I want to solve the following of the heat equation using separation of variable: But have one problem a the end of the method, Thx for your help.
$ \left\{\begin{matrix} u_{t}-u_{xx}=tx \; \; ,0<x<L, t>0 \\ u_{x}(0,t)= 0 \\ u(L,t)=0 \\ u(x,0)=1 \end{matrix}\right. $
First I will try to solve the homogeneous
$ \left\{\begin{matrix} u_{t}-u_{xx}=0 \; \; ,0<x<L, t>0 \\ u_{x}(0,t)= 0 \\ u(L,t)=0 \\ u(x,0)=1 \end{matrix}\right. $
So we write $u(x,t)=X(x)T(t)$ and we get $$\frac{T_t}{T}=\frac{X_{xx}}{X}=-\lambda$$
it follows that: $$T(t)=Be^{-\lambda t}$$ and:
1) if $\lambda <0$ we have $X(x)=\alpha e^{\sqrt{-\lambda}x}+ \beta e^{-\sqrt{-\lambda}x}$
2)if $\lambda = 0$ we have $X(x) = \alpha + \beta x$
3)if $\lambda > 0$ we have $X(x)=\alpha cos({\sqrt{\lambda}x})+ \beta sin({\sqrt{\lambda}x})$
So 2) cannot be, because if we replace the boundary condition we get that $X(x)=0$
So with 3)its same thing we get $X(x)=0$ and this cannot be.
(edited)
So we only have 3) and from the boundary condition we get: $$X_n(x)=cos(\frac{(2n-1)\pi x}{L}) \\n=1,2,...$$
Those are them the eigenfunction with eigenvalue $$\lambda_n = {(\frac{(2n-1)\pi}{L})}^{2} \\ n=1,2,3...$$ (observation I think that $ n \neq 0$ because $\lambda$ cannot be zero )
So we get $$u_n(x,t)=B_ncos(\frac{(2n-1)\pi x}{L})e^{-(\frac{(2n-1)\pi}{L})^2 t}$$
Using the superposition we get :
$$u(x,t)=\sum_{n=1}^{\infty}B_ncos(\frac{(2n-1)\pi x}{L})e^{-(\frac{(2n-1)\pi}{L})^2 t}$$
Here is my problem: To get the $B_n$ I must introduce the boundary condition $u(x,0) = 1$ So if I must get the fourier expansion of the constant. But them using this expansion we have that $B_n=0 $ for $n =1,2,3...$
Mmmm I just dont know were I forgot the $B_0$ in my deduction process. After this I think I can solve the nonhomogeneous.
Thanks for your help!
