It is known that [here] the following PDE
has no solution. However, this is no longer true if we replace the whole space by a ball of radius , say . In this entry, we show that if is a solution of
To this purpose, let us recall the following
The Maximum Principle. Let assume be open and bounded. We consider an elliptic operator of the form
where coefficients are continuous and the standard uniform ellipticity condition holds.
Theorem (Weak MP for ).
A proof of this theorem can be found in, for example, Partial Differential Equation [chapter 6] by L.C. Evans.
Corollary. The sign of solutions to PDE depends on those signs on the boundary.
We are now in a position to solve the question above. In the proof, we will make use a so-called boundary blow-up argument.
written in the polar coordinates. It follows from the presentation formula of Laplacian in the polar coordinate
So by the maximum principle we have
Obviously, if we replace by the whole space , then by taking the limit as we deduce that which is impossible. This is why in the whole space, our PDE has no solution.