Past exam of the differential equations 1 course of the University of Oxford 2023 3 b i Solution 2026-10-25
Define the squared L2 norm . Multiplication of the semilinear partial differential equation by , followed by integration by parts, givesAt the Dirichlet boundary condition makes the boundary term zero. At the Robin boundary condition is , so subtraction of the lower endpoint gives . In particular the sign is dissipative. Thus the exact energy identity iswhich implies
Write and . Subtraction of the two boundary value problems gives , with . Multiply by and apply integration by parts:Consequently the required energy identity isThe nonlinear term is nonnegative pointwise, sinceThis is the monotonicity energy estimate for a semilinear Dirichlet problem for the monotone function . Apply Young's inequality for products, , inside the last integral and discard the nonnegative nonlinear term. Moving to the left givesIn particular, equal forcing gives , proving uniqueness and a quantitative continuous dependence on forcing estimate for this boundary value problem.
Let and , and define the scaled forward difference operator, backward difference operator and second-order central difference byThe two finite difference methods are the nonlinear systemswith the prescribed Dirichlet boundary conditions and . These equations determine the interior nodal values.
For clarity distinguish the interior and edge inner products and their norms:The bracketed norm includes the last edge . Let , so , and put for . Direct telescoping gives discrete summation by parts:Subtract the two nonlinear systems, multiply each equation by and sum. The resulting discrete energy identity isThe nonlinear summands are nonnegative by the factorization in (a). Using Young's inequality for products at every node givesThis also proves uniqueness of each discrete solution. Existence can be seen by minimizing, with the endpoints fixed, the strictly convex energy functionalIt is a coercive function of its finitely many interior values, so it attains a minimum. Its first derivatives are ; thus its minimizer solves the scheme. The same argument applies to .