Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 68 / 4

Take $L=-d^2/d{x}^2+x$. The Airy equation is $Lu=0$. To discuss positivity of this linear operator, use its homogeneous variation domain $D(L)=\{v\in H^2(0,1):v(0)=0,v^{\prime }(1)=0\}$. Integration by parts givesThe boundary term is zero at both ends. The same calculation with two different functions proves that the associated form is a symmetric bilinear form. Positivity here concerns the homogeneous domain; the condition $u(0)=1$ instead defines an affine set of admissible solutions.

For the variational problem defineand minimizeThe Sobolev trace at zero is meaningful in $H^1$. No derivative boundary value is imposed on this trial space: the Neumann boundary condition will emerge naturally. For $v\in V$, Cauchy-Schwarz inequality and integration giveso $a(v,v)\ge \frac{2}{3}\Vert v{\Vert }_{H^1}^2$. Thus $a$ is a bounded, symmetric, coercive bilinear form on the Hilbert space $V$. The Lax-Milgram theorem says that a bounded coercive bilinear form and a bounded linear functional determine a unique weak solution. Apply it toand set $u=1+v$. This is precisely the first variation condition $a(u,w)=0$ for $J$.

Compactly supported test functions first yield $u^{\prime \prime }=xu$ in the sense of distributional derivatives. Since $xu\in L^2$, the solution is in $H^2$. Integration by parts then leaves $u^{\prime }(1)w(1)=0$ for every $w\in V$, hence $u^{\prime }(1)=0$. The other boundary condition is built into $1+V$. Conversely a solution of this boundary value problem satisfies the first variation condition. Finally,This proves existence, uniqueness and the global minimum characterization, including the natural boundary condition. It is the mixed-boundary Airy energy principle.

Choose a partition $0=x_0<x_1<\cdots <x_N=1$, with the usual nodal piecewise-linear hat functions ${\varphi }_i$, including the half hats at the endpoints. Write $u_h={\sum }_{i=0}^N{U}_i{\varphi }_i$ with $U_0=1$. Differentiating the Ritz method energy with respect to $U_i$, $1\le i\le N$, givesOn an element $[a,b]$ of length $\ell$, its two basis functions are $(b-x)/\ell$ and $(x-a)/\ell$. Direct integration yields the element matrixFor example the first weighted diagonal is ${\int }_a^{b}x(b-x{)}^2/{\ell }^{2}dx=\ell (3a+b)/12$; the off-diagonal is ${\int }_a^{b}x(b-x)(x-a)/{\ell }^{2}dx=\ell (a+b)/12$. This gives the affine-weighted hat mass matrix and explicit equations on any partition by assembling adjacent elements.

For the uniform choice $h=1/N$, $x_i=ih$, the assembled interior equations areThe last node contributes only one element, so its equation isIn the first interior equation the known $U_0$ term moves to the right, giving $1/h-h^2/12$. For $N=1$ only the endpoint equation is needed. The last row enforces the weak Neumann boundary condition, so no artificial value outside the interval is introduced. The matrix on $U_1,\dots ,U_N$ is a symmetric positive-definite matrix: its quadratic form is $a(v_h,v_h)>0$ for every nonzero trial variation $v_h$. The discrete minimizer is therefore unique.