Let $\lambda < \lambda_1$ and $ 2 < q < 2^*$. I am trying solve this problem
$$(P) \begin{align} \begin{cases} -\Delta u &= \lambda u + |u|^{q-2} u \ \text{in} \ \Omega,\\ u &= 0 \ \text{on} \ \partial \Omega. \end{cases} \end{align}$$
I was able to find a non-trivial and positive solution for the auxiliary problem:
$$(P') \begin{align} \begin{cases} -\Delta u &= \lambda u + \theta |u|^{q-2} u \ \text{in} \ \Omega,\\ u &= 0 \ \text{on} \ \partial \Omega, \end{cases} \end{align}$$
where $\theta \in \mathbb{R}$ was obtained by the method of Lagrange's multipliers.
My question is how can I find a solution of the problem $(P)$ from the solution of the problem $(P')$?
Thanks in advance!
Let be $u$ the solution of the problem $(P')$ and let be $t > 0$. Define $\overline{u} := t u$. On the one hand,
$$- \Delta \overline{u} = t (- \Delta u) = t (\lambda u + \theta |u|^{q-2} u).$$
On the other hand,
$$\lambda \overline{u} + |\overline{u}|^{q-2} \overline{u} = \lambda t u + t^{q-1} |u|^{q-2} u.$$
We want to show that $\overline{u}$ is a solution of the problem $(P)$, i.e.,
$$- \Delta \overline{u} = \lambda \overline{u} + |\overline{u}|^{q-2} \overline{u}.$$
Suppose the equality above holds, then
$$t (\lambda u + \theta |u|^{q-2} u) = \lambda t u + t^{q-1} |u|^{q-2} u.$$
Thus, we need to have $t = \theta^{\frac{1}{q-2}}$ in order to $\overline{u}$ be the solution of the problem $(P)$.