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 $c$.

  • $\begingroup$ 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? $\endgroup$ – Leo Moos Dec 18 '20 at 21:58
  • $\begingroup$ @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 $u$? $\endgroup$ – Ma Joad Dec 18 '20 at 22:06
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    $\begingroup$ 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 $\mathbf{R}^n$. $\endgroup$ – Leo Moos Dec 19 '20 at 18:30
  • $\begingroup$ @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 $u$. $\endgroup$ – Ma Joad Dec 19 '20 at 18:38

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 $n = 5,6,8.$

  • $\begingroup$ Argh! You beat me by 30 seconds. $\endgroup$ – Willie Wong Dec 18 '20 at 22:32
  • $\begingroup$ 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? $\endgroup$ – Ma Joad Dec 19 '20 at 8:59
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    $\begingroup$ 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. $\endgroup$ – Connor Mooney Dec 19 '20 at 18:26
  • $\begingroup$ @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. $\endgroup$ – Ma Joad Dec 19 '20 at 18:46
  • $\begingroup$ 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 $(\log r)^{-1}$). $\endgroup$ – 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 $u\not\in H^2$.


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