Fundamental lemma of calculus of variations with second derivative

Question

Intense debate at work place around the solution for this:

Let $M \in C[a,b]$ be a continuous function on the closed interval $[a,b]$ that satisfies $$\int_{a}^{b}M(x)\eta^{\prime\prime}dx = 0,$$ for all $\eta \in C^{2}\left[a,b\right]$ satisfying $\eta(a) = \eta(b) = \eta^{\prime}(a) = \eta^{\prime}(b) = 0$. Prove that $M(x) = c_{0}+c_{1}x$ for suitable $c_{0}$ $c_{1}$. What can you say about $c_{0}$, $c_{1}$?

I tried to use integration by parts, and use the fundamental lemma of the calculus of variations, and the lemma of Du Bois and Reynolds to prove it, but that requires $M$ to be $C^1([a, b])$.

Answer 1

Define $M_\epsilon(x) = (M \ast \phi_\epsilon)(x)$ where $\phi_\epsilon$ is a standard mollifier supported on $[-\epsilon,\epsilon]$. Then for all $\eta \in C^\infty$ supported on $[a+\epsilon, b-\epsilon],$

$$\begin{align*} \int_{a+\epsilon}^{b-\epsilon} M_\epsilon''(x)\eta(x)\,dx &= \int_{a+\epsilon}^{b-\epsilon} M_\epsilon(x)\eta''(x)\,dx \\ &= \int_{a+\epsilon}^{b-\epsilon}\int_{-\epsilon}^\epsilon M(x-t)\phi_\epsilon(t)\eta''(x)\,dt\,dx \\ &= \int_{-\epsilon}^\epsilon \phi_\epsilon(t)\int_{a+\epsilon-t}^{b-\epsilon-t} M(x)\eta''(x+t)\,dx\,dt \\ &= \int_{-\epsilon}^\epsilon \phi_\epsilon(t)\int_a^b M(x)\eta''(x+t)\,dx\,dt \\ &= 0 \end{align*}$$

In particular, this shows that $M_\epsilon(x) = c_{0,\epsilon} + c_{1,\epsilon}x$ for $x\in [a+\epsilon, b-\epsilon].$

Since $M_\epsilon\to M$ pointwise on $(a,b)$ as $\epsilon \to 0^+,$ $M(x) = c_0 + c_1x$ for $a< x < b.$

Answer 2

This solution is more elementary than the one of Brian Moehring because it does not use mollificators. It also has the advantage of explicitly giving $c_0$ and $c_1$, as requested in the question.

Let me make a minor notation change and assume that $m\in C([0, 1])$ is the function with the property that $$ \int_0^1 m(t)\eta''(t)\, dt = 0, \qquad \forall \eta\in C^2, $$ where $$ \tag{*}\eta(0)=\eta(1)=\eta'(0)=\eta'(1)=0.$$ Define $$\tag{1} M(x):=\int_0^x (m(t)-c_0-c_1t)(x-t)\, dt, $$ where $c_0, c_1$ are chosen in such a way that $M$ satisfies the boundary conditions (*). (See Remark, below, for more information on (1)). This amounts to solving a system of 2 linear equations in 2 unknowns, which is not singular, and thus admits one and only one such solution regardless of $m$. (The solution is given in the Appendix, below).

Now notice that the assumption on $m$ implies that, for every polynomial $P$ of degree 1, and for every smooth $\eta$ satisfying (*), we have that $$\tag{2} 0=\int_0^1(m(t)-P(t))\eta''(t)\, dt.$$ Indeed, integration by parts shows that $\int_0^1 P(t)\eta''(t)\, dt=0$. Using (2), we compute $$ 0=\int_0^1 (m(t)-c_0-c_1 t)M''(t)\, dt = \int_0^1 (m(t)-c_0-c_1t)^2\, dt, $$ which finally implies $$ m(t)=c_0+c_1 t.$$

Remark. How did I come up with the formula for $M$? That is actually a special case of a general formula; the function $$ M(x)=\int_0^x \frac{(x-t)^{n-1}}{(n-1)!} m(t)\, dt $$ satisfies $M(0)=M'(0)=\ldots=M^{(n-1)}(0)=0$ and $$ M^{(n)}(x)=m(x), $$ and is therefore known as "a $n$-th antiderivative of $m$".

Appendix. The values of $c_0, c_1$ are: $$ c_0=2\int_0^1 m(t)\, dt -\frac12\left( m(1)-m(0)\right), \quad c_1=m(1)-m(0).$$