I'm trying to wrap my head around the ways to approach the heat problem. This time, with Neumann boundary conditions. Now, I have the theory down, but the only example I found leaves me behind by leaping to the solution without any mention of the process used to get there. So here it is:
I want the weak form of: $$- \biggl( {u' \over x^2+1} \biggl)'+u=e^x$$ $$u'(0)=u'(1)=0$$ defined in $(0,1)$
The solution simply whips out a formula that is basically the Lax-Milgram theorem $$\int_0^1 \biggl[ {1 \over x^2+1}u'v'+uv\biggl]dx=\int_0^1e^xvdx$$
Which of course guarantees the existence of one solution, assuming that the bilinear form is continuous and coercive and the functional is continuous. It makes sense, but what are the steps in between the starting and the ending points?
Thanks.
You multiply your equation formally by a test function $v\in H^1$ and integrate by parts $$ -\int_0^1\left(\frac{u'}{1+x^2}\right)'v\,dx+\int_0^1 uv\,dx=\int_0^1 e^xv\,dx, $$ $$ \int_0^1\left(\frac{u'}{1+x^2}\right)v'\,dx-\left(\frac{u'}{1+x^2}\right)v|_0^1+\int_0^1 uv\,dx=\int_0^1 e^xv\,dx. $$ Using the boundary condition the non-integral term cancels and you get your weak form of the equation.