Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 63 / 2 / 3 / Solution

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 $Lu$.