How to prove that solutions of semilinear differential equations is even function?
My question comes from the book Stable Solutions of Elliptic Partial Differential Equations Louis Dupaigne, pages 30-32.
Summary: Which uniqueness theorem to use for this differential equation ?
I am working with the following semilinear differential equation \begin{equation} -u^{\prime\prime}=\lambda e^u, \qquad u(-1)=u(1)=0. \end{equation} with $\lambda>0$ a parameter.
Uniqueness
I want to prove that the solutions are even. After verifying that $u(x)$ and $\tilde{u}(x) = u(-x)$ are solutions of the differential equation and that they yield the same value when they are evaluated at the boundary conditions, I need to use some uniqueness theorem to ensure the following equality: $u(x) = \tilde{u}(x) = u(-x)$.
I have been searching but have not found any theorem applicable to this case. In Dupaigne's book, one could use Proposition 1.3.1 on page 15, but the problem here is that the solutions, (I'm not sure), are minimal and maximal and that proposition is only valid for stable solutions, and since the maximal solution has a Morse index different from zero, it isn't stable.
Even
The verification that it is an even function is immediate.
Let $\tilde{u}(x) = u(-x)$. Let’s show that the function $\tilde{u}(x)$ satisfies the differential equation. Its first and second derivatives are $$ \frac{d\tilde{u}(x)}{dx} = \frac{du(-x)}{dx} = -u^{\prime}(-x), \qquad \frac{d^2\tilde{u}(x)}{dx^2} = \frac{d^2u(-x)}{dx^2} = u^{\prime\prime}(-x), $$ Given the equality $u(x) = \tilde{u}(x) = u(-x)$, we have that $$ -\tilde{u}^{\prime\prime}(x) = \lambda e^{\tilde{u}(x)} \ -u^{\prime\prime}(-x) = \lambda e^{u(-x)} \ -u^{\prime\prime}(x) = \lambda e^{u(x)}, $$ and since the boundary conditions are satisfied, the first for $x = 1$ and the second for $x = -1$, $$ \tilde{u}(1) = u(-1) = 0, \qquad \tilde{u}(-1) = u(1) = 0, $$ it is shown that $u(x)$ is an even function.
0 answers
0 comment threads