Chapter 6 Partial Differential Equations (PDE)

Chapter 6 Partial Differential Equations (PDE)6-1 Classification of Partial Differential EquationsThe first-order linear PDE: a ( x, y ) u u b ( x, y ) f ( x, y )u g ( x, y ) 0 x yThe second-order linear PDE: a ( x, y ) d ( x, y ) 2u 2u 2u b(x,y) c(x,y) x y x 2 y 2 u u e( x, y ) f ( x, y )u g ( x, y ) 0 x y hyperbolic at ( x0 , y 0 ) : ( x0 , y 0 ) b( x0 , y 0 ) 2 4a( x0 , y 0 )c( x0 , y 0 ) 0 elliptic at ( x0 , y 0 ) : ( x0 , y 0 ) 0 parabolic at ( x0 , y 0 ) : ( x0 , y 0 ) 0 𝟐 𝒖 𝟐 𝒖Eg. Which is the type of partial differential equation for 𝟒 𝒙𝟐 𝟑 𝒚𝟐 ? [2018 台大電研] 2 𝑢 2 𝑢(Sol.) 4 𝑥 2 3 𝑦 2 0, 0-4 4 (-3) 48 0, it is hyperbolic.Wave equation:2 2u 2u 2u u2 u 2 ) b Fa(222 t x y z t2 u 2u 2u2 u a ( 2 2 2) a 2 2uHeat equation or Diffusion equation: t x y zLaplace’s and Poisson’s equations: 2u Schrodinger’s equation: i 2 u 2 u 2 u 0 x 2 y 2 z 2 ρ ψ 2 2 ψ Vψ in quantum mechanics.2m t 52

6-2 Separation-of-Variable MethodEg. Solve 2θ 2θ θ θ 2 , θ(x,0) x, θ(0,t) 0, 0 , and 0.2 t t 0 x x 1 x t(Sol.) Let θ ( x, t ) X ( x)T (t ) , X ′′( x)T (t ) X ( x)T ′′(t ) ,X ′′( x) T ′′(t ) λX ( x)T (t ) (2n 1)π (2n 1)πX(0) 0, X’(1) 0 λ , X n C n sin 22 2 x T ′′(t ) (2n 1)π (2n 1)π, T(0) constant, T’(0) 0 Tn d n cos 2T (t )2 2 (2n 1)π θ ( x, t ) An cos 2 n 1 (2n 1)πt sin 2 x (2n 1)πx sin 2 (2n 1)π x An θ ( x,0) x An sin 12 2 ( 2n 1)πn 1 1 sin 21 Eg. Solve 1 t , x dxn 1 8( 1)(2n 1) 2 π 2 x dx 2u 2u u 2 , u(0,t) u(1,t) u(x,0) 0, and sin(πx) .2 t t 0 x t(Sol.) u ( x, t ) X ( x)T (t ) , X ′′( x)T (t ) X ( x)T ′′(t ) ,X ′′( x) T ′′(t ) λ,X ( x)T (t )X(0) 0 X(1) λ -(nπ)2 and X(x) C n sin(nπx),T ′′(t ) -(nπ)2 and T(0) 0, T’(0) constant T(t) d n sin(nπt),T (t ) u ( x, t ) An sin( nπx) sin( nπt ) ,n 1 u ( x, t ) nπAn sin( nπx) cos(nπt ) tn 1 u1 sin(πx) πA 1 1 A 1 1/π but A n 0 for n 1 u ( x, t ) sin(πx) sin(πt ) t t 0π u 2 u, u(x,0) 3sin(2πx), u(0,t) u(1,t) 0, 0 x 1, t 0. t x 2X ′′( x) T ' (t )(Sol.) u ( x, t ) X ( x)T (t ) , X ( x)T ' (t ) X " ( x)T (t ) , λ,X ( x) T (t )Eg. SolveX(0) 0 X(1) λ -(nπ)2 and X(x) C n sin(nπx), T ' (t ) -(nπ)2 and T(0) constantT (t ) T(t) d n e n π t , u ( x, t ) An sin( nπx) e n π222n 12 tand u(x,0) An sin( nπx)n 1u(x,0) 3sin(2πx) A 2 3 but A n 0 for n 2 u(x,t) 3e 4π t sin( 2πx)2 53

2x y y 1 2 , y(x,0) f(x) 0.1 1 2π t x 2Eg. Solve2π 0.2 x,0 x π2, xπ 1 0.2 , x π π 2 and y(x,t) b (t ) sin(nx) . (a) Find the Fourier sine series for f(x) on [0,π]. (b)n 1nFind the ordinary differential equation and the initial condition for b n (t). (c) Findy(x,t). [1990 中央電研] (Sol.) (a) f ( x) a n sinn 1an nπx nπx a n sin( nx) , 2 L 2π , L π , nxLLn 11 π2 πf ( x) sin( nx)dx f ( x) sin( nx)dx π LL 0π2 π 2 0.2 xx 0.8 nπ sin(nx )dx π 0.2 1 sin(nx )dx 2 2 sin π 0 π2 π 2 n π0. 8 nπ sin sin( nx)2 2 2 n 1 n π f ( x) n 1n 1(b) y ( x, t ) bn (t ) sin( nx) , y ( x,0) bn (0) sin( nx) f ( x) n 1n 1 bn′′ (t ) sin( nx) bn (t ) n 2 ( sin( nx)) bn′′ (t ) n 2 bn (t ) 0 is the ordinary differential equation. bn (t ) α n cos(nt ) b n sin( nt ) 0.8 nπ sin()sinnxbn (0) sin( nx) 2 2 2 n 1 n πn 10.8 nπ bn (0) 2 2 sin α n (the initial condition), β n 0nπ 2 y ( x ,0 ) f ( x ) n 1n 1(c) y ( x, t ) bn (t ) sin( nx) α n cos(nt ) sin( nx)0.8 nπ sin cos(nt ) sin( nx) .2 2 2 n 1 n π 54

6-3 Laplace Transform Solutions of Partial Differential EquationsEg. Solve θ θ y , θ(x,0) 0 for x 0, and θ(0,y) y for y 0. [1990 台大化工研] x y(Sol.) Method 1: By Laplace transform 1 L[θ (0, y )] Θ(0, s )s211dΘ ( x , s ) sΘ( x, s ) θ ( x,0) 2 Θ( x, s ) A( s ) e sx 3dxss1 11111 1Θ(0, s ) A( s ) 3 2 A( s ) 2 3 Θ( x, s ) 2 3 e sx 3sssss s sL[θ ( x, y )] θ ( x, y )e sy dy Θ( x, s ) , L[ y ] 01y2 θ ( x, y ) y x ( y x ) 2 u ( y x ) 22 y2 , y x 22 y x 1 ( y x) 2 y , y x 22u vMethod 2: Define u x y and v x-y, x ,2u vy ,2 θ θ u θ v θ θwe have x u x v x u v θ θ u θ v θ θ and y u y v y u vu v θ u v θ θ θ 1 u 2 y 2 ,, θ (u , v) ( uv) C(v) 2 u x y4 2 u41 θ(x,0) 0forx 0 θ(u vu v, 0) 0,22u vandwehave 1 v 2 v 2 ) C (v) 0,4 2θ (y2 1 u 2v2v 2 1for v 0 or x y θ ( uv) (u v) 2 for x y4 2288 8u vu v u v2 θ(0,y) y for y 0 θ( 0,) , u -v and we have222 C(v) 1 v 23v 2for v 0 or x y v 2 ) C (v) -v, C(v) v 4 28θ (( x y) 2v2 y2 1 u 23v 2 1 (u v) 2 -v (y-x)for x y( uv) v 82224 28Note: This partial differential equation cannot be solved by separation of variables. θ 55

Eg. Solve 2u 2u u 2 , u(0,t) u(1,t) u(x,0) 0, and sin(πx) .2 t t 0 x t (Sol.) L[u ( x, t )] u( x, t )e st dt U ( x, s )0d 2U ( x, s ) u s 2U ( x, s ) su( x,0) s 2U ( x, s ) sin(πx )2dx t t 0d 2U ( x, s )sin πx s 2U ( x, s ) sin(πx) U ( x, s ) c1e sx c2 e sx 22s π2dx u (0, t ) 0 U (0, s ) 0 c1 01 u ( x, t ) sin(πx) sin(πt ) π U (1, s ) 0 u (1, t ) 0 c 2 0Eg. Solve(Sol.) u 2 u, u(x,0) 3sin(2πx), u(0,t) u(1,t) 0, 0 x 1, t 0. t x 2 L[u( x, t )] u( x, t )e st dt U ( x, s ) sU ( x, s ) u ( x,0) 0d2U ( x, s )dx 2d23U ( x, s ) sU ( x, s ) 3 sin( 2πx) U ( x, s ) c1e s x c2 e s x sin(2πx )2s 4π 2dxL[u(0, t )] U (0, s ) 0 , L[u(1, t )] U (1, s ) 0 c1 0 , c 2 0U ( x, s ) 23 3 sin( 2πx) 3e 4π t sin( 2πx) sin(2πx ) u ( x, t ) L 1 22s 4π s 4π 56

6-4 Fourier Transform Solutions of Partial Differential Equations2 u 2 uEg. Solve 2 , u(x,0) e x , - x , t 0. [2011 成大電研、2001 台大電研類 t x似題] 2 u ( x, t ) 2(Sol.) ℑ[u ( x, t )] U (ω , t ) , ℑ ω U (ω , t )2 x 2dU (ω , t ) ω 2U (ω , t ) U (ω , t ) Ae ω tdtAccording to ℑ[e a2 2x] πaU (ω ,0) A ℑ[u( x,0)] ℑ[e e x2ω24a2,] π e ω24 A π e Let b2 t 1/4 and according to ℑ 1 [e b ω ] 2 u ( x, t ) ℑ 1 [U (ω , t )] ℑ 1 [ π eEg. Solve ω242e ω24x24b 22b π,e ω t ] π ℑ 1 [e21 ( t )ω 24] e x21 4 t1 4t 2u 2u u ( x,0), u(x,0) 4e-5 x , 9 0 , - x , t 0.22 t t x 2 u ( x, t ) 2(Sol.) ℑ[u ( x, t )] U (ω , t ) , ℑ 9 9ω U (ω , t )2 x d2U (ω , t ) 9ω 2U (ω , t ) U (ω , t ) A cos(3ωt ) B sin(3ωt )2dtdU (ω , t ) -3ωAsin(3ωt) 3ωBcos(3ωt)dt402aAccording to ℑ e a x 2, U (ω ,0) A ℑ[u ( x,0)] ℑ[4e 5 x ] 225 ω 2a ω40 u ( x,0)d cos(3ωt )] U (ω ,0) 0 3ωB B 0, U (ω , t ) 2ℑ[ tdtω 25By ℑ -1[ejaωF(ω)] f(x a) and ℑ -1[e-jaωF(ω)] f(x-a),[]4040e i 3ωt e i 3ωt 1cos(3t)] ω[ℑ ] 2ω 2 25ω 2 252 5 e i 3ωt )] 2ℑ 1 [ 2 (e i 3ωt e i 3ωt )] 2e-5 x-3t 2e-5 x 3t ω 25u(x,t) ℑ 1 [U (ω , t )] ℑ 1 [ ℑ 1 [20 (e i 3ωtω 252 57

6-5 Miscellaneous Methods of Solving Partial Differential Equations2 x u 2Error function: erf ( x) e duπ0Complementary error function: erfc( x) 1 erf ( x)Eg. Solve 𝟐 𝒖 𝟐 𝟐 𝒖 𝟐 𝒖 𝟐 𝟐 𝟎, u(0,y) 0 and u(x,1) x2 .(Sol.) Set u(x,y) f(y mx) and ε y mx m2f”(ε) 2mf”(ε) f”(ε) 0, m2 2m 1 0,m -1, -1 u(x,y) f(y-x) xg(y-x). u(0,y) 0 f(y-0) 0 g(y-0) f(y) f(1-x) 0 and f(y-x) 0u(x,1) x2 f(1-x) xg(1-x) xg(1-x) g(1-x) x, g(x) 1-x g(y-x) 1-y x u(x,y) x (1-y x) 2u u u 2 u . [2015 台大電子所工數 K]2 x y x(Sol.) u ( x, y ) X ( x)Y ( y ) , [ X ′′( x) 2 X ' ( x)] Y ( y ) X ( x) [Y ' ( y ) Y ( y )] ,X ′′( x) 2 X ' ( x) Y ' ( y ) Y ( y ) λ X”(x) 2X’(x)-λX(x) 0 and Y’(y) (1-λ)Y(y) 0X ( x)Y ( y)Eg. SolveX”(x) 2X’(x)-λX(x) 0, r2 2r-λ 0, r -1 1 λ1． If 1 λ 0,1 λ k R, λ k2-1 and r -1 k X(x) Ae(-1 k)x Be(-1-k)x, 1-λ 2-k2,Y’(y) (1-λ)Y(y) 0 Y(y) C’e-(1-λ)y C’ e ( k 2 2 ) y u(x,y) e x 2 y [ C (k )e kx k y dk D(k )e kx k y dk ]20202． If 1 λ 0, λ -1 and r -1, -1 X(x) Ae-x Bxe-x, Y’(y) 2Y(y) 0 Y(y) C’e-2y u(x,y) e x 2 y (C Dx)3． If 1 λ 0,1 λ iω, r -1 iω X(x) Ae-xcos(ωx) Be-xsin(ωx),λ -ω2-1, 1-λ 2 ω2, Y’(y) (2 ω2)Y(y) 0, Y(y) C’ e ( ω 2 2 ) y u(x,y) e x 2 y [ C (ω )e ω y cos(ωx)dω D(ω )e ω y sin(ωx)dω ]020 582

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52 Chapter 6 Partial Differential Equations (PDE) 6-1 Classification of Partial Differential Equations . The first-order linear PDE: ( , ) ( , ) ( , ) ( , ) 0