I am not sure how to handle inhomogeneous Neumann boundary conditions in a week formulation of a pde. The problem is:

Derive the variational formulation of
$$ -u''=-e^x \; \;\; in \; \Omega \in (0,1) \\ u(0) = 0 , \; u'(1) = -1 $$

First I multiply the equation with a test function v and integrate over the domain, which leads to

$$ \int_\Omega -u''v\;dx = \int_\Omega-e^xv\;dx $$

With integration by parts I get $$ \int_0^1-u''v\;dx=-[u'v]_0^1 + \int_0^1u'v'\;dx = -u'(1)v(1)+u'(0)v(0)+\int_0^1u'v'\;dx $$ where u'(0)v(0) disappears if the testspace of v yields v(0)=0 but -u'(1)v(1)=v(1) because of the boundary condition. So my week formulation would be $$ \int_0^1u'v'dx=\int_0^1-e^xvdx - v(1) $$ However all the basic examples in the lecture have the form $$ \int_0^1u'v'dx=\int_0^1fv\;dx \\ a(u,v)=F(v) $$

Am I missing something here?

  • $\begingroup$ what is your testspaces here? $\endgroup$ – Guy Fsone Jan 16 '18 at 9:44

You are missing the clarification on your test space

you should chose the following space

$$V=\{u\in H^1(0,1): u(0)=0\}$$

V is closed subspace of $H^1(0,1)$ and therein your variational formulation is given by $$a(u,v)= F(v)~~~~~v\in V$$

where $$a(u,v)=\int_0^1u'v'dx+ u'(0)v(0)~~~~~and ~~~~~F(v)=\int_0^1-e^xvdx - v(1)$$

  • $\begingroup$ Is your formulation correct? I am worried about the possiblity that there could be $u \in H^1$ or $u \in V$ such that you cannot force $u'(1)$ to exist or be defined with that formulation. Due to Sobolev-embedding, you can find a continuous representant of $u \in H^1$, but that doesn't say much about its derivative. $\endgroup$ – mdot Jan 16 '18 at 10:42
  • $\begingroup$ @max that was a typo $\endgroup$ – Guy Fsone Jan 16 '18 at 11:30

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