Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 63 / 2

In the variational convention needed here, a positive definite symmetric operator on a real Hilbert space satisfiesFor a bounded operator on the whole space, symmetry is self-adjointness. On a complex Hilbert space, the corresponding convention is Hermitian positivity; the quadratic expression is real. Strict positivity should be distinguished from uniformly positive definite symmetric operator coercivity, which requires a fixed $\gamma >0$ such that $⟨Lv,v⟩\ge \gamma \Vert v{\Vert }^2$.

Symmetry matters in a real space. If “positive definite” means only the inequality for the quadratic form, the next variational assertion is false: $L=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$ has $⟨Lv,v⟩=\Vert v{\Vert }^2>0$, but its skew part disappears from that form. Its Euler-Lagrange equation is $v=f$, not $Lv=f$. Thus the usual symmetric/Hermitian convention, or an explicit symmetry assumption, is required.

For a real Hilbert space, differentiate along an arbitrary direction $w$:The last equality uses the symmetry in the definition of a positive definite symmetric operator. Vanishing of this first variation for every $w$ is equivalent to $Lv=f$, givingIf $u$ solves it, expand the functional about $u$ to findThis is strictly positive for nonzero $w$, so $u$ is the unique minimizer. Conversely a minimizer has zero first variation and solves the equation. On a complex Hilbert space, use the real functional $⟨Lv,v⟩-2\mathrm{Re}⟨f,v⟩$ and real directional derivatives; varying $w$ and $iw$ yields the same result.

This proves the quadratic variational principle for a symmetric positive operator, conditional on existence. Strict positivity in infinite dimension does not by itself guarantee a minimizer for every $f$: on ${\ell }^2$, the operator $(Lv{)}_j=v_j/j$ is strictly positive, but $f_j=1/j$ would require the non-square-summable solution $v_j=1$. A coercive bilinear form supplies existence through the Lax-Milgram theorem under its usual boundedness assumptions.

Interpret the differential expression as $Lu=(p{u}^{\prime \prime }{)}^{\prime \prime }+qu$, with a suitable differential-operator domain and the stated clamped boundary conditions. For sufficiently smooth $u,v$ in that domain, twice integrating by parts givesThe boundary term vanishes because $v=v^{\prime }=0$ at both ends. For real coefficients this form is symmetric/Hermitian. In particularIndeed nonnegativity follows from $p>0$ and $q\ge 0$. If the integral were zero, $u^{\prime \prime }=0$ almost everywhere in the interval, so $u$ would be affine. Its endpoint values then force $u=0$. This proves strict positivity on the intended clamped domain and gives the weak formulation of a clamped fourth-order equation.

There is a genuine domain issue in the literal wording. The closure of clamped smooth functions in the stated L2 norm is all of $L^2(-1,1)$, since it contains the dense set $C_c^{\infty }(-1,1)$. Boundary traces are not retained by that closure, and a fourth-order expression does not define an operator on every element of $L^2$. For example $p=(1-x^2{)}^2$, $q=0$ satisfy the given sign conditions, but $u=1_{(-1/2,1/2)}$ belongs to this closure and makes $(p{u}^{\prime \prime }{)}^{\prime \prime }$ a distribution rather than an $L^2$ function. Even a general twice-smooth function need not have the four derivatives required by the classical expression.

A rigorous classical repair is to take smooth enough coefficients, for example $p\in C^2([-1,1])$, $q\in C([-1,1])$, and use a clamped domain such as $H^4\cap H_0^2$ in $L^2$, with the appropriate operator realization. A weak repair uses the clamped second-order Sobolev space $H_0^2$ and the form just displayed, with hypotheses ensuring its boundedness and closure; uniform positivity $p\ge p_0>0$ makes it coercive there. The integration-by-parts and strict-positivity proof does not need that stronger uniform bound, but existence and completion statements must not be inferred from the original $L^2$ closure alone. The printed $L[f]$ with an expression in $u$ is also a harmless variable-name mismatch; the calculation consistently uses $Lu$.