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Consider the diffusion equation as $$\frac{\partial c}{\partial t} = D\frac{\partial^2c}{\partial x^2}$$ with boundary conditions of

c(x,0) = 0 (x > 0)
c(0,t) = A
c(∞,t) = 0

the Laplace transform leads to $$c(x,t) = A erfc\Big(\frac{x}{2\sqrt{Dt}}\Big)$$ then, at x=0 $$J_{(x=0)} = D\frac{\partial c(x,t)}{\partial t} = \frac{A\sqrt{D}}{\sqrt{\pi t}}$$

Now, how the solution will be if adding middle boundary conditions for x as

c(0,t) = A
c(x,0) = B (x=0)
c(x,0) = 0 (x >= x1)
A > B

meaning that there is a pre-existing gradient in the domain of (0 < x < x1) where x1 is a given value.

Kama
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  • Where is $x/2$? what are you trying to do? – Philip Roe Aug 14 '17 at 21:06
  • Wait, you mean you have put a single nonzero value in the middle of the initial condition? The solution will not see this as a point has measure zero. – Ian Aug 14 '17 at 21:16
  • @ yes, exactly. I modified the question to be clear. – Kama Aug 14 '17 at 22:04
  • @PhilipRoe my bad, it was a bad notation. I meant a given value of x. – Kama Aug 14 '17 at 22:08
  • Do you see how there is a significant difficulty when $A \neq B$? The discontinuity in the initial condition itself is no big deal, that is mollified away immediately by the regularizing properties of the diffusion equation. But the discrepancy at the corner is a serious problem. – Ian Aug 14 '17 at 22:32
  • What do you expect for the behaviour of the function at $0<x\leq x_1$ ? Otherwise I think it is difficult to give you the perfect answer. – doraemonpaul Aug 16 '17 at 14:35
  • @doraemonpaul There is nothing special about x1. Just there is already a concentration gradient in the domain of 0 < x ≤ x 1. In other words, t=0 is the condition of =t1 if A=B. – Kama Aug 16 '17 at 18:16

1 Answers1

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Hint:

For $0\leq x\leq x_1$ ,

Ommit the condition $c(\infty,t)=0$ ,

According to Does solve PDE by combination of variables always cannot find the general solutions?,

$c(x,t)=A+(B-A)\text{erf}\left(\dfrac{x}{2\sqrt{Dt}}\right)$

Hence for $x\geq x_1$ ,

Replace the condition of $c(0,t)=A$ as $c(x_1,t)=A+(B-A)\text{erf}\left(\dfrac{x_1}{2\sqrt{Dt}}\right)$

doraemonpaul
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  • It was a typo in my boundary conditions. c(x,0)=B only at x=0. If fact there is a gradient between x=0 and x=x1, and c is not equal to B in the whole domain. Sorry! – Kama Aug 15 '17 at 18:54