Maximum of $u$ solution of $-\Delta u + u^2 = f(|x|)$ in $B(0,1)$

Question

Let $u \in C^2(B(0,1)) \cap C^0(B[0,1])$ solution of $$ \left\{ \begin{array}{rcl} -\Delta u + u^2 & = &f(|x|), \ \ \mbox{in} \ \ B(0,1);\\ u(x) & = & 1, \ \ \mbox{on} \ \ \partial B(0,1) \end{array}\right.$$ where $f(|x|) \geq 0$ is of class $C^1$. What is the maximum of $u$ in $B[0,1]$? This value depends on $f$ or not?

Answer 1

It's unlikely that you will be able to find the maximum of $u$ but you can certainly say somethings about it.

Firstly, we have that $$\Delta u = u^2-f\geqslant -f \qquad \text{in }B_1. $$ By the maximum principle for the Laplacian with right hand side (for example see Theorem 3.7 in Elliptic Partial Differential Equations of Second Order by Gilbarg and Trudinger) we have \begin{align*} \sup_{B_1} u \leqslant \sup_{\partial B_1} u^+ + C \sup_{B_1} \vert f^+ \vert \end{align*} for some $C>0$. Recall that $u^+ (x) = \max \{ u(x),0 \}$. (In fact, the proof in G&T gives that the constant above is $C=e^2-1$). Since $u=1$ on $\partial B_1$ and $f \geqslant 0$ it follows that $$ \sup_{B_1} u \leqslant 1 + C \sup_{B_1} f.\tag{$\ast$}$$

Secondly, since $f \geqslant 0$, we have $$\Delta u = u^2-f \leqslant u^2 \qquad \text{in }B_1.$$ Let $v=-u$ to find $$\Delta v \geqslant -u^2 \qquad \text{in } B_1. $$ Applying Theorem 3.7 in G&T again we have $$\sup_{B_1}v \leqslant \sup_{\partial B_1}v^+ + C \sup_{B_1} u^2 .$$ As $v=-1$ on $\partial B_1$ and $\sup_{B_1}v = - \inf_{B_1}u$ it follows that $$\inf_{B_1}u \geqslant -C (\sup_{B_1}u)^2 \tag{$\ast\ast$} $$

Equation ($\ast$) shows that $\sup_{B_1} f$ controls the $\sup_{B_1}u$. Equation ($\ast\ast$) is some kind of Harnack-type inequality.

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Equation ($\ast$) and ($\ast\ast$) don't directly show that $\sup_{B_1}u$ has to depend on $\sup_{B_1}f$ but they certainly hint at it. Regardless, it seems very unlikely to me that $\sup_{B_1} u$ would not depend on $\sup_{B_1}f$. If you think of $u$ as the density of a population of animals, $f$ as a distribution of resources in $B_1$ then $$ -\Delta u = f(x) -u^2$$ might be (a weird) steady state model where the population of animals grows according to the resources available but independently of the current population and has a constant mortality rate. In this model, $\sup_{B_1} u$ being independent of $\sup_{B_1}f$ would mean that you can have spots with arbitrarily large amounts of resources and this would have no effect on the maximum population density.

To prove $\sup_{B_1} u$ has to depend of $\sup_{B_1} f$ rigorously I would set $n=1$ and try choose a 1-parameter family of functions $f_\lambda$ such that you can solve $-u''+u^2 =f_\lambda$ explicitly, and $\sup_{B_1}f_\lambda \to \infty$ as $\lambda \to \infty$. Then, solve for $u$ in terms of $f_\lambda$ and you'll likely find that $\sup_{B_1} u \to \infty$ as $\lambda \to \infty$ as well.