Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 5

For a second-order linear partial differential equation written $A{u}_{xx}+2B{u}_{xy}+C{u}_{yy}+\cdots =0$, the discriminant is $B^2-AC$. HereIt is positive exactly when $x\ne 0$, proving that the equation is a hyperbolic partial differential equation on each half-plane. At $x=0$ its principal part degenerates and it is not hyperbolic.

The characteristic curves satisfyIntegrating suggests the characteristic coordinatesTheir Jacobian determinant is $2/x$, so this is an invertible local change of variables whenever $x\ne 0$. Write $u(x,y)=U(\xi ,\eta )$. The chain rule givesSubstitution cancels both unmixed second derivatives and both first derivatives:The canonical form of a second-order partial differential equation is thereforeThe cancellation of the first derivatives is the key feature of these logarithmic characteristic coordinates.

Integrate the canonical form of a second-order partial differential equation first in $\eta$ and then in $\xi$. The first integration gives $U_{\xi }=\eta +g^{\prime }(\xi )$, and the second givesThusThe two arbitrary twice continuously differentiable functions describe the two characteristic families. Conversely, substituting this expression into $U_{\xi \eta }=1$ verifies every such solution. The formula applies on $x>0$ or $x<0$; independent choices can be made on the two half-planes. Smooth extension across the degenerate line $x=0$ requires extra compatibility and is not supplied by the logarithmic formula alone.

Use the classical regularity setting $u\in C^2(D)\cap C(\overline{D})$, with $D$ bounded. Continuity on the compact set $\overline{D}$ guarantees existence of a minimum. The inequality $\Delta u=f\le 0$ means that $u$ is superharmonic function.

A strict inequality $\Delta u<0$ would immediately exclude an interior minimum, because its Hessian matrix there would be positive semidefinite and have nonnegative trace. To handle equality, introduce the quadratic perturbation proof of the minimum principle:Its minimum must lie on $\partial D$. Choose $R$ so that $x^2+y^2\le R^2$ on $\overline{D}$. For any $p\in D$,Letting $\epsilon \downarrow 0$ gives $u(p)\ge {\min }_{\partial D}u$. ThereforeThis asserts that a minimum is attained on the boundary, not that it cannot also be attained inside; a constant function attains it everywhere.

For a harmonic function, $\Delta u=0$ and $\Delta (-u)=0$. Apply the preceding minimum argument to $u$ and to $-u$:These are the weak minimum and maximum principle for harmonic functions.

The difference $w=v-u$ is a harmonic function, because the Laplacian is a linear differential operator:The boundary inequality says $w\ge 0$ on $\partial D$. The minimum principle from part (ii) givesThis is the comparison principle for harmonic functions. Applying it in both directions also shows why fixed Dirichlet boundary conditions determine at most one harmonic function.