H 0 0 We note that h0nm (0) = 1 for n = 0, m = 0, but otherwise h0nm (0) = ωnm .

2

7.3.6 (b) By separating the z-variable, u(x, y, z) = φ(x, y)h(z), ddzh2 = λh and ∇2 φ = −λφ. From Section 7.3, φ = 0 on the boundaries implies that λnm = (nπ/L)2 + (mπ/W )2 with the eigenfunctions being φnm = sin nπx/L sin mπy/W, n = 1, 2, 3, . . . , m = 1, 2, 3, . . .. Since dh/dz(0) = 0, it follows that √ h(z) = cosh λnm z. Thus by superposition u=

∞ X ∞ X

Anm sin

n=1 m=1

p nπx mπy sin cosh λnm z.

L W

The nonhomogeneous boundary condition, f=

∞ X ∞ X

Anm cosh

n=1 m=1

determines Anm Anm cosh

p

p nπx mπy λnm H sin sin , L W

RW RL λnm H = R W R0L 0

0

22

0

mπy f sin nπx L sin W dx dy

2 mπy LW sin2 nπx L sin W dx dy = 4

.

7.3.7 (c) A solution only exists if of variables

RH RW 0

0

u(x, y, z) =

f (y, z) dy dz = 0, the condition for an equilibrium. By separation ∞ X ∞ X

Anm cos

n=0 m=0

nπy mπz cos cosh βnm x, W H

2 where βnm = (nπ/W )2 + (mπ/H)2 . The nonhomogeneous boundary condition,

f=

∞ X ∞ X

Anm βnm sinh βnm L cos

n=0 m=0

determines Anm

RH RW Anm βnm sinh βnm L = R 0H R 0W 0

0

nπy mπz cos , W H

mπz f cos nπy W cos H dy dz 2 mπz dy dz cos2 nπy W cos H

.

RH RW RH RW Since β00 = 0, we note that if 0 0 f dy dz 6= 0, there are no solutions. However, if 0 0 f dy dz = 0, then A00 is arbitrary and the other coefficients uniquely determined.

7.3.7 (d) See the solution to 7.3.7(c). Instead the nonhomogeneous condition g(y, z) =

∞ X ∞ X

mπz nπy cos cosh βnm L, W H

Anm cos

n=0 m=0

always determines Anm (without difficulty) RH RW Anm cosh βnm L = R 0H R 0W 0

0

mπz g cos nπy W cos H dy dz 2 mπz dy dz cos2 nπy W cos H

.

Section 7.4 7.4.1 (a) Equation (7.3.l5) with f 0 (0) = f 0 (L) = 0 yields µ = (mπ/L)2 with f (x) = cos mπx/L, m = 0, 1, 2, . . .. Equation (7.3.l6) with g(0) = g(H) = 0 yields λ − µ = (nπ/H)2 with g(y) = sin nπy/H, n = 1, 2, . . .. Thus λnm = (nπ/H)2 + (mπ/L)2 , where n = 1, 2, 3, . . . and m = 0, 1, 2, . . ..

Section 7.7 7.7.1 The product solutions are represented in (7.7.45). The initial condition u(r, θ, 0) = 0 implies that all √ terms with cos c λnm t vanish. The other initial condition ∂u/∂t(r, θ, 0) = α(r) sin 3θ implies that only m = 3 is needed (and in fact only the sin 3θ terms). Thus, by superposition, u(r, θ, t) =

∞ X

An J3

³p

´ p λ3n r sin 3θ sin c λ3n t.

n=1

From the initial condition (cancelling the sin 3θ term) α(r) =

∞ X

³p ´ p An c λ3n J3 λ3n r .

n=1

Using the orthogonality of J3

¡√

¢ λ3n r with weight r:

ÁZ a ³p Z a ³p ´ ´ p An c λ3n = α(r)J3 λ3n r r dr J32 λ3n r r dr .

0

0

23

7.7.2 (d) The boundary condition in (7.7.35) should be f 0 (a) = 0. From the Rayleigh quotient (7.6.5), λ ≥ 0 with λ = 0 only if φ(r, θ) is constant. Thus from (7.7.34) √ λ = 0 only for m = 0 in which case φ(r, θ) = 1. Otherwise λ > 0 and the transformation z = λr may be used. Thus, first (7.7.37) 0 is valid, then f (r) satisfies (7.7.38). ¡√ ¢ The boundary condition f (a) = 0, then implies that the other 0 λmn are determined by Jm λmn a = 0. The time-dependent part yields (7.7.5). Thus, the initial √ condition u(r, θ, 0) = 0 implies that h(0) = 0 and hence h(t) = sin c λmn t. However, if λ = 0, h(t) = t from (7.7.5). Thus, by superposition, u(r, θ, t) =

∞ X ∞ X

(Amn cos mθ + Bmn sin mθ) Hmn (r, t),

m=0 n=1

where

½ Hmn (r, t) =

Jm

¡√

m = 0, n = 1 ¢t √ λmn r sin c λmn t otherwise.

The other initial condition becomes β(r, θ) =

∞ X ∞ X

(Amn cos mθ + Bmn sin mθ) φmn (r),

m=0 n=1

where

½ φmn (r) =

1¡√ ¢ √ λmn r c λmn Jm

m = 0, n = 1 otherwise.

From orthogonality of these two-dimensional eigenfunctions Rπ Ra β(r, θ)φmn (r) cos mθ rdr dθ Amn = −πR π0 R a 2 φ (r) cos2 mθ rdr dθ −π 0 mn and Bmn is the same as Amn with cos mθ above replaced by sin mθ.

7.7.3 (a) The boundary conditions for (7.7.l2) are g(0) = g(π/2) = 0. This is a standard eigenvalue problem with L = π/2. Thus, µ = (mπ/L)2 = 4m2 with eigenfunctions g(θ) = sin 2mθ, m = 1, 2, 3, . . .. Thus, the m2 in (7.7.34) must be replaced by 4m2 [i.e., by 2m in (7.7.37) - (7.7.38)]. Thus, the ¡√ m replaced ¢ √ frequencies of vibration are c λmn where J2m λmn a = 0.

7.7.5 The boundary conditions for (7.7.l2) are g(0) = g(π/2) = 0. This is a standard eigenvalue problem with L = π/2. Thus, µ = (mπ/L)2 = 4m2 with eigenfunctions g(θ) = sin 2mθ, m = 1, 2, 3, . . .. Thus, the m2 in (7.7.34) must be replaced by 4m2 [i.e., m replaced by 2m in (7.7.37)]. The boundary conditions for (7.7.37) are f (a) = f (b) = 0. Thus ³√ ´ ³√ ´ c1 J2m λa + c2 Y2m λa = 0 ³√ ´ ³√ ´ c1 J2m λb + c2 Y2m λb = 0 .

Nontrivial solutions can be obtained from the determinant (or by elimination): ³√ ´ ³√ ´ ³√ ´ ³√ ´ J2m λa Y2m λb − J2m λb Y2m λa = 0, √ where the frequencies are c λ.

7.7.8 For the heat equation, Section 7.7 is valid with (7.7.5) replaced by dh/dt = −λkh. The boundary condition introduces more substantial changes. The Rayleigh quotient shows λ ≥ 0 with λ = 0 only when φ(r, θ) is constant, which means m = 0. The ´ other eigenfunctions still satisfy (7.7.38) with the ³√ 0 0 λa = 0. Thus boundary condition f (a) = 0 yielding Jm ( φmn (r, θ) =

³√ 1´ λr cos mθ Jm 24

m = 0, n = 1 otherwise.

By superposition u(r, θ, t) =

∞ X ∞ X

Amn φmn (r, θ)e−λmn kt +

n=1 m=0

∞ X ∞ X

Bmn Jm

³p

´ λmn r sin mθe−λmn kt .

n=1 m=1

The initial conditions yield equations for Amn and Bmn . Using the two-dimensional orthogonality of the eigenfunctions Rπ Ra f (r, θ)φmn (r, θ)rdr dθ Amn = −πR π0 R a 2 φ (r, θ)rdr dθ −π 0 mn mn kt and a similar expression for Bmn . As t → ∞, e−λ = 0, n = 1 since then ³R R → 0 except for ´. m ¡ 2¢ a π f (r, θ)rdr dθ πa , the average of the λmn = 0. Thus u(r, θ, t) → A01 , where A01 = −π 0 initial temperature distribution. Using physical reasoning, the equilibrium should be a constant. By conservation of thermal energy that constant should be the above value, since the boundaries are insulated.

7.7.9 (b) For the heat equation, Section 7.7 is valid with (7.7.5) replaced by dh/dt = −λkh. The boundary conditions for (7.7.l2) are g 0 (0) = g 0 (π) = 0. This is a cosine series in θ with L = π, i.e. µ = (mπ/L)2 = m2 (as before) but with g = cos mθ only. Thus (7.7.18-20) is valid with the condition at r = a being f 0 (a) = 0. The Rayleigh quotient shows λ ≥ 0 with λ = 0 only when m = 0 and f (r) = 1. The other √ eigenfunctions satisfy (7.7.37) - (7.7.38), where now the boundary condition at r = a yields 0 Jm ( λa) = 0. Thus the eigenfunctions are ½ ¡√1 ¢ m = 0, n = 1 φmn (r) = Jm λmn r otherwise.

By superposition u(r, θ, t) =

∞ X ∞ X

Amn φmn (r) cos mθe−λmn kt .

n=1 m=0

The initial conditions determine the coefficients Amn (by orthogonality): RπRa f (r, θ)φmn (r) cos mθrdrdθ Amn = 0 R π0 R a 2 . φ (r) cos2 mθrdrdθ 0 0 mn As t → ∞, e−λmn kt → 0 except for m = 0, n = 1 since then λmn = 0. Thus u(r, θ, t) → A01 , where RπRa f (r, θ)rdrdθ , A01 = 0 0 1 2 2 πa the average of the initial temperature distribution. The temperature approaches an equilibrium. This can be obtained easily by conservation of thermal energy since the boundary is insulated.

7.7.10 Subsection 7.7.9 is valid with h(t) instead solving dh/dt = −λkh. Thus by superposition u(r, t) =

∞ X

an J0

³p

´ λn r e−λn kt ,

n=1

where (7.7.63) is satisfied. The initial condition yields (7.7.66) with α(r) replaced by f (r). As t → ∞, u → 0 since λn > 0. This can be shown to be the only equilibrium solution since u(a, t) = 0, see exercise 1.5.10.

7.7.12 (a) x2 y 00 − 6y ≈ 0. Substituting y = xp yields p(p − 1) − 6 = 0 or (p − 3)(p + 2) = 0 or p = −2, 3.

Thus y = c1 (x−2 + . . .) + c2 (x3 + . . .).

7.7.12 (c) x2 y 00 +xy 0 +4y ≈ 0. Substituting y = xp yields p(p−1)+p+4 = 0 or p = ±2i. Since x±2i = e±2i ln x , we obtain real solutions y = c1 [cos(2 ln x) + . . .] + c2 [sin(2 ln x) + . . .].

7.7.12 (e) x2 y 00 − 4xy 0 + 6y ≈ 0. Substituting y ¡= xp yields p(p 1) −¢4p + 6 = 0 or p2 − 5p + 6 = 0 or ¢ ¡ − 2 3 (p − 3)(p − 2) = 0 or p = 2, 3. Thus y = c1 x + . . . + c2 x + . . . .

25

Section 7.8 7.8.1 (b) Since λ > 0, (7.7.37) is valid. The boundary conditions yield ³√ ´ ³√ ´ c1 Jm λ + c2 Ym λ =0 ³ √ ´ ³ √ ´ c1 Jm 2 λ + c2 Ym 2 λ = 0 .

By the determinant condition for a nontrivial solution (or by elimination), the eigenvalues are determined by ³ √ ´ ³ √ ´ ³√ ´ ³√ ´ λ Ym 2 λ − Jm 2 λ Ym λ =0.

Jm 7.8.1 (d) Using the Rayleigh quotient minimization principle (5.6.5) with p = r, q = −m2 /r, and σ = r (and x = r) R 2 ¡ du ¢2 2 [r dr + m2 ur ]dr 1 λ1 = min .

R2 u2 r dr 1 The smallest occurs with m = 0. Upper and lower bounds can be obtained using the technique discussed in Section 5.7. Since 1 ≤ r ≤ 2, 1 min 2

R 2 ¡ du ¢2

R 2 ¡ du ¢2

1

1

dr 2 ≤ λ1 ≤ min R 2dr 2 1 u dr 1

dr .

R 2dr 2 u dr 1

R 2 du 2 ( ) dr ¡ π ¢2 = L , where L = 2 − 1 = 1. Thus But min 1R 2dr 1

u2 dr

1 2 π ≤ λ1 ≤ 2π 2 .

2 7.8.2 (d) Section 7.7 is valid with dh/dt = −λkh. The boundary conditions for (7.7.l2) are g(0) = g(π/2) = 0.

This is the standard sine series with L = π/2 so that λ = (mπ/L)2 = 4m2 with g(θ) = sin 2mθ. Thus (7.7.34) - (7.7.39) are valid with m replaced by 2m. Thus by superposition u(r, θ, t) =

∞ X ∞ X

cmn J2m

³p ´ λmn r sin 2mθe−λmn kt ,

n=1 m=1

where (7.7.39) is valid with m replaced by 2m. The initial condition determines cmn ¡√

R π/2 R a cmn =

0

¢

G(r, θ)J2m λmn r sin 2mθ rdr dθ 0 . ¢ R π/2 R a 2 ¡√ J λmn r sin2 2mθ rdr dθ 0 0 2m

7.8.8 If m = 12 , then f = z −1/2 y where y 00 + y = 0. Thus J1/2 (z) = z −1/2 (c1 cos z + c2 sin z)

and Y1/2 (z) = z −1/2 (c3 cos z + c4 sin z) ,

where

we cannot determine ci from differential equation. Equation (7.7.33) can be used as a definition. As z → 0 we have obtained J1/2 ∼ c1 z −1/2 and Y1/2 ∼ c3 z −1/2 Thus from (7.7.33), c1 = 0 so that ¡ ¢ ¡ ¢ J1/2 ∼ c2 z 1/2 resulting in c2 = 1/ 21/2 21 ! and c3 = −21/2 − 12 !/π but c4 cannot be determined from this type p show that (−1/2)! p = √ of consideration. Gamma function results√(see exercise 10.3.14) = Γ(3/2) = 1/2Γ(1/2) = 1/2 π.

Thus, c = 2/π and c = − 2/π. Γ(1/2) = π and (1/2)!

2 3 p p p Therefore J1/2 (z) = 2/πz sin z and Y1/2 (z) = − 2/πz cos z + c4 π/2J1/2 (z). From (7.8.3), c4 = 0.

26

Section 7.9 7.9.1 (b) This is√similar to the problem in subsection 7.9.3. To satisfy u(r, θ, H) = 0, instead we use h(z) = sinh λ(z − H). The condition at z = 0 implies m = 7 only (and sin 7θ only also). Thus by superposition ∞ ³p ´ X p u(r, θ, z) = Amn sinh λ7n (H − z) sin 7θJ7 λ7n r , n=1

where (7.9.18) is satisfied with m = 7. It is not necessary to use notation λ7n ; instead λn may be used.

The boundary condition at z = 0 (after cancelling sin 7θ) determines An ¡√ ¢ Ra p α(r)J7 λ7n r r dr 0R ¢ An sinh λ7n H = . a 2 ¡√ J λ7n r r dr 0 7 7.9.2 (b) This has eigenvalue problems in θ and z and so is similar to subsection 7.9.4. The boundary conditions in θ are g(0) = g(π) = 0 so that µ = (mπ/L)2 = m2 since L = π with g(θ) √ = sin mθ only. In z the boundary conditions are h(0) = 0 and h0 (H) = 0. Thus h(z) = sin −λz with λ = −(n − 1/2)2 (π/H)2 , n = 1, 2, . . .. Thus (7.9.31) is valid with n replaced by n − 1/2. In this way (7.9.37) is valid (n → n − 1/2 and sin mθ only). By superposition ¶ ¸ µ ¶ ·µ ∞ X ∞ X 1 π 1 πz u(r, θ, z) = r sin n − sin mθ.

Amn Im n − 2 H 2 H m=1 n=1 The boundary condition at r = a determines Amn ¡ ¢ ·µ ¶ ¸ RπRH β(θ, z) sin mθ sin n − 21 πz 1 π H dz dθ 0 0 .

Amn Im n − a = RπRH ¡ ¢ 2 2 2 H sin n − 12 πz H sin mθ dz dθ 0 0 The normalization integral equals πH/4.

7.9.3 (b) By separation of variables, time only first, ∇23 φ + λφ = 0 and dh/dt = −λkh. Here ∇23 φ is the same operator as in Section 7.7 with an extra term φzz . If we next separate z, we know Qzz = −νQ with ν = (`π/H)2 , because of the boundary conditions and Q(z) = cos `πz/H, ` = 0, 1, 2, . . .. Thus ∇2 φ + λ0 φ = 0 where λ0 = λ − (`π/H)2 and here the Laplacian is the two-dimensional one previously analyzed. We follow Section 7.7. The boundary condition for (7.7.l2) is g 0 (0) = g 0 (π/2). This is a cosine series in θ, where µ = [mπ/(π/2)]2 = 4m2 and g(θ) = cos 2mθ, m = 0, 1, 2, . . . [and thus m is replaced by 2m in (7.7.34) and (7.7.38)]. Note that λ0 = 0 only ¡p for m = ¢0 with f (r) = 1. Otherwise 0 (7.7.38) is valid. The boundary condition at r = a yields J2m λ0mn a = 0. In this manner, the three-dimensional eigenfunctions are ½ 1 ¡p ¢ m = 0, ` = 0, n = 1 φ`mn (r, θ, z) = 0 r cos `πz cos 2mθJ λ otherwise.

2m mn H By superposition u(r, θ, z, t) =

∞ X ∞ X ∞ X

A`mn φ`mn (r, θ, z)e−λ`mn kt ,

n=1 m=0 `=0

where λ`mn = (`π/H)2 + λ0mn . The initial condition determines A`mn ÁZ Z Z Z Z Z A`mn = f (r, θ, z)φ`mn (r, θ, z) rdr dθ dz φ2`mn rdr dθ dz .

As t → ∞, e−λ`mn kt → ∞ except for ` = 0, m = 0, n = 1 since λ001 = 0 : Z Z Z 1 f (r, θ, z)r dr dθ dz, u(r, θ, z, t) → A001 = Hπa2 /4 the average of the initial temperature distribution. We expect this because with insulated boundaries the equilibrium temperature ought to be a constant (with the same thermal energy as the initial condition).

27

7.9.4 (a) By separation of variables in time ∇2 φ + λφ = 0 and dh/dt = −λkh. Here ∇2 φ is the same operator as in Section 7.7 with an extra term φzz . If we next separate z, we know Qzz = −νQ where 2 from the boundary condition ν = (`π/H) with φ = sin `πz/H, ` = 1, 2, 3, . . . ∇2 φ + λ0 φ = 0 where λ0 = λ − ν and here the Laplacian is the two-dimensional one previously analyzed. Since there is no θ-dependence, we may ¢ ¡pfollow ¢ Section 7.7.9. The orthogonal eigenfunctions in the radial direction are ¡p J0 λ0n r where J0 λ0n a = 0 determines λ0n . By superposition u(r, z, t) =

∞ X ∞ X

An` J0

`=1 n=1

³p

´ `πz −λkt λ0n r sin e , H

where λ = λ0 + (`π/H)2 . The initial condition determines An` ¡p ¢ RR f (r, z)J0 λ0n r sin `πz rdr dz ¢ 2 `πz H R R 2 ¡p An` = .

0 J0 λn r sin H rdr dz

28

Chapter 8. Nonhomogeneuous Problems Section 8.2 8.2.1 (a) An equilibrium satisfies d2 uE /dx2 = 0 with uE (0) = A and duE /dx(L) = B. Thus uE (x) = A+Bx. We now introduce the displacement from equilibrium, defined by (8.2.9). v(x, t) satisfies (8.2.10) subject to the boundary conditions, (8.2.11)and ∂v/∂x(L, t) = 0, and the initial conditions (8.2.13).

Since the boundary conditions and pde are homogeneous, it may be solved by separation of variables, 2 2 v(x, t) = φ(x)h(t). Thus dh/dt √ = −λkh√and d φ/dx1 = −λφ subject to φ(0) = 0 and dφ/dx(L) = 0.

The eigenfunctions are sin λx, where λL = (n − 2 )π, n = 1, 2, 3, ... Thus by superposition v(x, t) =

∞ X

1 an sin(n − )πx/Le−λn kt .

2 n=1

The initial conditions determine the coefficients an Z 2 L 1 an = g(x) sin(n − )πx/L dx, L 0 2 where g(x) = f (x) − uE (x).

8.2.1 (d) An equilibrium satisfies kd2 uE /dx2 = −Q = −k or d2 uE /dx2 = −1 subject to uE (0) = A and uE (L) = B. Thus, uE (x) = −x2 /2 + A + γx, where B = −L2 /2 + A + γL. The displacement from equilibrium, defined by (8.2.9), satisfies (8.2.10-13). Consequently the solution is given by (8.2.17).

8.2.2 (a) A reference temperature distribution r(x, t) must satisfy the boundary conditions, rx (0, t) = A(t) and rx (L, t) = B(t). An example is the quadratic r(x, t) = A(t)x+[B(t) − A(t)]x2 /2L. By introducing the difference by (8.2.25), we have u(x, t) = v(x, t) + r(x, t). Thus ∂v ∂2v ∂r ∂2r −k 2 =Q− + k 2, ∂t ∂x ∂t ∂x subject to the homogeneous boundary conditions vx (0, t) = vx (L, t) = 0 and the initial condition v(x, 0) = f (x) − r(x, 0).

8.2.2 (c) A reference temperature distribution r(x, t) must satisfy the boundary conditions, rx (0, t) = A(t) and r(L, t) = B(t). The simplest example is the linear function r(x, t) = A(t)x + B(t) − LA(t). By introducing the difference by (8.2.25), we have u(x, t) = v(x, t) + r(x, t). Thus ∂v ∂2v ∂r −k 2 =Q− ∂t ∂x ∂t since ∂ 2 r/∂x2 = 0, subject to the homogeneous boundary conditions, vx (0, t) = 0 and v(L, t) = 0, and the initial condition v(x, 0) = f (x) − r(x, 0).

8.2.6 (a) An equilibrium satisfies d2 uE /dx2 = 0 with uE (0) = A and uE (L) = B. Thus uE (x) = A + (B − A)x/L. By introducing the displacement, defined by (8.2.9), we have 2 ∂2v 2∂ v = c ∂t2 ∂x2 subject to v(0, t) = v(L, t) = 0 and the initial conditions v(x, 0) = f (x) − uE (x) and vt (x, 0) = g(x).

From Section 4.4 ∞ X v(x, t) = (An cos nπct/L + Bn sin nπct/L) sin nπx/L, n=1

where from initial conditions RL R nπc 2 L nπx An = L2 0 [f (x) − uE (x)] sin nπx L dx and Bn L = L 0 g(x) sin L dx. For large t, v(x, t) oscillates so that u(x, t) uE (x).

8.2.6 (d) An equilibrium satisfies c2 d2 uE /dx2 = − sin πx/L with uE (0) = uE (L) = 0. Thus uE (x) = (L/πc)2 sin πx/L. By introducing the displacement, defined by (8.2.9), we have the same result as exercise 8.2.6(a).

29

Section 8.3 8.3.1 (c) A reference temperature distribution r(x, t) must satisfy r(0, t) = A(t) and rx (L, t) = 0. The simplest example is r(x, t) = A(t). We introduce the difference [v(x, t) = u(x, t) − r(x, t)], so that dA ∂v ∂2v − k 2 = Q(x, t) − , ∂t ∂x dt since ∂r/∂t = dA/dt and ∂ 2 r/ ∂x2 = 0. The boundary conditions are homogeneous v(0, t) = 0 and vx (L, t) = 0. The initial condition is v(x, 0) = f (x) − A(0). The related homogeneous eigenfunctions are sin(n − 21 )πx/L, n = 1, 2, 3... since d2 φ/dx2 + λφ = 0 subject to φ(0) = 0 and dφ/dx(L) = 0.

According to the method of eigenfunction expansion v(x, t) =

∞ X

1 Bn (t) sin(n − )πx/L. 2 n=1

This can be differentiated term-by-term since both v(x, t) and the eigenfunctions satisfy the same set of homogeneous boundary conditions. Thus ¸ ∞ · X (n − 12 )2 π 2 dBn dA 1 πx +k Bn sin(n − ) = Q(x, t) − .

2 dt L 2 L dt n=1 By orthogonality dBn 2 + kλn Bn = q̄n (t) = dt L

Z L· Q(x, t) − 0

¸ 1 πx dA sin(n − ) dx. dt 2 L

The solution of this is given by (8.3.10).

8.3.1 (f) The related homogeneous eigenfunctions yield a Fourier cosine series. Thus, by the method of eigenfunction expansion ∞ X nπx u(x, t) = An (t) cos .

L n=0 This can be differentiated term-by-term since both u(x, t) and cos nπx/L satisfy the same set of homogeneous boundary conditions. Thus An (t) satisfies (8.3.9) whose solution is given by (8.3.10).

8.3.3 The related homogeneous eigenfunctions satisfy L(φ) = λσφ = 0 (where σ = cρ) and φ(0) = φ(L) = 0.

According to the method of eigenfunction expansions u(x, t) =

∞ X

an (t)φn (x).

n=1

This can be differentiated term-by-term since both u(x, t) and φn (x) satisfy the same set of homogeneous boundary conditions. Thus cρ

∞ X dan n=1

dt

φn (x) =

∞ X

an (t)L[φn (x)] + f (x, t),

n=1

but L(φn ) = −λn cρφn . Thus ¸ dan cρ φn (x) + λn an = f (x, t). dt n=1 ∞ X

·

Since the eigenfunctions are orthogonal with weight σ = cρ ,Z Z L L dan + λn an = fn (t) = f (x, t)φn (x)dx φ2n c ρ dx. dt 0 0 30

Now the integrating factor may be used to derive an expression similar to (8.3.10). The initial conditions yield ,Z Z L L an (0) = g(x)φn (x)cρdx φ2n cρdx.

0

0

8.3.4 (a) An equilibrium solution satisfies £ ¤ duE d = 0 subject to uE (0) = A and uE (L) = B. dx K0 (x) dx

Rx By integration K0 (x)duE /dx = c1 . By integrating again uE (x) = c2 + c1 0 Kdx̄ .

0 (x̄) RL The boundary conditions yield c2 = A and B − A = c1 0 dx̄/K0 (x̄).

√ 8.3.5 Since there √ is no θ-dependence, the2 corresponding homogeneous eigenfunctions are J0 ( λn r) such that J0 ( λn a) = 0. We note that ∇ φ = −λφ. According to the method of eigenfunction expansion u(r, t) =

∞ X

p An (t)J0 ( λn r).

n=1

√ This can be differentiated term-by-term since both u(x, t) and J0 ( λn r) satisfy the same set of homogeneous boundary conditions. Thus ∞ · X dAn n=1

dt

¸ + kλn An

p J0 ( λn r) = f (r, t).

Since these Bessel functions are orthogonal with weight r Ra √ f (r, t)J0 ( λn r)rdr dAn 0 R . + kλn An = fn (t) = a 2 √ dt J ( λn r)rdr 0 0 This is in the form of (8.3.9) whose solution is given by (8.3.10) involving the initial conditons, An (0) = 0.

8.3.7 A reference temperature distribution r(x, t) must satisfy r(0, t) = 0 and r(L, t) = t. The simplest example is r(x, t) = x t/L. The difference [v(x, t) = u(x, t) − r(x, t)] satisfies ∂v ∂2v x − =− . ∂t ∂x2 L subject to homogeneous boundary conditions v(0, t) = v(L, t) = 0 and the initial condition v(x, 0) = 0.

This problem has an equilibrium solution vE (x) = x3 /6L+c1 x where 0 = L2 /6+c1 L. This is equivalent to starting with the reference temperature distribution r1 (x, t) = xt/L + x3 /6L − Lx/6 .

2

∂ v1 1 In this case [v1 (x, t) = u(x, t) − r1 (x, t)], the new difference satisfies ∂v ∂t − ∂x2 = 0 with v1 (0, t) = 3 v1 (L, t) = 0 and the initial condition v1 (x, 0) = Lx/6 − x /6L. The solution is

v1 (x, t) =

∞ X

an sin

n=1

where an = L2

RL 0

(Lx/6 − x3 /6L) sin nπx L dx .

31

nπx −(nπ/L)2 t e , L

Section 8.4 8.4.1. (b) The related homogeneous eigenfunctions are cosines. Thus u(x, t) =

∞ X

An (t) cos nπx/L.

n=0

From the pde

Z L

0

An (t) = qn + k In

uxx cos nπx/L dx 0

where I0 = 1/L, (n 6= 0)In = 2/L, and Z L qn = In

Q(x, t) cos nπx/L dx .

0

Using Green’s formula (8.4.11) with v = cos nπx/L, 0

An + k(nπ/L)2 An = qn + kIn [(−1)n B − A].

Using the integrating factor An (t) = An (0)e−λn kt + e−λn kt

Z t

{qn + kIn [(−1)n B − A]} eλn kτ dτ

0

where An (0) = In

RL 0

f (x) cos nπx/L dx .

Section 8.5 8.5.2 (b) The related eigenfunctions are sines. Thus u(x, t) =

∞ X

An (t) sin nπx/L .

n=1

By term-by-term differentiation

00

An + c2 λn An = qn cos ωt,

RL where λn = (nπ/L)2 and qn = L2 0 g(x) sin nπx/L dx. If ω 2 =6= c2 λn = (nπc/L)2 , then (8.5.29) may be used. From the initial conditions, c2 = 0 and Z 2 L 2 2 c1 + [qn /(c λn − ω )] = f (x) sin nπx/L dx .

L 0 If ω 2 = (nπc/L)2 , resonance, then only for that one value of n, (8.5.31) is valid. From the initial conditions, c2 = 0 and Z 2 L c1 = f (x) sin nπx/L dx .

L 0 √ √ 8.5.5 (c) The eigenfunctions for ∇2 φ = −λφ on a semi-circle are Jm ( λmn r) sin mθ, where Jm ( λmn a) = 0, by modifying Section 7.7 with the boundary conditions g(0) = g(π) = 0. Thus, the method of eigenfunction expansions yields u(r, θ, t) =

∞ X ∞ X

p Amn (t) Jm ( λmn r) sin mθ .

m=1 n=1

By substituting into the pde

00

Amn + c2 λmn Amn = Qmn , 32

where

Z Z Qmn =

Using (8.5.19)

ÁZ Z p p 2 Q Jm ( λmn r) sin mθ rd rd θ Jm ( λmn r) sin2 mθ rd rd θ .

√ Z t p sin c λmn (t − τ ) √ Amn = cmn cos c λmn t + Qmn (τ ) dτ , c λmn 0

√ where the sin c λmn t vanishes because of the initial condition and where ÁZ Z Z Z p p 2 cmn = f Jm ( λmn r) sin mθ rd rd θ Jm ( λmn r) sin2 mθ rd rd θ .

8.5.6 (a) Since weight 1, dA = rd rd θ, a0 (0) = 0 and ±R R with R Rφ are orthogonal 2 a(0) = Hφ rdrdθ φ rd rd θ . From the pde, ÁZ Z Z Z 1 00 a + λa = g φ rd rdθ φ2 rd rdθ . c2

Section 8.6 8.6.1 (b) Using two-dimensional eigenfunctions ∞ X ∞ X

u(x, y) =

Anm sin nπx/L sin mπy/H.

n=1 m=1

Using (8.6.17), we need the normal derivative ∇φ · n̂ = ∂φ/∂x |x=L = Thus

I

nπ nπx mπy nπ cos sin | = (−1)n sin mπy/H. L L H x=L L

nπ u∇φ · n̂ d s = (−1)n L

so that

Z H sin mπy/Hdy = 0

nH (−1)n [1 − (−1)m ], mL

½

¾ 4n n m (−1) [1 − (−1) ] / λnm mL2 RR mπ 2 2 where λnm = ( nπ Q sin nπx/L sin mπy/H. L ) + ( H ) , and where Qnm = Anm =

−Qnm −

8.6.1 (d) Using two-dimensional eigenfunctions u(x, y) =

∞ X ∞ X

Anm cos nπx/L cos mπy/H .

n=0 m=0

By term-by-term differentiation (valid since u and the eigenfunctions satisfy the same set of homogeneous boundary conditions): RR Q cos nπx/L cos mπy/H dx dy 2 2 −Anm [(nπ/L) + (mπ/H) ] = R R . cos2 nπx/L cos2 mπy/H dx dy RR RR Thus if Q dx dy = / 0, there is no solution. However, if Q dx dy = 0, then A00 is arbitrary and the others are determined above. ½ ¾ √ cos mθ 8.6.3 (a) The two-dimensional eigenfunctions for ∇2 φ = −λφ are the set Jm ( λmn r) , where sin mθ √ Jm ( λmn a) = 0. By the method of eigenfunction expansion u(r, θ) =

∞ X ∞ X

∞ X ∞ X √ √ Amn cos mθ Jm ( λr) + Bmn sin mθ Jm ( λr).

m=0 n=1

m=1 n=1

33

From (8.6.17) µ

Amn Bmn

µ ¶ √ cos mθ Q Jm ( λr)r dr dθ sin mθ ¶ µ . √ RR cos2 mθ 2 Jm ( λr)rd rd θ sin2 mθ

− λ1

¶ =

RR

8.6.6 Using one-dimensional eigenfunctions u(x, y) =

∞ X

an (y) sin n x .

n=1

Substituting into the pde yields (8.6.5) ∞ · 2 X d an n=1

dy

¸ 2 2y − n a n sin nx = e sin x .

2

Thus for n 6= 1, an (y) = αn sinh ny + βn cosh n y. However, for n = 1, d 2 a1 dy 2

− a1 = e2y and thus a1 (y) = 13 e2y + α1 sinh y + β1 cosh y ,

using the method of undeterminedR coefficients for a particular solution. The other boundary conditions π are an (0) = 0 and an (L) = π2 0 f (x) sin nxdx. Thus 13 + β1 = 0 and (n 6= 1)βn = 0. Also 1 2L + α1 sinh L − 13 cosh L = a1 (L) and (n 6= 1)αn sinh n L = an (L), determining all αn .

3e

34

Chapter 9. Time-Independent Green’s Functions Section 9.2 9.2.1 (d) Using the method of eigenfunction expansion with a cosine series ∞ X

u(x, t) =

an (t) cos nπx/L .

n=0

The pde becomes ∂u/∂t =

∞ X

a0n (t) cos nπx/L = k ∂ 2 u/∂x2 + Q(x, t) .

n=0

.R L Thus a0n (t) = qn (t) + 0 k ∂∂xu2 cos nπx/Ldx 0 cos2 nπx/Ldx , .R RL L where qn (t) = 0 Q(x, t) cos nπx/L dx 0 cos2 nπx/L dx . Using Green’s formula (8.4.12) with v = cos nπx/L RL

Z L 0

2

∂2u cos nπx/L dx = − (nπ/L)2 ∂x2

Z L

u cos nπx/L dx + (−1)n B(t) − A(t).

0

Thus a0n (t) + k(nπ/L)2 an = qn (t) + k[(−1)n B(t) − A(t)] /In , where I0 = L and (n 6= 0)In = L/2.

Using the usual integrating factor [see (9.2.16)], ¸ Z t· (−1)n B(t0 ) − A(t0 ) −k(nπ/L)2 (t−t0 ) −k(nπ/L)2 t an (t) = an (0)e + qn (t0 ) + k e dt0 , In 0 where an (0) =

RL 0

g(x) cos nπx L dx /In . Therefore Z L u(x, t) =

g(x0 ) 0

+

Z tX ∞ 0 n=0

k

Q(x0 , t0 ) 0

I n=0 n

cos

nπx nπx0 −k(nπ/L)2 t cos e dx0 L L

nπx −k(nπ/L)2 (t−t0 ) (−1)n B(t0 ) − A(t0 ) cos e d t0 In L

Z LZ t +

∞ X 1

0

∞ X 1

I n=0 n

cos

nπx nπx0 −k(nπ/L)2 (t−t0 ) cos e d t 0 d x0 .

L L

We introduce the Green’s function, G(x, t; x0 , t0 ) = so that

∞ X 1

I n=0 n

cos

nπx nπx0 −k(nπ/L)2 (t−t0 ) cos e , L L

Z L u(x, t) =

Z LZ t g(x0 )G(x, t; x0 , 0)dx0 +

Q(x0 , t0 )G(x, t; x0 , t0 )dt0 dx0

0

0

Z t

+

Z t

kB(t0 )G(x, t; L, t0 )dt0 − 0

0

kA(t)G(x, t; 0, t0 )dt0 .

0

9.2.3 The method of eigenfunction expansion yields u =

∞ X n=1

an (t) sin nπx/L. This can be term-by-term

differentiated with respect to x since both u and sin nπx/L satisfy the same set of homogeneous

35

2

2 2 boundary conditions. Therefore, ddta2n = −c2 ( nπ L ) an + qn (t), where qn (t) = L Using (8.5.19)

an (t) = c1 cos where an (0) = c1 = L2

RL 0

nπct L nπct + c2 sin + L L nπc

Z t qn (t0 ) sin 0

dan nπc 2 f (x) sin nπx L dx and dt (0) = L c2 = L

RL 0

RL 0

Q(x, t) sin nπx L dx.

nπc (t − t0 )dt0 , L

g(x) sin nπx L dx.

By substituting these back into the series, we obtain Z L u(x, t) =

f (x0 ) 0

n=1

Z L g(x0 )

+ 0

Q(x0 , t0 ) 0

∞ X 2

0

∞ X 2 n=1

L

sin

L

n=1

Z LZ t +

∞ X 2

sin

L

sin

nπx0 nπx nπct sin cos dx0 L L L

nπx0 nπx sin nπct L sin dx0 nπc L L L

nπx0 nπx sin nπc L (t − t0 ) sin dt0 dx0 . nπc L L L

We introduce a Green’s function G(x, t; x0 , t0 ) =

∞ X 2 n=1

so that

L

sin

nπx0 nπx sin nπc L (t − t0 ) sin , nπc L L L

Z LZ t u(x, t) = Z L +

Q(x0 , t0 )G(x, t; x0 , t0 )dt0 dx0 0

0

Z L

g(x0 )G(x, t; x0 , 0) dx0 + 0

f (x0 ) 0

∂G (x, t; x0 , 0) dx0 . ∂t

We must have some ”faith” in these calculations since the series for ∂G/∂t does not converge. This result will be obtained later in a different way [see (11.2.24)].

Section 9.3

Rx 9.3.5 (a) du/dx = L f (x̄)dx̄ using the boundary condition at x = L. Integrating again gives ¢ R x ¡R x u(x) = 0 L 0 f (x̄)dx̄ dx0 using u(0) = 0. Integration-by-parts, dw = dx0 w = x0

Rx v = L 0 f (x̄)dx̄ dv = f (x0 )dx0 ,

Z x

Z x

yields u(x) = x

f (x̄)dx̄ − L

x0 f (x0 )dx0 , 0

which is equivalent to answer given.

9.3.5 (b) Let u1 be a homogeneous solution satisfying u(0) = 0, i.e., u1 = x, while let u2 be an independent dv2 1 one satisfying du/dx(L) = 0, i.e. u2 = 1. Since p = 1 from (9.3.10-12) dv dx = f and dx = −xf . Thus from (9.3.9) Z Z x

u=x

x

f (x0 )dx0 − 0

x0 f (x0 )dx0 + c1 x + c2 .

0

RL The boundary condition u(0) = 0 yields c2 = 0, while du/dx(L) = 0 yields c1 = − 0 f (x0 )dx0 , which is equivalent to answer given.

36

9.3.5 (c) In order for u(x) =

RL 0

f (x0 )G(x, x0 )dx0 , ½ G(x, x0 ) =

−x0 −x

x0 < x x0 > x.

9.3.6 (a) If x < x0 , G(x, x0 ) = c1 x + c2 , where c2 = 0 from G(0, x0 ) = 0. If x > x0 , G(x, x0 ) = c3 x + c4 , where c3 = 0 from dG/dx(L, x0 ) = 0. From continuity at x = x0 : c1 x0 = c4 . By integrating from x0 − ¯x + dG ¯¯ 0 to x0 +, = 1 or 0 − c1 = 1 . Thus c1 = −1 and c4 = −x0 , yielding dx ¯x0 − ½ −x x < x0 G(x, x0 ) = −x0 x > x0 .

9.3.6 (b) See answer on p. 743.

9.3.9 (a) Let u1 be a homogeneous solution satisfying u(0) = 0, i.e., u1 = sin x, while u2 satisfies u(L) = 0, i.e. u2 = sin(x − L). Note these are independent if L 6= nπ. We also note that u1 du2 /dx − u2 du1 /dx = sin L 6= 0. From (9.3.10-12) dv1 /dx = −f sin(x − L)/ sin L and dv2 /dx = f sin x/ sin L. Thus from (9.3.9) Z Z − sin x x sin(x − L) x u= f (x0 ) sin(x0 − L)dx0 + f (x0 ) sin x0 dx0 + c1 sin x + c2 sin(x − L) . sin L L sin L 0 The boundary condition u(0) = 0 yields c2 = 0, while u(L) = 0 yields c1 = 0.

9.3.9 (b) In order to put the result of part (a) into the form (9.3.15), ½ sin x sin(x0 − L)/ sin L x < x0 G(x, x0 ) = sin x0 sin(x − L)/ sin L x > x0 .

9.3.11 (a) For x 6= x0 , d2 G/dx2 +G = 0. If x < x0 , G(x, x0 ) = b sin x using G(0, x0 ) = 0. If x > x0 , G(x, x0 ) = a sin(x − L) using G(L, x0 ) = 0. To be continuous at x = x0 : b sin x0 = a sin(x0 − L).

(1)

Integrating the defining differential equation from x0 − to x0 + yields ¯x + Z x 0 + dG ¯¯ 0 + Gdx = 1. dx ¯x0 − x0 − This integral vanishes since G is continous, and thus a cos(x0 − L) − b sin x0 = 1.

(2)

Equations (1) - (2) may be solved by elimination only if L 6= nπ, yielding a = sin x0 / sin L and b = sin(x0 − L)/ sin L, using a trigonometric addition formula. Thus ½ sin x sin(x0 − L)/ sin L x < x0 G(x, x0 ) = sin x0 sin(x − L)/ sin L x > x0 .

Alternate method: The constants a and b may be redefined so that G(x, x0 ) is automatically continuous at x = x0 : ½ c sin(x0 − L) sin x x < x0 G(x, x0 ) = c sin x0 sin(x − L) x > x0 , where c is a constant independent of x0 . The jump condition on the derivative (9.3.45) determines c, yielding the same answer.

37

9.3.13 (b) For x 6= x0 , d2 G/dx2 + k 2 G = 0. A particularly convenient choice of homogeneous solutions is ½ c1 eik(x−x0 ) + c2 e−ik(x−x0 ) x < x0 G(x, x0 ) = c3 eik(x−x0 ) + c4 e−ik(x−x0 ) x > x0 .

The corresponding solution of φ is obtained by multiplying u and thus G(x, x0 ) by e−iωt . For example ei(kx−ωt) is a right-going wave (assuming ω/k > 0). For x < x0 , we want a left-going wave in order to be outward propagating and thus c1 = 0. For x > x0 , it should be right-going and thus c4 = 0: ½ c2 e−ik(x−x0 ) x < x0 G(x, x0 ) = c3 eik(x−x0 ) x > x0 .

To be continuous at x = x0 , c2 = c3 . The jump condition is obtained by integrating th defining differential equation from x0 − to x0 +: ¯x + Z x0 + dG ¯¯ 0 2 Gdx = 1 . + k dx ¯x0 − x0 − Since G is continuous

¯x + dG ¯¯ 0 =1 dx ¯x0 −

or

1 c3 ik + c2 ik = 1, yielding c2 = c3 = 2ik . A particularly convenient form for the Green’s function is

G(x, x0 ) =

1 ik|x−x0 | e .

2ik

9.3.14 (d) Using Green’s formula with v = G(x, x0 ): ¯L dG du ¯¯ [uL(G) − GL(u)]dx = p(u − G )¯ . dx dx ¯ 0

Z L

0

We note that L(u) = f and L(G) = δ(x − x0 ). Thus Z L u(x0 ) = 0

¯L dG du ¯¯ f (x)G(x, x0 )dx + p(u − G )¯ . dx dx ¯ 0

The Green’s function satisfies the related homogeneous boundary conditions, G(0, x0 ) = 0 and dG dx (L, x0 ) + h G(L, x0 ) = 0, while u satisfies the given conditions. Thus Z L u(x0 ) =

f (x)G(x, x0 )dx + p(L)G(L, x0 )(−hu(L) + hu(L) − β) − p(0)α 0

dG (0, x0 ) . dx

Switching x and x0 and using symmetry yields the given answer.

9.3.15 (a) For x 6= x0 , L(G) = 0. This problem defines homogeneuous solutions needed to solve the boundary conditions for the Green’s function. Thus ½ c1 y1 (x) x < x0 G(x, x0 ) = c2 y2 (x) x > x0 .

Continuity at x = x0 yields c1 y1 (x0 ) = c2 y2 (x0 ) .

(3)

The jump condition for the derivative is obtained by integrating the defining differential equation, yielding 38

since

R x0 + x0 −

¯x + dG ¯¯ 0 p =1 dx ¯x0 − q G dx vanishes because G is continuous. In this manner c1

dy1 dy2 (x0 ) − c2 (x0 ) = −1/p(x0 ) . dx dx

(4)

Solving (3) - (4) simultaneously yields ½ G(x, x0 ) =

y1 (x)y2 (x0 )/c x < x0 y2 (x)y1 (x0 )/c x > x0 ,

where c is the constant defined on page 387 and is related to the Wronskian w = y1

dy1 dy2 − y2 = c/p(x) . dx dx

The constant c is determined from the calculations of y1 and y2 at either end: c/p(0) = −y2 (0) or c/p(L) = y1 (L) .

9.3.21 If x 6= x0 , then dG/dx = 0. If x < x0 , G(x, x0 ) = 0 in order to satisfy G(0, x0 ) = 0. If x > x0 , the general solution is G(x, x0 ) = c. The Green’s function is not continuous, since its derivative is a Dirac delta function. The jump condition is obtained by integrating the defining differential equation ¯ G ¯xx00 + − =1 .

Therefore c − 0 = 1. We conclude that ½ G(x, x0 ) =

0 1

x < x0 x > x0 ,

which is not symmetric.

0

00

9.3.25 (b) The simplest particular solution corresponds to the initial condition u(0) = u (0) = u (0) = 000 u (0) = 0. Taking the Laplace transform of the differential equation yields s4 ū(s) = F (s) and thus ū(s) = F (s)/s4 .

Using the convolution theorem,

Z x u(x) =

f (x̄)g(x − x̄)dx̄ , 0

where g(x) is a function whose Laplace transform is 1/s4 . From the table inside the front cover, g(x) = x3 /3!. Thus Z x

u(x) =

f (x̄)(x − x̄)3 /3!dx̄.

0

Section 9.4 9.4.2 (a) Using Green’s formula Z L 0

µ ¶¯L dφh du ¯¯ [uL(φh ) − φh L(u)] dx = p u − φh ¯ . dx dx ¯ 0

Since L(u) = f, u(0) = α, u(L) = β and L(φh ) = 0, φh (0) = 0, φh (L) = 0, it follows that Z L dφh (0) dφh (L) − p(0)α .

0= f φh dx + p(L)β dx dx 0 39

9.4.3 (b) Homogeneous solutions satisfy d2 φ/dx2 + φ = 0 with dφ/dx(0) = dφ/dx(π) = 0. Thus φh = cos x is a nontrivial homogeneous solution. According to the Fredholm alternative, solutions to the nonhomogeneous problem (subject to homogeneous boundary conditions) exist only if the right hand side is orthogonal to φ(x). Since Z π

cos x sin x dx = 0, 0

there are an infinite number of solutions.

9.4.6 (a) The general solution of the differential equation is u = 1 + c1 cos x + c2 sin x .

The boundary condition u(0) = 0 implies 0 = 1 + c1 , while u(π) = 0 yields 0 = 1 − c1 . These are inconsistent so that no solutions exist. To apply the Fredholm alternative, we note that φh = sin x is a homogeneous solution since φ(0) = φ(π) = 0. Furthermore, Z π 1 · sin x dx = 2 6= 0 .

0

Thus, there are no solutions since the right hand side f (x) = 1 is not orthogonal to φh = sin x.

9.4.6 (b) The general solution of the differential equation is u = 1 + c1 cos x + c2 sin x .

The boundary condition du/dx(0) = 0 implies 0 = c2 , while du/dx(π) = 0 yields 0 = −c2 . This is possible with c2 = 0 and c1 arbitrary: u = 1 + c1 cos x .

There are an infinite number of solutions. To apply the Fredholm alternative, we note that φh = cos x 0 0 is a homogeneous solution since φ (0) = φ (π) = 0. Since Z π 1 · cos x dx = 0 , 0

the right-hand side f (x) = 1 is orthogonal to all homogeneous solutions φh = cos x. Thus there should be an infinite number of solutions.

9.4.6 (c) The general solution of the differential equation is u = 1 + c1 cos x + c2 sin x .

The boundary condition u(−π) = u(π) implies 1 − c1 = 1 − c1 , while u0 (−π) = u0 (π) yields −c2 = −c2 .

Thus the above soltuion for u is valid with both c1 and c2 arbitrary. There are an infinite number of solutions. To apply the Fredholm alternative, we note that both φh = sin x and φh = cos x are homogeneous solutions satisfying the homogeneous boundary conditions. Here Z π Z π 1 · cos x dx = 0 and 1 · sin x dx = 0 . −π

−π

Thus, there are an infinite number of solutions.

9.4.8 (a) To obtain a particular solution, we substitute up = A x sin x into the differential equation in order 00 to determine the constant A. Here u0 p = A(x cos x + sin x) and up = A(−x sin x + 2 cos x).

Therefore A(−x sin x + 2 cos x + x sin x) = cos x or A = 12 . Thus the general solution is u = x sin x/2 + c1 cos x + c2 sin x .

40

The boundary condition u(0) = 0 yields 0 = c1 , while u(π) = 0 also yields 0 = c1 . Thus u=

1 x sin x + c2 sin x , 2

with c2 arbitrary, which is an infinite number of solutions. To apply the Fredholm alternative, we note that φh = sin x is a homogeneous solution satisfying the boundary conditions. Since Z π cos x sin x dx = 0 , 0

there are an infinite number of solutions.

9.4.10 Since φh = sin x is a solution of both the corresponding homogeneous differential equation and boundary condition, a modified Green’s function Gm (x, x0 ) must be introduced satisfying d2 Gm + Gm = δ(x − x0 ) + c sin x , dx2 subject to Gm (0, x0 ) = Gm (π, x0 ) = 0. The constant c is chosen so that the right-hand side is orthogonal to φh = sin x: Z π 2 0= sin x[δ(x − x0 ) + c sin x]dx or c = − sin x0 . π 0 If x 6= x0 , G00m +Gm = c sin x. The method of undetermined coefficients can be used to find a particular solution, Gm = − 2c x cos x. Thus ½ 1 c1 cos x + c2 sin x x < x0 Gm (x, x0 ) = sin x0 x cos x + c3 cos x + c4 sin x x > x0 . π The boundary conditions yield 0 = c1

and

0 = − sin x0 − c3 ,

while continuity at x = x0 yields c1 cos x0 + c2 sin x0 = c3 cos x0 + c4 sin x0 , which simplifies to c2 = − cos x0 + c4 . The jump condition x = x0 is ¯x + dGm ¯¯ 0 =1 dx ¯x0 − since Gm (x, x0 ) is continuous. The jump condition will be satisfied since c was picked so that Gm (x, x0 ) exists. Thus there are an infinite number of modified Green’s functions ½ 1 (c4 − cos x0 ) sin x x < x0 Gm (x, x0 ) = sin x0 x cos x + c4 sin x − sin x0 cos x x > x0 , π where c4 is an arbitrary function of x0 , c4 (x0 ). We chose Gm (x, x0 ) to be symmetric, as is guaranteed

by Exercise 9.4.9, so that (9.4.17) may be derived. For example, for x < x0 Gm (x, x0 ) = Gm (x0 , x) yields 1 sin x0 x cos x + [c4 (x0 ) − cos x0 ] sin x π 1 = sin x x0 cos x0 + c4 (x) sin x0 − sin x cos x0 . π As a function of x, c4 (x) is some combination of sin x and x cos x: c4 (x) = a sin x + b x cos x cos x : 41

where a and b are constants, independent of x and x0 . Equating the coefficients of sin x and x cos x yields sin x : x cos x :

a sin x0 + b x0 cos x0 − cos x0 = π1 x0 cos x0 + a sin x0 − cos x0 1 π sin x0 = b sin x0 .

Thus b = 1/π and a is arbitrary, yielding an infinite number of symmetric modified Green’s functions:  1 (x cos x sin x0 + x0 cos x0 sin x)    π − cos x0 sin x x < x0 Gm (x, x0 ) = a sin x sin x0 + 1 (x cos x sin x + x cos x sin x )  0 0 0   π − cos x sin x0 x > x0 .

To obtain a formula for u(x), we use Green’s formula ¯π Z π dGm du ¯¯ . [uL(Gm ) − Gm L(u)] dx = u − Gm dx dx ¯0 0 Using the differential equations for u(x) and Gm (x, x0 ), we obtain Z π u(x0 ) = 0

2 f (x)Gm (x, x0 )dx + sin x0 π

Z π u(x) sin x dx + β 0

dGm dGm (π) − α (0) . dx dx

We can use any modified Green’s function. With a = 0 dGm 1 (π) = − (x0 cos x0 + sin x0 ) dx π 1 dGm (0) = (sin x0 + x0 cos x0 ) − cos x0 . dx π Switching x and x0 , using the symmetry of Gm (x, x0 ), we obtain Z π β 1 u(x) = f (x0 )Gm (x, x0 )dx0 − (x cos x + sin x) − α[ (sin x + x cos x) − cos x] + k sin x, π π 0 Rπ where the constant k = π2 0 u(x0 ) sin x0 dx0 . This constant is arbitrary, representing an arbitrary mulRπ tiple of the homogeneous solution. This solution is consistent (based on calculating 0 u(x0 ) sin x0 dx0 ) is the original problem actually has a solution. From exercise 9.4.2 a solution only exists if Z π 0= f (x) sin x dx − α − β , 0

since φh = sin x .

9.4.11 (a) By integrating the differential equation from 0 to L, ¯L dGa ¯¯ = 1. dx ¯0 Applying the boundary conditions yields c = 1.

9.4.11 (b) The modified Green’s function represents a steady-state heat flow problem with heat source −δ(x − x0 ) insulated at x = 0 with heat flow to the right (out) at x = L of −c. The total thermal RL energy generated inside per unit time is 0 −δ(x − x0 )dx = −1. For equilibrium, the heat generated inside must equal the heat flowing out. Thus −1 = −c .

42

9.4.11 (c) From the differential equation ½ Ga (x, x0 ) =

c1 x + c2 c3 x + c4

x < x0 x > x0 .

dGa /dx(0) = 0 implies c1 = 0, while dGa /dx(L) = c = 1 implies c3 = c = 1 .

To be continous at x = x0 , c1 x0 + c2 = c3 x0 + c4 or c2 = x0 + c4 . Thus ½ x0 + c4 (x0 ) x < x0 Ga (x, x0 ) = x + c4 (x0 ) x > x0 , which automatically satisfies the jump condition dGa /dx |x0 + − dGa /dx |x0 − = 1 . In general, this different modified Green’s function is not symmetric.

9.4.11 (d) To be symmetric, for example Ga (x, x0 ) = Ga (x0 , x) for x < x0 : x0 + c4 (x0 ) = x0 + c4 (x) .

This is only valid if c4 (x0 ) is an arbitrary constant α, independent of x0 : ½ x0 x < x0 Ga (x, x0 ) = α + x x > x0 .

There is an infinite number of symmetric modified Green’s funciton of this type because if there is one, any multiple of the homogeneous solution φh (x0 ) can be added.

9.4.11 (e) Using Green’s formula with v = Ga (x, x0 ), ¯L ¶ Z Lµ 2 d Ga d2 u dGa du ¯¯ u − Ga (x, x0 ) 2 dx = u − Ga (x, x0 ) ¯ . dx2 dx dx dx ¯ 0 0

Using the differential equations for u and Ga (x, x0 ) and the fundamental property of the Dirac delta function yields ¯ ¯ Z L ¯ ¯ dGa dGa ¯ ¯ u(x0 ) = (x, x0 )¯ (x, x0 )¯ f (x)Ga (x, x0 )dx + u(L) − u(0) . ¯ ¯ dx dx 0 x=L

¯ a ¯ From part (d), dG dx (x, x0 ) x=L = 1

x=0

¯

dGa ¯ dx (x, x0 ) x=0 = 0, and thus

and

Z L u(x0 ) =

f (x)Ga (x, x0 )dx + u(L) .

0

Switching x and x0 using the symmetry of Ga (x, x0 ) yields Z L u(x) =

f (x0 )Ga (x, x0 )dx0 + u(L) .

0

To show this is consistent, we substitute x = L, in which case we obtain Z L f (x0 )dx0 = 0 0

since Ga (L, x0 ) is independent of x0 . This is the condition for the solution to exist from the Fredholm alternative since φh (x) = 1. Thus u(L) is arbitrary above (and may be replaced by an arbitrary constant), representing an arbitrary multiple of the homogeneuous solution φh (x) = 1.

43

Section 9.5 9.5.3 (c) On the semi-circle √ with G = 0 on the boundary, the eigenfunctions of ∇2 φ = −λφ are the √ family φmn = sin mθJm ( λmn r) where λmn is determined from Jm ( λmn a) = 0. By the method of eigenfunction expansion, G(r, θ; r0 , θ0 ) =

∞ X ∞ X

Amn φmn (r, θ) .

(5)

n=1 m=1

Since G and φmn satisfy the same set of homogeneous boundary conditions, 2

∇ G=

∞ X ∞ X

2

Amn ∇ φmn =

n=1 m=1

∞ X ∞ X

Amn (−λmn φmn ) .

n=1 m=1

Using the orthogonality of φmn and the differential equation for G: RR φ ∇2 G dx dy φmn (r0 , θ0 ) R Rmn −λmn Amn = =RR 2 .

2 φmn dx dy φmn dx dy Substituting this into (5) yields G(r, θ; r0 , θ0 ) =

∞ X ∞ X φmn (r, θ)φmn (r0 , θ0 ) RR . −λmn φ2mn r̄dr̄dθ̄ n=1 m=1

→ − 9.5.4 Using Green’s formula with v = G(− x,→ x 0 ): ZZZ ZZ (u∇2 G − G∇2 u)dx dy dz = ° (u∇G − G∇u) · n̂ dS . → − Substituting ∇2 G = δ(− x −→ x 0 ) with G = 0 on the boundary yields ZZZ ZZ → − → − → u(− x 0) = G(→ x,− x 0 )f (→ x )dx dy dz + ° h(− x )∇ G · n̂ dS . − → By switching → x and − x 0 and using the symmetry of G, we obtain ZZZ ZZ → → − → → u(− x) = G(− x,→ x 0 )f (− x 0 )dx0 dy0 dz0 + ° h(− x 0 )∇− → x 0 G · n̂dS0 .

9.5.9 (b) The θ−dependent eigenfunctions are sin mθ, m = 1, 2, ... since G = 0 on θ = 0 and θ = π. Thus, the method of eigenfunction expansion yields G=

∞ X

Am (r) sin mθ .

m=1

By taking the Laplacian and using the differential equation for the Green’s function, we obtain → → ∇2 G = δ(− x −− x 0) = Since

ZZ

µ ¶ ∞ · X 1 d dAm m2 r − 2 Am ] sin mθ . r dr dr r m=1

→ − f (r, θ)δ(− x −→ x 0 )dx dy =

ZZ

it is helpful to note that

44

− → f (r, θ)δ(→ x −− x 0 )r dr dθ = f (r0 , θ0 ),

(6)

1 − → δ(→ x −− x 0 ) = δ(r − r0 )δ(θ − θ0 ) . r0 Thus from (6), using the orthogonality of sines, Z 2 π 1 d dAm m2 sin mθ 2 sin mθ0 (r ) − 2 Am = δ(r − r0 )δ(θ − θ0 ) dθ = δ(r − r0 ) . r dr dr r π 0 r0 πr0 This may be solved by the methods of Section 9.3 to obtain one-dimensional Green’s functions. For r 6= r0 , the equation is equidimensional, whose solutions are rp with p = ±m. For r < r0 , the solution must be bounded, and thus it is proportional to rm . For r > r0 , it satisfies the zero boundary condition at r = a. In this manner  i h  r < r0  c rm ( ra0 )m − ( ra0 )m Am (r) =  ¤ £  r > r0 , c r0m ( ar )m − ( ar )m where we have made this function automatically continuous at r = r0 . The constant c is chosen to ¯r + 2 m¯ 0 satisfy the jump-condition r dA dr r0 − = π sin mθ0 . Thus · ¸ r0 m−1 1 1 a m+1 1 r0 m 1 a m 2 + ( ) sin mθ0 , c ( ) − ( ) + ( ) mr0m = a a a r0 r0 a r0 r0 πr0 m θ0 so that c = a−m sinmπ , yielding the answer.

9.5.10 (a)

ZZZ

£ ¤ u(∇2 + k 2 )v − v(∇2 + k 2 )u dV =

ZZZ

ZZ (u∇ v − v∇ u)dV = ° (u∇v − v∇u) · n̂ d S .

2

2

→ → x −− x 0 | in three dimensions as in (9.5.28b). Then (ρ 6= 0)∇2 G + k 2 G = 0. Spherically 9.5.10 (b) Let ρ = |− symmetric solutions will satisfy µ ¶ 1 d 2 dG ρ + k2 G = 0 . ρ2 dρ dρ Introduce the change of variable, G = y/ρ. This is admittedly motivated by the answer. Since dG/dρ = ρ1 dy/dρ − y/ρ2 , it follows that d2 y + k2 y = 0 . dρ2 Thus y = c1 eikρ + c2 e−ikρ or G = (c1 eikρ + c2 e−ikρ )/ρ .

The time-dependent problem is obtained by multiplying by e−iωt . We note that ei(kρ−ωt) is outgoing if ω/k > 0, while e−i(kρ+ωt) is incoming. Thus, to be outgoing c2 = 0 G = c1 eikρ /ρ. → → To determine c1 we integrate the defining differential equation over a small sphere centered at − x =− x0 with radius ρ: ZZZ (∇2 G + k 2 G)d V = 1 .

ZZZ GdV = 0 and thus from the divergence theorem

It can be shown that lim

ρ→0

ZZ lim °∇G · n̂ d S = 1 .

ρ→0

ZZ ∂G ° d S = 1. However, the ρ→0 ∂ρ

On the surface of the sphere, ∇G · n̂ = ∂G/∂ρ is constant. Hence lim surface area is 4πρ2 . Thus lim 4πρ2 ρ→0

∂G = 1, which implies c1 = −1/4π. ∂ρ 45

− → 9.5.10 (c) Let r = |→ x −− x 0 | in two dimensions as in (9.5.28a). For r 6= 0, ∇2 G+k 2 G = 0. Circularly symmetric d 2 2 2 2 2 2 solutions will satisfy 1r dr (r dG dr ) + k G = 0, which is equivalent to r d G/dr + rdG/dr + k r G = 0.

This can be related to Bessel’s differential equation, so that G = c1 J0 (kr) + c2 Y0 (kr).

To investigate whether or not this solution is outgoing for large r, we use the asymptotic expansions for the Bessel functions (7.8.3). In this way, for large r G ∼ (2/πkr)1/2 [c1 cos(kr − π/4) + c2 sin(kr − π/4)] .

By using Euler’s formulas, we obtain an equivalent expression G ∼ (2πkr)−1/2 [(c1 − ic2 )ei(kr−π/4) + (c1 + ic2 )e−i(kr−π/4) ] .

The time-dependent problem is obtained by multiplying by e−iωt . We note that ei(kr−ωt) is outgoing if ω/k > 0, while e−i(kr+ωt) is incoming. Thus, to be outgoing c1 + ic2 = 0 .

(7)

To determine the remaining constant we must investigate the singularity at r = 0. If we integrate → → the defining differential equation over a circle of radius r centered at − x =− x 0 , we obtain using the divergence theorem I ZZ ∇G · n̂ds + k 2

GdA = 1 .

H Since G is only a function of r, ∇G · n̂ = dG/dr, and ds = 2πr, we have Z r dG 2πr + 2πk 2 Gr̄ d r̄ = 1 . dr 0 Since rG is integrable, we take the limit as r → 0 and obtain the simpler expression for the singularity condition dG lim 2πr =1. r→0 dr We use the asymptotic expansions for Bessel functions for small arguments, (7.7.33). In this manner G = c1 [1 − (kr/2)2 + ...] + c2 [2/π log(kr) + ...] and hence

dG = 2c2 /πr + ... dr Thus, c2 = 1/4 to satisfy the singularity condition. By also using (7), we obtain G=

1 [Y0 (kr) − iJ0 (kr)] .

4

9.5.13 (a) To satisfy the boundary condition that ∂G/∂y = 0 at y = 0, we introduce a positive image source at (x0 , −y0 , z0 ). The Green’s function for this half-space is the superposition of the two infinite space Green’s functions ¤−1/2 1 ³£ → → (x − x0 )2 + (y − y0 )2 + (z − z0 )2 G(− x,− x 0) = − 4π £ ¤−1/2 ´ . + (x − x))2 + (y + y0 )2 + (z − z0 )2 It can be verified that ∂G/∂y = 0 at y = 0.

46

9.5.13 (b) From the three-dimensional version of Green’s formula ZZZ ZZ → − → → u(− x 0) = f (→ x )G(− x,− x 0 )dV − ° G∇u · n̂ dS , → → since ∂G/∂y = 0 on y = 0. We note that n̂ = −ĵ. Thus, after switching − x and − x 0 and using the symmetry of G ¯ ¯ ZZZ ZZ ¯ ¯ ∂u ¯ − → − → → − − → → − − → u( x ) = f ( x 0 )G( x , x 0 )dV0 + G( x , x 0 )¯¯ dx0 dz0 , ¯ ¯ ∂y 0 y0 =0 y=0

where ∂u/∂y0 |y0 =0 = h(x0 , z0 ) and − → G(→ x,− x 0 )|y0 =0 = −

1 [(x − x0 )2 + y 2 + (z − z0 )2 ]−1/2 .

2π

9.5.14 To satisfy the boundary conditions that G = 0 at x = 0, we use the method of images. We introduce three image sources, negative ones at (−x0 , y0 ) and (x0 , −y0 ) and a positive one at (−x0 , −y0 ). The Green’s function in the first quadrant is the superposition of the four infinite-space two-dimensional Green’s functions each satisfying (9.5.31a). By using properties of logarithms, we obtain ¾ ½ 1 [(x − x0 )2 + (y − y0 )2 ][(x + x0 )2 + (y + y0 )2 ] − → → − G( x , x 0 ) = . ln 4π [(x − x0 )2 + (y + y0 )2 ][(x + x0 )2 + (y − y0 )2 ] It is easy to verify that G = 0 along x = 0 and along y = 0.

9.5.19 The Green’s function inside a circle of radius a is given by (9.5.57), where φ is the angle between − → → → → x and − x 0 . We use polar coordinates with − x being represented by (r, θ) and − x 0 by (r0 , θ0 ). Thus φ = θ − θ0 . For a semi-circle with G = 0 on the boundaries, image sources may be introduced. There are negative ones located at (r0 , −θ0 ) and the corresponding point outside of the circle. In this case φ = θ + θ0 . Thus ¸ · 1 r 2 + r02 − 2rr0 cos(θ − θ0 ) 2 − → − → G( x , x 0 ) = ln a 2 2 4π r r0 + a 4 − 2rr0 a 2 cos(θ − θ0 ) ¸ · 1 r 2 + r02 − 2rr0 cos(θ + θ0 ) 2 − ln a 2 2 .

4π r r0 + a 4 − 2rr0 a 2 cos(θ + θ0 ) At y = 0, θ = 0 or θ = π in which case the above formula yields G = 0 as desired. → − 9.5.22 (c) In three dimensions the response to a Dirac delta function is G = −1/4πρ, where ρ = |− x −→ x 0| − → is the distance from the source and where x 0 = (x0 , y0 , z0 ). To satisfy G = 0 at x = 0 and at x = L, image sources must be introduced. To satisfy G = 0 at x = 0, a negative source must be introduced at (−x0 , y0 , z0 ). Note: to satisfy G = 0 at x = L, a negative image source at (−x0 + 2L, y0 , z0 ) and a positive image source at (x0 + 2L, y0 , z0 ) exists. This process must continue indefinitely. There will be positive sources at x0 + 2Ln and negative sources at −x0 + 2Ln, where n is an arbitrary integer (positive or negative). Thus à ! ∞ 1 X 1 1 − → → − G( x , x 0 ) = − − , − → → → → 4π |− x −− α n | |− x − β | n=−∞

− → → where − α n = (x0 + 2Ln, y0 , z0 ) and β n = (−x0 + 2Ln, y0 , z0 ) .

47

n

Chapter 10. Fourier Transform Section 10.2

R∞ R0 2 2 10.2.1 u = 0 c(ω)e−iωx e−kω t dω + −∞ c(ω)e−iωx e−kω t dω. In the latter integral, let ω 0 = −ω. Then R∞ R 2 0 02 ∞ u = 0 c(ω)e−iωx e−kω t dω + 0 c(−ω 0 )eiω x e−kω t dω 0 . Since ω 0 is a 00 dummy00 variable, it may be called ω. Now Euler’s formula yields Z ∞ Z ∞ 2 −kω 2 t u= [c(ω) + c(−ω)] cos ωxe dω + i [−c(ω) + c(−ω)] sin ωxe−kω t dω.

0

0

Comparing this with (10.2.9) yields (for ω > 0) A(ω) = c(ω) + c(−ω) and B(ω) = i[−c(ω) + c(−ω)] .

By adding and subtracting these formulas (for ω > 0), c(ω) = 21 [A(ω) + iB(ω)] and c(−ω) = 12 [A(ω) − iB(ω)].

If A and B are real, then c̄(ω) = 21 [A(ω) − iB(ω)] = c(−ω).

Section 10.3 10.3.6 The Fourier transform is obtained from (10.3.6) with γ = 1: F (ω) =

1 2π

Z ∞

f (x)eiωx dx =

−∞

1 2π

Z a

eiωx dx =

−a

¯a eiωx ¯¯ .

2πiω ¯−a

1 1 Thus, F (ω) = 2πiω (eiωa − e−iωa ) = πω sin ωa .

10.3.7 The inverse Fourier transform is obtained from (10.3.7) with γ = 1: Z ∞ Z ∞ f (x) = F (ω)e−iωx dω = e−|ω|α e−iωx dω −∞

Z ∞ =

−∞

e−ωα e−iωx dω +

Z 0

0

Thus

eωα e−iωx dω .

−∞

¯0 ¯∞ eω(−α−ix) ¯¯ eω(α−ix) ¯¯ f (x) = + ¯ −α − ix ¯0 α − ix ¯

−∞

=

1 2α 1 + = 2 . α + ix α − ix α + x2

10.3.10 (a) By separation of variable (u = h(t)φ(r)) 1 dh 1 d dφ = (r ) = −s2 . kh dt rφ dr dr 2

2 2 Thus r2 ddrφ2 + r dφ dr + s r φ = 0, which is Bessel’s differential equation of order 0. Thus

φ = c1 J0 (sr) + c2 Y0 (sr) .

Since u is bounded at r = 0, c2 = 0. By the superposition principle (see p. 447), Z ∞ 2 u(r, t) = B(s)J0 (sr)e−s kt ds .

0

It is convenient (due to the weight function for Bessel functions) to let B(s) = sA(s) .

48

10.3.10 (b) One or two-dimensional Green’s formula may be used. In one-dimension ¯b Z b dv du ¯¯ [uL(v) − vL(u)]dr = p(u − v )¯ , dr dr ¯ a a

d d where L = dr (r dr ), so that p = r and a = 0 and b = L. We note that

L[J0 (sr)] = −s2 rJ0 (sr)

and L[J0 (s1 r)] = −s21 rJ0 (s1 r) .

Thus, letting u = J0 (sr) and v = J0 (s1 r) in Green’s formula yields ¶¯ µ Z L du ¯¯ dv 2 2 −v (s − s1 ) J0 (sr)J0 (s1 r)r d r = L u ¯ dr dr ¯ 0

.

L

For large L, the right-hand side may be approximated using (7.8.3): r r 2 du 2 u = J0 (sr) ∼ cos(sr − π/4) and thus ∼ −s sin(sr − π/4) πsr dr πsr and a similar expression for v (with s replaced by s1 ). Thus r ·r ¸ Z L 2 s π π s1 π π 2 2 (s −s1 ) J0 (sr)J0 (s1 r)rdr ∼ L sin(sL − ) cos(s1 L − ) − sin(s1 L − ) cos(sL − ) , πL s1 4 4 s 4 4 0 RL so that 0 J0 (sr)J0 (s1 r)rdr q p s1 s π π π π s sin(s1 L − 4 ) cos(sL − 4 ) 2 s1 sin(sL − 4 ) cos(s1 L − 4 ) − ∼ . π s2 − s21 10.3.10 (c) In order to use the results of part (b), we multiply equation on bottom of page 456 by J0 (s1 r)r and integrate from 0 to L: Z L Z ∞ Z L f (r)J0 (s1 r)rdr = A(s) J0 (sr)J0 (s1 r)rdr sds .

0

0

0

In the above expressions, we note that A(s) = A(s1 ) + A(s) − A(s1 ) . The contribution to the integral from A(s) − A(s1 ) vanishes as L → ∞ by the Riemann-Lebesgue lemma (see first edition of this text) since [A(s) − A(s1 )]/(s2 − s21 ) is continuous. This result can be shown as in exercise 10.3.9 (b) by integration-by-parts. Thus q p s1 s π π π π Z ∞ Z ∞ s1 sin(sL − 4 ) cos(s1 L − 4 ) − s sin(s1 L − 4 ) cos(sL − 4 ) 2 f (r)J0 (s1 r)rdr = A(s1 ) lim sds .

L→∞ 0 π s2 − s21 0 q q q p s1 p s1 s We also note that ss1 = 1 + ss1 − 1 and s = 1+ s − 1 , where again terms s1 − 1 and p s1 2 2 s − 1, do not contribute due to the Riemann-Lebesgue lemma. Similarly, s − s1 = (s − s1 )(s + s1 ) = 2 2s1 (s − s1 ) + (s − s1 ) , where the latter term will not contribute as L → ∞, as well as s = s1 + s − s1 .

Thus, Z ∞ Z ∞ sin(sL − π4 ) cos(s1 L − π4 ) − sin(s1 L − π4 ) cos(sL − π4 ) 2 f (r)J0 (s1 r)rdr = A(s1 ) lim s1 ds.

L→∞ 0 π 2s1 (s − s1 ) 0 Z ∞ R∞ sin L (s − s1 ) ds .

Using the trigonometric addition formulas, 0 f (r)J0 (s1 r)rdr = π1 A(s1 ) lim L→∞ 0 s − s1 The change of variables w = L(s − s1 ) shows that Z ∞ Z A(s1 ) ∞ sin w f (r)J0 (s1 r)rdr = dw = A(s1 ) π 0 −∞ w 49

for s1 > 0, using the sine integral (see exercise 10.3.9 (b) again). This gives a symmetric inversion formula Z ∞ A(s) =

f (r)J0 (sr)rdr .

0

10.3.16 Let t = ky n , so that dt = kny n−1 dy. Thus Z ∞ Z ∞ Z ∞ n 1 dt t 1+p−n 1 y p e−ky dy = y p e−t n−1 = ( ) n e−t dt . kn 0 y kn 0 k 0 Using the definition of the gamma function in exercise 10.3.14, Z ∞ n 1 y p e−ky dy = k −(1+p)/n Γ(x) , n 0 where x − 1 = 1+p n − 1 or x = (1 + p)/n .

Section 10.4 10.4.3 (a) Taking the Fourier transform yields ∂ ū = −kω 2 ū − iωcū . ∂t The solution of this initial value problem is 2

t −iωct ū(ω, t) = F (ω) e|−kω {z e } .

G(ω)

Inverting using the convolution theorem yields Z ∞ 1 u(x, t) = f (x̄)g(x − x̄)dx̄ , 2π −∞ where the shift theorem determines g(x), g(x) = u(x, t) = √

1 4πkt

pπ

Z ∞

−(x+ct)2 /4kt . Thus kt e 2

f (x̄)e−(x−x̄+ct) /4kt dx̄ .

−∞

10.4.5 (a) Taking the Fourier transform yields ∂ ū = −kω 2 ū + Q̄ . ∂t A particular solution is obtained using the usual integrating factor.

10.4.5 (b) The general solution is a particular solution plus an arbitrary multiple of the homogeneous solution −kω 2 t

ū = c e

−kω 2 t

Z t

+e

2

Q̄(ω, τ )ekω τ dτ .

0

From the initial conditions, c = ū(ω, 0) = F (ω), the Fourier transform of the initial conditions.

10.4.5 (c) Using the convolution theorem, r Z ∞ Z tZ ∞ r 2 π −(x−x̄)2 /4kt 1 π 1 f (x̄) e dx̄ + Q(x̄, τ ) e−(x−x̄) /4k(τ −t) dx̄dτ . u(x, t) = 2π −∞ kt 2π 0 −∞ k(τ − t)

50

10.4.6 Taking the Fourier transform yields −ω 2 Y + idY /dω = 0 , since F[xf (x)] = −idF/dω. the general solution of this first-order differential equation is 3

Y (ω) = ce−iω /3 .

By using the formula for the inverse Fourier transform, we obtain Z ∞ Z ∞ 3 ω3 y(x) = c e−iω /3 e−iωx dω = 2c cos( + ωx)dω , 3 −∞ 0 where Euler’s formula and symmetry has been used. The constant c can be determined from the initial conditon √ Z ∞ . 2 ω3 3 −2/3 1 −2/3 cos( )dω = 2c 3 3 Γ( ) = y(0) = 2c Γ( ) , 3 3 2 3 0 from exercise 10.3.17. Thus, c= √

1 3 Γ( 13 )Γ( 23 )

=

1 sin π/3 √ = 2π π 3

from exercise 10.3.15. In this manner y(x) ≡ Ai (x) =

1 π

Z ∞ cos( 0

ω3 + ωx)dω .

3

10.4.7 (a) By taking the Fourier transform of the pde, we obtain ∂ ū = k(−iω)3 ū . ∂t The solution of this ordinary differential equation is 3

ū(ω, t) = c(ω)eikω t , where c(ω) is determined from the initial conditions, ū(ω, 0) = F (ω) = c(ω). Thus Z ∞ 3 u(x, t) = F (ω)eikω t e−iωx dω . −∞

10.4.7 (b) If g(x) =

R∞ −∞

3

eikω t e−iωx dω, then by the convolution theorem u(x, t) =

1 2π

Z ∞ f (x̄)g(x − x̄)dx̄ . −∞

10.4.7 (c) Using symmetry and the change of variables kω 3 t = s3 /3 [or ω = s/(3kt)1/3 ] yields µ 3 ¶ Z ∞ Z ∞ s 2 s cos g(x) = cos(kω 3 t − ωx)dω = − x ds .

3 (3kt)1/3 0 (3kt)1/3 −∞ h i −x 2π , so that By using exercise 10.4.6, we obtain g(x) = (3kt) 1/3 Ai (3kt)1/3 u(x, t) =

1 (3kt)1/3

·

Z ∞ f (x̄)Ai −∞

51

¸ x̄ − x dx̄ .

(3kt)1/3

Section 10.5 2

10.5.3 Only the Fourier cosine transform of e−αx has a simple expression: Z Z 2 2 ∞ −αx2 1 ∞ −αx2 iωx C[e−αx ] = e cos ωxdx = e e dx , π 0 π −∞ using symmetry. Thus

2 2 2 1 C[e−αx ] = 2 F [e−αx ] = √ e−ω /4α . πα

transform of e−ωα . From exercise 10.5.1 (see the table of Fourier cosine 10.5.10 We use the inverse cosine R ∞ −ωα α transforms), x2 +α2 = 0 e cos ωxdω . Differentiating this with respect to α yields · ¸ Z ∞ ∂ α = (−ω)e−ωα cos ωxdω . ∂α x2 + α2 0 Thus

· ¸ £ ¤ ∂ α 2α2 − (x2 + α2 ) C −1 ωe−ωα = − = .

2 2 ∂α x + α (x2 + α2 )2

10.5.11 (a) We take the Fourier sine transform and obtain 2 ∂ ū = k( ω − ω 2 ū) , ∂t π where ū is the sine transform and where (10.5.27) has been used. The general solution of this ordinary differential equation is 2 2 ū(ω, t) = + A(ω)e−kω t . πω 2 By using the initial condition, ū(ω, 0) = F (ω) = A(ω) + πω , · ¸ 2 2 2 ū(ω, t) = + F (ω) − e−kω t . πω πω

To invert this, we note from the tables that S[1] = 2/πω. Furthermore, the remaining term is the form of a product, one the sine transform √ of a known function and the other the cosine transform of 2 2 a known function, C[e−αx ] = e−ω /4α / πα, where we will let 4α = 1/kt or α = 1/4kt. Thus by the convolution theorem for sine transforms [see the table or exercise (10.5.6)] r ·r ¸ Z 1 ∞ π −(x−x̄)2 /4kt π −(x+x̄)2 /4kt 1 u(x, t) = 1 + [f (x̄) − 1] e − e dx. π 0 2 kt kt 2

∂ v 10.5.11 (b) By letting v(x, t) = u(x, t), we obtain ∂v ∂t = k ∂x2 subject to v(0, t) = 0 and v(x, 0) = f (x) − 1.

This problem has been solved in Section 10.5. From (10.5.39) Z ∞ 2 2 1 v(x, t) = √ [f (x̄) − 1][e−(x−x̄) /4kt − e−(x+x̄) /4kt ]dx̄ .

4πkt 0

10.5.11 (c) The answers are the same.

10.5.17 (a) By taking the sine transform of the pde, we obtain · ¸ 2 ∂ ū iσ0 t 2 =k ωAe − ω ū , ∂t π

52

where (10.5.27) has been used and where ū(ω, t) is the sine transform of u(x, t). Homogeneous solution 2 of this equation is e−kω t , while a particular solution is obtained by the method of undetermined coefficients (judiciuous guessing). In this manner 2

ū(ω, t) = c(ω)e−kω t + The initial condition implies ū(ω, 0) = 0 = c(ω) +

2 π ωA eiσ0 t . ω 2 + iσk0

2 π ωA iσ

ω 2 + k0

2

10.5.17 (b) As t → ∞, e−kω t → 0. Thus for large t ū(ω, t) ∼

.

2 π ωA iσ

ω 2 + k0

eiσ0 t .

Section 10.6 10.6.1 (a) By taking the Fourier transform in x, we obtain −ω 2 ū +

d2 ū = 0. dy 2

The boundary conditions for this differential equation are ū(ω, 0) = F1 (ω) and ū(ω, H) = F2 (ω). The solution of this equation, solving these boundary conditions, is ū(ω, y) = F2 (ω)

sinh ωy sinh ω(H − y) + F1 (ω) . sinh ωH sinh ωH

The desired solution u(x, y) is the inverse Fourier transform obtained by the defining integral, (10.3.7).

10.6.2 (b) The boundary conditions suggest the use of a cosine transform in y. By applying this transform, Laplace’s equation becomes d2 ū − ω 2 ū = 0 , dx2 where ū(x, ω) is the cosine transform in y of u(x, y) and where (10.5.26) has been used, simplified by ∂u/∂y(x, 0) = 0. The solution of the differential equation is a linear combination of hyperbolic (or exponential) functions. Since ∂u/∂x(L, y) = 0, ∂ ū/∂x(L, ω) = 0. Thus ū(x, ω) = A(ω) cosh ω(L − x) , where the other boundary condition u(0, y) = g1 (y) becomes ū(0, ω) = G1 (ω), the cosine transform of g1 (y). Thus G1 (ω) = A(ω) cosh ω L , determining A(ω). The solution u(x, y) is the inverse cosine transform, determined by the defining integral, of ū(x, ω).

10.6.4 (a) By taking the sine transform, in x, we obtain −ω 2 ū + d2 ū/dy 2 = 0 , where ū(ω, y) is the sine transform in x of u(x, y). The general solution of this differential equation is ū(ω, y) = a(ω)e−ωy + b(ω)eωy , where only ω > 0 is needed. Since we assume u → 0 as y → ∞, it follows that ū(ω, y) → 0 as y → ∞.

Thus b(ω) = 0, so that ū(ω, y) = a(ω)e−ωy .

The nonhomogeneous boundary condition ∂u/∂y(x, 0) = f (x), becomes ∂ ū/∂y(ω, 0) = F (ω) when the sine transform is introduced, where F (ω) is the sine transform of f (x), F (ω) = S[f (x)]. Thus −ωa(ω) = S[f (x)], 53

so that

1 ū(ω, y) = − S[f (x)]e−ωy . ω The apparent singularity at ω = 0 may cause some difficulty in calculating the inverse transform. The simplest way to determine u(x, y) is to first eliminate the singularity. Note that ∂ ū(ω, y) = S[f (x)]e−ωy . ∂y The convolution theorem for sine transforms may be used since h(x) = C −1 [e−ωy ] =

y x2 + y 2

from the tables or exercise 10.5.1. Thus · ¸ Z 1 ∞ ∂u y y (x, y) = f (x̄) dx̄ . − ∂y π 0 (x − x̄)2 + y 2 (x + x̄)2 + y 2 By integrating this with respect to y, we obtain Z ∞ 1 (x − x̄ )2 + y 2 u(x, y) = f (x̄)ln dx̄ + Q(x ) , 2π 0 (x + x̄ )2 + y 2 where Q(x) is an arbitrary function of x. Since u → 0 as y → ∞, we obtain Q(x) = 0.

10.6.11 (a) By taking the sine transform in x and the sine transform in y, we obtain ∂ ū = −k (ω12 + ω22 )ū , ∂t − where ū(ω1p , ω2 , t) is the appropriate 00 double00 transform of u(x, y, t). We introduce → ω ≡ (ω1 , ω2 ) and 2 2 − → ω ≡ | ω | = ω1 + ω2 . Then 2 − → ū(→ ω ) = c(− ω )e−kω t , → → where c(− ω ) = F (− ω ), the double transform of the initial conditions. Thus Z ∞Z ∞ u(x, y, t) = 0

→ F (− ω )e−kω t sin ω1 x sin ω2 y dω1 dω2 , 2

0

R∞R∞ 4

− − where F (→ ω ) = π2 0 0 f (x, y) sin ω1 x sin ω2 y dx dy . We note that F (→ ω ) is odd in ω1 and odd in ω2 . Thus the integrand in the double integral for u(x, y, t) is even in both ω1 and ω2 . Furthermore we may extend f (x, y) as an odd function of x and y. If we replace sin ω1 x by e±iω1 x / ± i in both double integrals, no error is made since the cosine term makes no contribution due to its symmetry. Similarly sin ω2 x may be replaced by e±iω2 x / ± i. Thus Z ∞Z ∞ → − − → 2 − −F (→ ω )e−kω t e−i ω · r dω1 dω2 u(x, y, t) = −∞

and

−∞

Z ∞Z ∞ −F (ω) = −∞

→ − − → f (x, y)ei ω · r d x d y .

−∞

These are related to the formulas for double Fourier transforms, where the convolution theorem may be used so that u(x, y, t) satisfies (10.6.95). However, this answer depends on f (x0 , y0 ) for −∞ < x0 < ∞ and −∞ < y0 < ∞ while it is only naturally given in the first quadrant. We note that we have extended f (x, y) as an odd function in both x and y. If we add up contributions from all four quadrants, we obtain the answer given.

54

10.6.12 (a) If we introduce first a sine transform in y, then µ 2 ¶ ∂ ū ∂ ū 2 =k − ω ū ∂t ∂x2 with ū(0, ω, t) = ū(L, ω, t) = 0 and the initial conditions ū(x, ω, 0) = F (x, ω). Here ū(x, ω, t) is the sine transform in y of u(x, y, t). The pde for ū(x, ω, t) is suited for the usual Fourier series in x ū(x, ω, t) =

∞ X

An (ω, t) sin

n=1

nπx , L

(8)

where the initial conditions yield ∞ X

2 or An (ω, 0) = L

nπx F (x, ω) = An (ω, 0) sin L n=1

Z L F (x, ω) sin 0

nπx dx.

L

By substituting (8) into the pde, we obtain the differential equation satisfied by An (ω, t): dAn /dt = −k[(nπ/L)2 + ω 2 ]An .

Thus

2

2

An (ω, t) = An (ω, 0)e−k[(nπ/L) +ω ]t .

By inversion of the sine transform u(x, y, t) =

Z ∞ ÃX ∞

nπx An (ω, t) sin L n=1

0

! sin ω y d ω ,

where An (ω, t) is given above with 4 An (ω, 0) = Lπ

Z LZ ∞ f (x, y) sin ωy sin 0

0

nπx dy dx , L

since F (x, ω) is the sine transform in y of f (x, y). → 10.6.15 (a) We introduce ū(− ω , z) the double Fourier transform in x and y of u(x, y, z): −ω 2 ū + where ω =

d2 ū =0, dz 2

p

ω12 + ω22 > 0 . The general solution of this equation is → → − ū(− ω , z) = A(− ω )e−ωz + B(→ ω )eωz .

→ Since u → 0 as z → ∞, if follows that ū → 0 as z → ∞, and hence B(− ω ) = 0. Furthermore, the − → boundary condition at z = 0 is that ū equals F ( ω ) there, the double Fourier transform of f (x, y): − → ū(→ ω,z) = F (− ω )e−ωz .

10.6.15 (b) By using the convolution theorem for double Fourier transforms ZZ 1 → → u(x, y, z) = f (x, y)g(− r −− r 0 )dx0 dy0 , (2π)2 → where the double Fourier transform of g(− r ) is known to equal e−ωz : − F[g(→ r )] = e−ωz .

55

− Determining g(→ r ) is not easy. From the general inversion formula ZZ Z 2π Z ∞ − →→ − → g(− r)= e−ωz e−i ω · r dω1 dω2 = e−ωz e−iωr cos(φ−θ) ω dω dφ , 0

0

where the latter integral is obtained using polar coordiantes. If polar coordinates were chosen oriented → − → in the direction of − r (which is θ) then → ω ·− r = ωr cos φ, which is equivalent to replacing θ by 0 in the double integral. The ω -integral (and what follows) is easier by integrating from ∞ to z: Z z Z 2π Z ∞ → g(− r )dz = − e−ωz e−iωr cos φ dωdφ , since

Rz ∞

∞

e

−ωz

0

Z z

− g(→ r )dz = −

∞

Thus

0

ωdz = −e−ωz |z∞ = −e−ωz . The ω integral is now easy Z 2π 0

¯∞ Z 2π dφ e−ω(z+ir cos φ) ¯¯ dφ = − . ¯ −(z + ir cos φ) 0 z + ir cos φ 0

Z 2π z − ir cos φ dφ dφ = −z , 2 + r 2 cos2 φ 2 + r 2 cos2 φ z z ∞ 0 0 since the cos φ integral is zero by symmetry. The resulting integral can be put in the given form Z 2π Z z dφ −2πz − → g( r )dz = −z = , 2 2 2 2 2 2 z(z + r2 )1/2 (z + r ) cos φ + z sin φ ∞ 0 Z z

− g(→ r )dz = −

Z 2π

− for z > 0. Differentiating this yields g(→ r ) = 2πz(z 2 + r2 )−3/2 , so that Z ∞Z ∞ z f (x0 , y0 ) u(x, y, z) = dx0 dy0 .

2π −∞ −∞ [(x − x0 )2 + (y − y0 )2 + z 2 ]3/2 10.6.16 (b) We want the radial problem to be an eigenvalue problem. By separation of variables of Laplace’s equation in polar coordinates, u(r, θ) = φ(r)h(θ), d2 h = λh dθ2

and

r

d dφ (r ) + λφ = 0 ; dr dr

see (2.5.37) and (2.5.40). The eigenvalue problem consists of the radial equation, to be solved subject to φ(0) bounded and φ(a) = 0. this is in the form of a singular Sturm-Liouville problem, singular at r = 0. In some sense, though, the eigenfunctions will be orthogonal with weight 1/r. The radial differential equation is equidimensional r2

d2 φ dφ +r + λφ = 0 , dr2 dr

√ √ √ so that solutions are in the form φ = rp with p(p−1)+p+λ = 0√ or p = ±i λ . Since r±i λ = e±i λ ln r , √ real solutions are linear combinations of sin( λ ln r) and cos( λ ln r). However, the boundary condition at r = a is most easily satisfied using the following set of independent solutions: √ √ r r φ = c1 cos ( λ ln ) + c2 sin ( λ ln ) . a a

Here φ(a) = 0 implies that c1 = 0, so that √ r φ(r) = c2 sin( λ ln ) . a The other condition, φ(0) is bounded, is satisfied for all √ λ > 0. We thus use the generalized superposition principle of integrating over λ. We introduce ω = λ and integrate over ω instead: Z ∞ r u2 (r, θ) = A(ω) sinh ω θ sin (ω ln )dω , a 0 56

where we have noted that the θ-dependent part which satisfies h(0) = 0 is sinh ω θ. The nonhomogeneous boundary condition becomes Z ∞ π r g2 (r) = A(ω) sinh ω sin (ω ln )dω .

2 a 0 In order to determine A(ω), it is convenient to consider the variable ρ = − ln(r/a). As 0 < r < a, it follows that 0 < ρ < ∞, a semi-infinite variable similar to the type that transforms might be used for: Z ∞ π [−A(ω) sinh ω ] sin(ωρ)dω . g2 (r) = 2 0 Thus −A(ω) sinh ω π/2 is the Fourier transform in ρ of g2 (r), and hence by definition of its transform Z π 2 ∞ −A(ω) sinh ω = g2 (r) sin ω ρ d ρ .

2 π 0 We may return to the physically meaningful variable r, by noting that dρ = −dr/r in which case Z π 2 a dr −A(ω) sinh ω = − g2 (r) sin [ω ln r/a] .

2 π 0 r Note the appearance of the 00 weight00 1/r .

10.6.18 By taking the Fourier transform in x, we obtain ∂ 2 ū = −c2 ω 2 ū , ∂t2 where ū(ω, t) is the Fourier transform in x of u(x, t). The general solution of this differential equation is ū(ω, t) = A(ω) cos cωt + B(ω) sin c ω t .

By transforming the initial conditions, we obtain ū(ω, 0) = 0

and

∂ ū(ω, 0) = G(ω) , ∂t

where G(ω) is the Fourier transform in x of g(x). Hence 0 = A(ω) and G(ω) = c ω B(ω). Thus ū(ω, t) = G(ω)

sin c ω t π = G(ω)F (ω) . cω c

We note that in the tables (or by exercise 10.3.6) that if ½ 0 |x| > ct f (x) = 1 |x| < ct , then F[f (x)] = F (ω) = sin cωt/πω. This enables us to use the convolution theorem for Fourier transforms: Z ∞ π 1 g(x̄)f (x − x̄)dx̄] , u(x, t) = [ c 2π −∞ where

½ f (x − x̄) =

Thus u(x, t) =

1 2c

57

0 |x − x̄| > ct 1 |x − x̄| < ct .

Z x+ct g(x̄)dx̄ . x−ct

Chapter 11. Green’s Functions for Wave and Heat Eqns Section 11.2 11.2.6 (a) From (11.2.24),

Z tZ ∞ u(x, t) = ½

where from (11.2.31-32) G =

G(x, t; x0 , t0 )Q(x0 , t0 )dx0 dt0 , 0

0

−∞

|x − x0 | > c(t − t0 ) |x − x0 | < c(t − t0 ) .

1 2c

Contributions only occur if x − c(t − t0 ) < x0 < x + c(t − t0 ). Thus 1 u(x, t) = 2c

Z t Z x+c(t−t0 ) Q(x0 , t0 )dx0 dt0 .

0

x−c(t−t0 )

11.2.6 (b) If x = x1 , t = t1 , then x1 − c(t1 − t0 ) < x0 < x1 + c(t1 − t0 ), as sketched on page 746 for 0 < t0 < t.

11.2.7 (a) From exercise 11.2.6, u(x, t) =

1 2c

Z t Z x+c(t−t0 ) Q(x0 , t0 )dx0 dt0 .

0

x−c(t−t0 )

If Q(x, t) = g(x)e−iωt , then u(x, t) =

Z t Z x+c(t−t0 )

1 2c

0

g(x0 )e−iωt0 dx0 dt0 .

x−c(t−t0 )

In order to do the t0 -integration first, the order of integration must be interchanged, requiring an analysis of the triangular region (see page 746). If x0 is first fixed, there appears to be two cases, 0 x0 < x in which case t0 ranges from 0 to t − x−x and x0 > x in which case t0 ranges from 0 to c x0 −x 0| t − c . Both correspond to t0 ranging from 0 to t − |x−x . Thus c u(x, t) =

1 2c

Z x+ct g(x0 ) x−ct

Z t− |x−x 0| c

e−iωt0 dt0 dx0 .

0

By performing the t0 -integration, we obtain Z x+ct |x−x0 | 1 dx0 u(x, t) = g(x0 )[1 − e−iω(t− c ) ] , 2c x−ct iω

or

Z ∞ u(x, t) =

g(x0 )I(x, x0 )dx0 , −∞

where the influence function I(x, x0 ) satisfies ( I(x, x0 ) =

0 1−e

|x − x0 | > ct −iω(t−

|x−x0 | ) c

2icω

|x − x0 | < ct . x0 −x

x−x0 )

If x > x0 , then e−iω(t− c occurs which is right-going (outward), while if x < x0 , then e−iω(t− c ) occurs which is left-going (also outward). This outward propagating influence function does not have an infinite spatial extent. Instead, it is cut-off at a distance corresponding to the propagation velocity ±c.

58

11.2.8 (a) From (11.2.24) − u(→ x , t) =

Z t ZZZ

→ → → G(− x , t; − x 0 , t0 )Q(− x 0 , t0 )dV0 dt0 ,

0

→ − → → where Q(− x 0 , t0 ) = g(→ x 0 )e−iωt0 and the Green’s function satisfies (11.2.47) with ρ = |− x −− x 0 |. Thus Z t ZZZ 1 → → u(− x , t) = g(− x 0) δ[ρ − c(t − t0 )]e−iωt0 dV0 dt0 .

4πcρ 0 → − − → 1 The singularity occurs at t0 = t − | x −c x 0 | . However, note that δ(ct) = |c| δ(t) [see (9.3.34)]. In order → − |− x −→ x | 0 for t > 0, t > . Thus 0

c

→ u(− x , t) = where − → I(→ x,− x 0 , t) =

ZZZ

→ − → g(− x 0 )I(→ x,− x 0 , t)dV0 ,

  0 

1 e − → 4πc2 |→ x −− x 0|

− → → − | x − x 0| −iω(t− ) c

→ − − → t < | x −c x 0 | → − − → , t > | x −c x 0 | .

→ − Notice that the region of influence is |− x −→ x 0 | < ct, the inside of a sphere expanding (outward) as t increases.

11.2.10 (a) By symmetry, a negative image source at (−x0 , t0 ) may be used so that G(x, t; x0 , t0 ) = G∞ (x, t; x0 , t0 ) − G∞ (x, t; −x0 , t0 ) , where G∞ is given by (11.2.31-32).

11.2.10 (b) We use the one-dimensional version of (11.2.16). At x = 0, n̂ = −î, and at x = ∞, n̂ = i. Thus ¯∞ Z t ∂G ∂u ¯¯ 2 u(x, t) = −c (u −G ) dt0 . ∂x0 ∂x0 ¯0 0 Since G = ∂G/∂x0 = 0 at x0 = ∞ and G(x, 0) = 0 because G(0, x) = 0. Thus 2

u(x, t) = c

¯ ∂G ¯¯ h(t0 ) dt0 , ∂x0 ¯x0 =0 0

Z t

since u(0, t) = h(t). We evaluate ∂G/∂x0 from part (a): 1 {−H[x − x0 − c(t − t0 )] + H[x − x0 + c(t − t0 )] 2c +H[x + x0 − c(t − t0 )] − H[x + x0 + c(t − t0 )]} , ∂G 1 = {δ[x − x0 − c(t − t0 )] − δ[x − x0 + c(t − t0 )] ∂x0 2c +δ[x + x0 − c(t − t0 )] − δ[x + x0 + c(t − t0 )]} , ¯ ∂G ¯¯ 1 1 = {δ[x − c(t − t0 )] − δ[x + c(t − t0 )]} = δ[x − c(t − t0 )], ¯ ∂x0 x0 =0 c c G(x, t; x0 , t0 ) =

since δ[x + c(t − t0 )] = 0 because x > 0 and t > t0 . Thus from (9) Z t 1 h(t0 ) δ[x − c(t − t0 )] dt0 u(x, t) = c2 c 0 Z t x = h(t0 )δ(t0 + − t) dt0 , c 0 because of (9.3.34). Thus ½ 0 if x > ct u(x, t) = h(t − xc ) if 0 < x < ct .

59

(9)

11.2.12 (a) The solution is obtained from (11.2.24) using the three-dimensional infinite space Green’s function for the wave equation (11.2.39). In this manner 1 G(x, y, t; x1 , y1 , t1 ) = 4πc

Z t ZZZ 0

1 δ[ρ − c(t − t0 )] · δ(x0 − x1 )δ(y0 − y1 )δ(t0 − t1 )dx0 dy0 dz0 dt0 , ρ

where ρ2 = [(x − x0 )2 + (y − y0 )2 + (z − z0 )2 ] and where we will need to show our answer does not depend on z. The x0 , y0 , t0 integrals are easy due to the delta functions. If t > t1 , then Z ∞ 1 1 G(x, y, t; x1 , y1 , t1 ) = δ[ρ1 − c(t − t1 )]dz0 , 4πc −∞ ρ1 where ρ21 = r2 + (z − z0 )2 and r is the two-dimensional distance from source point to response point r = [(x − x1 )2 + (y − y1 )2 ]1/2 .

The integrand is symmetric around z0 = z, z0 = z ± (ρ21 − r2 )1/2 . Thus Z ∞ 1 1 G(x, y, t; x1 , y1 , t1 ) = δ[ρ1 − c(t − t1 )]dz0 .

2πc z ρ1 It is now convenient to change Rvariables (fix r), ρ21 = r2 + (z − z0 )2 so that 2ρ1 dρ1 = 2(z − z0 )(−dz0 ). ∞ 1 1 Thus G(x, y, t; x1 , y1 , t1 ) = 2πc δ[ρ1 − c(t − t1 )]dρ1 . There is an impulse at ρ1 = c(t − t1 ). Thus r z0 −z ( G(x, y, t; x1 , y1 , t1 ) =

0 1 √ 1 2πc c2 (t−t1 )2 −r 2

if r > c(t − t1 ) if r < c(t − t1 ) .

Section 11.3 11.3.2 (a)

¯b ¯b R b du d2 u d d2 d +r + qu] dx = u[rv − dx (pv)]¯a + du pv ¯a + a u[qv + dx v[p 2 (pv) − dx (rv)]dx , dx a 2 dx dx | {z }

Rb

L(u) 2

2

d du d du d d d d du d (vp du since vp ddxu2 = dx dx ) − dx dx (vp) , dx dx (vp) = dx (u dx (vp)) − u dx2 (vp) , and vr dx = dx (vru) − d u dx (rv) . By letting

d2 d d2 v dp dv d2 p dr (pv) − (rv) + qv = p 2 + (2 − r) +( 2 − + q)v , 2 dx dx dx dx dx dx dx ¯b Rb ¯ we obtain a [vL(u) − uL∗ (v)]dx = −H ¯ , where L∗ (v) =

a

H(x) = u[ If L = L*, then p

d du dv du dp (vp) − vr] − pv = p(u − v ) + uv( − r) . dx dx dx dx dx

d2 dp d d2 p dr d2 d + (2 − r) + ( − + q) = p +r +q .

2 2 2 dx dx dx dx dx dx dx

This requires 2dp/dx − r = r and d2 p/dx2 − dr/dx = 0. Consequently r = dp/dx, which results in L=

d d (p ) + q . dx dx

11.3.6 (c) We use the one-dimensional version of (11.3.21). Since n̂ = −î at x = 0, Z t u(x, t) = k

[u(0, t0 ) 0

∂u ∂G (x, t; 0, t0 ) − G(x, t; 0, t0 ) (0, t0 )]dt0 . ∂x0 ∂x0 60

Since the Green’s fucntion satisfies the homogeneous boundary condition, G(0, t; x0 , t0 ) = 0, which because of symmetry yields G(x0 , t0 ; 0, t) = 0. Thus, Z t u(x, t) = k

u(0, t0 ) 0

∂G (x, t; 0, t0 )dt0 . ∂x0

For the Green’s function we use (11.3.34). Thus · ¸ · ¸¾ ½ ∂G 1 (x − x0 ) (x − x0 )2 (x + x0 ) (x + x0 )2 (x, t; x0 , t0 ) = p exp − + exp − ∂x0 4k(t − t0 ) 2k(t − t0 ) 4k(t − t0 ) 4πk(t − t0 ) 2k(t − t0 ) and

· ¸ ∂G x2 x/k(t − t0 ) exp − (x, t; 0, t0 ) = p , ∂x0 4k(t − t0 ) 4πk(t − t0 )

so that in general since u(0, t) = A(t), x u(x, t) = √ 4πk

·

Z t A(t0 )(t − t0 )

−3/2

0

¸ −x2 exp dt0 .

4k(t − t0 )

In the given case where A(t0 ) = 1 x u(x, t) = √ 4πk

·

Z t (t − t0 )

−3/2

0

¸ −x2 exp dt0 .

4k(t − t0 )

This can be simplified by the change of variables η = √ this manner, we obtain

2 u(x, t) = √ π

x and thus dη = 2√x4k (t − t0 )−3/2 dt0 . In 4k(t−t0 )

Z ∞

2 e−η dη . √ x/ 4kt

11.3.7 The Green’s function satisfies the related homogeneous boundary condition ∂G/∂x = 0 at x = 0. This boundary condition can be satisfied using one-dimensional infinite space Green’s functions (11.3.27), if we introduce a positive image source at (−x0 , t0 ). In this manner, the desired Green’s function is ½ · ¸ · ¸¾ 1 (x − x0 )2 (x + x0 )2 G(x, t; x0 , t0 ) = p exp − + exp − .

4k(t − t0 ) 4k(t − t0 ) 4πk(t − t0 )

61

Chapter 12. Method of Characteristics Section 12.2 12.2.2 The general solution of the pde is w(x, t) = P (x + 3t) using (12.2.12) since c = −3. In order to satisfy the initial condition, w(x, 0) = P (x) = cos x. Therefore w(x, t) = cos(x + 3t) .

12.2.5 (b) Using the ideas of subsection 12.2.2, if dx/dt = x, then dw/dt = 1. The characteristics satisfy x = x0 et ,

(10)

where the characteristics start (t = 0) at x = x0 . Along the characteristics w(x, t) = t + w0 , where w0 is the value of w at t = 0 where x = x0 , i.e., w0 = f (x0 ). Thus w(x, t) = t + f (x0 ) = t + f (xe−t ), using (10) to solve for x0 as a function of x and t.

12.2.5 (d) Using the ideas of subsection 12.2.2, if dx/dt = 3t, then dw/dt = w. The characteristics satisfy x=

3 2 t + x0 , 2

(11)

where the characteristics start (t = 0) at x = x0 . Along the characteristics, w(x, t) = w0 et , where w0 is the value of w at t = 0, where x = x0 , i.e., w0 = f (x0 ). Thus 3 w(x, t) = f (x0 )et = f (x − t2 )et , 2 using (11) to solve for x0 as a function of x and t.

12.2.6 Using the ideas of subsection 12.2.2, if dx/dt = 2u, then du/dt = 0. The latter equation states that u is constant along the characteristic. If the characteristic starts (t = 0) at x0 , then u = f (x0 ) there and everywhere on this characteristic. Thus x = 2f (x0 )t + x0 , since x = x0 at t = 0. These characteristics are straight lines with velocity 2f (x0 ). However, they are not necessarily parallel to each other, as different lines have different slopes.

12.2.8 The characteristics satisfy x = 2f (x0 )t + x0 . If x0 > 0, then f (x0 ) = 2 : x = 4t + x0 .

Along these characteristics (x0 > 0) u = 2. Thus u=2

for

x > 4t .

If x0 < 0, then f (x0 ) = 1 : x = 2t + x0 .

62

Along these characteristics (x0 < 0) u = 1. Thus ½ 1 for x < 2t u= 2 for x > 4t .

We have not determined what occurs for 2t < x < 4t. In the limit as L → 0 of exercise 12.2.7, all values of u between 1 and 2 occur at the discontinuity at x0 = 0. Thus there are an infinite number of characteristics starting at x0 = 0, each satisfying x = 2f (x0 )t u(x, t) = f (x0 ) .

By eliminating f (x0 ) in this ”fan-shaped” region u(x, t) = x/2t .

Thus

  u(x, t) =

1 x/2t  2

x < 2t 2t < x < 4t x > 4t .

This is similar to the result sketched in Figures 12.6.4 and 12.6.5.

Section 12.3

¯ ¯ dF (x−ct) 12.3.4 (a) ∂u = (−c) = −c dF ¯ ∂t d(x−ct) dx (x) . t=0 ¯ dF (x−ct) ¯ (−ct) 1 d 12.3.4 (b) ∂u = dF ∂x = d(x−ct ¯ d(−ct) = − c dt F (−ct) . x=0

Section 12.4 12.4.1 The general solution of the wave equation is given by (12.3.4) u(x, t) = F (x − ct) + G(x + ct) .

(12)

Since for x > 0 u(x, 0) = 0 and ∂u/∂t(x, 0) = 0, it follows from (12.3.10-11) that F (x) = 0

and G(x) = 0

only for

x>0.

The boundary condition, u(0, t) = h(t) for t > 0, yields h(t) = F (−ct) + G(ct) if

t>0.

(13)

We now can evaluate (12). If x > ct, then the arguments of both F and G are positive and there F = G = 0. Thus u(x, t) = 0 if x > ct .

However if x < ct, then the argument of F only is negative. From (13), letting z = −ct, F (z) = h(−z/c) − G(−z) for Thus F (x − ct) = h(

z<0.

x − ct ) − G(ct − x), −c

resulting in

x ) − G(ct − x) + G(x + ct) c if x < ct. In this case both arguments of G are positive, and hence both G equal zero. Thus ½ 0 x > ct u(x, t) = h(t − xc ) 0 < x < ct . u(x, t) = h(t −

It is very helpful in understanding this result to sketch characteristics (as for example in Figures 12.4.1 and 12.4.2).

63

12.4.2 The general solution of the wave equation is given by (12.3.4) u(x, t) = F (x − ct) + G(x + ct) .

(14)

The initial conditons are only valid for x < 0. Thus (12.3.10-11) implies that G(x) =

1 cos x 2

and F (x) =

1 cos x 2

for

x<0.

(15)

The boundary condition, u(0, t) = e−t for t > 0 yields e−t = F (−ct) + G(ct)

for

t>0.

(16)

We now can evaluate (14). If x < −ct, then the arguments of both F and G are negative, and there F = G = 1/2 cos x. Thus u(x, t) =

1 1 cos(x − ct) + cos(x + ct) = cos x cos ct , x < −ct .

2 2

If −ct < x < 0, then the argument of G only is positive. From (16), letting z = ct G(z) = e−z/c − F (−z) Thus for x + ct > 0 which yields

for

z>0.

G(x + ct) = e−(x+ct)/c − F (−x − ct) , u(x, t) = e−(x+ct)/c − F (−x − ct) + F (x − ct)

if x + ct > 0. In this case both arguments of F are negative, and hence (15) may be used: u(x, t) = e−(t+x/c) + In conclusion

½ u(x, t) =

1 1 cos(x − ct) − cos(−x − ct) .

2 2

cos x cos ct x + ct < 0 e−(t+x/c) + sin x sin ct x + ct > 0 .

Figure 12.4.1 reflected around the t-axis, i.e., x replaced by −x, shows the important characteristic, x + ct = 0.

12.4.6 The general solution of the wave equation is given by (12.3.4) u(x, t) = F (x − ct) + G(x + ct) .

The zero initial conditions from (12.3.10-11) imply that F (x) = 0

and G(x) = 0

only for

x>0.

Since x + ct > 0, it follows that u(x, t) = F (x − ct) .

To apply the nonhomogeneous boundary condition, we calculate: dF (x − ct) ∂u (x, t) = ∂x d(x − ct and thus

dF (−ct) 1 d ∂u (0, t) = =− F (−ct) = h(t) ∂x d(−ct) c dt 64

for t > 0. Consequently

Z t F (−ct) = −c

h(t̄)dt̄ , 0

where we have used F (0) = 0 since we have assumed F (x) is continuous and we already know F (x) = 0 for x > 0. We note that (z = −ct) for z < 0 Z −z/c F (z) = −c

h(t̄)dt̄ 0

and hence

Z t−x/c h(t̄)dt̄ if x − ct < 0 .

F (x − ct) = −c 0

In summary,

½ u(x, t) =

0 x > ct R t− xc h(t̄)dt̄ x < ct . −c 0

12.4.7 The general solution of the wave equation is given by (12.3.4) u(x, t) = F (x − ct) + G(x + ct) .

(17)

The initial conditions (valid for x > 0) are applied in (12.3.10-11). Since it is initially at rest, F (x) = G(x) =

1 f (x) 2

for

x>0.

The boundary condition requires ∂u/∂x: ∂u dF (x − ct) dG(x + ct) = + . ∂x d(x − ct) d(x + ct) Evaluating this at x = 0 yields h(t) = −

1 d 1 d F (−ct) + G(ct), c dt c dt

for t > 0 .

By integrating this, we obtain for t > 0 Z t c h(t̄)dt̄ = −F (−ct) + G(ct)

(18)

0

since G(0) − F (0) = 12 f (0) − 12 f (0) = 0 by continuity. In (17) G is known for positive arguments, but F is not known for negative arguments, needed if x < ct. However, from (18), letting z = −ct Z −z/c F (z) = G(−z) − c

h(t̄)dt̄

for z < 0 .

0

Consequently, for x − ct < 0 (i.e., 0 < x < ct) Z t−x/c F (x − ct) = G(ct − x) − c

h(t̄)dt̄ .

0

We thus can calculate u, ½ 1 u(x, t) =

ct 2 [f (x + ct) + f (x − ct)] x > R t−x/c 1 [f (x + ct) + f (ct − x)] − c h(t̄)dt̄ 2 0

65

x < ct .

Section 12.5 12.5.1 (b) By separation of variables, u(x, t) is given by (4.4.11) where the conditions for An and Bn are given in (4.4.12) and (4.4.13). If g(x) = 0, Bn = 0 and ∞ X

u(x, t) =

An sin

n=1

nπx nπct cos , L L

P∞

where f (x) = n=1 An sin nπx L for 0 < x < L. However, the series does not equal f (x) outside of 0 < x < L. Instead, from the theory of Fourier sine series. ∞ X

An sin

n=1

nπx = f¯(x) , L

(19)

where f¯(x) is the odd periodic extension of f (x). Since sin a cos b = 21 sin (a + b) + 12 sin(a − b), it follows that ∞ ∞ 1X 1X nπ nπ u(x, t) = An sin (x + ct) + An sin (x − ct) .

2 n=1 L 2 n=1 L Thus from (19) u(x, t) =

1 ¯ [f (x + ct) + f¯(x − ct)] , 2

where f¯(x) is the odd periodic extension of f (x). This is the result found on page 545.

Section 12.6 12.6.1 (a) By integration, ρ(x, t) = c(x), where c(x) is an arbitrary function of x. From the initial condition, ρ(x, 0) = f (x) = c(x). Thus the solution is ρ(x, t) = f (x) .

12.6.1 (c) This is an ordinary differential equation, keeping x fixed, whose general solution is ρ(x, t) = A(x)e−3xt , where A(x) is an arbitrary function of x. From the initial condition ρ(x, 0) = f (x) = A(x). Therefore ρ(x, t) = f (x)e−3xt .

The method of characteristics can also be used (if dx/dt = 0, then dρ/dt = −3xρ).

12.6.2 This is an ordinary differential equation, keeping x fixed, whose general solution is ρ(x, t) = A(x)et , where A(x) is an arbitrary function of x. The condition says that 1 + sin x = A(x)et = A(x)e−x/2 since this is given along x = −2t (equivalent to t = −x/2). Thus A(x) = ex/2 (1 + sin x) , yielding the solution

x

ρ(x, t) = ex/2 (1 + sin x)et = (1 + sin x)et+ 2 .

66

12.6.3 (a) The general solution of this pde is ρ(x, t) = A(x − c0 t) , where A is an arbitrary function. From the initial condition, ρ(x, 0) = A(x) = sin x. Therefore, ρ(x, t) = sin(x − c0 t) .

By the method of characteristics, if dx/dt = c0 , then dρ/dt = 0. The characteristics satisfy x = c0 t + x0 , where x(0) = x0 . Also ρ(x, t) = ρ(x0 , 0) = sin x0 , from the initial condition. Thus, ρ(x, t) = sin(x − c0 t) .

12.6.3 (b) By the method of characteristics, if dx/dt = c0 , then dρ/dt = 0. The characteristics that start at x(0) = x0 for x0 > 0 are x = c0 t + x0 .

From the initial condition ρ(x, t) = ρ(x0 , 0) = f (x0 ) for

x0 > 0 .

Thus ρ(x, t) = f (x − c0 t) for x0 = x − c0 t > 0. However, there are characteristics that start at x = 0 at t = t0 < 0. For these x = c0 (t − t0 ) .

From the boundary condition ρ(x, t) = ρ(0, t0 ) = g(t0 )

for

t0 = t −

x >0. c0

t0 = t − x/c0 is called the retarded time. In summary, ½ ρ(x, t) =

f (x − c0 t) g(t − cx0 )

x > c0 t c0 t > x > 0 .

12.6.4 (a) The condition u(0) = umax yields α = umax , while u(ρmax ) = 0 yields 0 = α + βρmax or β = −umax /ρmax . Thus u(ρ) = umax (1 − ρ/ρmax ) . The flow satisfies q = ρu, and hence q(ρ) = umax ρ(1 − ρ/ρmax ) . This is a parabola which is concave down with q(0) = q(ρmax ) = 0.

12.6.4 (b) The flow is maximum where q 0 (ρ) = umax (1 − 2ρ/ρmax ) = 0 .

Thus ρ = ρmax /2, u = umax /2, and hence q = ρmax umax /4 is the maximum flow.

12.6.8 (a) By the method of characteristics, if dx/dt = c, then dρ/dt = e−3x . The characteristics satisfy x = ct + x0 ,

(20)

where x(0) = x0 . Along the characteristics dρ/dt = e−3(ct+x0 ) .

Since x0 is constant, by integration ρ(x, t) =

1 −3(ct+x0 ) e +k , −3c 67

(21)

where k is determined from the initial condition ρ(x0 , 0) = f (x0 ) = − Using (20) - (22), ρ(x, t) = −

1 −3x0 e +k .

3c

(22)

1 −3x 1 e + e−3(x−ct) + f (x − ct) .

3c 3c

12.6.8 (c) By the method of characteristics, if dx/dt = t, then dρ/dt = 5. The characteristics satisfy x=

1 2 t + x0 , 2

(23)

where x(0) = x0 . Along the characteristics, ρ(x, t) = 5t + ρ(x0 , 0) , since at t = 0, x = x0 . From the initial condition and (23) 1 ρ(x, t) = 5t + f (x − t2 ) .

2 12.6.8 (e) By the method of characteristics, if dx/dt = −t2 , then dρ/dt = −ρ. The characteristics satisfy 1 x = − t3 + x0 , 3

(24)

where x(0) = x0 . Along the characteristics, ρ satisfies an ordinary differential equation whose general solution is ρ(x, t) = ρ(x0 , 0)e−t , since at t = 0, x = x0 . Furthermore, ρ(x0 , 0) = f (x0 ). Therefore, from (24) 1 ρ(x, t) = f (x + t3 )e−t .

3 12.6.8 (g) By the method of characteristics, if dx/dt = x, then d ρ/dt = t. The characteristics satisfy a differential equation whose solution is x = x0 et , (25) since x(0) = x0 . Along these characteristics, ρ(x, t) =

1 2 t + ρ(x0 , 0) , 2

since at t = 0, x = x0 . Furthermore, ρ(x0 , 0) = f (x0 ). Therefore, from (25) ρ(x, t) =

1 2 t + f (xe−t ) .

2

12.6.9 (a) By the method of characteristics, if dx/dt = −ρ2 , then dρ/dt = 3ρ. Along the characteristics ρ(x, t) = ρ(x0 , 0)e3t , since we choose x = x0 at t = 0. thus the characteristics satisfy dx = −ρ2 (x0 , 0)e6t . dt Since x0 is constant,

1 x = − (e6t − 1)ρ2 (x0 , 0) + x0 .

6 From the initial condition ρ(x0 , 0) = f (x0 ), we have a parametric representation of the solution: ρ(x, t) = f (x0 )e3t , where x0 satisfies x = − 16 (e6t − 1)f 2 (x0 ) + x0 .

68

12.6.9 (c) By the method of characteristics, if dx/dt = t2 ρ, then dρ/dt = −ρ. Along the characteristics ρ(x, t) = ρ(x0 , 0)e−t , since we choose x = x0 at t = 0. The characteristics satisfy dx = t2 e−t ρ(x0 , 0) . dt Since x0 is constant and ρ(x0 , 0) = f (x0 ) , Z t x = f (x0 )

τ 2 e−τ dτ + x0 ,

(26)

0

where the integral can be evaluated (by tables, symbolic integration routines, or repeated integrationby-parts). A parametric representation is ρ(x, t) = f (x0 )e−t , where x0 satisfies (26).

12.6.14 (a) Since ρ = f (x − V t), ρt = −V f 0 , ρx = f 0 , etc.. Therefore 2f

−V f 0 + umax (1 −

ρmax

)f 0 = νf 00 .

12.6.14 (c) By integration −V f + umax (f −

f2 ρmax

) = νf 0 + c. ρ2

2 ) = c. Similarly as x → −∞ If f → ρ2 as x → ∞ such that f 0 → 0, then −V ρ2 + umax (ρ2 − ρmax

ρ2

1 −V ρ1 + umax (ρ1 − ρmax ) = c. Subtracting these yields

V (ρ1 − ρ2 ) + umax (ρ2 − Thus

ρ22 ρmax

) − umax (ρ1 −

ρ2

V =

ρ21 ρmax

)=0.

ρ2

2 1 umax (ρ2 − ρmax ) − umax (ρ1 − ρmax )

ρ2 − ρ1

,

where long division yields the given answer. However, in this form V = [q]/[ρ] since q = ρu and u = umax (1 − ρ/ρmax ) if ν = 0. This is the same as the shock velocity derived in (12.6.20).

12.6.17 (a) The pde is (12.6.12) where c(ρ) = q 0 (ρ), where q = ρu = umax (ρ − ρ2 /ρmax ). Thus c(ρ) = umax (1 − 2ρ/ρmax ) .

By the method of characteristics, if dx/dt = c(ρ), then dρ/dt = 0. Since ρ is constant along the characteristics, the characteristics are straight lines. We use the notation x(0) = x0 . We need to calculate the density wave velocity for ρ = ρmax /5 and ρ = 3ρmax /5: c(ρmax /5) = 3umax /5

and

c(3ρmax /5) = −umax /5 .

Thus, if x0 > 0, from the initial condition ρ(x, t) = 3ρmax /5

along x = −umax t/5 + x0 ,

while if x0 < 0, ρ(x, t) = ρmax /5 along 69

x = 3umax t/5 + x0 .

These two families of characteristics intersect as in Fig. 12.6.9 (a), though if x > x0 , the slopes of the lines should be negative here. Thus a shock exists separating ρ = ρmax /5 from ρ = 3ρmax /5, starting at xs = 0 at t = 0. The shock velocity satisfies (12.6.20): 3 ρmax · 25 ρmax − 15 ρmax · 54 ρmax dxs [q] umax = = umax 5 , = 3 1 dt [ρ] 5 5 ρmax − 5 ρmax

since q = ρu and u( 35 ρmax ) = 25 umax and u( 15 ρmax ) = 45 ρmax . Figure 12.6.9 (c) is valid with xs = umax 5 t, and thus ½ ρmax /5 x < umax 5 t ρ(x, t) = 3ρmax /5 x > umax 5 t.

In addition Fig. 12.6.10 is valid with the changes reflected by the above formula.

12.6.18 (b) The pde is (12.6.12), where

u = umax (1 − ρ2 /ρ2max ) q = ρu = umax (ρ − ρ3 /ρ2max ) c(ρ) = q 0 (ρ) = umax (1 − 3ρ2 /ρ2max ).

Since q 00 (ρ) < 0, q(ρ) is concave down. Furthermore q 0 (ρ) is a decreasing function of ρ. Thus, since initially ½ ρ1 x < 0 ρ(x, 0) = ρ2 x > 0 with ρ2 < ρ1 , the faster density wave velocity is ahead of the slower density wave velocity. Thus density will spread out necessitating fan-like characteristics. By the method of characteristics, if dx/dt = c(ρ), then dρ/dt = 0. The characteristics are straight lines. We use the notation x(0) = x0 . Thus, if x0 > 0 from the initial condition ρ(x, t) = ρ2 along x = c(ρ2 )t + x0 , while if x0 < 0 ρ(x, t) = ρ1

along x = c(ρ1 )t + x0 .

Since these diverge, Figure 12.6.4 is valid. So far ½ ρ2 ρ(x, t) = ρ1

x > c(ρ2 )t x < c(ρ1 )t ,

but we have not determined ρ(x, t) for c(ρ1 )t < x < c(ρ2 )t, since c(ρ2 ) > c(ρ1 ). There, by the method of characteristics, x = c(ρ)t, since at t = 0, x = x0 = 0. We solve for ρ(x, t) there as follows: x = c(ρ) = umax (1 − 3ρ2 /ρ2max ) t x and hence 3ρ2 /ρ2max = 1 − umax t or

ρmax ρ(x, t) = √ 3

r 1−

x . umax t

This is valid in the fan-shaped region, c(ρ1 )t < x < c(ρ2 )t.

70

Chapter 13. Laplace Transform Section 13.2 13.2.4 Using the convolution theorem with g(t − t̄) = 1 (and hence g(t) = 1), ·Z t ¸ L f (t̄)dt̄ = F (s)G(s) = F (s)/s , 0

since g(t) = 1 implies that G(s) = 1/s. An alternate method to derive this result is to let Z t u(t) =

f (t̄)dt̄ ,

so that

du/dt = f (t) .

0

By taking the Laplace transform, we obtain sU (s) = F (s) or

U (s) = F (s)/s, since u(0) = 0.

13.2.5 (b) We use L[−tf (t)] = dF/ds .

Here f (t) = − sin 4t, so that F (s) = −4/(s2 + 16). Thus L[t sin 4t] = 13.2.5 (d) We use

d −4 8s = 2 . ds s2 + 16 (s + 16)2

L[eat f (t)] = F (s − a)

with a = 3 and f (t) = sin 4t. Since from the tables F (s) =

4 , s2 + 16

we have L[e3t f (t)] =

4 4 = 2 . (s − 3)2 + 16 s − 6s + 25

13.2.5 (f) It is possible to express f (t) in terms of step functions: f (t) = t2 [H(t − 5) − H(t − 8)] .

Now we use the formula

L[f (t − b)H(t − b)] = e−bs F (s) .

with b = 5. We want f (t − 5) = t2 , and thus f (t) = (t + 5)2 = t2 + 10t + 25. From the tables, 25 F (s) = s23 + 10 s2 + s . Thus 2 10 25 L[t2 H(t − 5)] = e−5s ( 3 + 2 + ) . s s s 64 Also f (t − 8) = t2 , and thus f (t) = (t + 8)2 = t2 + 16t + 64. From the tables, F (s) = s23 + 16 s2 + s .

Thus 2 10 25 2 16 64 L{t2 [H(t − 5) − H(t − 8)]} = e−5s ( 3 + 2 + ) − e−8s ( 3 + 2 + ) . s s s s s s

13.2.5 (h) We use the formula

L[H(t − b)f (t − b)] = e−bs F (s)

with b = 1 and f (t) = t4 . Consequently from the tables 4!

L[(t − 1)4 H(t − 1)] = e−s 5 . s

71

13.2.6 (e) We use the method of partial fractions s s a b = = + , s2 + 8s + 7 (s + 7)(s + 1) s+7 s+1 where from (13.2.4)

s

a = lim

s→−7 s + 1

Thus

· L

−1

13.2.6 (j) First we let F (s) =

=

7 6

and b = lim

s

s→−1 s + 7

=

−1 .

6

¸ s 7 1 = e−7t − e−t .

2 s + 8s + 7 6 6

s+2 a b c = + + , 2 s(s + 9) s s + 3i s − 3i

using complex roots, where a = lim

s+2

s→0 s2 + 9

Thus f (t) =

=

2 s+2 2 − 3i s+2 2 + 3i , b = lim = , c = lim = . s→−3i s(s − 3i) s→+3i s(s + 3i) 9 −18 −18

2 2 + 3i 3it 2 − 3i −3it 2 2 3i + e + e = − cos 3t + (−2i sin 3t), 9 −18 −18 9 9 18

c1 s+c2 using Euler’s formulas. This can also be obtained by noting F (s) = 2/9 s + s2 +9 and therefore

s + 2 − 29 (s2 + 9) s − 29 s2 1 − 92 s c1 s + c2 s+2 2/9 = − = = = . s2 + 9 s(s2 + 9) s s(s2 + 9) s(s2 + 9) s2 + 9 In this manner using the tables f (t) =

2 1 2 + sin 3t − cos 3t .

9 3 9

We also need to invert −5e−4s F (s) . From the table the inversion of this is −5H(t − 4)f (t − 4) . Thus · ¸ · ¸ s+2 2 1 2 2 1 2 −1 −4s L (1 − 5e ) = + sin 3t− cos 3t−5H(t−4) + sin 3(t − 4) − cos 3 (t − 4) . s(s2 + 9) 9 3 9 9 3 9 13.2.7 (b) By taking the Laplace transform of the differential equation, we obtain sY (s) − 2 + Y (s) = 1/s , since y(0) = 2. Thus Y (s) =

2 + 1/s 1 + 2s a b = = + , s+1 s(s + 1) s s+1

using partial fractions, where from (13.2.4) a = lim

s→0

Consequently

1 + 2s =1 s+1

and b = lim

s→−1

y(t) = 1 + e−t .

72

1 + 2s =1. s

13.2.7 (d) The right-hand side can be expressed in terms of a step function f (t) = e−t H(t − 3) .

Its Laplace transform is

F (s) = G(s)e−3s

where g(t − 3) = e−t or g(t) = e−(t+3) , so that G(s) = e−3 /s + 1. Thus F (s) = e−3(s+1) / s + 1 , which can be obtained in other ways. By taking the Laplace transform of the differential equation, we obtain s2 Y (s) − s3 − 7 + 5[sY (s) − 3] − 6Y (s) = e−(s+1) /s + 1 .

Consequently, Y (s) =

3s + 22 s2 + 5s − 6

+

e−3(s+1) . (s + 1)(s2 + 5s − 6)

We invert these separately: Y1 (s) =

3s + 22 3s + 22 a b = = + , s2 + 5s − 6 (s + 6)s − 1) s+6 s−1

using partial fractions, where a = lim

s→−6

3s + 22 4 =− s−1 7

and b = lim

s→1

3s + 22 25 = . s+6 7

4 −6t t . For the other term we first need Thus y1 (t) = 25 7 e − 7e

G1 (s) =

1 1 c1 c2 c3 = = + + , (s + 1)(s2 + 5s − 6) (s + 1)(s + 6)(s − 1) s+1 s+6 s−1

where using partial fractions 1

c1 = lim

s→−1 (s + 6)(s − 1)

=

−1 1 1 1 1 , c2 = lim = , c3 = lim = . s→−6 s→1 10 (s + 1)(s − 1) 35 (s + 1)(s + 6) 14

Thus g1 (t) = − We use the formula Thus y(t) =

1 −t 1 1 e + e−6t + et .

10 35 14

L−1 [e−3s G1 (s)] = H(t − 3)g1 (t − 3) .

25 t 4 −6t 1 1 1 e − e + e−3 H(t − 3)[− e−(t−3) + e−6(t−3) + et−3 ] .

7 7 10 35 14

Note that for t > 3

· t

y(t) = e

¸ e15 4 1 25 e−6 + + e−6t [ − ] + e−t .

7 14 35 7 10

13.2.7 (f) By taking the Laplace transform of the differential equation, we obtain s2 Y (s) + 4Y (s) = and thus Y (s) =

1 , s2 + 1

1 . (s2 + 1)(s2 + 4) 73

Partial fractions may be utilized. Alternatively Y (s) =

a b + s2 + 1 s2 + 4

because Y (s) is even is s. Thus by multiplying by (s2 + 1)(s2 + 4) a(s2 + 4) + b(s2 + 1) = 1 .

Consequently a + b = 0 and 4a + b = 1. Thus a = −b = 1/3. By inversion, we obtain y(t) =

1 1 1 sin t − ( sin 2t).

3 3 2

Section 13.3 13.3.2 By taking the Laplace transform, we obtain s2 Y (s) + Y (s) = F (s) or Y (s) = sF2(s) +1 . Using the convolution theorem Z t

y(t) =

f (t̄)g(t − t̄)dt̄ , 0

where G(s) = 1/s2 + 1. Thus g(t) = sin t so that Z t y(t) =

f (t̄) sin(t − t̄)dt̄ .

0

If f (t) = δ(t − t0 ), then y(t) = G(t, t0 ): Z t G(t, t0 ) =

δ(t̄ − t0 ) sin(t − t̄)dt̄ = sin(t − t0 ), 0

for t > t0 . Thus

Z t y(t) =

f (t̄)G(t, t̄)dt̄ .

0

Section 13.4 13.4.3 By taking the Laplace transform in t, we obtain s2 U (x, s) − s sin x = c2 ∂ 2 U/∂x2 .

The general solution of this ordinary differential equation is U (x, s) =

s s2 + c2

−sx

sx

sin x + A(s)e c + B(s)e c ,

using the method of undetermined coefficients for a particular solution. Since u(x, t) is bounded as x → ±∞, its Laplace transform must also be bounded as x → ±∞. Thus A(s) = B(s) = 0: U (x, s) =

s sin x . s2 + c2

From the inverse Laplace transform (using the table), we obtain u(x, t) = sin x cos ct .

13.4.4 By taking the Laplace transform in t, we obtain sU = k∂ 2 U/∂x2

74

for x > 0

since u(x, 0) = 0. The general solution of this differential equation is √

√

√

√

U (x, s) = A(s)e− sx/ k + B(s)e sx/ k .

The problems is defined for 0 < x < ∞. The boundary conditions are u(0, t) = f (t) and u(x, t) → 0 as x → ∞. These are Laplace transformed in t, yielding U (0, s) = F (s)

and U (x, s) → 0 as

x→∞,

where F (s) is the Laplace transform of f (t). The condition as x → ∞ yields B(s) = 0, while the condition at x = 0 yields A(s) = F (s): √

√

U (x, s) = F (s)e− sx/ k . √

√

To apply the convolution theorem, we √ must invert G(s) = e− sx/ k . From the table 2 a −3/2 −a /4t g(t) = 2√π t e , where a = x/ k. Thus Z t 2 x f (t̄) √ (t − t̄)−3/2 e−x /4k(t−t̄) dt̄ . u(x, t) = 2 kπ 0

Section 13.5 13.5.3 By taking the Laplace transform in t, we obtain s2 Ū = c2 ∂ 2 Ū /∂x2 ∂ since u(x, 0) = ∂t u(x, 0) = 0. We also Laplace transform the boundary conditions

Ū (0, s) = 0

and

Ūx (L, s) = B(s) ,

where B(s) is the Laplace transform of b(t). The general solution of this ordinary differential equation is s s Ū (x, s) = c1 (s) sinh x + c2 (s) cosh x . c c The boundary conditions yield 0 = c2 (s) and B(s) = c1 sc cosh sc L , and thus sinh sc x c Ū (x, s) = B(s) . s cosh sc L From the convolution theorem

Z t u(x, t) =

b(t̄)f (t − t̄)dt̄ 0

sinh s x

where F (s) = sc cosh sc L ≡ G(s) s . Thus from exercise 13.2.4, c Z t f (t) = g(t̄)dt̄, 0

where

s s ∞ h s iX sinh sc x s e c x − e− c x (x−L) − sc (x+L) c G(s) = c = c sL =c e −e (−1)n e− c 2Ln s cosh sc L e c + e− c L n=0

=c

∞ X

h s i s (−1)n e− c (L−x+2nL) − e− c (x+L+2Ln) .

n=0

Each term in this series is inverted easily: · ¸ ∞ X L − x + 2Ln x + L + 2Ln g(t) = c (−1)n δ(t − ) − δ(t − ) . c c n=0 By integration f (t) = c

∞ X

· (−1)

n=0

n

¸ L − x + 2Ln x + L + 2Ln H(t − ) − H(t − ) . c c

75

Section 13.6 13.6.4 (a) For the Green’s function, q(x, t) is replaced by δ(x − x0 )δ(t − t0 ). The corresponding homogeneous boundary condition must be satisfied G(0, t; x0 , t0 ) = 0 and G → 0 as x → ∞, along with the causality condition G(x, t; x0 , t0 ) = 0 for t < t0 . by taking the Laplace transform in t, the pde becomes sḠ = k∂ 2 Ḡ/∂x2 + δ(x − x0 )e−st0 subject to the boundary conditions Ḡ(0, s; x0 , t0 ) = 0 and Ḡ → 0 as x → ∞. The differential . We equation for Ḡ may be solved using techniques for Green’s functions. For x 6= x0 , sḠ = k∂ 2 Ḡ/∂x2p choose independent homogeneous solutions that each satisfy one of the boundary conditions (sinh ks x √ and e− s/kx ). Thus ( √ p x < x0 A e− s/k x0 sinh √ks x Ḡ(x, s; x0 , t0 ) = p − s/k x A sinh s/k x0 e x > x0 , where we have chosen coefficients such that Ḡ is automatically continuous at x = x0 . Here A is a constant, independent of x0 , obtained from the jump condition on the derivative: ¯x + dḠ ¯¯ 0 + e−st0 , 0=k dx ¯x0 − since Ḡ is continuous. In this way, we obtain √ p p e−st0 e− s/k x0 e−st0 √ A= (sinh s/k x0 + cosh s/k x0 ) = √ . sk sk

Section 13.7 13.7.1 (b) Here F (s) = 1/(s2 + 9) has simple poles at s = ±3i. The inversion integral is Z γ+i∞ 1 est f (t) = ds , 2 2πi γ−i∞ s + 9 for t > 0, where γ > 0 to be to the right of all singularities. Closing the contour to the left for t > 0 and neglecting the large semi-circle (as can be justified) yields (13.7.11) X f (t) = res(sn ) , n

where these are the residues of est /s2 + 9. Since these are simple poles at s = ±3i f (t) =

X e sn t n

2sn

=

e3it e−3it 1 e3it − e−3it 1 + = = sin 3t , 6i −6i 3 2i 3

using Euler’s formula.

Section 13.8 13.8.1 One way to obtain this result is to use (11.2.24) in which case Z L ∂ u(x, t) = − f (x0 ) G(x, t; x0 , 0)dx0 . ∂t 0 0 This can also be obtained using Laplace transforms. Here we evaluate ∂t∂0 G(x, t; x0 , 0) using Laplace transforms. From (13.6.11-12) ½ −1 ∂ sinh sc (L − x0 ) sinh sc x x < x0 Ḡ(x, t; x0 , 0) = s sinh sc x0 sinh sc (L − x) x > x0 . ∂t0 c sinh c L 76

To invert this, we note as in (13.8.5) that the singularities are only simple poles sn , located at the zeros of the denominator sinh Lc sn = 0 . Again s = 0 is not a pole. There are an infinite number of poles located at L sn = i n π n = ±1, ±2, ±3, ... . c The residue at each pole may be evaluated res(sn ) =

R(sn ) P (sn )esn t = , Q0 (sn ) L cosh( Lc sn )

where if x < x0

sn sn (L − x0 ) sinh x , c c and the result for x > x0 can be obtained by symmetry. Since sinh ix = i sin x P (sn ) = − sinh

P (sn ) = sin

nπ nπx nπx0 nπx (L − x0 ) sin = − cos nπ sin sin , L L L L

which is thus also the result for x > x0 . Consequently ∞ X ∂ G(x, t; x0 , 0) = ∂t0 n=−∞

(n6=0)

∞ nπx inπct/L 0 − cos nπ sin nπx 2 X nπx0 nπx nπct L sin L e =− sin sin cos .

L cos(nπ) L n=1 L L L

Thus

Z L

∞

nπx0 2 X nπx nπct sin sin cos dx0 .

L n=1 L L L 0 P∞ nπx nπct This is the result of separation of variables since u(x, t) = n=1 an sin L cos L , where an = R nπx0 2 L L 0 f (x0 ) sin L dx0 . u(x, t) =

f (x0 )

77

Chapter 14. Dispersive Waves Section 14.2 14.2.1 (c) u = ei(kx−ωt) implies i(−iω) = (ik)2 or ω = k 2 .

2

2

14.2.5 Assume φ = A(y)ei(kx−ωt) and free surface s = Bei(kx−ωt) . From Laplace’s equation ∂∂xφ2 + ∂∂yφ2 = 0, we have −k 2 A + A00 (y) = 0.

To solve the bc at the bottom y = −h, it is easier to use solutions which are functions of y + h rather than y. The general solution is a linear combination of exponentials but hyperbolic functions are simpler: A(y) = c1 cosh k(y + h) + c2 sinh k(y + h).

The bc at y = −h becomes A0 (−h) = 0 or c2 = 0. With c1 = A, φ = A cosh k(y + h)ei(kx−ωt) .

The free surface conditions (at y = 0) become ∂φ + gs = 0 ⇒ −iωA cosh kh + gB = 0 ∂t ∂φ ∂s −g = 0 ⇒ kA sinh kh − iωB = 0 ∂y ∂t This linear combination (in A and B) has a nontrivial solution only if its determinant is zero ¯ ¯ ¯ −iω cosh kh g ¯¯ ¯ ¯ k sinh kh iω . ¯ Thus ω 2 cosh kh − gk sinh kh = 0 or ω 2 = gk tanh kh. This can also be obtained by elimination.

2

2

14.2.7 Assume φ = A(y)ei(kx−ωt) and free surface s = Bei(kx−ωt) . From Laplace’s equation ∂∂xφ2 + ∂∂yφ2 = 0, we have −k 2 A + A00 (y) = 0, whose general solution is A(y) = c1 eky + c2 e−ky .

0 i(kx−ωt) Since ∂φ , the bc becomes A0 (y) → 0 as y → −∞. Thus c2 = 0 (assuming k > 0). ∂y = A (y)e With c1 = A,

φ = Aeky ei(kx−ωt) .

The free surface conditions (at y = 0) become ∂φ + gs = 0 ⇒ −iωA + gB = 0 ∂t

78

∂φ ∂s −g = 0 → kA − iωB = 0 ∂y ∂t This linear combination (in A and B) has a nontrivial solution only if its determinant is zero ¯ ¯ −iω ¯ ¯ k

¯ g ¯¯ iω. ¯

Thus ω 2 − gk = 0 or ω 2 = g k. This can also be obtained by elimination.

Section 14.3 14.3.9 (a) Let u = G(x)e−iωf t so that −iωf G = G000 + δ(x).

The jump conditions are that G and G0 are continuous at x = 0, but [G00 ] = −1. Homogeneous solutions √ 1/3 (G = erx ) satisfy −iωf = r3 . One root is r = iωf ≡ ik. The other two roots are r = (± 3 − i)k/2.

1/3

dω

Since k = ωf or ωf = k 3 , in this problem the group velocity dkf = 3k 2 > 0. The radiation condition implies that the oscillatory term exists only for x > 0. Thus n G=

√ i

3

b2 e− 2 kx e 2 a1 e

ikx

¯ ¯ ¯k ¯x

+ b1 e

√

¯ ¯

3 − 2i kx − 2 ¯k¯x

e

x<0 x > 0,

where we have insisted that the non-oscillatory solutions decay. The conditions at x = 0 are (letting x =sgnk) G continuous: b2 = a1 + b1

√

√

G0 continuous: b2 (− 2i + 23 s) = ia1 + b1 (− 2i − 23 s) [G00 ] = −1: −a1 + b1 (− 14 + 34 s2 ) − b2 (− 14 + 34 s2 ) = −1 √ √ Solving this system yields a1 = 2/3, 3b1 = i 3s − 1, and 3b2 = i 3s + 1.

Section 14.5 14.5.3 (b) ω 0 = k 2 − k. The group velocity is a parabola with minimum when ω 00 = 0 or k = 21 . Here ω 0 ( 12 ) = − 14 . There are two waves if xt > − 14 and there are no waves if xt < − 14 .

Section 14.6 ∂k0 ∂θ ∂θ 0 0 0 = k0 + (x − ω 0 (k0 )t) ∂k 14.6.1 k = ∂x ∂x = k0 and −ω = ∂t = −ω0 + (x − ω (k0 )t) ∂t = −ω0 since x = ω (k0 )t.

Section 14.7 14.7.8 From exercise (14.7.7), the consistency equation is ut = − 12 Pxxx + 2Px (u − λ) + P ux . Here we assume P = A + Bλ + Cλ2 with C assumed constant. Substituting this into the consistency equation yields 1 ut = − (Axxx + Bxxx λ) + 2(Ax + Bx λ)(u − λ) + (A + Bλ + Cλ2 )ux .

2 Since we do not wish u(x, t) to depend on λ, the coefficients of each power of λ must be identical. The O(λ2 ) terms are 0 = −2Bx + Cux which can be immediately integrated to yield 2B = Cu + 2B0 (t).

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The O(λ) terms are 0 = − 12 Bxxx +2uBx −2Ax +Bux . In order to integrate to solve for A, we eliminate B to show that 2uBx + Bux = Cuux + ux ( 12 Cu + B0 (t)) may in fact be integrated. Thus, 1 3 2A = − Bxx + Cu2 + B0 u + 2A0 (t).

2 4 Then the O(1) terms yield the following generally nonlinear partial differential equations that satisfy Lax’s equations 1 ut = − Axxx + 2uAx + Aux .

2

Section 14.8 14.8.6 (b) Substituting u = eikx+(σ−iω)t yields σ − iω = −1 + Rk 2 − k 4 − ik 3 . Collecting real and imaginary parts yields the answer.

14.8.7 (c) Substituting u = eikx+(σ−iω)t yields σ − iω = −i(ik) = k. Thus, σ = k and ω = 0. This problem is ill-posed because the growth rate σ is unbounded as k → ∞.

Section 14.9 14.9.3 In the method of multiple scales, (14.9.22) is substituted into the equation and we obtain one addition term to (14.9.23) due to the nonlinear damping: ω2

2 ∂2u ∂ ∂u ∂u ∂u ∂u 3 0 2∂ u + ²[2ω(T ) + ω (T ) ] + ² + ω 2 u = −²(ω +² ) . ∂θ2 ∂T ∂θ ∂θ ∂T 2 ∂θ ∂T

When the perturbation expansion (14.9.24) is introduced, to leading order we obtain ω2 (

∂ 2 u0 + u0 ) = 0 ∂θ2

whose general solution is u0 (θ, T ) = A(T )eiθ + A∗ (T )e−iθ . The O(²) equation is ω2 (

∂ 2 u1 ∂ ∂u0 ∂u0 ∂u0 3 + u1 ) = −[2ω(T ) + ω 0 (T ) ] − ω3 [ ] ∂θ2 ∂T ∂θ ∂θ ∂θ

= −[2ω(T )A0 (T ) + ω 0 (T )A(T )]ieiθ + iω 3 [A3 e3iθ − 3A2 A∗ eiθ ] + (∗) after substituting the expression for u0. The third harmonic terms e±3iθ are not secular. However the terms e±iθ correspond to the forcing frequency equaling the natural frequency, are said to be secular, and must be eliminated: −2ω(T )A0 (T ) − ω 0 (T )A(T ) − 3ω 3 A2 A∗ = 0, which is a differential equation the complex amplitude A(T ) must satisfy. To obtain equations for the amplitude and phase, we substitute A(T ) = r(T )eiψ(T ) and obtain 2ωr0 (T ) + ω 0 (T )r(T ) + 3ω 3 r3 = 0, and ψ 0 (T ) = 0 which implies that the slow phase ψ(T ) is a constant.

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14.9.5 We use the method of multiple scales with a phase θ satisfying dθ dt = ω(²t) where the slowly varying frequency is the usual frequency for a pendulum of length L r ω(²t) =

g .

L(²t)

2

2 dL du We substitute (14.9.22) into ddt2u + ω 2 (²t)u = −² L(T ) dT dt and obtain

ω2

2 ∂ ∂u ∂u 2 dL ∂u ∂u ∂2u 0 2∂ u + ²[2ω(T ) + ω (T ) ] + ² + ω 2 u = −² (ω +² ).

2 2 ∂θ ∂T ∂θ ∂θ ∂T L(T ) dT ∂θ ∂T

instead of (14.9.23). When the perturbation expansion (14.9.24) is introduced, to leading order we obtain ω2 (

∂ 2 u0 + u0 ) = 0 ∂θ2

whose general solution is u0 (θ, T ) = A(T )eiθ + A∗ (T )e−iθ . The O(²) equation is ω2 (

∂ 2 u1 ∂ ∂u0 ∂u0 2 dL ∂u0 + ω 0 (T ) ]−ω + u1 ) = −[2ω(T ) ∂θ2 ∂T ∂θ ∂θ L(T ) dT ∂θ

= −[2ω(T )A0 (T ) + ω 0 (T )A(T )]ieiθ − iω

2 dL iθ Ae + (∗) L(T ) dT

after substituting the expression for u0. All the terms on the right hand side, corresponding to the forcing frequency equaling the natural frequency, are said to be secular and must be eliminated: 2ω(T )A0 (T ) + ω 0 (T )A(T ) + ω

2 dL A = 0, L(T ) dT

which is a differential equation the amplitude A(T ) must satisfy. This equation is easily solved by first dividing by ωA: 2

A0 ω0 2L0 + + = 0.

A ω L

Integrating yields 2 lnA+lnω + 2 lnL =constant, so that the amplitude satisfies A(T ) = c1 ω −1/2 L−1 = c2 L1/4 L−1 = c2 L−3/4 , q g using the expression for the frequency ω. Since dθ = ω(²t) = dt L(²t) , Rq g it follows that θ = L(²t) dt + ψ. Thus Z r g u0 = cL−3/4 cos( dt + ψ).

L(²t) 14.9.8 In the method of multiple scales, we assume the wave amplitude and frequency are slowly varying and ∂u ∂3u ∂θ and ω = − ∂θ satisfy k = ∂x ∂t . Using (14.9.67) and (14.9.68), our equation ∂t = ∂x3 becomes −ω

∂u ∂ ∂ 3 ∂3u ∂3u ∂2u ∂u +² = (k +² ) u = k 3 3 + ²(3k 2 + 3kkX 2 ) + O(²2 ).

2 ∂θ ∂T ∂θ ∂X ∂θ ∂X∂θ ∂θ

We claim that the perturbation method will not work unless the frequency satisfies the dispersion relation 81

ω = k3 .

Conservation of waves kT + ωX = 0 becomes kT + 3k 2 kX = 0. The perturbation expansion u = ∂ 3 u0 0 u0 + ²u1 + ..., yields the O(1) equation −k 3 ( ∂u ∂θ + ∂θ 3 ) = 0, so that the slowly varying plane wave satisfies u0 = A(X,T )eiθ + A∗ (X,T )e−iθ . The O(²) equation is −k 3 (

∂ 2 u0 ∂u1 ∂ 3 u1 ∂u0 ∂ 3 u0 + )=− + 3k 2 + 3kkX 3 2 ∂θ ∂θ ∂T ∂X∂θ ∂θ2 = −AT eiθ − 3(k 2 AX + kkX A)eiθ + (∗).

All terms on the right-hand side are secular (with forcing frequency equaling natural frequency) and must be eliminated, so that AT + 3k 2 AX + 3kkX A = 0.

Section 14.10 14.10.3 In this problem (as we will show), there can be a boundary layer at both boundaries x = 0 and x = 1.

The leading-order solution of the outer expansion (u = u0 + ...) is x u0 = − .

4 Inner (left): Near x = 0, we introduce the boundary layer scaling x = ²1/2 XL , so that the differential equation becomes d2 u − 4u = ²1/2 XL . dXL2 2

The inner (left) expansion (u = U0 (XL ) + ...) yields to leading-order ddXU20 − 4U0 = 0, whose general L

solution is U0 = c1 e2XL + c2 e−2XL . In matching the left boundary layer solution to the outer solution, the limit XL → +∞ is involved, so that to avoid exponential growth on the boundary layer scale, c1 = 0 and thus U0 = c2 e−2XL . The boundary condition at x = 0 (XL = 0) is satisfied if c2 = 1 so that the leading-order solution in the left boundary layer is U0 = e−2XL .

Inner (right): Near x = 1, we introduce the boundary layer scaling x − 1 = ²1/2 XR , so that the differential equation becomes d2 u 1/2 XR .

2 − 4u = 1 + ² dXR 2

The inner (right) expansion (u = V0 (XR ) + ...) yields to leading-order ddXV20 − 4V0 = 1, whose general R

solution is V0 = − 41 + c3 e2XR + c4 e−2XR . In matching the right boundary layer solution to the outer solution, the limit XR → −∞ is involved, so that to avoid exponential growth on the boundary layer scale, c4 = 0 and thus V0 = − 14 + c3 e2XR . The boundary condition at x = 1 (XR = 0) is satisfied if 2 = − 14 + c3 or c3 = 94 so that the leading-order solution in the right boundary layer is 1 9 V0 = − + e2XR .

4 4 82

In this problem, the inner and outer solutions have been obtained without matching. However, one should always do at least a leading-order match. In this problem the matching of the leading-order inner (left) to the leading-order outer solution is limXL →+∞ U0 = limx→0 u0 = 0 since U0 = e−2XL and u0 = − x4 . In this problem the matching of the leading-order inner (right) to the leading-order outer solution is limXR →−∞ V0 = limx→1 u0 = − 14 since V0 = − 14 + 94 e2XR and u0 = − x4 .

2

14.10.6 For the second-order differential equation ² ddxu2 + (2x + 1) du dx + 2u = 0, the differential equation is reduced to first-order when ² = 0. Thus, we expect a boundary layer to be necessary. To determine the location of the boundary layer (either x = 0 or x = 1), we note that derivatives will be large in d2 u du a boundary layer so that roughly the differential equation will be dX 2 + (positive) dX = 0 (because 2x + 1 > 0) whose solution is u = A + Be− (positive)X . The limit X → −∞ must be not permitted so that a boundary layer can only occur at x = 0. The outer solution (away from the boundary layer) 0 has an outer expansion u = u0 (x) + .... The leading-order outer equation is (2x + 1) du dx + 2u0 = 0. This first-order linear ordinary differential equation can be solved by many methods, but since it is an Euler equation perhaps the simplest method is to observe that some power u0 = c(2x + 1)r satisfies this equation, where by direct substitution 2r + 2 = 0 or equivalently r = −1. The constant c can be determined from the boundary condition at x = 1 since we have shown that a boundary layer exists in this problem at x = 0. In this way c = 6 and the leading-order outer solution (which solves the boundary condition at x = 1) is u0 = 6(2x + 1)−1 .

The boundary layer at x = 0 has the scaling x = ²X so that the differential equation in the boundary d2 u du layer is dX 2 + (2²X + 1) dX + 2²u = 0. The inner expansion is u = U0 (X) + ... and the leading-order 2 −X 0 inner solution satisfies ddXU20 + dU . The boundary dX = 0, whose general solution is U0 (X) = A + Be condition at x = 0 (X = 0) yields 1 = A + B. Thus, we may express the leading-order inner solution as U0 (X) = 1 − B + Be−X .

The constant B can be determined by matching the leading-order inner and outer solutions. Here, limX→+∞ U0 = limx→0 u0 , which yields 1 − B = 6 or B = −5. Thus, the leading-order inner solution is U0 (X) = 6 − 5e−X where x = ²X .

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