Explicit solutions to $-\triangle u = k f(u)$ Does anyone know any examples of $f$'s for which $-\triangle u(x) = k f(u(x))$ has an explicit solution (i.e. a formula for the solution, not a numerical approximation scheme) in terms of $k$?
I am interested in examples where $f\geq 0$ is neither constant nor linear. Optimally I would be interested in a smooth $f$ with $\int_0^\infty 1/f<\infty$. Any non-pathological open $U \subset \mathbb{R}^d$ domain is fine and $d=1,2,3$ are all of interest, with $d=2$ being the most interesting to me.
Edit: Boundary conditions $u=0$ on boundary of domain.
 A: In $d=1$ with $f(u)=u^2$, we get
$$
u''(x)+k u^2 = 0
$$
which has solutions in terms of the "Weierstrass elliptic function". Finding the constants that match the boundary conditions on a given interval is not trivial to do in a closed form, but you should be able to numerically find two constants that do that trick.
In $d=1$ with $f(u)=-\exp(u)$, the family of solutions is
$$
u(x)=\log \left(\frac{c_1-c_1 \tanh ^2\left(\frac{1}{2} \sqrt{c_1 \left(c_2+x\right){}^2}\right)}{2 k}\right)
$$
I found these by trying some functions for $f$ and using WolframAlpha.
A: When $d=1$ , it is just an autonomous ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0301.pdf and the solution can only express in implicit form for most $f(u)$ .
Starting from $d=2$ , even no known nice form of the solution for most $f(u)$ . Perhaps the only known exception is when $d=2$ and $f(u)=ae^{bu}$ , which is very lucky to have nice explicit form of the solution according to http://eqworld.ipmnet.ru/en/solutions/npde/npde3103.pdf.
