Question

Solve the following initial value problem

1. U_t(x,t)=10U_xx(x,t) -1<x<1, t>0
2. U(-1,t)=U(1,t) U_x(-1,t)=U_x(1,t) t>0
3. U_x(x,0)=x+1 -1<x<1

Answer 1

solution of heat equation:

$u(x,t)=\sum B_n sin(n\pi x/L)e^{-k(n\pi /L)^2t}$

where

$B_n=\frac{2}{L}\int^L_0f(x)sin(n\pi x/L)dx$

$f(x)=u(x,0)$

we have:

$k=10,L=1$

then:

$f(x)=\int^1_{-1}u_x(x,0)dx=\int^1_{-1}(x+1)dx=2$

$B_n=4\int^1_0sin(n\pi x)dx=-\frac{4}{\pi n}cos(n\pi x)|^1_0=-\frac{4}{\pi n}cos(n\pi )+\frac{4}{\pi n}$

$B_n=0$ for even n

$B_n=\frac{8}{\pi n}$ for odd n

$u_n(x,t)=0$ for even n

$u_n(x,t)=\frac{8}{\pi n}sin(n\pi x)e^{-10(n\pi )^2t}$ for odd n

for $u(-1,t)=u(1,t)$ :

$u(-1,t)=\frac{8}{\pi n}sin(-n\pi )e^{-10(n\pi )^2t}=0$

$u(1,t)=\frac{8}{\pi n}sin(n\pi )e^{-10(n\pi )^2t}=0$

for $u_x(-1,t)=u_x(1,t)$ :

$u_x(x,t)=8cos(n\pi x)e^{-10(n\pi )^2t}$

$u_x(-1,t)=8cos(-n\pi )e^{-10(n\pi )^2t}=-8e^{-10(n\pi )^2t}$

$u_x(1,t)=8cos(n\pi )e^{-10(n\pi )^2t}=-8e^{-10(n\pi )^2t}$