AB2.14: Heat Equation: Solution by Fourier Series

Transcription

1 AB2.14: Heat Equation: Solution by Fourier Series Consider the boundary value problem for the one-dimensional heat equation describing the temperature variation in a bar with the zero-temperature ends: = t c2 2 u x 2, u = u(x, t), t >, < x <, u(x, ) = f(x), (1) u(, t) =, u(, t) =, t. The solution is determined by the separation of variables (the Fourier method): u(x, t) = F (x)g(t). Then t = F 2 u G, x = F G 2 Substituting this into one-dimensional heat equation and separating variables, G F G = c 2 F G c 2 G = F = const = p2 F we obtain the differential equations for G(t) and F (x) G + c 2 p 2 G =, Satisfy the boundary conditions: F + p 2 F =. u(, t) = F ()G(t) =, u(, t) = F ()G(t) =, t. Thus, The general solution for F is F () =, F () =. F = A cos px + B sin px. and which yields F () = : A = ; F () = : B sin p = sin p = (B ) p = nπ, p = p n = nπ (n = 1, 2,...). F = F n = sin p n x = sin nπ x (n = 1, 2,...). The equation for G becomes G + λ 2 ng =,.

2 The general solution of this equation is G(t) = G n (t) = B n e λ2 nt (n = 1, 2,...). Hence the solutions of satisfying are t = 2 u c2 x, 2 < x < u(, t) =, u(, t) =, t. u n (x, t) = F n (x)g n (t) = B n e λ2nt sin nπ x (n = 1, 2,...). These functions are called eigenfunctions and are called eigenvalues. Now we can solve the entire problem by setting Satisfy the initial conditions: u(x, t) = u n (x, t) = B n e λ2nt sin nπ x. u(x, ) = B n sin nπ x = f(x). Thus, B n = 2 f(x) sin nπ xdx, n = 1, 2,.... EXAMPE 1 Sinusoidal initial temperature. Find the solution to the boundary value problem (1) for the one-dimensional heat equation with the initial temperature f(x) = 1 sin π 8 x. Solution. Satisfy the initial conditions: Then the solution is u(x, ) = B n sin nπ 8 x = f(x) = 1 sin π 8 x. λ n = λ 1 = cπ 8. u(x, t) = u n (x, t) = B 1 e λ2 1 t sin π 8 x = 1e λ2 1 t sin π 8 x.

3 EXAMPE 2 Triangular initial temperature in a bar. Find the solution to the boundary value problem (1) for the one-dimensional heat equation with the triangular initial temperature f(x) = Solution. Satisfy the initial conditions: { x if < x < /2, x if /2 < x <. u(x, ) = B n sin nπ x = f(x). For the odd periodic extension of f(x) the Fourier coefficients are B n = 2 f(x) sin nπ xdx = 2 ( /2 x sin nπ xdx + ( x) sin nπ ) /2 xdx = 2 nπ x cos nπ x /2 2 nπ ( x) cos nπ x + 2 /2 cos nπ nπ xdx /2 2 cos nπ nπ /2 xdx = πn cos nπ π 2 n 2 sin nπ 2 + πn cos nπ π 2 n 2 sin nπ 2 = (1) 4 n 2 π 2 sin nπ 2 = 4 (2l 1) 2 π 2 ( 1)l+1 = 4 ( 1) l+1 π (2l 1) = 2 We have and the solution is u(x, t) = { 4 u n (x, t) = 4 π 2 n 2 π 2 if n = 1, 5, 9,..., 4 n 2 π 2 if n = 3, 7, 11,...., ( sin π ( cπ xe ) 2t 1 9 sin 3π ( 3cπ xe ) 2t ) EXAMPE 3 A bar with insulated ends. Find the solution to the boundary value problem for the one-dimensional heat equation = t c2 2 u x 2, u = u(x, t), t >, < x <, u(x, ) = f(x), (1) u x (, t) =, u x (, t) =, t.

4 Solution. The solution is determined by the Fourier method: u(x, t) = F (x)g(t). Then t = F G, 2 u x 2 = F G, G + c 2 p 2 G =, Satisfy the boundary conditions: F + p 2 F =. u x (, t) = F ()G(t) =, u x (, t) = F ()G(t) =, t. Thus, The general solution for F is F x () =, F x () =. F = A cos px + B sin px. and which yields F () = : B = ; F () = : Ap sin p = sin p = (A ) p = p n = nπ F = F n = cos nπ (n = 1, 2,...). x (n = 1, 2,...). The solution for G is G(t) = G n (t) = A n e λ2 nt (n = 1, 2,...). Hence the solutions of satisfying are t = 2 u c2 x, 2 < x < u x (, t) =, u x (, t) =, t. u n (x, t) = F n (x)g n (t) = A n e λ2 n t cos nπ x (n = 1, 2,...). These functions are eigenfunctions and are eigenvalues.

5 Now we can solve the entire problem by setting u(x, t) = u n (x, t) = A n e λ2nt cos nπ x. n= n= Satisfy the initial conditions: u(x, ) = A n cos nπ x = f(x). Thus, A = 1 f(x)dx, A n = 2 f(x) cos nπ xdx, n = 1, 2,.... EXAMPE 4 Triangular initial temperature in a bar with insulated ends. Find the solution to the boundary value problem for the one-dimensional heat equation in a bar with insulated ends = t c2 2 u with the triangular initial temperature x 2, u = u(x, t), t >, < x <, u(x, ) = f(x), (1) u x (, t) =, u x (, t) =, t. f(x) = { x if < x < /2, x if /2 < x <. Solution. Satisfy the initial conditions: u(x, ) = A n cos nπ x = f(x). For the even periodic extension of f(x) the Fourier coefficients are A = 1 f(x)dx = 1 ( /2 ) xdx + ( x)dx = /2 4 ; A n = 2 f(x) cos nπ xdx = 2 ( /2 x cos nπ xdx + ( x) cos nπ ) /2 xdx = 2 nπ x sin nπ x /2 2 nπ ( x) sin nπ x 2 /2 sin nπ nπ xdx+ /2 + 2 sin nπ nπ /2 xdx = πn sin nπ π 2 n (cos nπ 2 2 1) πn sin nπ π 2 n (cos nπ 2 2 cos nπ) = (2)

6 We have and the solution is u(x, t) = n= u n (x, t) = 4 8 π 2 2 (2 cos nπ ) n 2 π 2 2 cos nπ 1., ( 1 2 cos 2π ( 2cπ 2 xe ) 2t cos 6π ( 6cπ 2 xe ) 2t ] The boundary value problem for the two-dimensional wave equation Consider the boundary value problem describing vibrations of a planar rectangular membrane fixed at its edges and excited by means of a certain initial displacement with a given initial velocity: u tt = a 2 u, u = u(x, y, t), t >, < x < a, < y < b, u(x, y, ) = ϕ(x, y) u t (x, y, ) = ψ(x, y) (1) u(, y, t) =, u(a, y, t) =, u(x,, t) =, u(x, b, t) =, t. The partial differential equation in (1) is called the two-dimensional wave equation. Here, u t = t, u tt = 2 u t 2, u = 2 u x u y 2. The solution is determined by the separation of variables (the Fourier method): u(x, y, t) = v(x, y)t (t). Substituting this into the differential equation in (1) we obtain the equation for T (t) T + a 2 λt = (11), and the equation and the boundary value problem for v(x, y) v xx + v yy + λv =, < x < a, < y < b v(, y) =, v(a, y) = (12) v(x, ) =, v(x, b) = The solution v(x, y) of (12) is also determined by the separation of variables v(x, y) = X(x)Y (y). Substituting this into the differential equation in (12) we obtain the equations and the boundary value problems for X(x) and Y (y) { X + νx =, < x < a X() =, X(a) = (13);

7 { Y + µy =, < y < b Y () =, Y (b) = (14); Here µ and ν are number parameters such that µ + ν = λ. v. (13) and (14) are called the Sturm iouville eigenvalue problems. We have X n (x) = sin nπ a x, ν n = ( nπ a )2 ; Y m (y) = sin mπ b y, λ = ( nπ a )2 + ( mπ b )2 v n,m = A n,m sin nπ a µ m = ( mπ b )2. x sin mπ b y. Coefficients A n,m A n,m are determined from the orthogonality conditions Finally, and where a b The solution to (1) is u(x, y, t) = v 2 n,mdxdy = A 2 n,m v n,m (x, y) = m=1 a (C n,m cos A n,m = sin 2 nπ a xdx b 4 ab, 4 ab sin nπ a x sin mπ b y. λ n,m at + D n,m sin sin 2 mπ ydy = 1. b λ n,m at)v n,m (x, y), C n,m = a b = 4 ab ϕ(x, y)v n,m (x, y)dxdy = a b ϕ(x, y) sin nπ a x sin mπ b ydxdy, D n,m = 1 a2 λ n,m 4 ab a b ψ(x, y) sin nπ a x sin mπ b ydxdy.

8 PROBEM Find the solution to the boundary value problem (1) for the one-dimensional heat equation with the given c and the initial temperature f(x) = sin.1πx. Solution. We have f(x) = sin.1πx = sin π 1 x. Thus = 1. Satisfy the initial conditions: Then the solution is u(x, ) = B n sin nπ x = f(x) = sin.1πx. λ n = λ 1 = cπ 1. For the data of the problem, u(x, t) = u n (x, t) = B 1 e λ2 1 t sin π 1 x = e λ2 1 t sin π 1 x. c 2 = K σρ = =