Apologies for my overly simple problem.

I am looking at the generic diffusion-decay PDE

$$u_t=D\nabla^2u-\delta u(x,y,t),~u(0,0,t)=u_0,$$

and I am interested in the steady-state profile of $u(x,y,t)$, i.e. a solution to

$$0=D\nabla^2u-\delta u(x,y,t),~u(0,0)=u_0.$$

Using the ansatz

$$u(x,y,t) = F e^{\lambda x} + G e^{-\lambda x} + J e^{\lambda y} + K e^{-\lambda y},$$

I find one possible solution with


The problem I have now is that this solution grows away from the origin which I find puzzling as I expected to find a solution that has a peak in the origin and decays away from it.

Could anyone point me in the right direction please?

Thank you.


Apologies for my slow response but @Andrew's answer was a little intimidating (and still is, although I don't seem to see it now - did they delete their answer?) and so I had to do a little background reading.

Thanks @Willie Wong for your answer and for giving me some intuition as to why my expectation is wrong. Also thanks to @Andrew for pointing me in the direction of fundamental solutions and so forth.

Following @Andrew I found these lecture notes: http://www.stanford.edu/class/math220b/handouts/laplace.pdf

Where they construct a radial solution for the Laplace equation using the ansatz


and knowing the derivative of the absolute value function and defining radius $r=|\mathbf{x}|$ I get:

$$v_t=D(v_{rr} + \frac{1}{r} v_r) - \delta v~;~D,\delta>0~;~r>0$$

For the steady state equation

$$0=D(v_{rr} + \frac{1}{r} v_r) - \delta v~;~D,\delta>0~;~r>0$$

I use the ansatz

$$v(r)=v_0 \exp{\lambda r},$$

which gives me

$$0=D v_0 \exp{(\lambda r)} (\lambda^2 + \frac{\lambda}{r} - \delta)$$

so using $D>0$, $v_0> 0$ I get

$$\lambda_{1,2}=-\frac{1}{2r} \mp \frac{1}{2} \sqrt{r^{-2}+4 \delta / D}.$$

Of course this solution blows up towards the origin so I am again a little puzzled and would appreciate any advice / help!

In those notes I linked above, they construct the fundamental solution (starting around Eqn 3.2) which is probably what I want but I don't yet understand how some of the steps work.


1 Answer 1


If $D$ and $\delta$ are both positive, as indicated by the nomenclature diffusion-decay, that is the solution you should find. At a critical point $\nabla u = 0$ you have that

$$ D\nabla^2 u = \delta u $$

suggesting that if $u$ is positive at the critical point, it cannot be a maximum (as other wise $\nabla^2 u < 0$ and leads to a contradiction) and if $u$ is negative at the critical point, it cannot be a minimum.

This means that your expectation to have a peak in the origin and decays away from it is unrealistic and untenable.

Physically speaking, your equation has two processes, a diffusion and a decay, both are trying to drive the function toward 0. If you don't continuously inject more stuff in from infinity (which is represented by the growing solutions), the only reasonable steady state is $u(x,y) = 0$.

  • $\begingroup$ Thanks for your answer! Do you know if the other person deleted their answer? I don't seem to be able to see the answer they gave. Do you have any thoughts on the steps I have taken trying to get a radial solution (see my edits above)? $\endgroup$
    – Name
    May 29, 2012 at 15:10
  • $\begingroup$ The fundamental solution has to blow up near the origin: in your case it should formally solve the equation $D\nabla^2 u - \delta u = \delta_0$, where $\delta_0$ is the Dirac delta function. I am not sure if that is in fact what you want. May I ask why you choose an exponential ansatz, and why you need to think about the steady-state problem? $\endgroup$ May 29, 2012 at 15:55
  • $\begingroup$ What you are doing is not quite right: $\lambda$ should be constant in your ansatz (I think), yet what you derived has $\lambda$ depending on $r$. Solutions of the radial problem should be in the form of Bessel functions. $\endgroup$ May 29, 2012 at 15:59
  • $\begingroup$ Yea OK, I think I misunderstood the lecture notes I linked to. Do I understand what you wrote correctly when I say: we want to find a radial solution $v$ such that $\int_{-\infty}^{\infty}{(v(s) D \nabla^2 v - \delta v(s))ds} = v(0)$? This is how I understand the Dirac delta notation, but I don't think I understand the intuition behind this. I choose an exponential ansatz because I have a process where material is released at $r=0$ and diffuses and decays everywhere else in the two-dimensional domain. So in my head an exponential seemed reasonable. $\endgroup$
    – Name
    May 29, 2012 at 16:38
  • $\begingroup$ I updated my edit to show more steps. I can also solve the homogeneous equation of the radial ODE I get to find $v_{\text{hom}} = -c_1 \log{r}+c_2$. However, this homogeneous solution also blows up towards the origin so that doesn't help me? $\endgroup$
    – Name
    May 29, 2012 at 16:50

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