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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 $Lu$.

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 $q$.