Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 72 / 1

Multiply by a test function vanishing at the endpoints and apply integration by parts. The positive operator is $-u^{\prime \prime }+xu$, so the forcing changes sign. The weak formulation of this forced Airy boundary-value problem is: find $u\in V=H_0^1(0,1)$ such thatHere $V$ is the zero-boundary Sobolev space and can be equipped with $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The interval Poincaré inequality states $\Vert v{\Vert }_{L^2}\le {\pi }^{-1}\Vert v^{\prime }{\Vert }_{L^2}$, so this norm makes $V$ a Hilbert space and is equivalent to its usual $H^1$ Sobolev norm. The form is symmetric and satisfiesThe Lax-Milgram theorem says that a bounded coercive bilinear form on a Hilbert space and a bounded linear functional determine exactly one weak solution. All its hypotheses have just been verified. Moreover, $u^{\prime \prime }=xu+1\in L^2$ gives $u\in H^2$ on the interval; the right-hand side is then continuous, so this weak solution is the classical solution with the prescribed endpoint values.

Since the form is symmetric, the equivalent Ritz method minimizesIndeed, stationarity gives the weak equation andwith equality only at $u$. This proves the minimization equivalence, including the sign of the linear term.

Restricting to a conforming finite element space $V_h\subset V$ gives a unique discrete solution and Galerkin orthogonality, $a(u-u_h,v_h)=0$ for every $v_h\in V_h$. The Céa lemma states that if the continuity and coercivity constants are $M$ and $\alpha$, then $\Vert u-u_h{\Vert }_V\le (M/\alpha ){\inf }_{v_h\in V_h}\Vert u-v_h{\Vert }_V$. For this symmetric problem the exact best-approximation constant in the energy norm is one. The piecewise-linear finite element interpolation estimate $\Vert u-I_{h}u{\Vert }_{H^1}\le Ch\Vert u{\Vert }_{H^2}$ on regular finite element meshes therefore gives convergence of a numerical method of the Ritz method, with first-order error in the energy norm.

Take a uniform finite element mesh $x_i=ih$, $0\le i\le J$, $h=1/J$, and interior piecewise-linear hat functions ${\varphi }_i$. Write $u_h={\sum }_{i=1}^{J-1}{U}_i{\varphi }_i$. The Ritz method gives $KU=F$ withThe derivative part has diagonal $2/h$, neighboring entries $-1/h$ and zero other entries. For the affine-weighted hat mass matrix, symmetry around $x_i$ givesOn $[x_i,x_{i+1}]$, the product of the two hats is symmetric about the midpoint, integrates to $h/6$, and therefore has weighted integralConsequently the explicit tridiagonal equations arefor $1\le i\le J-1$, with $U_0=U_J=0$. The reaction coefficient has been integrated exactly; replacing it by an unweighted constant mass matrix would give a different system. For a nonzero coefficient vector, the represented function is nonzero and $U^{T}KU=a(u_h,u_h)>0$, so the matrix is symmetric positive-definite and the algebraic solution is unique.

Uniform spacing is a choice, not a requirement of the method. On a general finite element mesh with $h_i=x_i-x_{i-1}$, the corresponding entries arewith symmetric lower entries. These reduce to the displayed uniform-grid system.