A classic problem I'm having problems with (problem requires to use Laplace transform)
$\frac{\partial ^2}{\partial x^2} u(x,t)=\frac{\partial}{\partial t} u(x,t) $
with conditions:
$u(0,t)=0;\,\,u(1,t)=u_0;\,\,\,t\gt 0$
$u(x,0)=0; \,\,0\lt x\lt 1$
I started proposing: $u(x,t)=w(x,t)+v(x)$ then following with:
$v(x)=u_0x$ and $w(0,t)=w(1,t)=0; w(x,0)=-v(x)=-u_0x$
$\frac{\partial ^2}{\partial x^2} w(x,t)=\frac{\partial}{\partial t} w(x,t) $
Now transforming t :
$\mathcal{L}(w'')=\mathcal{L}(\dot w) \rightarrow W''(x,s)=sW(x,s)-w(x,0)$
$W''(x,s)=sW(x,s)+u_0x$
Which leads me to:
$W(x,s)=Wh+Wp=c_1e^{-x\sqrt{s}}+c_2e^{x\sqrt{s}} - \frac{u_0x}{s}$
and lastly: setting $c_2=0$
$$\mathcal{L}^{-1}(W(x,s))=c_1\frac{xe^{-x^2/4t}}{2\sqrt{\pi}t^{3/2}} -u_0x =w(x,t)$$
But these solution is wrong as $w(1,t)\neq0$
and the problem states that solutions of the form: $erfc(\frac{2n+1\pm x}{2\sqrt{t}})$ should appear.
I think the given boundaries makes this problem have a solution which obtained by the separation of variables method is:
$u(x,t)=\sum_{n=1}^{\infty}[\frac{u_0(-1)^{n+1}}{\pi n}] \sin{(\pi n x)}e^{(\pi n)^2t/4} + u_0x$
this solution is valid for the given boundaries.
What am i doing wrong???
Just figure it out, my error was to set $C_2$ 0, i thuought it would make imposible the inverse transform to be computed as $C_2 e^{\sqrt s x}$ is not dominated by an exponencial.
For the solution i just had to transform all $u(x,t)$ and not separate it into 2 functions.
$\mathcal{L}(u'')=\mathcal{L}(\dot u) \rightarrow U''(x,s)=sU(x,s)-u(x,0)$ so we have: $U''(x,s)=sU(x,s)\rightarrow U(x,s)=c_1e^{x\sqrt{s}}+c_2e^{-x\sqrt{s}}$
Verifying (the transformed) first boundary condition: $U(0,s)=0=c_1e^{0\sqrt{s}}+c_2e^{-0\sqrt{s}}\rightarrow c_2=-c_1$ $\rightarrow U(x,s)=c_1(e^{x\sqrt{s}}-e^{-x\sqrt{s}})$
Now verifying the second boundary condition:
$U(1,s)=\frac {u_0}s=c_1(e^{1\sqrt{s}}-e^{-1\sqrt{s}})\rightarrow c_1=\frac {u_0} {s(e^{\sqrt{s}}-e^{-\sqrt{s}})}$
Thus $U(x,s)=\frac {u_0(e^{x\sqrt{s}}-e^{-x\sqrt{s}})} {s(e^{\sqrt{s}}-e^{-\sqrt{s}})}\rightarrow U(x,s)=(\frac {u_0}{se^{\sqrt s}})(\frac {1}{(1-e^{-2\sqrt s})})(e^{x\sqrt{s}}-e^{-x\sqrt{s}})$
Now using a series representation for: $ \frac {1}{1-q}=\sum_{n=0}^\infty\,q^n$, where: $q=e^{-2\sqrt s}$
$ \frac {1}{1-e^{-2\sqrt s}}=\sum_{n=0}^\infty \, (e^{-2\sqrt s})^n=\sum_{n=0}^\infty \, e^{-2n\sqrt s}$
$U(x,s)=\sum_{n=0}^\infty(\frac {u_0}{s} e^{-(1+2n+x)\sqrt s} -\frac {u_0}{s} e^{-(1+2n-x)\sqrt s})$
Finally using the inverse tranform:
$u(x,t)=\sum_{n=0}^\infty erfc(\frac {1+2n\pm x}{2\sqrt t})$
and that's what i wanted. Did i miss something??