Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2014 / 5

The maximum principle for subharmonic functions here states that if $u\in C^2(D)\cap C(\overline{D})$ and $\Delta u=f\ge 0$ in $D$, thenIf $f\le 0$, apply this to $-u$ to obtain the corresponding minimum principle. In particular a harmonic function has both its maximum and minimum on the boundary. Without a sign condition on $f$, this conclusion need not hold.

First suppose $\Delta u>0$. By compactness and continuity, $u$ attains a maximum on $\overline{D}$. At an interior maximum, the first derivatives vanish and the Hessian matrix is negative semidefinite. Its trace is then nonpositive, contradicting $\Delta u>0$. Thus a maximum lies on the boundary.

For the non-strict case put $u_{ϵ}=u+ϵ(x^2+y^2)$, with $ϵ>0$. Then $\Delta u_{ϵ}=f+4ϵ>0$, so the strict case yields, for $z\in \overline{D}$,Let $ϵ\downarrow 0$. This gives $u(z)\le {\max }_{\partial D}u$, proving the theorem. The argument proves the weak maximum principle; it does not claim that every maximizer must be on the boundary when $u$ is constant.

Write the elliptic differential operator as $L=\Delta +r{\partial }_x+s{\partial }_y$. At an interior maximum, $u_x=u_y=0$ and $u_{xx}+u_{yy}\le 0$. Therefore $Lu\le 0$ there, contradicting $Lu=f>0$.

The extreme value theorem ensures that a maximum exists on the compact closed disk. It cannot lie in the interior, so the maximum is attained on $\partial D$. The drift terms disappear at a stationary point, which is why the strict-sign proof is so short.

Use an exponential perturbation whose image under $L$ is strictly positive everywhere. The coefficient $r$ is bounded on the closed disk; put $R={\max }_{\overline{D}}|r|$ and choose $k>R$, $k>0$. For $v(x,y)=e^{kx}$,Consequently $u_{ϵ}=u+ϵv$ satisfies $L{u}_{ϵ}=f+ϵLv>0$. Part (i) applies, so for every $z\in \overline{D}$,Letting $ϵ\downarrow 0$ givesThus the maximum is attained on the boundary also when $f\ge 0$. This is the exponential perturbation proof of the weak maximum principle with drift. It does not require differentiating the continuous coefficients $r,s$.

If $u_1,u_2$ satisfy the same boundary value problem, their difference $w=u_1-u_2$ satisfies $Lw=0$ in $D$ and $w=0$ on the boundary, by linearity. Part (b)(ii) givesApply it again to $-w$, which also satisfies $L(-w)=0$ and zero Dirichlet boundary conditions. This gives $-w\le 0$. Hence $w=0$ throughout $\overline{D}$, provingThis proves uniqueness whenever a solution exists; existence is the separately supplied assumption. When the drift coefficients vanish, the same argument uses part (a).

For a real solution, $u^4\ge 0$, so $\Delta u\ge 0$. The maximum principle for subharmonic functions gives $u\le {\max }_{\partial D}u=0$. The additional nonnegativity condition gives $u\ge 0$. ThereforeThis function satisfies both the equation and boundary condition, so it is exactly the full set of nonnegative solutions. The sign restriction is used alongside the maximum principle, rather than treating the nonlinear equation as a linear uniqueness problem.