Past exam of the approximation theory and finite element methods course of the University of Oxford / 2008 / B / 8

Use the full continuous piecewise linear space $V_h\subset H^1(0,1)$, including endpoint basis functions, because the boundary conditions are Neumann. DefineChoose $u_h^0$ as the $L^2$ projection of $u_0$ onto $V_h$. For each time step seek $u_h^{m+1}\in V_h$ withThis is the conforming Crank-Nicolson method. The reaction coefficient and the diffusion are both treated at the midpoint; integration by parts has removed the boundary flux through the homogeneous Neumann condition.

Test the discrete equation with $v_h=u_h^{m+1/2}$. The identity $(a-b,a+b)=\Vert a{\Vert }^2-\Vert b{\Vert }^2$ yieldsSince $e^x\ge 1$ on the interval, $a(v,v)\ge \Vert v{\Vert }_{H^1}^2$. Therefore the requested inequality holds:Discarding the nonnegative second term and iterating proves $\Vert u_h^m{\Vert }_2\le \Vert u_h^0{\Vert }_2$ for every positive $\Delta t$, with no restriction relating it to $h$. This is unconditional $L^2$ stability; the initial projection also satisfies $\Vert u_h^0{\Vert }_2\le \Vert u_0{\Vert }_2$.

Write $u_h^m={\sum }_i{U}_i^m{\phi }_i$ and define the mass matrix and reaction-diffusion stiffness matrixThen each step isBoth component matrices are symmetric, so $A$ is symmetric. For a nonzero coefficient vector $v$, its finite element function $v_h$ is nonzero andThus $A$ is also positive definite and every update has a unique solution. Because nonneighboring hats have disjoint interior supports, the matrix is tridiagonal in nodal order.