A pde via the Theorem of Lagrange Multipliers

Question

The following is an exercise of a text (in portuguese) on Critical Point Theory I am reading:

Use the Theorem of Lagrange Multipliers to obtain a weak solution to the problem $$ (P) \quad \begin{cases} - \Delta u + u = h(x) |u|^{q - 2}u \quad \text{ in } \Bbb{R}^N \\ u \in H^1(\Bbb{R}^N) \end{cases} $$ where $2 < q < 2^*$, $h \in L^{\frac{2^*}{2^* - q}}(\Bbb{R}^N) \cap L^\infty (\Bbb{R}^N)$, $h \geq 0$ and $h \neq 0$.

Here is what I tried:

We will assume $N \geq 3$. Consider the functionals $J, F: H^1 (\Bbb{R}^N) \longrightarrow \Bbb{R}$ defined by $$ J(u) = \frac12 ||u||^2, \quad F(u) = \frac1q \int_{\Bbb{R}^N} h(x) |u|^q \ dx. $$ We know that these functionals are both of class $C^1$, with $$ J'(u)v = \int_{\Bbb{R}^N} \nabla u \cdot \nabla v \ dx + \int_{\Bbb{R}^N} uv \ dx, \quad F'(u)v = \int_{\Bbb{R}^N} h(x)|u|^{q - 2}uv \ dx $$ for $v \in H^1(\Bbb{R}^N)$.

Let $$ M = F^{-1}(\{1\}) = \{ u \in H^1 (\Bbb{R}^N) \ : \ F(u) = 1\}. $$

To apply the Theorem of Lagrange Multipliers we need to show that $F'(u) \neq 0$ for all $u \in M$. Indeed, for all such $u$ we have that $$ F'(u)u = \int_{\Bbb{R}^N} h(x) |u|^q \ dx = q. $$

Note that $J \geq 0$, so in particular it is bounded from below on $M$. Let $$ c = \inf_M J. $$ Then there exists a sequence $(u_n) \subset M$ such that $$ J(u_n) = \frac12 ||u_n||^2 \to c \geq 0, $$ hence $(u_n)$ is bounded. Since $H^1(\Bbb{R}^N)$ is a reflexive Banach space there exists $u_0 \in H^1(\Bbb{R}^N)$ such that $u_n \rightharpoonup u_0$ in $H^1(\Bbb{R}^N)$, up to a subsequence.

Now, since we don't have compact Sobolev embeddings, I don't know how to proceed. Any hints will be the most appreciated.

Thanks in advance and kind regards.

Answer 1

Besides the semicontinuity of the norm mentioned in your answer, you need to show that $F(u_n) \to F(u_0)$ modulo subsequences.

Indeed, on any ball $B$, by Rellich–Kondrachov, one has $\|u_n - u_0\|_{L^q(B)} \to 0$. Therefore,

$$ \left|\int_B h \left(|u_n|^{q} - |u_0|^{q} \right)\right| \lesssim \|h\|_{L^\infty}\left(\|u_n\|_{L^q}^{q-1}+\|u_0\|_{L^q}^{q-1} \right) \|u_n - u_0\|_{L^p(B)} \to 0, $$ since $u_n$ is bounded in $H^1$.

Now, given $\varepsilon >0$, $B$ can be chosen big enough such that, $$ \|h\|_{L^\frac{2^*}{2^*-q}(B^c)} < \varepsilon. $$

This gives

$$ \left|\int_{B^c} h \left(|u_n|^{q} - |u_0|^{q} \right)\right| \lesssim \|h\|_{L^\frac{2^*}{2^*-q}(B^c)}\left(\|u_n\|_{L^{2^*}}^{q}+\|u_0\|_{L^{2^*}}^{q} \right) \lesssim \left(\|u_n\|_{H^1}^{q}+\|u_0\|_{H^1}^{q} \right) \varepsilon, $$

so that the proof of the convergence of $F(u_n)$ is done.

One finishes the way you mentioned, noting that $u_0 \in M$ and that the lower semicontinuity gives you $J(u_0) \leq c$ and therefore $J(u_0) = c$. Your scaling argument reduces the multiplier to 1.

Answer 2

I think I got this with some hints by Luccas Campos. What is your opinion?

Since the norm is weakly lower semicontinuous, it follows that $$ J(u_0) = \frac12 ||u_0||^2 \leq \liminf_{n \to \infty}\frac12 ||u_n||^2 = c. $$

On the other hand, it holds that $F(u_n) \to F(u)$ and therefore $u \in M$. Indeed, note that, for every $\phi \in C_c^\infty(\Bbb{R}^N)$, by the compact Sobolev embeddings, $$ u_n|_{\text{supp} \phi} \to u_0|_{\text{supp} \phi} \quad \text{ in } L^q(\text{supp} \phi). $$ Then, by Vainberg's theorem, there is a subsequence and a function $g \in L^q(\text{supp}\phi)$ such that $$ u_n(x) \to u_0(x) \quad \text{a.e. in supp}\phi $$ and $$ |u_n(x)| \leq g(x) \quad \text{ a.e. in supp}\phi. $$ Then $$ h(x)|u_n|^{q - 2}u_n \phi \to h(x) |u_0|^{q - 2}u_0\phi \quad \text{ a.e. in supp}\phi $$ and $$ |h(x)|u_n|^{q - 2}u_n \phi| \leq ||\phi||_\infty h(x) |g|^{q - 1} \in L^1(\text{supp}\phi). $$ Then, by the Dominated Convergence Theorem, $$ \int_{\text{supp}\phi} h(x) |u_n|^{q - 2}u_n \phi \ dx \to \int_{\text{supp}\phi} h(x) |u_0|^{q - 2}u_0 \phi \ dx. $$ But this holds for every $\phi \in C_c^\infty(\Bbb{R}^N)$, so by density it holds for every $v \in H^1(\Bbb{R}^N)$. In particular, it holds for $u_0$, so it follows that $1 = F(u_n) \to F(u_0)$, which in turn implies that $u_0 \in M$. We conclude that $c \leq J(u_0)$.

It therefore follows that $J(u_0) = c$, hence $u_0$ is a local minimum of $J$ constrained to $M$. Since $F'(u_0) \neq 0$, from the Theorem of Lagrange Multipliers it follows that $$ J'(u_0)v = \lambda F'(u_0) v \quad \forall v \in H^1(\Bbb{R}^N). $$ Hence $u_0$ is a solution of $$ \begin{cases} - \Delta u + u = h(x) |u|^{q - 2}u \quad \text{ in } \Bbb{R}^N \\ u \in H^1(\Bbb{R}^N) \end{cases}. $$ Note that $$ 0 < ||u_0||^2 = \lambda qF(u_0) = \lambda q, $$ hence $\lambda > 0$. Now, let $w = \alpha u_0$ for some $\alpha > 0$ to be chosen later on. Then $$ - \Delta w + w = \alpha (-\Delta u_0 + u_0) = \alpha \lambda h(x) |u_0|^{q - 2} u_0 = \frac{\alpha \lambda}{\alpha^{q - 1}}h(x) |w|^{q - 2}w. $$ It then suffices to chose $\alpha > 0$ such that $\frac{\alpha \lambda}{\alpha^{q - 1}} = 1$.