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I am trying to solve the conduction problem for the temperatures $\theta(x,t)$. \begin{align} \frac{\partial\theta}{\partial t}&=k\frac{\partial^2\theta}{\partial x^2} \\ \theta(0,t)&=T_0e^{-bt}, \ \ \ t>0, \ b>0 \tag{1}\\ \theta(x,0)&=0, \ \ \ x>0. \end{align}

My attempt:

I took the Laplace transform with respect to t of the PDE. \begin{align} \mathcal{L}_t(\theta_t(x,t))&=k\mathcal{L}_t(\theta_{xx}(x,t)) \\ s\mathcal{L}_t(\theta(x,t))&=k\frac{d^2}{dx^2}\mathcal{L}_t(\theta(x,t)) \\ s\bar{\theta}&=k\frac{d^2}{dx^2}\bar{\theta}. \end{align}

Solving this ODE, I get $$\bar{\theta}(x,t)=Ae^{x\sqrt{\frac{s}{k}}}+Be^{-x\sqrt{\frac{s}{k}}}, \ \ A,B\in\mathbb{R}.$$ To ensure $\bar{\theta}$ is finite, take $A=0$ as $|\bar{\theta}|\rightarrow\infty$ as $|s|\rightarrow\infty.$ Taking the Laplace transform of $(1)$ and imposing this boundary condition, I get $$\bar{\theta}(x,t)=\frac{T_0}{s+b}e^{-x\sqrt{\frac{s}{k}}}.$$ Assuming this is correct, how can I invert? A hint would be appreciated in (I have tried convolution theorem). I expect the result to be in terms of error functions.

Update:

$$\mathcal{L^{-1}}\left(\frac{1}{s+b}\times e^{-x\sqrt{\frac{s}{k}}}\right)=e^{-bt}\ast\frac{kxe^{\frac{x^2}{4tk}}}{2\sqrt{\pi (kt)^3}}.$$ I have used the property $$\mathcal{L}(f(ct))=\frac{1}{c}F\left(\frac{s}{c}\right).$$

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  • $\begingroup$ You might find this helpful: math.stackexchange.com/questions/1779581/… $\endgroup$
    – Paul
    Apr 7 '19 at 3:04
  • $\begingroup$ @Paul Thanks, I was on the right track. I have updated my attempt $\endgroup$
    – user557493
    Apr 7 '19 at 3:12
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Let $\theta(x,t)=X(x)T(t)$ ,

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

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

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

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

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

$\theta(0,t)=T_0e^{-bt}$ :

$\int_0^\infty C_2(s)e^{-kts^2}~ds=T_0e^{-bt}$

$C_2(s)=T_0\delta\left(s-\sqrt{\dfrac{b}{k}}\right)$

$\therefore\theta(x,t)=\int_0^\infty C_1(s)e^{-kts^2}\sin xs~ds+\int_0^\infty T_0\delta\left(s-\sqrt{\dfrac{b}{k}}\right)e^{-kts^2}\cos xs~ds$

$\theta(x,t)=\int_0^\infty C_1(s)e^{-kts^2}\sin xs~ds+T_0e^{-bt}\cos\sqrt{\dfrac{b}{k}}x$

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  • $\begingroup$ The answer given in my textbook is $$\theta(x,t)=\frac{T_0e^{-bt}}{2}\left(e^{x\sqrt{-b/k}}\text{erfc}\left(\frac{x+2\sqrt{-b/k}kt}{2\sqrt{kt}}\right)+e^{-x\sqrt{-b/k}}\text{erfc}\left(\frac{x-2\sqrt{-b/k}kt}{2\sqrt{kt}}\right)\right).$$ Here, erfc is the complementary Error function. Is this equivalent to your result? $\endgroup$
    – user557493
    Apr 7 '19 at 11:01

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