1.3.1 Left end at zero temperature, 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 ) & =T_{1}\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 \(x=0\,,u\left ( 0,t\right ) =T_{1}\), some constant value. Therefore \(U\left ( 0,s\right ) =\mathcal{L}\left ( u\left ( 0,t\right ) \right ) =\mathcal{L}\left ( T_{1}\right ) =\frac{1}{s}T_{1}\). Hence at \(x=0\) the above gives\[ \frac{1}{s}T_{1}=c_{2}\] Therefore (2) becomes\begin{equation} U\left ( x,s\right ) =\frac{T_{1}}{s}e^{-\sqrt{\frac{s}{k}}x} \tag{3} \end{equation} From tables, the inverse Laplace transform of the above is (since \(x>0,k>0\))\begin{align*} u\left ( x,t\right ) & =T_{1}\operatorname{erfc}\left ( \frac{x}{2\sqrt{kt}}\right ) \\ & =T_{1}\left ( 1-\operatorname{erf}\left ( \frac{x}{2\sqrt{kt}}\right ) \right ) \end{align*}

This completes the solution.

Animation

This example was solved also numerically to verify the above result. Let \begin{align*} k & =\frac{1}{10}\frac{m^{2}}{\sec }\\ u\left ( x,0\right ) & =0\\ u\left ( 0,t\right ) & =T_{1}=60^{0} \end{align*}

Solution

\begin{align*} u\left ( x,t\right ) & =60\operatorname{erfc}\left ( \frac{x}{2\sqrt{\frac{t}{10}}}\right ) \\ & =60\left ( 1-\operatorname{erf}\left ( \sqrt{\frac{5}{2}}\frac{x}{\sqrt{t}}\right ) \right ) \end{align*}

This is an animation of the above.

Code used to the above