Summary of proof Let be an open subset of Euclidean space. Let u:M\to\mathbb{R} be a twice-differentiable function which attains its maximum value . Suppose that :a_{ij}\frac{\partial^2u}{\partial x^i\,\partial x^j}+b_i\frac{\partial u}{\partial x^i}\geq 0. Suppose that one can find (or prove the existence of): • a compact subset of , with nonempty interior, such that for all in the interior of , and such that there exists on the boundary of with . • a continuous function h:\Omega\to\mathbb{R} which is twice-differentiable on the interior of and with ::a_{ij}\frac{\partial^2h}{\partial x^i\,\partial x^j}+b_i\frac{\partial h}{\partial x^i}\geq 0, : and such that one has on the boundary of with Then on with on the boundary of ; according to the weak maximum principle, one has on . This can be reorganized to say :-\frac{u(x)-u(x_0)}\geq \frac{h(x)-h(x_0)} for all in . If one can make the choice of so that the right-hand side has a manifestly positive nature, then this will provide a contradiction to the fact that is a maximum point of on , so that its gradient must vanish.
Proof The above "program" can be carried out. Choose to be a spherical annulus; one selects its center to be a point closer to the closed set than to the closed set , and the outer radius is selected to be the distance from this center to ; let be a point on this latter set which realizes the distance. The inner radius is arbitrary. Define :h(x)=\varepsilon\Big(e^{-\alpha|x-x_{\text{c}}|^2}-e^{-\alpha R^2}\Big). Now the boundary of consists of two spheres; on the outer sphere, one has ; due to the selection of , one has on this sphere, and so holds on this part of the boundary, together with the requirement . On the inner sphere, one has . Due to the continuity of and the compactness of the inner sphere, one can select such that . Since is constant on this inner sphere, one can select such that on the inner sphere, and hence on the entire boundary of . Direct calculation shows :\sum_{i=1}^n\sum_{j=1}^na_{ij}\frac{\partial^2h}{\partial x^i\,\partial x^j}+\sum_{i=1}^nb_i\frac{\partial h}{\partial x^i}=\varepsilon \alpha e^{-\alpha|x-x_{\text{c}}|^2}\left(4\alpha\sum_{i=1}^n\sum_{j=1}^n a_{ij}(x)\big(x^i-x_{\text{c}}^i\big)\big(x^j-x_{\text{c}}^j\big)-2\sum_{i=1}^n a_{ii}-2 \sum_{i=1}^n b_i\big(x^i-x_{\text{c}}^i\big)\right). There are various conditions under which the right-hand side can be guaranteed to be nonnegative; see the statement of the theorem below. Lastly, note that the
directional derivative of at along the inward-pointing radial line of the annulus is strictly positive. As described in the above summary, this will ensure that a directional derivative of at is nonzero, in contradiction to being a maximum point of on the open set .
Statement of the theorem The following is the statement of the theorem in the books of Morrey and Smoller, following the original statement of Hopf (1927): {{blockquote|Let be an open subset of Euclidean space . For each and between 1 and , let and be continuous functions on with . Suppose that for all in , the symmetric matrix is positive-definite. If is a nonconstant function on such that :\sum_{i=1}^n\sum_{j=1}^na_{ij}\frac{\partial^2u}{\partial x^i\,\partial x^j}+\sum_{i=1}^nb_i\frac{\partial u}{\partial x^i}\geq 0 on , then does not attain a maximum value on .}} The point of the continuity assumption is that continuous functions are bounded on compact sets, the relevant compact set here being the spherical annulus appearing in the proof. Furthermore, by the same principle, there is a number such that for all in the annulus, the matrix has all eigenvalues greater than or equal to . One then takes , as appearing in the proof, to be large relative to these bounds. Evans's book has a slightly weaker formulation, in which there is assumed to be a positive number which is a lower bound of the eigenvalues of for all in . These continuity assumptions are clearly not the most general possible in order for the proof to work. For instance, the following is Gilbarg and Trudinger's statement of the theorem, following the same proof: {{blockquote|Let be an open subset of Euclidean space . For each and between 1 and , let and be functions on with . Suppose that for all in , the symmetric matrix is positive-definite, and let denote its smallest eigenvalue. Suppose that \textstyle\frac{a_{ii}}{\lambda} and \textstyle\frac{\lambda} are bounded functions on for each between 1 and . If is a nonconstant function on such that :\sum_{i=1}^n\sum_{j=1}^na_{ij}\frac{\partial^2u}{\partial x^i\,\partial x^j}+\sum_{i=1}^nb_i\frac{\partial u}{\partial x^i}\geq 0 on , then does not attain a maximum value on .}} One cannot naively extend these statements to the general second-order linear elliptic equation, as already seen in the one-dimensional case. For instance, the ordinary differential equation has sinusoidal solutions, which certainly have interior maxima. This extends to the higher-dimensional case, where one often has solutions to "eigenfunction" equations which have interior maxima. The sign of
c is relevant, as also seen in the one-dimensional case; for instance the solutions to are exponentials, and the character of the maxima of such functions is quite different from that of sinusoidal functions. == See also ==