# Regularity of the quasi-linear PDE $-\Delta u + c(u) = f$

Let $$K$$ be a compact set in $$\mathbb R^n,$$ $$f\in C^\infty(K),$$ and $$c:\mathbb R \to \mathbb R$$ be a smooth function. Consider an element $$u\in H_0^1(K),$$ satisfying the PDE $$-\Delta u + c(u) = f$$ in weak sense, i.e.

$$\int_K \nabla u\cdot \nabla v+\int_K vc(u)=\int_K fv.$$

I think we can establish something similar to $$u \in C^\infty(K).$$ Indeed, we can consider a standard elliptic regularity argument: let $$v=-D^{-h}_kD^{h}_ku$$, where $$D^{h}_ku(x)=\frac{u(x+he_k)-u(x)}{h}$$. Then, we can estimate: $$\int \nabla u\cdot \nabla v=\|D^{h}_k \nabla u\|_{L^2(K)}^2.$$ The difficult part is the estimate on the nonlinearity. Firstly I try to get $$u\in H^2.$$ We can easily write $$\int_K (f-c(u))D^{-h}_kD^{h}_ku\leq (\|f\|_{L^2} + \|c\circ u\|_{L^2}) \|D^{-h}_kD^{h}_k u\|_{L^2}$$ If we can show that $$\|c\circ u\|_{L^2}<\infty,$$ then it is just a matter of simple calculations to confirm the gain in regularity. However, this is difficult. In dimensions $$n=1,2$$, we have the continuous Sobolev embedding $$H^1 \to L^\infty$$, so we can treat $$u$$ as a bounded function and hence $$\|c\circ u \|_{L^\infty}<\infty,$$ and since $$K$$ has finite measure, $$c\circ u\in L^2$$. So the desired regularity is obtained. However, in higher dimensions, we do not have such nice embedding.

Do we have (at least) $$H^2$$ regularity in higher dimensions?

Apparently it is more realistic to have a bound on the growth of $$c$$. However, I find out that this question is asked in Lawrence C. Evan's Partial Differential Equations book on page 366 without any bounds on $$c$$.

• Is there some bound on the growth of $c$ available? Without this, if say $c(t) = t^p$ with some large exponent, how do you ensure your weak identity is well-defined? – Leo Moos Dec 18 '20 at 21:58
• @LeoMoos Oh it should be $u\in H_0^1(K)$. Yes and I am thinking about this. Do we require the bilinear form to be defined for all functions in the space? Is it okay to just have it defined for a subspace including $u$? – Ma Joad Dec 18 '20 at 22:06
• The problem is on a different page in my edition, but I'm guessing you're referring to the question where Evans additionally imposes that $c' \geq 0$ and $c(0) = 0$? The hypotheses there are a bit different, notably $u$ satisfies the PDE on $\mathbf{R}^n$. – Leo Moos Dec 19 '20 at 18:30
• @LeoMoos That seems to be the question I see. Yes, its different because we need in addition have the equation satisfied outside the support of $u$. – Ma Joad Dec 19 '20 at 18:38

## 2 Answers

Not always. Consider the case $$n \geq 5$$, $$K = B_1$$, and $$u = |x|^{\frac{4-n}{2}} - 1$$. Then $$u \in H^1_0(B_1)$$ but $$u \notin H^2(B_1)$$, and $$\Delta u = \frac{n(4-n)}{4}(u+1)^{\frac{n}{n-4}} := c(u),$$ so $$c$$ is smooth when $$n = 5,6,8.$$

• Argh! You beat me by 30 seconds. – Willie Wong Dec 18 '20 at 22:32
• What's interesting is that the example works only for $n=5,6,8$, dues to the fraction $n/(n-4).$ But apparently we can find examples on other dimensions as well? – Ma Joad Dec 19 '20 at 8:59
• Yes, there are counterexamples of the form $u = |x|^{-2/p} - 1$, $K = B_1$, and $c(s) = (1+s)^{p+1}$ (up to multiplying by a constant) in any dimension $n \geq 3$ (see Willie's answer), because we can modify $c$ however we like in $\{s < 0\}$. When $n = 2$ one can take $u = \log(\log(1/|x|))$, $K = B_{1/e}$, and $c(s) = -e^{2(e^s-s)}$. In the case $n = 1$, $H^1$ functions are bounded, so solutions are smooth. – Connor Mooney Dec 19 '20 at 18:26
• @ConnorMooney Is $u= \ln(\ln(1/r))$ really in $H^1$? I find that the integral of the square of its derivative (gradient) $\int -(r^2 \ln r)^{-1}$ diverges. – Ma Joad Dec 19 '20 at 18:46
• Yes, $|\nabla u|^2 = r^{-2}(\log r)^{-2}$, the area element is $rdr$, and $r^{-1}(\log r)^{-2}$ is integrable near zero (as the derivative of $(\log r)^{-1}$). – Connor Mooney Dec 19 '20 at 18:54

Locally if you take

$$u = r^{-2/p}$$

you see that

$$\Delta u = - (n - 3 - \frac2p) \frac2p \frac{u}{r^2} = - (n-3-\frac2p)\frac2p u^{p+1}$$

Set your $$c$$ to be the function on the RHS.

If I did my back-of-envelop computations correct. If $$n > 2$$ and $$p > \frac{4}{n-2}$$ you will have (near the origin) $$u \in H^1$$ and $$c(u) \in L^1$$.

If $$n = 3,4$$, or if $$n > 4$$ and $$p \leq \frac{4}{n-4}$$, you see that $$c(u) \not\in L^2$$ and hence $$u\not\in H^2$$.