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$\mathbf{u_t=ku_{xx}; u)0,t)=u_0 , u(x,0)=0; \text{u bounded for} x>0,t>0:}$

Take LT wrt $t$:

$\frac{\partial ^2 \bar{u}}{\partial x^2}=\bar{u} \frac{s}{k}$

$\frac{\partial ^2 \bar{u}}{\partial x^2}-\bar{u} \frac{s}{k}=0$

$\rightarrow \bar{u}=Ae^{\sqrt{\frac{s}{k}}}+Be^{-\sqrt{\frac{s}{k}}}$

How should I proceed?

The answer is

$u=u_o$ $erfc$ $[\frac{x}{2\sqrt {kt}}]$

  • 0
    What do you mean? – 2017-01-04
  • 0
    How do I invert the $\bar{u}$. I need hints for proceeding from the roots of $\bar{u}$. – 2017-01-04
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    First you should also LT the boundary values and use them to find A and B – 2017-01-04
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    I seem to be lacking confidence in this. Could someone please post an answer, brief and precise. – 2017-01-04
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    If $u(x,0) = 0$ then the solution is $u(x,t) = 0 \ \ \forall x \in \mathbb{R}, t > 0$. I think you mean $u(x,0) = u_{0}$, $u(0, t) = 0$. Also, see [here](http://www.robots.ox.ac.uk/~jmb/lectures/pdelecture4.pdf) for a solution (the last 3 slides, set $c = 1$). – 2017-01-04

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Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=kX''(x)T(t)$

$\dfrac{T'(t)}{kT(t)}=\dfrac{X''(x)}{X(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-ks^2\\X''(x)+s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-kts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-kts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-kts^2}\cos xs~ds$

$u(0,t)=u_0$ :

$\int_0^\infty C_2(s)e^{-kts^2}~ds=u_0$

$C_2(s)=u_0\delta(s)$

$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-kts^2}\sin xs~ds+\int_0^\infty u_0\delta(s)e^{-kts^2}\cos xs~ds=\int_0^\infty C_1(s)e^{-kts^2}\sin xs~ds+u_0$

$u(x,0)=0$ :

$\int_0^\infty C_1(s)\sin xs~ds+u_0=0$

$\mathcal{F}_{s,s\to x}\{C_1(s)\}=-u_0$

$C_1(s)=\mathcal{F}^{-1}_{s,x\to s}\{-u_0\}=-\dfrac{2u_0}{\pi s}$

$\therefore u(x,t)=u_0-\dfrac{2u_0}{\pi}\int_0^\infty\dfrac{e^{-kts^2}\sin xs}{s}~ds=u_0~\text{erfc}\left(\dfrac{x}{2\sqrt{kt}}\right)$