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I'm trying to understand how to apply the Galerkin method to $u_t - \Delta u = u^3$. I understand how to obtain all of the a-priori estimates using Sobolev embeddings and such but my question concerns the actual discretization procedure where we project onto the finite dimensional subspace spanned by the eigenfunctions of $-\Delta$.

In the linear case we may simply set $u_N = \sum \phi_n c_n$ and plug this in to obtain a set of $N$ O.D.Es which we then show satisfy the same energy bounds. In the non-linear case though we may not just substitute directly because of the $u^3$ term. How can this be dealt with? Is there perhaps a better way to do the approximation?

Addition: In this example if we let $u_N = \sum \phi_n c_n^N(t)$ then when we plug this into the weak form of our PDE we obtain and then choose our test function to be one of the basis elements $w_k$ we obtain $d/dt c_k^N(t) + \sum_{i=1}^n e^{ki}(t) c_k^N(t) = \int (\sum_{n=1}^N c_n^N(t) \phi_n)^3w_k$, for some coefficients $e^{ki}(t)$. My question is, how do I deal with the integral on the right? I would like to be able to solve this ODE and then say I have a solution.

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There is no easy way. You have to honestly open the parentheses, compute the integrals, and deal with the resulting trilinear form if you pursue this approach. –  fedja Sep 10 '10 at 23:47

1 Answer 1

The solution of this problem will blow up in finite times, which can be understood from the ODE your posted. In fact, the nonlinear term in the right hand of the ODE is not globally lipchitz, you can only get a local solution by Picard existence theorem. To see that, one solves the following ODE $$ y' = y^3, y(0)=1 $$ you will the solution is $y(t) = -2(t - \sqrt[3]{2})^{-2}$ blow up as $t \rightarrow \sqrt[3]{2}$.

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Will all solutions blow up no matter what (nontrivial) initial and boundary conditions you have? –  timur Dec 1 '11 at 1:10

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