Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 71 / 5

Work with real functions and $f\in L^2(0,1)$. To give the variational problem a definite domain, choose the homogeneous Dirichlet boundary conditions used in the next part. The natural form space is the zero-trace Sobolev space $H=H_0^1(0,1)$, with norm $\Vert v{\Vert }_H=\Vert v^{\prime }{\Vert }_2$. The Sobolev trace operator is well defined because one-dimensional $H^1$ functions have absolutely continuous representatives. For example $v(0)={\int }_0^{1}v(x)dx-{\int }_0^1(1-x)v^{\prime }(x)dx$, so its endpoint trace is bounded in the $H^1$ norm; the other endpoint has the analogous estimate. The zero-trace space is closed in the Hilbert space $H^1$, and the Poincaré inequality $\Vert v{\Vert }_2\le {\pi }^{-1}\Vert v^{\prime }{\Vert }_2$ makes the chosen derivative norm equivalent to its full norm. Thus $H$ is a Hilbert space.

Differentiability of $p,q$ implies continuity on the compact interval. Hence there are finite constants $p_{\max },q_{\max }$ and a positive $p_{\min }$ with $p_{\min }\le p\le p_{\max }$ and $0\le q\le q_{\max }$. Define the symmetric bilinear formThe Cauchy-Schwarz inequality and Poincaré inequality give all the estimates required by the Lax-Milgram theorem:That theorem states that a continuous coercive bilinear form on a Hilbert space and a bounded linear functional determine a unique weak solution. Completeness, continuity, coercivity and boundedness have now all been verified. Therefore there is a unique $u\in H$ with $a(u,v)=⟨f,v⟩$ for every $v\in H$.

The energy version of the given variational problem is $I(v)=a(v,v)-2⟨f,v⟩$. Its exact variation isBoundedness of $a$ makes the last term quadratic in $\Vert w{\Vert }_H$, so the Fréchet derivative is $DI(v)[w]=2a(v,w)-2⟨f,w⟩$. Its vanishing for every test function is preciselyTaking compactly supported smooth tests and integrating in the distributional sense shows that this is the Euler-Lagrange equation $-(p{u}^{\prime }{)}^{\prime }+qu=f$. Conversely that weak equation gives the displayed stationary condition. The stationary point is actually the unique global minimizer, becauseEquality occurs only when $v=u$. This proves the variational equivalence, including existence and uniqueness rather than only a formal derivative calculation.

For precision about the literal L2 inner product $⟨Lv,v⟩$, use the flux-domain realization of a one-dimensional elliptic operatorFor such $v$, weak integration by parts and its zero endpoint values give $⟨Lv,v⟩=a(v,v)$. On the larger form space $H$, the same energy is interpreted by the weak operator pairing. The minimizer belongs to $D(L)$ automatically: its weak equation gives $(p{u}^{\prime }{)}^{\prime }=qu-f\in L^2$, while $p{u}^{\prime }\in L^2$. Thus its equation also holds as an equality of $L^2$ functions. This domain does not assume bounded $p^{\prime }$, which mere differentiability would not supply. If $p\in C^1$ one may equivalently use $D(L)=H^2\cap H_0^1$, but that stronger coefficient assumption is not needed for the proof here.

Choose linearly independent basis functions ${\varphi }_1,\dots ,{\varphi }_N$ in the zero-trace Sobolev space, and put $V_N=\mathrm{span}\{{\varphi }_i\}$. A concrete conforming Galerkin method uses continuous piecewise-linear hat functions on an interval mesh, with zero endpoint values. Seek $u_N={\sum }_j{c}_j{\varphi }_j$ satisfyingTesting against each basis function givesThe matrix is symmetric, and for a nonzero coefficient vector $c$, its represented function is nonzero by basis independence. The coercivity estimate gives $c^{T}Kc=a(u_N,u_N)>0$. Thus $K$ is a positive-definite matrix, the finite-dimensional problem has a unique solution, and it minimizes the same energy over $V_N$. This is Ritz-Galerkin equivalence for a symmetric coercive form.

Subtracting the continuous and discrete equations proves Galerkin orthogonality: $a(u-u_N,v_N)=0$ for all $v_N\in V_N$. Let $\Vert v{\Vert }_a=\sqrt{a(v,v)}$. For any comparison $v_N$, orthogonality and the Cauchy-Schwarz inequality in this inner product giveThereforeWith $C_a=p_{\max }+q_{\max }/{\pi }^2$, the same argument in the derivative norm gives the Céa lemma estimate $\Vert u-u_N{\Vert }_H\le (C_a/p_{\min }){\inf }_{v_N\in V_N}\Vert u-v_N{\Vert }_H$.

Convergence now follows from approximation, not from a bare assertion that the residual is small. The defining density of smooth compactly supported functions in $H_0^1$ lets one approximate $u$ by such a smooth function. Its piecewise-linear interpolant has zero endpoints and derivative error tending to zero as the largest mesh interval tends to zero. For a smooth $w$, the derivative on each element is the average of $w^{\prime }$ there, and the interval Poincaré inequality bounds its error by the element length times $\Vert w^{\prime \prime }{\Vert }_2$. Density and the boxed best-approximation estimate consequently give $u_N\to u$ in $H_0^1$, and hence in $L^2$, without assuming a bounded derivative of $p$. If the stronger regularity $u\in H^2$ is available, the same interpolation estimate gives first-order convergence in the derivative norm.