The solution of diffusion equation $$ \partial_t\rho=D\nabla^2\rho$$ with a point source $$ \rho(0,z)=\delta(z)$$

is in 1 dimension $$ \rho(t,z)=\frac{1}{\sqrt{4\pi Dt}}e^{-\frac{z^2}{4Dt}}$$

My question is : is there any analytical solution of the same problem with a boundary solution $$v(z_0)=0=-D\partial_z\rho(z_0)$$

meaning if you put a wall somewhere ?

My feeling is that the problem would be overconstrained. Is it right ?

If it's possible, what would be the way to achieve it ?

Thanks in advance

Edit 1 :

And if it's possible is it also possible to add a force towards the wall ? I mean for example if you have diffusion + gravity : $$v=-v_g-\frac{D\nabla\rho}{\rho} $$ which associate with : $$\partial_t \rho +\nabla.(\rho v)=0 $$ to

$$\partial_t \rho=D\nabla^2\rho + v_g \nabla\rho $$

and in 1d : $$\partial_t \rho= D\frac{d^2\rho}{dz^2} + v_g \frac{d\rho}{dz} $$

This can be simplified thanks to a change of variable : $z\rightarrow\tilde{z}=z+v_g t$ into : $$\partial_t \rho= D\frac{d^2\rho}{d\tilde{z}^2} $$

and the solution is $$ \rho(t,z)=\frac{1}{\sqrt{4\pi Dt}}e^{-\frac{(z-vt)^2}{4Dt}}$$

So I was wondering how this solution was changed if there is a wall.

Edit 2

I found this link in wikipedia about the Mason-Weaver equation where a solution for my second equation in the edit seems to be derived.


1 Answer 1


An analytical solution exists. The idea is to but an instantaneous "image source" behind the wall, so that assuming $z_0 > 0$, the problem is replaced with one where there is no wall, but the initial condition is $$ \rho(0, z) = \delta(z) + \delta(z - 2z_0). $$ Because of the symmetry about $z=z_0$, $\partial_z\rho$ will be zero here at all times. This technique can be applied to a wide range of problems. The solution is $$\rho(t, z) = \frac{1}{\sqrt{4\pi D t}}\left[e^{-\frac{z^2}{4Dt}} + e^{-\frac{(z - 2z_0)^2}{4Dt}} \right]. $$

  • $\begingroup$ @Pulk Thanks that's a very nice answer. I tried your solution on a modified diffusion equation that I described in the new Edit but I don't get satisfying result, by changing the exponentials $z^2\rightarrow (z-t)^2$ and $(z-2z_0)^2\rightarrow (z-2z_0+t)^2$ but the waves go away... Would you have a way to solve that complication plz ? $\endgroup$
    – J.A
    Jul 23, 2019 at 20:10
  • 1
    $\begingroup$ Unfortunately this technique can't be used (at least not directly) to the new problem because the drift term breaks the symmetry. Adding two Gaussian solutions no longer works, so i don't know of a simple analytical solution. $\endgroup$
    – Puk
    Jul 23, 2019 at 21:35

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