How to understand the unique continuation result

Question

Let $E$ be the closure of $C_c^{\infty}\left(\mathbb{R}^N\right)$ ($N \geqq 3)$ under the norm $$ \|u\|_E=\left(\int_{\mathbb{R}^N}|\nabla u|^2\right)^{1 / 2}. $$

Suppose $K(x) \in C^1\left(\mathbf{R}^3\right), K(x)$ and $\nabla K(x)$ are bounded in $\mathbf{R}^3, K_{x_2}(x)$ is non-negative but not identically zero ($K_{x_2}(x)$ denotes the partial derivative of $K(x)$ in $x_2$-direction).

Then the only solution of $$ -\Delta u=K(x) u^5 \text { in } \mathbf{R}^3 $$ in $E$ is the trivial solution $u \equiv 0$.

Proof: Let $u$ be any solution in $E$. Multiply the equation $$ -\Delta u=K(x) u^5 \text { in } \mathbf{R}^3 $$ by $u_{x_2}$ and integrate by parts. We obtain $$ \int_{\mathbf{R}^3} K_{x_2}(x) u(x)^6=0. $$

The hypotheses on $K(x)$ imply that $u$ is identically zero in an open set. My question is how to understand the unique continuation result to show $u \equiv 0$.

Answer 1

If I understand things correctly, $u$ is vanishing on some non-empty open subset. Moreover, you have pointwisely on $\mathbb R^3$ the differential inequality $$ \lvert \Delta u\rvert\le C\lvert u\rvert^5. $$ Also, you know that the function $u$ is in $\smash{\dot H}^1(\mathbb R^3)$, which is a subset of $L^6(\mathbb R^3)$. This implies that $$ \lvert u\rvert^5=\lvert u\rvert\underbrace{\lvert u\rvert^4}_{L^{3/2}}, $$ implying that you have a differential inequality $ \lvert \Delta u\rvert\le V\lvert u\rvert, \ V \in L^{d/2} $ in $\mathbb R^d$, $d\ge 3$. Then the strong unique continuation result due to Jerison & Kenig [MR0794370] entails that $u\equiv 0$. Note that $d/2$ is critical and that $L^{d/2}_{\text{loc}}$ is enough.