Derive weak formulation of a PDE from the integral form

Question

I am taking a class about theory of elasticity. Therefore we cover some PDEs.

There is one exercise that puzzles my mind. We shall derive the weak/variational formulation from the following DE:

$\begin{align}-(a(x)u'(x)))' + b(x)u(x) &= f(x) \qquad x \in (0,l) \\ \alpha u'(0) - \beta u(0) &= 0 \\ \gamma u'(l) +\delta u(l) &= 0\end{align}$

but we shall do it from the integral form:

$a(x)u'(x) - a(0)u'(0) + \int_0^x (f(s) -b(s)u(s))ds = 0$

I really have no idea how to do that. Normally I would just multiply with some test function $v \in V$, integrate and then do partial integration.

But here ?! No idea.

I would be so thankful for any help.

EDIT: It seems that my question is misunderstood. I try it once again.

Given: $-\int_0^x(a(s)u'(s))'ds + \int_0^x f(s)ds + \int_0^x b(s)u(s)ds = 0$ how can I derive the weak formulation?

Answer 1

we look for a weak solution $u\in H^1(0, \ell)$ satisfying the integral identity (weak formulation of the problem) $$\int_0^\ell u\bigg(-(a\phi')'+b\phi\bigg)dx=\int_0^\ell f\phi dx$$(obtained by twice integration by parts on the first term of the equation) for all $\phi\in C_0^\infty(0, \ell)$ and by density $$\int_0^\ell u\bigg(-(a\phi')'+b\phi\bigg)dx=\int_0^\ell f\phi dx$$ for all $\phi\in H_0^1(0, \ell)$.

Answer 2

I am going to write something obvious (too long for a comment), please do not misinterpret my aim.

Apart from elegance, what is fundamentally wrong with first going from the integral form to the differential form, and then to the weak formulation (via the usual proceudre you are familiar with)? It still fulfills the requirements of the exercise.

The integral to differential conversion is trivial, as $$a(x)u´(x) - a(0)u´(0) = \int_0^x (a(s) u´(s)) \mathrm{d}s$$ and your differential expression is only a few passages away, waiting for the integration by parts machinery.

ADDENDUM Ok I just saw your edit in which you performed the integration on the first term. Well is that not all is there to do? If the integral expression is zero, will not the integrand be? And you have the DE.