Lecture 24. Scott Pauls 5/21/07

Transcription

1 Lecture 24 Department of Mathematics Dartmouth College 5/21/07

2 Material from last class The heat equation α 2 u xx = u t with conditions u(x, 0) = f (x), u(0, t) = u(l, t) = Separate variables to get X X = T α 2 T = λ or X λx = 0, T α 2 λt = 0 2. Translation of boundary conditions X(0) = 0 = X(L)

3 Eigenvalue For the X equation, we have a two point boundary value X λx = 0, X(0) = 0 = X(L) Solutions: X n (x) = sin(nπx/l), n = 1, 2, 3,... with eigenvalues λ n = n2 π 2 L 2.

4 Eigenvalue For the X equation, we have a two point boundary value X λx = 0, X(0) = 0 = X(L) Solutions: X n (x) = sin(nπx/l), n = 1, 2, 3,... with eigenvalues λ n = n2 π 2 L 2.

5 T equation For these eigenvalues (λ n = n2 π 2 ), solve for T via L 2 T + α2 n 2 π 2 L 2 T = 0 T n (t) = e α2 n 2 π 2 L 2 t So, for each n we have a product solution to α 2 u xx = u t with u(0, t) = u(l, t) = 0: u n (x, t) = e α2 n 2 π 2 L 2 t sin(nπx/l)

6 T equation For these eigenvalues (λ n = n2 π 2 ), solve for T via L 2 T + α2 n 2 π 2 L 2 T = 0 T n (t) = e α2 n 2 π 2 L 2 t So, for each n we have a product solution to α 2 u xx = u t with u(0, t) = u(l, t) = 0: u n (x, t) = e α2 n 2 π 2 L 2 t sin(nπx/l)

7 T equation For these eigenvalues (λ n = n2 π 2 ), solve for T via L 2 T + α2 n 2 π 2 L 2 T = 0 T n (t) = e α2 n 2 π 2 L 2 t So, for each n we have a product solution to α 2 u xx = u t with u(0, t) = u(l, t) = 0: u n (x, t) = e α2 n 2 π 2 L 2 t sin(nπx/l)

8 Superposition Now, it is highly unlikely that u n (x, 0) = sin(nπx/l) = f (x) so we are unable to impose the last condition on a single choice of solution. So, we use the principle of superposition u(x, t) = = u n (x, t) n=1 n=1 c n e α2 n 2 π 2 L 2 t sin(nπx/l) This is a Fourier sine series when t = 0 so we need to represent f as such a series.

9 Superposition Now, it is highly unlikely that u n (x, 0) = sin(nπx/l) = f (x) so we are unable to impose the last condition on a single choice of solution. So, we use the principle of superposition u(x, t) = = u n (x, t) n=1 n=1 c n e α2 n 2 π 2 L 2 t sin(nπx/l) This is a Fourier sine series when t = 0 so we need to represent f as such a series.

10 Fourier sine series representation Steps: 1. Make f odd an periodic with period 2L 2. Compute coefficients 3. Then c n = b n b n = 2 L L 0 f (x) sin ( nπx ) L dx

11 Example L = π, f (x) = (π x)x 3 bterrell/h1/heat1.html

12 Other types of boundary conditions Insulated ends: α 2 u xx = u t, u x (0, t) = u x (L, t) = 0, u(x, 0) = f (x) Also called Neumann boundary conditions. Two point boundary value : X λx = 0, X (0) = X (L) = 0 and T α 2 λt = 0

13 Other types of boundary conditions Insulated ends: α 2 u xx = u t, u x (0, t) = u x (L, t) = 0, u(x, 0) = f (x) Also called Neumann boundary conditions. Two point boundary value : X λx = 0, X (0) = X (L) = 0 and T α 2 λt = 0

14 Insulated ends Solutions: λ n = n 2 π 2 /L, X n = cos(nπx/l), T n = e n2 π 2 α 2 L 2 t So, u 0 (x, t) = 1, u n (x, t) = e n2 π 2 α 2 L 2 t cos(nπx/l) And where u(x, t) = 1 + c n = 2 L n=1 L 0 c n e n2 π 2 α 2 L 2 t cos(nπx/l) f (x) cos(nπx/l) dx

15 Work for next class Read 10.5,10.7,10.8 Homework 9 assigned but is not due! These are practice s for the final.