Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 69 / 1

Let $I=(-1,1)$ and use the clamped second-order Sobolev spaceHere weak derivatives up to order two are square integrable, with Sobolev norm $\Vert v{\Vert }_{H^2}^2=\Vert v{\Vert }_2^2+\Vert v^{\prime }{\Vert }_2^2+\Vert v^{\prime \prime }{\Vert }_2^2$. The boundary values are the continuous traces given by the Sobolev trace theorem, not arbitrary pointwise representatives. This space is the $H^2$ closure of $C_c^{\infty }(I)$, so it is a closed Hilbert space. The derivative boundary conditions are essential, and $H_0^1(I)$ alone is insufficient.

Define the symmetric bilinear form and bounded linear functionalThen the variational problem isFor complex data replace $a$ by its sesquilinear version, take the real part of the forcing term and use absolute squares in $J$; the same argument applies.

Its first variation is $J^{\prime }(u)[v]=a(u,v)-\ell (v)$. Thus stationary points satisfy $a(u,v)=\ell (v)$ for every $v\in H$, the weak formulation of a clamped fourth-order equation. For smooth coefficients and functions, twice integrating the highest term and once the middle term gives the differential equation, with all boundary terms zero because $v=v^{\prime }=0$ at both ends. With only the stated continuous coefficients, this manipulation is instead understood distributionally: compactly supported test functions give $(p{u}^{\prime \prime }{)}^{\prime \prime }-(q{u}^{\prime }{)}^{\prime }+ru=f$ as a distribution. This avoids assuming nonexistent classical derivatives of $p$ or $q$.

By the extreme value theorem, $p_0={\min }_{[-1,1]}p$ is strictly positive, and $p,q,r$ are bounded. The Cauchy-Schwarz inequality therefore givesFor completeness, the clamped interval derivative norm bound follows from $v^{\prime }(-1)=v(-1)=0$ and the one-dimensional Sobolev representative: integrate $v^{\prime \prime }$ to recover $v^{\prime }$, and then $v^{\prime }$ to recover $v$. Applying Cauchy-Schwarz and integrating over an interval of length two givesConsequently $\Vert v{\Vert }_{H^2}^2\le 21\Vert v^{\prime \prime }{\Vert }_2^2$. Nonnegativity of $q,r$ yields the precise coercive bilinear form boundThe Lax-Milgram theorem states that a bounded bilinear form on a real Hilbert space satisfying $a(v,v)\ge c\Vert v{\Vert }^2$ with $c>0$ represents every bounded linear functional uniquely: there is exactly one $u$ with $a(u,v)=\ell (v)$ for all $v$. All of its hypotheses have now been verified on $H$.

This weak solution is the unique global minimum, rather than merely a stationary point. The Ritz-Galerkin equivalence for a symmetric coercive form here is the exact identityIt is strictly positive for every nonzero $w\in H$. Conversely, a minimum has zero first variation in every direction, so it satisfies the same weak equation. Density of compactly supported test functions and boundedness of $a,\ell$ extend the distributional equation back to all $H$ test vectors. There is exactly one minimum, and it is exactly the clamped weak solution. No additional coefficient smoothness or classical fourth derivative is needed.