Deriving the Euler-Lagrange equation using partial integration

Question

From our homework sheet (Germany):

Consider the following variational problem $$ S [u] = \int_{0}^{1} \left( \frac{1}{2}u^{''}(x)^2-\lambda u^{'}(x) \right) {\rm d} x \to \min \quad \text{with} \quad u \in C^2(0;1) \cap C[0;1], u(0) = u(1) = 0 $$ Derive the Euler-Lagrange equation using partial integration.

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So I set $\phi(t) = S[u + tv]$ with $v \in C_{0}^{\infty}(0;1)$

Considering $u$ is a solution of the variation problem: $$\phi'(t) = \int_{0}^{1}(u''(x)+tv''(x))v''(x)-\lambda v'(x)dx = 0$$ for $t = 0$. So $$0 = \phi'(0) = \int_{0}^{1} u''(x) v''(x) - \lambda v'(x)dx$$

Integrating $ -v'(x) \lambda$ gives you $$\int_{0}^{1}u''(x)v''(x)dx = 0$$

But how do I continue from here? If $u \in C^4(0;1)$ one could get

$$\int_{0}^{1}u^{(4)}(x)v(x)dx = 0$$

and thus $u^{(4)}(x) = 0$. Does anybody have an idea how to continue from here?

Answer 1

For what it's worth, we can find the minimum configuration for $u$ without knowing the EL equations as follows: The total derivative term in $S[u]$ always vanishes due to the boundary conditions (BCs), so we may write $S[u]$ with a non-negative integrand. The minimum of $S[u]$ is then clearly when the integrand is zero, i.e. $u^{\prime\prime}\equiv 0$. Together with the BCs, this leads to $u\equiv 0$.

Answer 2

EDIT: THIS IS WRONG.

The Euler--Lagrange equation in this case is $u''=0$ (this is actually not true, see comments). Indeed, as you correctly derived in the question, we must have $$ \int_{0}^{1}u''(x)v''(x)dx = 0, $$ and $v\in C^\infty_0$ is arbitrary, so $v''$ is a general element of $C^\infty_0$. We conclude $u''=0$.

The only solutions to this Euler--Lagrange equation are affine linear functions $u(t)=at+b$, but we also need to impose boundary conditions $u(0)=u(1)=0$, which means that the only critical point is $u=0$ and we find the same result as in the other answer.