I have the solution of a PDE of the form: $$ \Delta \Psi(r,\theta, \phi) = k \Psi(r,\theta,\phi)$$ on a set $\mathbb{R}^3 \backslash B(0,R)$.

Hence, the actual solution is known there! Regarding this, you might have a look at: Related thread Notice, that my solution is bounded and vanishes at infinity! Now I want to consider an additional small perturbation $$ \Delta \Psi(r,\theta, \phi) = k \Psi(r,\theta,\phi) + \epsilon \Psi(r,\theta,\phi)^2.$$

Of course, I suspect that you cannot analytically solve this PDE anymore, but is it possible to come up with some kind of perturbation theory(if it is difficult to answer this question with mathematical rigour, you are invited to argue with handwaving).

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    $\begingroup$ Naively (as a physicist), one would write $\Psi=\Psi_0+\epsilon \Psi_1$. Replace in original equation assuming $\Psi_0$ solves original equation. Get a new (linear) equation for $\Psi_1$ involving r.h.s. with (known) $\Psi_0^2$… iterate? $\endgroup$
    – chris
    Mar 23, 2014 at 14:40
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    $\begingroup$ But your question claims actual solution is known, so I it is fair to assume the square is known too? $\endgroup$
    – chris
    Mar 23, 2014 at 14:48
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    $\begingroup$ @Lipschitz: yes it does. $\endgroup$
    – mookid
    Mar 23, 2014 at 15:12
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    $\begingroup$ @Lipschitz it is rigorous! At least for a physicist :-) $\endgroup$
    – chris
    Mar 23, 2014 at 15:21
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    $\begingroup$ You can treat this equation by means of the method of sub- and super-solutions. I don't know if this can be referred to as a "perturbative method", but you can judge by yourself by looking at Evans's book on PDEs, §9.3 $\endgroup$ Mar 23, 2014 at 17:39

1 Answer 1


Let me simplify your problem to a purely radial one for the sake of simplicity.

Let us write $\Psi=\Psi_0(r) + \epsilon \Psi_1(r)$ and first solve for $\Psi_0$.

The solution is (If I believe mathematica) $\Psi_0(r)=\frac{c_1 e^{-\sqrt{k} r}}{r}+\frac{c_2 e^{\sqrt{k} r}}{2 \sqrt{k} r}$

Now I can plug this solution into the linearized first order (in $\epsilon$) equation

$\nabla^2 \Psi_1- k\Psi_1 =\Psi_0^2$, which again if I trust mathematica,

$\frac{e^{-\sqrt{k} r} \left(-4 c_1^2 k \text{Ei}\left(-\sqrt{k} r\right)-4 c_2 c_1 \sqrt{k} \text{Ei}\left(\sqrt{k} r\right)+e^{2 \sqrt{k} r} \left(4 c_1^2 k \text{Ei}\left(-3 \sqrt{k} r\right)+4 c_2 c_1 \sqrt{k} \text{Ei}\left(-\sqrt{k} r\right)+c_2^2 \text{Ei}\left(\sqrt{k} r\right)+4 c_4 k\right)-c_2^2 \text{Ei}\left(3 \sqrt{k} r\right)+8 c_3 k^{3/2}\right)}{8 k^{3/2} r}$

where $Ei(z)$ gives the exponential integral function. The first order solution is therefore $\Psi_0(r)+\Psi_1(r)$ given above (it seems to obey the original equation at order $\epsilon^2$).

Now one would follow the same path to solve the more general problem, while properly accounting for the boundary conditions, order by order.


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