Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is somewhat of a curiosity that can hide somewhat deeper. For a Green function of a nonlinear PDE I mean something like $$ \partial^2\phi+V(\phi)=\delta^D(x). $$ I do not know if a real meaning can be attached to something like this. But I can think to a gradient expansion. E.g., consider the nonlinear Klein-Gordon equation $$ \partial^2\phi+\lambda\phi^3=\delta^D(x). $$ We can rewrite this equation as $$ \partial_t^2\phi+\lambda\phi^3=\delta(t)\delta^{D-1}(x) + \epsilon\Delta\phi. $$ and to treat the Laplacian as a perturbation (for this aim I introduced an arbitrary $\epsilon$ that I will set to 1 at the end of computation). Then, I have a gradient expansion in term of powers $\epsilon$. This has for the leading order $$ \partial_t^2\phi_0+\lambda\phi_0=\delta^D(x) $$ and I know the exact solution to $\partial_t^2\phi'+\lambda\phi'=\delta(t)$. Can I attach a meaning to something like this from the exact solution of this latter equation maybe using Coulombeau functions?

There is also a less singular approach. I just rescale time as $\tau=\sqrt{\lambda}t$ and I get $$ \lambda\partial_\tau^2\phi+\lambda\phi^3=\sqrt{\lambda}\delta(\tau)\delta^{D-1}(x) + \Delta\phi $$ that is $$ \partial_\tau^2\phi+\phi^3=\frac{1}{\sqrt{\lambda}}\delta(\tau)\delta^{D-1}(x) + \frac{1}{\lambda}\Delta\phi. $$ This is again a gradient expansion in powers of $\frac{1}{\sqrt{\lambda}}$ but the leading order has the form $$ \partial_\tau^2\phi_0+\phi_0^3=0 $$ that has a known exact solution. Next-to-leading order is no more singular giving a linear equation.

This is more than an exercise as can have some applications in physics. But what I am really interested to is if all this can have a mathematical meaning.

Thanks.

share|improve this question
    
In the third equation you have $\partial_t^2$ but in the first two equations you have $\partial^2$. Could you clarify what you mean by the latter? Also, could you explain how you get the third equation from a "gradient expansion"? –  Deane Yang Jun 6 '13 at 22:17
    
@DeaneYang: Yes, I will clarify this in the question. –  Jon Jun 7 '13 at 6:06
2  
Depending on the nonlinearity, it may or may not be straightforward to give a meaning to this. Since there is no superposition principle, I would guess a fundamental solution might not be that useful. Out of curiosity, what use would it have? –  timur Jun 7 '13 at 7:03
    
@timur: This is quite interesting. If you consider such kind of "nonlinear" Green functions, you can get a good approximation to the problem $\partial^\phi+V(\phi)=j$ with a leading order $\phi\approx\int d^Dx'G(x-x')j(x')$, similarly to the linear case. This represents the leading term of a strong coupling series and, for the simplest case of an ODE, it is just a small time expansion. So, the rescaling $\sqrt{\lambda}t$ accounts for the range of validity of this approximation. –  Jon Jun 7 '13 at 7:18
    
Wouldn't you also get a good approximation by using a fundamental solution of the linear part only? Or perhaps the nonlinear fundamental solution gives a better result? –  timur Jun 7 '13 at 7:20
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.