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Solution Manual for Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 5th Edition

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    S OLUTIONS M ANUAL A PPLIED P ARTIAL D IFFERENTIAL E QUATION WITH F OURIER S ERIES AND B OUNDARY V ALUE P ROBLEMS Richard Haberman Southern Methodist University

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    Chapter 1. Heat Equation Section 1.2 1.2.9 (d) Circular cross section means that P = 2 πr, A = πr 2 , 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 cρ du dt = − 2 h r u . The solution of this first-order linear differential equation with constant coefficients, which satisfies the initial condition u (0) = u 0 , is u ( t ) = u 0 exp [ − 2 h cρr t ] . Section 1.3 1.3.2 ∂u/∂x is continuous if K 0 ( x 0 − ) = K 0 ( x 0 +), that is, if the conductivity is continuous. Section 1.4 1.4.1 (a) Equilibrium satisfies (1.4.14), d 2 u/dx 2 = 0, whose general solution is (1.4.17), u = c 1 + c 2 x . The boundary condition u (0) = 0 implies c 1 = 0 and u ( L ) = T implies c 2 = T /L so that u = T x/L . 1.4.1 (d) Equilibrium satisfies (1.4.14), d 2 u/dx 2 = 0, whose general solution (1.4.17), u = c 1 + c 2 x . From the boundary conditions, u (0) = T yields T = c 1 and du/dx ( L ) = α yields α = c 2 . Thus u = T + αx . 1.4.1 (f) In equilibrium, (1.2.9) becomes d 2 u/dx 2 = − Q/K 0 = − x 2 , whose general solution (by integrating twice) is u = − x 4 / 12 + c 1 + c 2 x . The boundary condition u (0) = T yields c 1 = T , while du/dx ( L ) = 0 yields c 2 = L 3 / 3. Thus u = − x 4 / 12 + L 3 x/ 3 + T . 1.4.1 (h) Equilibrium satisfies d 2 u/dx 2 = 0. One integration yields du/dx = c 2 , the second integration yields the general solution u = c 1 + c 2 x . x = 0 : c 2 − ( c 1 − T ) = 0 x = L : c 2 = α and thus c 1 = T + α. Therefore, u = ( T + α ) + αx = T + α ( x + 1). 1.4.7 (a) For equilibrium: d 2 u dx 2 = − 1 implies u = − x 2 2 + c 1 x + c 2 and du dx = − x + c 1 . From the boundary conditions du dx (0) = 1 and du dx ( L ) = β, c 1 = 1 and − L + c 1 = β which is consistent only if β + L = 1. If β = 1 − L , there is an equilibrium solution ( u = − x 2 2 + x + c 2 ). 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: d dt ∫ L 0 cρu dx = − du dx (0) + du dx ( L ) + ∫ L 0 Q 0 dx = − 1 + β + L. If β + L = 1, then the total thermal energy is constant and the initial energy = the final energy: ∫ L 0 f ( x ) dx = ∫ L 0 ( − x 2 2 + x + c 2 ) dx, which determines c 2 . If β + L 6 = 1, then the total thermal energy is always changing in time and an equilibrium is never reached. 1

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    Section 1.5 1.5.9 (a) In equilibrium, (1.5.14) using (1.5.19) becomes d dr ( r du dr ) = 0. Integrating once yields rdu/dr = c 1 and integrating a second time (after dividing by r ) yields u = c 1 ln r + c 2 . An alternate general solution is u = c 1 ln( r/r 1 ) + c 3 . The boundary condition u ( r 1 ) = T 1 yields c 3 = T 1 , while u ( r 2 ) = T 2 yields c 1 = ( T 2 − T 1 ) / ln( r 2 /r 1 ). Thus, u = 1 ln( r 2 /r 1 ) [( T 2 − T 1 ) ln r/r 1 + T 1 ln( r 2 /r 1 )]. 1.5.11 For equilibrium, the radial flow at r = a , 2 πaβ , must equal the radial flow at r = b , 2 πb . Thus β = b/a . 1.5.13 From exercise 1.5.12, in equilibrium d dr ( r 2 du dr ) = 0. Integrating once yields r 2 du/dr = c 1 and integrat- ing a second time (after dividing by r 2 ) yields u = − c 1 /r + c 2 . The boundary conditions u (4) = 80 and u (1) = 0 yields 80 = − c 1 / 4 + c 2 and 0 = − c 1 + c 2 . Thus c 1 = c 2 = 320 / 3 or u = 320 3 ( 1 − 1 r ) . 2

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    Chapter 2. Method of Separation of Variables Section 2.3 2.3.1 (a) u ( r, t ) = φ ( r ) h ( t ) yields φ dh dt = kh r d dr ( r dφ dr ) . Dividing by kφh yields 1 kh dh dt = 1 rφ d dr ( r dφ dr ) = − λ or dh dt = − λkh and 1 r d dr ( r dφ dr ) = − λφ . 2.3.1 (c) u ( x, y ) = φ ( x ) h ( y ) yields h d 2 φ dx 2 + φ d 2 h dy 2 = 0. Dividing by φh yields 1 φ d 2 φ dx 2 = − 1 h d 2 h dy 2 = − λ or d 2 φ dx 2 = − λφ and d 2 h dy 2 = λh . 2.3.1 (e) u ( x, t ) = φ ( x ) h ( t ) yields φ ( x ) dh dt = kh ( t ) d 4 φ dx 4 . Dividing by kφh , yields 1 kh dh dt = 1 φ d 4 φ dx 4 = λ . 2.3.1 (f) u ( x, t ) = φ ( x ) h ( t ) yields φ ( x ) d 2 h dt 2 = c 2 h ( t ) d 2 φ dx 2 . Dividing by c 2 φh , yields 1 c 2 h d 2 h dt 2 = 1 φ d 2 φ dx 2 = − λ . 2.3.2 (b) λ = ( nπ/L ) 2 with L = 1 so that λ = n 2 π 2 , n = 1 , 2 , . . . 2.3.2 (d) (i) If λ > 0 , φ = c 1 cos √ λx + c 2 sin √ λx. φ (0) = 0 implies c 1 = 0, while dφ dx ( L ) = 0 implies c 2 √ λ cos √ λL = 0. Thus √ λL = − π/ 2 + nπ ( n = 1 , 2 , . . . ). (ii) If λ = 0 , φ = c 1 + c 2 x . φ (0) = 0 implies c 1 = 0 and dφ/dx ( L ) = 0 implies c 2 = 0. Therefore λ = 0 is not an eigenvalue. (iii) If λ < 0, let λ = − s and φ = c 1 cosh √ sx + c 2 sinh √ sx. φ (0) = 0 implies c 1 = 0 and dφ/dx ( L ) = 0 implies c 2 √ s cosh √ sL = 0. Thus c 2 = 0 and hence there are no eigenvalues with λ < 0. 2.3.2 (f) The simpliest method is to let x ′ = x − a . Then d 2 φ/dx ′ 2 + λφ = 0 with φ (0) = 0 and φ ( b − a ) = 0. Thus (from p. 46) L = b − a and λ = [ nπ/ ( b − a )] 2 , n = 1 , 2 , . . . . 2.3.3 From (2.3.30), u ( x, t ) = ∑ ∞ n =1 B n sin nπx L e − k ( nπ/L ) 2 t . The initial condition yields 2 cos 3 πx L = ∑ ∞ n =1 B n sin nπx L . From (2.3.35), B n = 2 L ∫ L 0 2 cos 3 πx L sin nπx L dx . 2.3.4 (a) Total heat energy = ∫ L 0 cρuA dx = cρA ∑ ∞ n =1 B n e − k ( nπ L ) 2 t 1 − cos nπ nπ L , using (2.3.30) where B n satisfies (2.3.35). 2.3.4 (b) heat flux to right = − K 0 ∂u/∂x total heat flow to right = − K 0 A∂u/∂x heat flow out at x = 0 = K 0 A ∂u ∂x ∣ ∣ x =0 heat flow out ( x = L ) = − K 0 A ∂u ∂x ∣ ∣ x = L 2.3.4 (c) From conservation of thermal energy, d dt ∫ L 0 u dx = k ∂u ∂x ∣ ∣ ∣ L 0 = k ∂u ∂x ( L ) − k ∂u ∂x (0) . Integrating from t = 0 yields ∫ L 0 u ( x, t ) dx ︸ ︷︷ ︸ heat energy at t − ∫ L 0 u ( x, 0) dx ︸ ︷︷ ︸ initial heat energy = k ∫ t 0 [ ∂u ∂x ( L ) ︸ ︷︷ ︸ integral of flow in at x = L − ∂u ∂x (0) ] dx ︸ ︷︷ ︸ integral of flow out at x = L . 2.3.8 (a) The general solution of k d 2 u dx 2 = αu ( α > 0) is u ( x ) = a cosh √ α k x + b sinh √ α k x . The boundary condition u (0) = 0 yields a = 0, while u ( L ) = 0 yields b = 0. Thus u = 0. 3

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    2.3.8 (b) Separation of variables, u = φ ( x ) h ( t ) or φ dh dt + αφh = kh d 2 φ dx 2 , yields two ordinary differential equations (divide by kφh ): 1 kh dh dt + α k = 1 φ d 2 φ dx 2 = − λ . Applying the boundary conditions, yields the eigenvalues λ = ( nπ/L ) 2 and corresponding eigenfunctions φ = sin nπx L . The time-dependent part are exponentials, h = e − λkt e − αt . Thus by superposition, u ( x, t ) = e − αt ∑ ∞ n =1 b n sin nπx L e − k ( nπ/L ) 2 t , where the initial conditions u ( x, 0) = f ( x ) = ∑ ∞ n =1 b n sin nπx L yields b n = 2 L ∫ L 0 f ( x ) sin nπx L dx . As t → ∞ , u → 0, the only equilibrium solution. 2.3.9 (a) If α < 0, the general equilibrium solution is u ( x ) = a cos √ − α k x + b sin √ − α k x . The boundary condition u (0) = 0 yields a = 0, while u ( L ) = 0 yields b sin √ − α k L = 0. Thus if √ − α k L 6 = nπ, u = 0 is the only equilibrium solution. However, if √ − α k L = nπ , then u = A sin nπx L is an equilibrium solution. 2.3.9 (b) Solution obtained in 2.3.8 is correct. If − α k = ( π L ) 2 , u ( x, t ) → b 1 sin πx L , the equilibrium solution. If − α k < ( π L ) 2 , then u → 0 as t → ∞ . However, if − α k > ( π L ) 2 , u → ∞ (if b 1 6 = 0). Note that b 1 > 0 if f ( x ) ≥ 0. Other more unusual events can occur if b 1 = 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). (a) A 0 = 1 L ∫ L L/ 2 1 dx = 1 2 , A n = 2 L ∫ L L/ 2 cos nπx L dx = 2 L · L nπ sin nπx L ∣ ∣ ∣ L L/ 2 = − 2 nπ sin nπ 2 (b) by inspection A 0 = 6 , A 3 = 4, others = 0. (c) A 0 = − 2 L ∫ L 0 sin πx L dx = 2 π cos πx L ∣ ∣ ∣ L 0 = 2 π (1 − cos π ) = 4 /π, A n = − 4 L ∫ L 0 sin πx L cos nπx L dx (d) by inspection A 8 = − 3, others = 0. 2.4.3 Let x ′ = x − π . Then the boundary value problem becomes d 2 φ/dx ′ 2 = − λφ subject to φ ( − π ) = φ ( π ) and dφ/dx ′ ( − π ) = dφ/dx ′ ( π ). Thus, the eigenvalues are λ = ( nπ/L ) 2 = n 2 π 2 , since L = π, n = 0 , 1 , 2 , ... with the corresponding eigenfunctions being both sin nπx ′ /L = sin n ( x − π ) = ( − 1) n sin nx = > sin nx and cos nπx ′ /L = cos n ( x − π ) = ( − 1) n cos nx = > cos nx . Section 2.5 2.5.1 (a) Separation of variables, u ( x, y ) = h ( x ) φ ( y ), implies that 1 h d 2 h dx 2 = − 1 φ d 2 φ dy 2 = − λ . Thus d 2 h/dx 2 = − λh subject to h ′ (0) = 0 and h ′ ( L ) = 0. Thus as before, λ = ( nπ/L ) 2 , n = 0 , l, 2 , . . . with h ( x ) = cos nπx/L . Furthermore, d 2 φ dy 2 = λφ = ( nπ L ) 2 φ so that n = 0 : φ = c 1 + c 2 y , where φ (0) = 0 yields c 1 = 0 n 6 = 0 : φ = c 1 cosh nπy L + c 2 sinh nπy L , where φ (0) = 0 yields c 1 = 0. The result of superposition is u ( x, y ) = A 0 y + ∞ ∑ n =1 A n cos nπx L sinh nπy L . The nonhomogeneous boundary condition yields f ( x ) = A 0 H + ∞ ∑ n =1 A n sinh nπH L cos nπx L , so that A 0 H = 1 L ∫ L 0 f ( x ) dx and A n sinh nπH L = 2 L ∫ L 0 f ( x ) cos nπx L dx. 4

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    2.5.1 (c) Separation of variables, u = h ( x ) φ ( y ), yields 1 h d 2 h dx 2 = − 1 φ d 2 φ dy 2 = 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 ) = ∞ ∑ n =1 A n cosh nπx H sin nπy H . The nonhomogeneous boundary condition at x = L yields g ( y ) = ∑ ∞ n =1 A n cosh nπL H sin nπy H , so that A n is determined by A n cosh nπL H = 2 H ∫ H 0 g ( y ) sin nπy H dy. 2.5.1 (e) Separation of variables, u = φ ( x ) h ( y ), yields the eigenvalues λ = ( nπ/L ) 2 and corresponding eigenfunctions φ = sin nπx/L, n = 1 , 2 , 3 , ... The y -dependent differential equation, d 2 h dy 2 = ( nπ L ) 2 h , satisfies h (0) − dh dy (0) = 0. The general solution h = c 1 cosh nπy L + c 2 sinh nπy L obeys h (0) = c 1 , while dh dy = nπ L ( c 1 sinh nπy L + c 2 cosh nπy L ) obeys dh dy (0) = c 2 nπ L . Thus, c 1 = c 2 nπ L and hence h n ( y ) = cosh nπy L + L nπ sinh nπy L . Superposition yields u ( x, y ) = ∞ ∑ n =1 A n h n ( y ) sin nπx/L, where A n is determined from the boundary condition, f ( x ) = ∑ ∞ n =1 A n h n ( H ) sin nπx/L, and hence A n h n ( H ) = 2 L ∫ L 0 f ( x ) sin nπx/L dx . 2.5.2 (a) From physical reasoning (or exercise 1.5.8), the total heat flow across the boundary must equal zero in equilibrium (without sources, i.e. Laplace’s equation). Thus ∫ L 0 f ( x ) dx = 0 for a solution. 2.5.3 In order for u to be bounded as r → ∞ , c 1 = 0 in (2.5.43) and ¯ c 2 = 0 in (2.5.44). Thus, u ( r, θ ) = ∞ ∑ n =0 A n r − n cos nθ + ∞ ∑ n =1 B n r − n sin nθ. (a) The boundary condition yields A 0 = ln 2 , A 3 a − 3 = 4, other A n = 0 , B n = 0 . (b) The boundary conditions yield (2.5.46) with a − n replacing a n . Thus, the coefficients are determined by (2.5.47) with a n replaced by a − n 2.5.4 By substituting (2.5.47) into (2.5.45) and interchanging the orders of summation and integration u ( r, θ ) = 1 π ∫ π − π f ( ¯ θ ) [ 1 2 + ∞ ∑ n =1 ( r a ) n ( cos nθ cos n ¯ θ + sin nθ sin n ¯ θ ) ] d ¯ θ. Noting the trigonometric addition formula and cos z = R e [ e iz ], we obtain u ( r, θ ) = 1 π ∫ π − π f ( ¯ θ ) [ − 1 2 + Re ∞ ∑ n =0 ( r a ) n e in ( θ − ¯ θ ) ] d ¯ θ. Summing the geometric series enables the bracketed term to be replaced by − 1 2 + Re 1 1 − r a e i ( θ − ¯ θ ) = − 1 2 + 1 − r a cos( θ − ¯ θ ) 1 + r 2 a 2 − 2 r a cos( θ − ¯ θ ) = 1 2 − 1 2 r 2 a 2 1 + r 2 a 2 − 2 r a cos( θ − ¯ θ ) . 5

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    2.5.5 (a) The eigenvalue problem is d 2 φ/dθ 2 = − λφ subject to dφ/dθ (0) = 0 and φ ( π/ 2) = 0. It can be 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 = (2 n − 1) 2 . The radially dependent term satisfies (2.5.40), and hence the boundedness condition at r = 0 yields G ( r ) = r 2 n − 1 . Superposition yields u ( r, θ ) = ∞ ∑ n =1 A n r 2 n − 1 cos(2 n − 1) θ. The nonhomogeneous boundary condition becomes f ( θ ) = ∞ ∑ n =1 A n cos(2 n − 1) θ or A n = 4 π ∫ π/ 2 0 f ( θ ) cos(2 n − 1) θ dθ. 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 = (2 n ) 2 and φ = sin nπθ L = sin 2 nθ . The radial part that remains bounded at r = 0 is G = r √ λ = r 2 n . By superposition, u ( r, θ ) = ∞ ∑ n =1 A n r 2 n sin 2 nθ . To apply the nonhomogeneous boundary condition, we differentiate with respect to r : ∂u ∂r = ∞ ∑ n =1 A n (2 n ) r 2 n − 1 sin 2 nθ . The bc at r = 1, f ( θ ) = ∑ ∞ n =1 2 nA n sin 2 nθ , determines A n , 2 nA n = 4 π ∫ π/ 2 0 f ( θ ) sin 2 nθ 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 = n 2 and corresponding eigenfunctions φ = sin nπθ/L = sin nθ, n = 1 , 2 , 3 , ... The radial part which is bounded at r = 0 is G = r √ λ = r n . Thus by superposition u ( r, θ ) = ∞ ∑ n =1 A n r n sin nθ . The bc at r = a , g ( θ ) = ∑ ∞ n =1 A n a n sin nθ, determines A n , A n a n = 2 π ∫ π 0 g ( θ ) sin nθ dθ . 2.5.7 (b) The boundary conditions of (2.5.37) must be replaced by φ ′ (0) = 0 and φ ′ ( π/ 3) = 0. This will yield a cosine series with L = π/ 3 , λ = ( nπ/L ) 2 = (3 n ) 2 and φ = cos nπθ/L = cos 3 nθ, n = 0 , 1 , 2 , . . . . The radial part which is bounded at r = 0 is G = r √ λ = r 3 n . Thus by superposition u ( r, θ ) = ∞ ∑ n =0 A n r 3 n cos 3 nθ . The boundary condition at r = a , g ( θ ) = ∑ ∞ n =0 A n a 3 n cos 3 nθ, determines A n : A 0 = 3 π ∫ π/ 3 0 g ( θ ) dθ and ( n 6 = 0) A n a 3 n = 6 π ∫ π/ 3 0 g ( θ ) cos 3 nθ 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 r n and r − n (1 and ln r for n = 0), we choose φ 1 ( r ) such that φ 1 ( a ) = 0 and φ 2 ( r ) such that φ 2 ( b ) = 0 : φ 1 ( r ) = { ln( r/a ) n = 0 ( r a ) n − ( a r ) n n 6 = 0 φ 2 ( r ) = { ln( r/b ) n = 0 ( r b ) n − ( b r ) n n 6 = 0 . 6

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    Then by superposition u ( r, θ ) = ∞ ∑ n =0 cos nθ [ A n φ 1 ( r ) + B n φ 2 ( r )] + ∞ ∑ n =1 sin nθ [ C n φ 1 ( r ) + D n φ 2 ( r )] . The boundary conditions at r = a and r = b , f ( θ ) = ∞ ∑ n =0 cos nθ [ A n φ 1 ( a ) + B n φ 2 ( a )] + ∞ ∑ n =1 sin nθ [ C n φ 1 ( a ) + D n φ 2 ( a )] g ( θ ) = ∞ ∑ n =0 cos nθ [ A n φ 1 ( b ) + B n φ 2 ( b )] + ∞ ∑ n =1 sin nθ [ C n φ 1 ( b ) + D n φ 2 ( b )] easily determine A n , B n , C n , D n since φ 1 ( a ) = 0 and φ 2 ( b ) = 0 : D n φ 2 ( a ) = 1 π ∫ π − 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 = (2 n ) 2 and the eigenfunctions are φ = sin nπθ/L = sin 2 nθ, n = 1 , 2 , 3 , . . . . The radial part which is zero at r = a is G = ( r/a ) 2 n − ( a/r ) 2 n . Thus by superposition, u ( r, θ ) = ∞ ∑ n =1 A n [ ( r a ) 2 n − ( a r ) 2 n ] sin 2 nθ. The nonhomogeneous boundary condition, f ( θ ) = ∑ ∞ n =1 A n [ ( b a ) 2 n − ( a b ) 2 n ] sin 2 nθ, determines A n : A n [ ( b a ) 2 n − ( a b ) 2 n ] = 4 π ∫ π/ 2 0 f ( θ ) sin 2 nθ dθ. 2.5.9 (b) The two homogeneous boundary conditions are in r , and hence φ ( r ) must be an eigenvalue problem. By separation of variables, u = φ ( r ) G ( θ ) , d 2 G/dθ 2 = λG and r 2 d 2 φ dr 2 + r dφ dr + λφ = 0 . The radial equation is equidimensional (see p.78) and solutions are in the form φ = r p . Thus p 2 = − λ (with λ > 0) so that p = ± i √ λ . r ± i √ λ = e ± i √ λ ln r . 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 = c 1 cos[ √ λ ln( r/a )] + c 2 sin[ √ λ ln( r/a )] . The homogeneous condition φ ( a ) = 0 yields 0 = c 1 , while φ ( b ) = 0 implies sin[ √ λ ln( r/a )] = 0 . Thus √ λ ln( b/a ) = nπ, n = 1 , 2 , 3 , ... and the corresponding eigenfunctions are φ = sin [ nπ ln( r/a ) ln( b/a ) ] . The solution of the θ -equation satisfying G (0) = 0 is G = sinh √ λθ = sinh nπθ ln( b/a ) . Thus by superposition u = ∞ ∑ n =1 A n sinh nπθ ln( b/a ) sin [ nπ ln( r/a ) ln( b/a ) ] . The nonhomogeneous boundary condition, f ( r ) = ∞ ∑ n =1 A n sinh nπ 2 2 ln( b/a ) sin [ nπ ln( r/a ) ln( b/a ) ] , will determine A n . 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 A n sinh nπ 2 2 ln( b/a ) = 2 ∫ 1 0 f ( r ) sin [ nπ ln( r/a ) ln( b/a ) ] dz. But dz = dr/r ln( b/a ). Thus A n sinh nπ 2 2 ln( b/a ) = 2 ln( b/a ) ∫ 1 0 f ( r ) sin [ nπ ln( r/a ) ln( b/a ) ] dr/r. 7

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    Chapter 3. Fourier Series Section 3.2 3.2.2 (a) x ∼ a 0 + ∑ ∞ n =1 a n cos nπx/L + ∑ ∞ n =1 b n sin nπx/L . From (3.2.2), a 0 = 0 since f ( x ) is odd, ( n 6 = 0) a n = 0 since f ( x ) is odd, and b n = 1 L ∫ L − L x sin nπx/L dx = 2 L nπ ( − 1) n +1 . 3.2.2 (c) sin πx/L ∼ a 0 + ∑ ∞ n =1 a n cos nπx/L + ∑ ∞ n =1 b n sin nπx/L . By inspection, b 1 = 1, all others = 0. 3.2.2 (f) From (3.2.2), a 0 = 1 2 L ∫ L 0 dx = 1 / 2 ( n 6 = 0) a n = 1 L ∫ L 0 cos nπx/L dx = 0 b n = 1 L ∫ L 0 sin nπx/L dx = − 1 nπ cos nπx L ∣ ∣ ∣ L 0 = 1 − cos nπ nπ Thus b n = 2 /nπ, n odd, but b n = 0, n even. 3.3.2 (d) From (3.3.6), B n = 2 L ∫ L/ 2 0 sin nπx/L dx = − 2 nπ cos nπx/L ∣ ∣ ∣ L/ 2 0 = 2 nπ ( 1 − cos nπ 2 ) . 3.3.10 f ( − x ) = { x 2 − x < 0 (or x > 0) e x − x > 0 (or x < 0) . Thus . f e ( x ) = 1 2 [ f ( x ) + f ( − x )] = 1 2 { x 2 + e x x < 0 x 2 + e − x x > 0 f 0 ( x ) = 1 2 [ f ( x ) − f ( − x )] = 1 2 { x 2 − e x x < 0 e − x − x 2 x > 0 3.3.13 b n = 2 L ∫ L 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 b n = 0 for n even. Section 3.4 3.4.1 (a) ∫ b a = ∫ c − a + ∫ b c + . Thus ∫ b a u dv dx dx = uv ∣ ∣ ∣ ∣ ∣ c − a + uv ∣ ∣ ∣ ∣ ∣ ∣ b c + − ∫ b a v du dx dx = uv ∣ ∣ ∣ ∣ ∣ ∣ ∣ b a + uv ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ c − c + − ∫ b a v du dx dx. 3.4.3 (a) We want to determine the sine coefficients of df /dx : df dx = ∑ ∞ n =1 b n sin nπx L , where the cosine coefficients of f are given f = ∞ ∑ n =0 a n cos nπx L ( n 6 = 0) a n = 2 L ∫ L 0 f cos nπx L dx. Here by integration by parts b n = 2 L ∫ L 0 df dx sin nπx L dx = 2 L [ f sin nπx L ∣ ∣ ∣ x 0 − 0 + f sin nπx L ∣ ∣ ∣ ∣ L x 0 + − nπ L ∫ L 0 f cos nπx L dx ] . Thus b n = 2 L sin nπx 0 L ( α − β ) − nπ L a n . 8

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    3.4.9 ∂u ∂x = ∑ ∞ n =1 ( nπ L ) b n cos nπx L since u (0) = 0 and u ( L ) = 0. ∂ 2 u ∂x 2 ∼ − ∑ ∞ n =1 ( nπ L ) 2 b n sin nπx L . Thus from p. 119, ∑ ∞ n =1 db n dt sin nπx L ∼ − k ∑ ∞ n =1 ( nπ L ) 2 b n sin nπx L + q . Thus db n dt + k ( nπ L ) 2 b n = 2 L ∫ L 0 q sin nπx L dx . 3.4.12 The eigenfunctions of the related homogeneous problem are cos nπx/L, n = 0 , 1 , 2 , . . . . Thus u ∼ ∑ ∞ n =0 A n ( t ) cos nπx/L , which can be differentiated (if u is continuous) since it is a cosine series: ∂u/∂x ∼ ∑ ∞ n =0 A n ( − nπ/L ) sin nπx L . This can be differentiated again (if ∂u/∂x is continuous) only because ∂u/∂x (0) = 0 and ∂u/∂x ( L ) = 0: ∂ 2 u/∂x 2 ∼ − ∑ ∞ n =0 A n ( nπ/L ) 2 cos nπx/L . Thus from p. 119 ∞ ∑ n =0 [ dA n dt + k ( nπ L ) 2 A n ] cos nπx L = e − t + e − 2 t cos 3 πx L . The right hand side is a simple cosine series (with only two non-zero terms). Thus dA n dt + k ( nπ L ) 2 A n =    e − t n = 0 e − 2 t n = 3 0 otherwise . The initial conditions are A 0 (0) = 1 L ∫ L 0 f ( x ) dx and ( n 6 = 0) A n (0) = 2 L ∫ L 0 f ( x ) cos nπx L dx. The solution of the differential equations are n 6 = 0 , 3 A n ( t ) = A n (0) e − k ( nπ/L ) 2 t A 0 ( t ) = A 0 (0) + 1 − e − t , obtained by integration A 3 ( t ) = A 3 (0) e − k ( nπ/L ) 2 t + e − 2 t − e − k ( nπ/L ) 2 t k ( 3 π L ) 2 − 2 , obtained by using the method of undetermined coefficients (judicious guessing) for the particular solution. This works if e − 2 t is not a homogeneous solution, i.e., − 2 6 = − k (3 π/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 ) , sinh x ∼ 1 L (cosh L − 1) + ∑ ∞ n =1 [ nπ L b n + 2 L (( − 1) n cosh L − 1) ] cos nπx L . Since this is a cosine series, it may be differentiated cosh x ∼ − ∞ ∑ n =1 [ ( nπ L ) 2 b n + 2 nπ L 2 (( − 1) n cosh L − 1) ] sin nπx L . Thus b n = − ( nπ L ) 2 b n − 2 nπ L 2 [( − 1) n cosh L − 1] or b n = 2 nπ L 2 1 − ( − 1) n cosh L 1+( nπ/L ) 2 . 3.5.4 (b) Integrating yields sinh x = A 0 + ∑ ∞ n =1 − L nπ b n cos nπx/L , where A 0 = 1 L ∫ L 0 sinh x dx = 1 L (cosh L − 1). Integrating again yields cosh x − 1 = A 0 x + ∑ ∞ n =1 − ( L nπ ) 2 b n sin nπx/L . Thus ∞ ∑ n =1 b n [ 1 + ( L nπ ) 2 ] sin nπx/L = 1 + A 0 x. Using (3.3.8) and (3.3.l2) b n [ 1 + ( L nπ ) 2 ] = 2 nπ [1 − ( − 1) n ] + 1 L (cosh L − 1) 2 L nπ ( − 1) n +1 or b n = 2 nπ 1 − ( − 1) n cosh L 1 + ( L nπ ) 2 . 3.5.7 Evaluate (3.5.6) at x = L/ 2: L 2 8 = L 2 4 − 4 L 2 π 3 ( 1 − 1 3 3 + 1 5 3 − . . . ) or 1 − 1 3 3 + 1 5 3 − 1 7 3 = L 2 / 8 4 L 2 /π 3 = π 3 / 32 . 9

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    Section 3.6 3.6.1 The complex Fourier coefficient is defined by (3.6.7): c n = 1 2 L ∫ L − L f ( x ) e inπx/L dx = 1 2 L 4 ∫ x 0 + 4 x 0 e inπx/L dx . Thus c n = 1 2 L 4 L inπ e inπx/L ∣ ∣ ∣ ∣ x 0 + 4 x 0 = 1 2 inπ 4 e inπx 0 /L ( e inπ 4 /L − 1) . Equivalently, c n = 1 2 inπ 4 e inπ ( x 0 + 4 / 2) /L ( e inπ 4 / 2 L − e − inπ 4 / 2 L ) or c n = 1 nπ 4 e inπ ( x 0 + 4 / 2) /L sin( nπ 4 / 2 L ) . 10

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    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 are c √ λ . The boundary condition φ (0) = 0 implies φ = c 1 sin √ λx , while dφ/dx ( H ) = 0 yields √ λH = ( m − 1 2 ) π with m = 1 , 2 , 3. Thus the frequencies are ( m − 1 2 ) πc/H and the eigenfunctions are sin( m − 1 2 ) πx/H . 4.4.2 (c) By separation of variables, u = φ ( x ) h ( t ) , d 2 h dt 2 = − λh and T 0 d 2 φ dx 2 + ( α + λρ 0 ) φ = 0. With φ (0) = 0 and φ ( L ) = 0 , ( α + λρ 0 ) /T 0 = ( nπ/L ) 2 , 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 ) = ∞ ∑ n =1 A n sin √ λ n t sin nπx/L, where the frequencies of vibration are √ λ n = √ ( nπ/L ) 2 T 0 − α ρ 0 . The other initial condition, f ( x ) = ∑ ∞ n =1 A n √ λ n sin nπx/L, determines A n A n √ λ n = 2 L ∫ L 0 f ( x ) sin nπx/L dx. 4.4.3 (b) By separation of variables, u = φ ( x ) h ( t ) , ρ 0 h ′′ + βh ′ hT 0 = φ ′′ φ = − λ. The boundary conditions φ (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, ρ 0 h ′′ + βh ′ + ( nπ L ) 2 T 0 h = 0 , and hence can be solved by substitution h = e rt . This yields the quadratic equation ρ 0 r 2 + βr + ( nπ L ) 2 T 0 = 0 , whose roots are r = − β ± √ β 2 − 4 ρ 0 T 0 ( nπ/L ) 2 2 ρ 0 . Since β 2 < 4 ρ 0 T 0 ( π/L ) 2 , the discriminant is < 0 for all n : r = − β 2 ρ 0 + iw n , where w n = √ T 0 ρ 0 ( nπ L ) 2 − β 2 4 ρ 2 0 . Real solutions are h = e − βt/ 2 ρ 0 (sin w n t, cos w n t ). Thus by superposition u = e − βt/ 2 ρ 0 ∞ ∑ n =1 ( a n cos w n t + b n sin w n t ) sin nπx L . The initial condition u ( x, 0) = f ( x ) determines a n , a n = 2 L ∫ L 0 f ( x ) sin nπx L dx, while ∂u ∂t ( x, 0) = g ( x ) is a little more complicated, g ( x ) = ∑ ∞ n =1 b n w n sin nπx L − β 2 ρ 0 ∞ ∑ n =1 a n sin nπx L ︸ ︷︷ ︸ f ( x ) , and thus b n w n = βa n 2 ρ 0 + 2 L ∫ L 0 g ( x ) sin nπx L dx. 11

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    Chapter 5. Sturm-Liouville Eigenvalue Problems Section 5.3 5.3.1 By separation of variables, u = φ ( x ) h ( t ) , 1 h d 2 h dt 2 = 1 ρ 0 φ ( x ) [ T 0 d 2 φ dx 2 + αφ ] = − λ. Thus, d 2 h dt 2 = − λh and T 0 d 2 φ dx 2 + αφ + λρ 0 φ = 0 . For the time-dependent equation (if λ > 0), h = c 1 cos √ λt + c 2 sin √ λt . The spatial equation is in Sturm-Liouville form: ρ = T 0 constant, q = α ( x ) , σ = ρ 0 ( x ). 5.3.3 Hd 2 φ/dx 2 = d dx ( Hdφ/dx ) − dH/dx dφ/dx . Thus d dx ( Hdφ/dx )+ dφ/dx ( Hα − dH/dx )+( λβ + γ ) Hφ = 0. To be in standard Sturm-Liouville form, dH dx = αH or H = c 1 exp [∫ x α ( t ) dt ] , let c 1 = 1 . Then ρ ( x ) = H, q ( x ) = γH , and σ ( x ) = βH . 5.3.4 (b) By separation of variables, u = φ ( x ) h ( t ) , 1 h dh dt = 1 φ ( k d 2 φ dx 2 − v 0 dφ dx ) = − λ . The boundary value problem is k d 2 φ dx 2 − v 0 dφ dx + λφ = 0. φ = e rx implies kr 2 − v 0 r + λ = 0 or r = v 0 ± √ v 2 0 − 4 λk 2 k = v 0 ± i √ 4 λk − v 2 0 2 k . To satisfy the boundary conditions φ (0) = 0 and φ ( L ) = 0: √ 4 λk − v 2 0 2 k = nπ L and φ ( x ) = e v 0 2 k x sin nπx L . Thus, by superposition, u = ∞ ∑ n =1 A n e v 0 2 k x sin nπx L e − λ n t = e v 0 2 k x ∞ ∑ n =1 A n sin nπx L e − λ n t . The initial value problem can be solved f ( x ) = e v 0 2 k x ∞ ∑ n =1 A n sin nπx L so that A n = 2 L ∫ L 0 f ( x ) e − v 0 2 k x sin nπx L dx. Note that λ = v 2 0 4 k + k ( nπ L ) 2 . 5.3.9 (c) Since it is equidimensional, φ = x r , which implies r ( r − 1)+ r + λ = 0 or r 2 = − λ . If λ > 0 , r = ± i √ λ , where x ± i √ λ = e ± i √ λ ln x . Thus the general solution is φ = c 1 cos ( √ λ ln x ) + c 2 sin ( √ λ ln x ) . The boundary condition φ (1) = 0 implies c 1 = 0, and hence φ ( b ) = 0 yields sin ( √ λ ln b ) = 0 or √ λ ln b = nπ, n = 1 , 2 , . . . . Thus λ = ( nπ/ ln b ) 2 . If λ = 0 the differential equation becomes d dx ( xdφ/dx ) = 0. Thus, the general solution is φ = c 1 + c 2 ln x . The condition φ (1) = 0 yields c 1 = 0, and then φ ( b ) = 0 implies c 2 = 0. Thus λ = 0 is not an eigenvalue. 12

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    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 = ∞ ∑ n =1 a n φ n ( x ) e − λ n t (with λ 1 = 0 , φ 1 = 1) . The initial conditions, f ( x ) = ∞ ∑ n =1 a n φ n ( x ) , yield a n = ∫ L 0 f φ n cρ dx ∫ L 0 φ 2 n cρ dx , since the eigenfunctions are orthogonal with weight cρ . As t → ∞ , u → a 1 = ∫ L 0 f cρ dx ∫ L 0 cρ dx , 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 φ n ( r ), are orthogonal with weight r . The time-dependent part satisfies (5.2.10), and hence h ( t ) = e − λkt . By superposition u = ∞ ∑ n =1 a n φ n ( r ) e − λkt . The initial condition, f ( r ) = ∞ ∑ n =1 a n φ n ( r ) , determines a n , a n = ∫ a 0 f ( r ) φ n ( r ) r dr ∫ a 0 φ 2 n ( r ) r dr . 5.4.6 By separation of variables, u = φ ( x ) h ( t ), T 0 d 2 φ dx 2 + λρ 0 φ = 0 d 2 h dt 2 + λh = 0 . The boundary conditions are of the Sturm-Liouville type, and therefore the eigenfunctions denoted φ n ( x ) are orthogonal with weight ρ 0 ( x ). We call the eigenvalues λ n . Then the time-dependent problem has solutions cos √ λ n t and sin √ λ n t . Initially at rest means ∂u/∂t ( x, 0) = 0, so that only cosines are needed. Thus by superposition u = ∞ ∑ n =1 A n φ n ( x ) cos √ λ n t. The initial position yields f ( x ) = ∞ ∑ n =1 A n φ n ( x ) , 13

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    determining A n , A n = ∫ L 0 f ( x ) φ n ( x ) ρ 0 ( x ) dx ∫ L 0 φ 2 n ρ 0 dx . Section 5.5 5.5.1 (g) To be self-adjoint (if p is constant), u ( L ) dV dx ( L ) − v ( L ) du dx ( L ) = u (0) dv dx (0) − v ((0) du dx (0) . The derivatives at x = L may be eliminated from the boundary conditions. After some algebra, we obtain ( αδ − βγ ) [ u (0) dv dx (0) − v (0) du dx (0) ] = u (0) dv dx (0) − v (0) du dx (0) . Thus, it is self-adjoint only if αδ − βγ = 1. 5.5.9 Multiply the differential equation by φ and integrate: ∫ 1 0 φ d 4 φ dx 4 dx + λ ∫ 1 0 e x φ 2 dx = 0 . Note that φ d 4 φ dx 4 = d dx ( φ d 3 φ dx 3 ) − dφ dx d 3 φ dx 3 = d dx ( φ d 3 φ dx 3 ) − d dx ( dφ dx d 2 φ dx 2 ) + ( d 2 φ dx 2 ) 2 . Thus φ d 3 φ dx 3 ∣ ∣ ∣ 1 0 − dφ dx d 2 φ dx 2 ∣ ∣ ∣ ∣ 1 0 + ∫ 1 0 ( d 2 φ dx 2 ) 2 dx + λ ∫ 1 0 e x φ 2 dx = 0. From the boundary conditions, the boundary contributions vanish. Thus λ = − ∫ 1 0 ( d 2 φ dx 2 ) 2 dx ∫ 1 0 e x φ 2 dx , 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, λ 1 ≤ u 2 T (1) + ∫ 1 0 ( du T dx ) 2 dx ∫ 1 0 u 2 T dx . Any trial function should satisfy the boundary conditions (and be continuous with no zeroes). Ge- ometrically, a simple example is a parabola ( u T (0) = 0 , u T (1) = 1 , du T /dx (1) = − 1, from the boundary conditions. To obtain this parabola, we substitute u T = ax + bx 2 into the condition at x = 1 : a + 2 b + a + b = 0, a simple choice being a = 3 , b = − 2 so that u T = 3 x − 2 x 2 and u T /dx = 3 − 4 x . Thus, λ 1 ≤ 1 + ∫ 1 0 (3 − 4 x ) 2 dx ∫ 1 0 (3 x − 2 x 2 ) 2 dx = 1 − 1 12 (3 − 4 x ) 3 ∣ ∣ ∣ 1 0 ∫ 1 0 (9 x 2 − 12 x 3 + 4 x 4 ) dx = 1 − 1 12 ( − 1 − 27) 3 − 3 + 4 5 = 40 / 12 4 / 5 = 4 1 6 . 14

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