Answer to Question #204415 in Calculus for smi

Question

Q. Construct the solution of the heat equation using separation of variables method:

u_xx=4u_t , 0<x<40, t>0

u(0, t)=0, u(40, t)=0, t>0

u(x,0)= x, 0 less than or equal x less than or equal 20, u(x,0)=40-x, 20 less than or equal x less than or equal 40

Answer 1

$\frac{T'}{T}=\frac{X''}{X}=\lambda$

$0=u(0,t)=X(0)T(t)=X(0)=0$

$0=u_x(40,t)=X'(40)T(t)=X'(40)=0$

$T'(t) = λT(t)$

$X(0) = X' (40) = 0.$

$X'(40)=0=\mu C_1cos(40\mu )=0$

$λn =-(n-\frac{1}{2})^2$

$X_n(x)=sin(n- \frac{1}{2})^x$

$u_n(x,t)=X_n(x)T_n(t)=sin(n-\frac{1}{2})x )exp(-(n-\frac{1}2)^2t)$

$u_n(x,t)=\sum_{n-1}^\infty sin(n-\frac{1}{2})x )exp(-(n-\frac{1}2)^2t)$

$f(x)=\sum^\infty_{n-1}b_n sin((n-\frac{1}{2})x)$

$b_n=\frac{2}L \int_0^Lf(x) sin((n-\frac{1}2)x)dx=\int_0^{40} f(x)sin((n- \frac{1}2)x)dx$

$\int_0^{40} sin((n-\frac{1}2)x)sin((m-\frac{1}2)x)dx=[_{20}^0$

$b_n=\frac{2}{40} \int _0^{40}f(x) sin ((n-\frac{1}2)x)dx=\frac{6}{40}\int_0^{40} sin(\frac{5x}2)sin((n- \frac{1}2)x)dx$

$b_n=\frac{6}{40} \int_0^{40} sin(100x)sin((n-\frac{1}2)x)dx=[_0^3$

$f(x)=\sum_{n=1}^\infty b_nsin((n-\frac{1}2)x)$

$u(x,t)=3 sin (\frac{5x}{2})exp (-(\frac{1}{2})t)$