1.3.2 Left end at temperature that depends on time. No source term, zero initial conditions

\begin{align} u_{t} & =ku_{xx}\tag{1}\\ t & >0\nonumber \\ x & >0\nonumber \\ u\left ( x,0\right ) & =0\nonumber \\ u\left ( 0,t\right ) & =\sin \left ( t\right ) \nonumber \end{align}

And \(u\left ( x,t\right ) <\infty \) as \(x\rightarrow \infty \). This means \(u\left ( x,t\right ) \) is bounded. This conditions is always needed to solve these problems.

Solution

Let \(U\left ( x,s\right ) \) be the Laplace transform of \(u\left ( x,t\right ) \). Defined as \[\mathcal{L}\left ( u,t\right ) =\int _{0}^{\infty }e^{-st}u\left ( x,t\right ) dt \] Applying Laplace transform to the original PDE (1) gives\[ sU\left ( x,s\right ) -u\left ( x,0\right ) =kU_{xx}\left ( x,s\right ) \] But \(u\left ( x,0\right ) =0\), therefore the above becomes\[ U_{xx}-\frac{s}{k}U=0 \] The solution to this differential equation is\[ U\left ( x,s\right ) =c_{1}e^{\sqrt{\frac{s}{k}}x}+c_{2}e^{-\sqrt{\frac{s}{k}}x}\] Since \(u\left ( x,t\right ) \) is bounded in the limit as \(x\rightarrow \infty \) and \(k>0\), therefore it must be that \(c_{1}=0\) to keep the solution bounded. The above simplifies to\begin{equation} U\left ( x,s\right ) =c_{2}e^{-\sqrt{\frac{s}{k}}x} \tag{2} \end{equation} At \(x=0\,,u\left ( 0,t\right ) =\sin \left ( t\right ) \). Therefore \(U\left ( 0,s\right ) =\mathcal{L}\left ( u\left ( 0,t\right ) \right ) =\mathcal{L}\left ( \sin \left ( t\right ) \right ) =\frac{1}{1+s^{2}}\). Hence at \(x=0\) the above gives\[ \frac{1}{1+s^{2}}=c_{2}\] Therefore (2) becomes\begin{equation} U\left ( x,s\right ) =\frac{1}{1+s^{2}}e^{-\sqrt{\frac{s}{k}}x} \tag{3} \end{equation} Need to find inverse Laplace transform of the above. Using convolution. Since \begin{align*} e^{-\sqrt{\frac{s}{k}}x} & \Longleftrightarrow \frac{xe^{-\frac{x^{2}}{4k\tau }}}{2\sqrt{\pi }\sqrt{k\tau ^{3}}}\\ \frac{1}{1+s^{2}} & \Longleftrightarrow \sin t \end{align*}

Then\begin{align*} u\left ( x,t\right ) & =\int _{0}^{t}\sin \left ( t-\tau \right ) \frac{xe^{-\frac{x^{2}}{4k\tau }}}{2\sqrt{\pi }\sqrt{k\tau ^{3}}}d\tau \\ & =\frac{x}{2\sqrt{\pi k}}\int _{0}^{t}\sin \left ( t-\tau \right ) \frac{e^{-\frac{x^{2}}{4k\tau }}}{\sqrt{\tau ^{3}}}d\tau \end{align*}

This completes the answer. The above integral is evaluated numerically in the animation below since a closed form inverse Laplace transform was not found at this time.

Animation

This is an animation of the above for \(20\) seconds using \(k=\frac{1}{10}\).

Code used to the above