#### 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$$U\left ( x,s\right ) =c_{2}e^{-\sqrt{\frac{s}{k}}x} \tag{2}$$ 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$$U\left ( x,s\right ) =\frac{1}{1+s^{2}}e^{-\sqrt{\frac{s}{k}}x} \tag{3}$$ 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