I am attempting to understand the Lyapunov-Schmidt method with a simple example but I am running into trouble. Here is the example I am considering. Suppose $ v>0$ satisfies $ -\Delta v - v=0 $ in $ \Omega$ with $ v=0 $ on $ \partial \Omega$. I would like to find solutions of

$$ -\Delta u - u = \lambda f(u) \qquad \Omega, \qquad \qquad u=0 \quad \partial \Omega $$ for small $ \lambda$ where $ f$ is some fixed smooth nonlinearity. Lets assume the space we are working in can be decomposed as $ span\{ v\} + X$ and so given $t \in R$ there is some $ \phi=\phi(\lambda,t) \in X$ and $C=C(\lambda,t) \in R$ such that $$ \Delta \phi + \phi + \lambda f(tv + \phi) = C v.$$ So we hope that for small $ \lambda$ we can pick $t$ such that $ C(\lambda,t)=0$ and then $ u=tv+\phi$ satisfies our original problem.

Now of course one must impose some restrictions on $f$ otherwise we cannot expect to have a solution. For instance if $u$ a solution then we must have $\int_\Omega f(u)v dx=0$ and hence $f$ cannot be of one sign. Now multiplying both sides of the PDE by $v$ (and assume $v$ is $L^2$ normalized) one arrives at $$C(\lambda,t)=\lambda \int_\Omega v f(tv+ \phi(\lambda,t)) dx.$$

So the part I am struggling with is to show that $C$ is zero for some $t$. I suspect the idea is to shows its positive for some $ t$ and negative for antoher $t$ and then use some continuity of $ \phi$ in $ t$ and $ \lambda$ and the intermediate value theorem to find a zero of $C$. One case I was almost able to prove $C$ had a zero was when $f$ was given by $ f(u)=-u^3 + h(u)$ where $h$ is bounded.

I realize my question is not very precise. Any suggestions would be greatly appreciated. thanks Craig

  • $\begingroup$ You need to distinguish two cases $\int f(v) v \neq 0$ and $\int f(v) v =0$. The first case is simpler and can be dealt with using the implicit function theorem. The second case is more complicated. You need to fix a function $w$ such that $-\Delta w-w +f(v)=0$, $w=0$ on the boundary. The bifurcation picture can be tricky and you need more info on $f$ to be able to same something meaningful. $\endgroup$ – Liviu Nicolaescu May 13 '14 at 8:54
  • $\begingroup$ It looks to me that if $\int f'(v) v^2 \neq 0$ and $\int f(v)v =0$ then you have a bifurcation at $\lambda =0$. $\endgroup$ – Liviu Nicolaescu May 13 '14 at 9:17
  • $\begingroup$ thank you very much for the tips. I was playing around with the implicit function theorem but i wasn't using $ \partial_t \phi(\lambda,t)=0$ when $ \lambda =0$. $\endgroup$ – Craig May 14 '14 at 1:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.