Past exam of the numerical solutions of partial differential equations course of the University of Oxford / 2023 / 1

Write $w=u-v$ and $r=f-g$. Subtraction of the two boundary value problems gives $-w^{\prime \prime }+w+u^3-v^3=r$, with $w(0)=w(1)=0$. Multiply by $w$ 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 $s\mapsto s^3$. Apply Young's inequality for products, $rw\le (r^2+w^2)/2$, inside the last integral and discard the nonnegative nonlinear term. Moving $\frac{1}{2}{\int }_0^1{w}^{2}dx$ to the left givesIn particular, equal forcing gives $w=0$, proving uniqueness and a quantitative continuous dependence on forcing estimate for this boundary value problem.

Let $f_i=f(ih)$ and $g_i=g(ih)$, 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 $U_0=V_0=A$ and $U_N=V_N=B$. 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 $i=N$. Let $W=U-V$, so $W_0=W_N=0$, and put $q_i=D_x^{-}{W}_i$ for $1\le i\le N$. Direct telescoping gives discrete summation by parts:Subtract the two nonlinear systems, multiply each equation by $h{W}_i$ 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 $h(-D_x^{+}{D}_x^{-}{W}_i+W_i+W_i^3-f_i)$; thus its minimizer solves the scheme. The same argument applies to $g$.

Use $u_i=u(x_i)$ for the sampled exact solution, and let $e_i=U_i-u_i$. The Dirichlet boundary conditions imply $e_0=e_N=0$. The definition of the truncation error $\phi$ gives the exact discrete equationSubtracting it from the numerical equation therefore givesThe sign on the right is positive because ${\phi }_i$ was defined as the difference approximation minus the exact second derivative. Multiply by $h{e}_i$, sum over the interior nodes and apply discrete summation by parts:Again $(U_i^3-u_i^3)e_i=e_i^2(U_i^2+U_i{u}_i+u_i^2)\ge 0$. By Young's inequality for products, $(\phi ,e{)}_h\le \frac{1}{2}\Vert \phi {\Vert }_h^2+\frac{1}{2}\Vert e{\Vert }_h^2$. HenceNo fourth derivative is needed for this step: the assumed twice continuous differentiability suffices to define $\phi$ and obtain the discrete energy estimate.

On the vector space of nodal values with zero endpoints define the discrete Sobolev normIt is a norm because it is the Euclidean norm of the scaled list of nodal values and first finite differences, and vanishes only for the zero list. It is the discrete counterpart of the Sobolev space $H^1$ norm. Here $W=U-u$ has zero endpoints, so this definition applies directly.

Set $M_4={\max }_{0\le x\le 1}|u^{(4)}(x)|$, finite by the assumed continuity. Taylor's theorem with fourth-order remainders gives, for suitable ${\xi }_{i,+}\in (x_i,x_i+h)$ and ${\xi }_{i,-}\in (x_i-h,x_i)$,Adding the two expansions cancels the odd derivatives, soConsequently $\Vert \phi {\Vert }_h^2\le h(N-1)M_4^2{h}^4/144\le M_4^2{h}^4/144$. Doubling the bound in (c) gives $2\Vert D_x^{-}e]{\Vert }_h^2+\Vert e{\Vert }_h^2\le \Vert \phi {\Vert }_h^2$, and thereforeThus $M_4/12$ is a valid nonnegative constant. To satisfy the request for a strictly positive constant even when $u^{(4)}\equiv 0$, take $C=\max \{1,M_4\}/12$, independent of $h$. This proves second-order convergence of a numerical method in a discrete Sobolev norm, including control of the backward difference operator applied to the error.