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This sort of problem is well documented, but, I can not see how to get this solution;

The problem is:

$$\frac{\partial c}{\partial t}=D\frac{\partial^2 c}{\partial x^2}, 0 < x < \infty, t > 0$$

with boundary conditions $c(0,t)= C_0, c(x,0) = 0$ and possibly $\lim_{x \rightarrow \infty}c(x,t) = 0$ as the capillary that the model is modelling is infinitely long.

whose solution is

$$c(x,t) = 2C_0\bigg( 1-\frac{1}{\sqrt{2\pi}}\int^z_{-\infty}\exp(-\frac{s^2}{2}) ds \bigg), z = \frac{x}{\sqrt{2Dt}}$$

Now I have let $c(x,t) = X(x)T(t)$ and then obtained

$$\frac{T'}{DT} = - s^2 = \frac{X''}{X}$$

we have

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

so

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

but my initial conditions give, first

$$\int_0^\infty C_2(s)e^{-Dts^2}ds = C_0$$

then changing variables $s = \sqrt{\frac{r}{D}}$

$$\int_0^\infty \frac{1}{2}\sqrt{\frac{D}{r}}C_2(r)e^{-tr}ds = C_0$$

$$ L[ \frac{1}{2}\sqrt{\frac{D}{r}}C_2(r)] = C_0$$

the inverse Laplace transformation of the right and side is a multiple of the delta function.... I feel like I have taken the wrong way.

  • The expressions $X(x)$ and $T(t)$ you found are both false. Even if you correct them, the particular solution with respect to the boundary conditions will require not only one function $c(x,t)=X(x)T(t)$ but the sum of an infinity of such functions with an infinity of values of $k$. This approach would be theoretically possible, but very arduous. – JJacquelin Sep 23 '16 at 09:21
  • Note that the method of separation of variables $X(x)T(t)$ is convenient in cases of some "educational" problems where the boundary conditions are especially chosen to make successful the calculus with this method. – JJacquelin Sep 23 '16 at 09:32
  • Right. Is there a method you can suggest. Or maybe an initial step to take? –  Sep 23 '16 at 16:27
  • Since the solution is known I suppose that the calculus behind is published or at least the references about the method used. – JJacquelin Sep 23 '16 at 17:16
  • The key idea is to remplace a discrete infinite sum of terms $\sum_sC_sX(x,s)T(t,s)$ by an integral $\int C(s)X(x,s)T(t,s)ds$. For example see http://math.stackexchange.com/questions/1486114/steps-to-solve-semi-infinite-ibvp – JJacquelin Sep 23 '16 at 17:47
  • Brilliant, thanks for that I will take a look through it. –  Sep 23 '16 at 17:58
  • would I be correct in saying that the constants depend on $s$ because $s$ was arbitrarily chosen? Also, I am slightly confused when it comes to the sum part. Why, when solving PDE's do we take a sum? –  Sep 23 '16 at 19:30
  • $D\frac{T'}{T}=\frac{X''}{X}=-s^2$ any $s$. Each value of $s$ generates a solution $C_sX(x,s)T(t,s)$ of the PDE. $C_s$ is any constant. The subscript $s$ is there only to indicate that all constants can be different (or not). In adding the particular solutions we obtain a more general solution. This is possible because the PDE is linear. The integral instead of discret sum allows to the general solution to be fitted to the boundary conditions. In fact, all this is done in order to obtain a solution satisfying the boundary conditions : this is the most difficult part of the task. – JJacquelin Sep 23 '16 at 20:55
  • Hi there, I have tried to work through the link, the problem in the link I can follow. Using a similar method to his I have encountered a problem in finding the value of $C_2$, I have updated my original post with some details of this. –  Sep 24 '16 at 14:24
  • Your $X(x)$ and $T(t)$ continue to be false. – JJacquelin Sep 24 '16 at 14:33
  • But the general solution in EDIT is the same as that in the link? the only way they differ is the boundary conditions. I am sorry about this JJacquelin –  Sep 24 '16 at 14:37
  • You set $=\frac{1}{k}$ instead of $=-s^2$. So, your $X(x)$ isn't sinusoidal as you wrote. If you want sinusoidal terms, you have to set $=-\frac{1}{k}$. Also your $T(t)$ is false because the solution of a first order ODE involves only one constant, not two. – JJacquelin Sep 24 '16 at 14:49
  • Yes, but if i proceeded as in the first answer in the link I would have arrived at the first statement in the EDIT section? i.e if I started the working from scratch, using $=-s^2$ I would arrive at the first statement in the EDIT section. –  Sep 24 '16 at 14:52
  • To be clear, correct your equation with $=-\frac{1}{k}$ and $T(t)$. – JJacquelin Sep 24 '16 at 14:56
  • I have updated to how I have now, after reading the link you suggested, attempted the problem. –  Sep 24 '16 at 15:15
  • Meanwhile doraemonpaul gives the full answer. The function $\delta(s)$ is not a problem since it leads to the second integral equal to $C_0$ – JJacquelin Sep 25 '16 at 05:27

1 Answers1

1

Of course use separation of variables:

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

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

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

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

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

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

$c(0,t)=C_0$ :

$\int_0^\infty C_2(s)e^{-kts^2}~ds=C_0$

$C_2(s)=C_0\delta(s)$

$\therefore c(x,t)=\int_0^\infty C_1(s)e^{-Dts^2}\sin xs~ds+\int_0^\infty C_0\delta(s)e^{-Dts^2}\cos xs~ds=\int_0^\infty C_1(s)e^{-Dts^2}\sin xs~ds+C_0$

$c(x,0)=0$ :

$\int_0^\infty C_1(s)\sin xs~ds+C_0=0$

$\mathcal{F}_{s,s\to x}\{C_1(s)\}=-C_0$

$C_1(s)=\mathcal{F}^{-1}_{s,x\to s}\{-C_0\}=-\dfrac{2C_0}{\pi s}$

$\therefore c(x,t)=C_0-\dfrac{2C_0}{\pi}\int_0^\infty\dfrac{e^{-Dts^2}\sin xs}{s}~ds=C_0~\text{erfc}\left(\dfrac{x}{\sqrt{4Dt}}\right)$

It luckily satisflies $\lim\limits_{x\to\infty}c(x,t)=0$ .

$\therefore c(x,t)=C_0~\text{erfc}\left(\dfrac{x}{\sqrt{4Dt}}\right)$

doraemonpaul
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  • Very nice : The Fourier sin transform to find $C_1(s)$. – JJacquelin Sep 25 '16 at 05:23
  • Could you expand a little on the inverse Fourier equality. I understand that you have used the Fourier transform. I am trying to follow what is on wikipedia on the transform. –  Sep 26 '16 at 15:50
  • I get $C_1(s) = -\frac{2}{\pi} \int_{0}^{\infty}C_0 \sin(xs)dx = \frac{2C_0}{\pi s} \bigg([\frac{\cos(ys)}{s}]_{y \rightarrow \infty}-1 \bigg)$ how do i deal with the first term? –  Sep 26 '16 at 17:46
  • @HMPARTICLE When $\int_0^\infty f(x)\sin sx~dx$ diverges, we shouldn't simply recognize $\mathcal{F}^{-1}_{s,x\to s}{f(x)}=\dfrac{2}{\pi}\int_0^\infty f(x)\sin sx~dx$ and should consider at other suitable approach. – doraemonpaul Sep 28 '16 at 12:19
  • Could you hint at another approach? –  Sep 28 '16 at 21:00