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
$$\lambda^2=\delta/D,~F=G=J=K=u_0/4.$$
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.
Edit:
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
$$u(\mathbf{x},t)=v(|\mathbf{x}|,t),$$
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.
Best Answer
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 $u(x,y) = 0$.