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Solve the Heat-Eqn. $$u_t=ku_{xx}$$ where $x,t>0$ and $u_x(0,t) =0$ and $u(x,0)=\begin{cases} 1, & 0 < x <2 \\ 0, & 2\leq x \end{cases} $

What are the solution methods?

  • Separation of variables?

  • Fourier Transform?

I can' t solve it by the separation of variables since $u_x(L,t)$ is ungiven and since $u(x,0)$ is a piecewise function.

Or can we solve by the Fourier Transform? It is a difficult question for me. Please help me.

My attempt:

Using separation of variables, we have $X''(x)-\lambda X(x)=0$ and $T'(t)-\lambda T(t)=0$.

If we select $\lambda=-\mu^2$, we get

$X(x)=Acos\mu x+B sin \mu x$ and $T(t)=C e^{-kt(\dfrac{n \pi}{L})^2}$

HD239
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  • kindly include your attempt. – Siong Thye Goh Oct 11 '17 at 16:34
  • The right boundary condition isn't given because, as you wrote yourself, the constraint on $x$ is $x>0$—this is a heat equation on a semi-infinite rod. – Nap D. Lover Oct 11 '17 at 16:39
  • However, i don't think that necessarily precludes separation of variables from working—where did you get stuck? – Nap D. Lover Oct 11 '17 at 16:40
  • I edited it. Please look again dear friends. – HD239 Oct 11 '17 at 16:50
  • Okay. Again, where does this quantity $L$ magically appear from? It isn't in the problem statement so how does it show up in your solution? Are you perhaps going through the formal steps of the analogous problem on $0<x<L$ without thinking of the context of the current problem? – Nap D. Lover Oct 11 '17 at 16:53
  • you need to consider all 3 cases for $\lambda$ (negative, zero and positive) – Vasili Oct 11 '17 at 16:56
  • Additionally: it seems you've made a typo: you give a piecewise function for the initial condition $u(x,0)$ but then refer to it as $u_x(x,0)$. Which one is it? – Nap D. Lover Oct 11 '17 at 16:57
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    You are right, it is $u(x,0)$ – HD239 Oct 11 '17 at 17:00
  • Wait a minute, are you saying $ku$ dropped its $x$? Great thing I rolled as fast as I could then! – mathreadler Oct 11 '17 at 17:14
  • I believe the domain $L$ is (0,2) since it is zero everywhere else in the initial condition. – David Oct 11 '17 at 17:22
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    @David No, stuff will diffuse out of (0,2) shortly thereafter. – Ian Oct 11 '17 at 17:33
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    The usual approach to the heat equation on the semi-infinite rod is to use symmetry to convert to the heat equation on the full infinite rod. You can find this online, for example https://ocw.mit.edu/courses/mathematics/18-303-linear-partial-differential-equations-fall-2006/lecture-notes/fourtran.pdf Another way to go is to look at a truncated domain $[0,M]$ and then send $M \to \infty$, but to do that you need to be careful to choose the correct boundary condition on the other end of the truncated domain. I think the correct boundary condition will be $u_x(M,t)=0$. – Ian Oct 11 '17 at 17:35
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    See this reference for a solution – xidgel Oct 11 '17 at 21:29

1 Answers1

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For $0<x<2$ , the PDE has conditions $u_x(0,t)=0$ and $u(x,0)=1$ only.

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

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

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

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

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

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

$u_x(x,t)=\int_0^\infty sC_1(s)e^{-kts^2}\cos xs~ds-\int_0^\infty sC_2(s)e^{-kts^2}\sin xs~ds$

$u_x(0,t)=0$ :

$\int_0^\infty sC_1(s)e^{-kts^2}~ds=0$

$C_1(s)=0$

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

$u(x,0)=1$ :

$\int_0^\infty C_2(s)\cos xs~ds=1$

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

$\therefore u(x,t)=\int_0^\infty\delta(s)e^{-kts^2}\cos xs~ds=1$

$u(2,t)=1$

Now for $x\geq2$ , the PDE has conditions $u(2,t)=1$ and $u(x,0)=0$ only.

Similarly, $u(x,t)=\int_0^\infty C_3(s)e^{-kts^2}\sin((x-2)s)~ds+\int_0^\infty C_4(s)e^{-kts^2}\cos((x-2)s)~ds$

$u(2,t)=1$ :

$\int_0^\infty C_4(s)e^{-kts^2}~ds=1$

$C_4(s)=\delta(s)$

$\therefore u(x,t)=\int_0^\infty C_3(s)e^{-kts^2}\sin((x-2)s)~ds+\int_0^\infty \delta(s)e^{-kts^2}\cos((x-2)s)~ds=1+\int_0^\infty C_3(s)e^{-kts^2}\sin((x-2)s)~ds$

$u(x,0)=0$ :

$\int_0^\infty C_3(s)\sin((x-2)s)~ds=-1$

$\mathcal{F}_{s,s\to x-2}\{C_3(s)\}=-1$

$C_3(s)=\mathcal{F}^{-1}_{s,x-2\to s}\{-1\}=-\dfrac{2}{\pi s}$

$\therefore u(x,t)=1-\dfrac{2}{\pi}\int_0^\infty\dfrac{e^{-kts^2}\sin((x-2)s)}{s}~ds=\text{erfc}\left(\dfrac{x-2}{2\sqrt{kt}}\right)$

doraemonpaul
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    This is totally wrong, because $u(2,t)$ will not be $1$ forever. Material will diffuse between the two subdomains and in fact $\lim_{t \to \infty} u(x,t)$ will be $0$ for each fixed $x$. It is possible to proceed in this general manner but then the interior boundary condition at $x=2$ is precisely this "continuous flux between subdomains" boundary condition which is not $u(2,t)=1$. – Ian Oct 12 '17 at 15:38
  • I think there should be some simple mistakes somewhere, though I can't immediately tell where it is. I've solved the problem with Mathematica here, you can use that solution for comparison. – xzczd Oct 20 '17 at 12:57