A regularity result for semilinear PDE of the form $\Delta u=f(x, u)$ in Michael E. Taylor's book Partial Differential Equations III.

Question

Under the assumption that

$$\partial_{u} f(x, u) = 0 \text{ for } |u| \geq K \quad（1.6）$$

Michael E. Taylor said that (proposition $(1.3)$)

For $k=1,2,..$, if $g \in H^{k+1 / 2}(\partial M)$ then any solution $u \in V$ to

\begin{align} \Delta u &= f(x, u) \text{ on $M$} \\ u &= g \qquad \text{ on $\partial M$} \end{align}

belongs to $H^{k+1}(M)$. Hence, if $g \in C^{\infty}(\partial M)$, then $u \in$ $C^{\infty}(\bar{M})$ where $V = \{u \in H^{1}(M): u = g \text{ on } \partial M \}$ and $f \in C^{\infty}(\bar{M} \times \mathbb{R})$.

Proof

We start with $u \in H^{1}(M)$. Then the right side of $(1.1)$ belongs to $H^{1}(M)$ if $f(x, u)$ satisfies $(1.6)$. This gives $u \in H^{2}(M)$, provided $g \in H^{3 / 2}(\partial M)$. Additional regularity follows inductively.

I don't get it, can someone give me some hints?

Answer 1

Note that the assumption (1.6) implies that there exists a constant $L$ such that $$ |\partial_u f(x,u)|\leq L\quad \text{on } \overline{M}\times \mathbb{R}.$$ Suppose that $u \in H^1(M)$ and define $$ h(x)=f(x,u(x)).$$ Since $|\partial_u f(x,u)|$ is bounded, it follows from the chain rule that $$ \frac{\partial h}{\partial x_j} = \frac{\partial f}{\partial x_j}(x,u(x))+f(x,u(x)) \frac{\partial u}{\partial x_j}(x) \in L^2(M). $$ Hence $h$ is in $H^1$. Therefore, by $L^2$-theory, $u \in H^3$. Then continue this process.

---

November 28th, 2021. After receiving the comment, I think that it is necessary to explain several steps. I will focus on the case when $M=\Omega$, $\Omega$ is a bounded smooth domain in $\mathbb{R}^n$. Also, by the trace theorem, we may assume that $g=0$. Recall the following regularity theorem.

Theorem A. Let $k\in \mathbb{N}\cup \{0\}$ and $h\in H^k(\Omega)$. Then there exists a unique $u\in H^1_0(\Omega)\cap H^{k+2}(\Omega)$ satisfying $$ -\Delta u=h\quad \text{in } \Omega,\quad u=0\quad \text{on } \partial\Omega.$$

Suppose that $u\in H^1_0(\Omega)$. By the previous argument, we see that $h(x)=f(x,u(x))$ belongs to $H^1(\Omega)$. By Theorem A, there exists a unique $v\in H^1_0(\Omega)\cap H^3(\Omega)$ satisfying $$ -\Delta v=h\quad \text{in }\Omega\quad v=0\quad \text{on } \partial\Omega.$$ Define $w=v-u$. Then $w\in H^1_0(\Omega)$ and $w$ satisfies $$ -\Delta w=0\quad \text{in } \Omega\quad w=0\quad \text{on } \partial\Omega.$$ Hence by the uniqueness assertion in Theorem A, $w$ is identically zero. This implies that $u\in H_0^1(\Omega)\cap H^3(\Omega)$. Continue this process.