# Stability for quasi-linear ODE

1. Mar 22, 2006

### ReyChiquito

I have the following quasi-linear ode (1)

$$-\frac{d^2 u}{dx^2}+\frac{\lambda}{(1+u)^2}=0$$

with boundary conditions $u(\pm \frac{1}{2})=0$ and $\lambda>0$.

I've proven this equation to have two solutions for $\lambda<\lambda^*$, one for $\lambda=\lambda^*$ and none for $\lambda*<\lambda$.
Also, i know that $-1< u(0) \le u(x) \le 0$ and that $u$ is convex. I can plot the bifurcation diagram ($\|u\|_\infty \quad vs \quad \lambda$) and find out that in $\lambda=\lambda^*$ I have a subcritical bifurcation.

With all this information, I need to find out wich branch of the bifurcation diagram is the stable one. In order to do that, I state the linealized eigenvalue problem (2) around a solution $u_s(x)$ of (1)

$$-\frac{d^2 u}{d x^2}-\left[\frac{2\lambda}{(1+u_s(x))^3}+\mu\right]u=0$$

and here is where i get stuck.

The eigenvalue $\mu=0$ will tell me where i loose stability, so I need to solve (2) with $\mu=0$ right? And if I found a $\lambda$ where the solution is nontrivial then I'll know wich branch is the unstable one. Is that correct?

How can I calculate such $\lambda$?

I've read that if I take the operator

$$A=-\frac{d^2}{d x^2}-\frac{2\lambda}{(1+u_s(x))^3}$$,

since

$$\frac{2\lambda}{\left[1+u_s(x)\right]^3} \le \frac{2\lambda}{\left[1+u_s(0)\right]^3}=c$$

then i can define the operator

$$A_c=A+(c+1)$$

wich is strongly positive and has eigenvalues $\mu_0-c-1,\mu_1-c-1,...,\mu_n-c-1,...$ where $\mu_0,\mu_1,...,\mu_n,...$ are the eigenvalues of (2). If I calculate the eigenvalues of $A_s$ then i can calculate $\lambda$. Is this a better way or am I just complicating things?

Other apporach i did was to use the WKB method in (2) but i get stuck trying to apply boundary conditions in my integrals.

I need someone to help me clarify my mind and tell me if im getting somewhere or im just waisting everyones time...

sorry for bad english and thx for help

Last edited: Mar 22, 2006