I am taking a class on FE PDE and I just have some hopefully easy questions on some specifics of going from the strong or differential form to the weak or variational form.
The equation I am evaluating is:
-u"(x) + u(x) = f(x) , u(0) = u(1) = 0
to go to the weak form I believe I should multiply by a test function, then integrate by parts and simplify what I can. Doing this with v(x) being the test function and dropping the function of x bit
-u"v + uv = fv
IBP
∫(u'v') - |vu' + ∫(uv) = ∫(fv)
the solution given in class to this problem is
(u',v') + (u,v) = (f,v)
My question is first I assume this is inner product notation? Where
(u,v) = ∫ ( u(x)*v(x) dx)
secondly how does the value v*u' drop out when evaluated over the domain (0-1) ?
and thirdly can someone just kind of walk me through how to derive this? Or if its just a condition on v(x) the weighting function?
∀ v(x)∈ H_0^1 (0,1)
H_0^1 (0,1)= {(v(x) , v(0)=v(1)=0 , ∫v^2 dx < ∞)}
Thanks for any help you can offer!!
1) Yes, you are right: $$\left( {u,v} \right) = \int\limits_0^1 {uv} dx$$
2) It drops out due to the boundary conditions. When you have homogeneous boundary conditions you choose the test functions $v$ so they also match the homogeneous boundary conditions. In your case: $$v \in {V^0} = \left\{ {\omega :\left\| \omega \right\| + \left\| {\omega '} \right\| < \infty ,\omega \left( 0 \right) = 0,\omega \left( 1 \right) = 0} \right\}$$ so when you integrate by parts $$\int\limits_0^1 { - u''v} dx = \left[ { - u'v} \right]_{x = 0}^{x = 1} + \int\limits_0^1 {u'v'} dx = - u'\left( 1 \right)\underbrace {v\left( 1 \right)}_{ = 0} + u'\left( 0 \right)v\underbrace {\left( 0 \right)}_{ = 0} + \int\limits_0^1 {u'v'} dx = \int\limits_0^1 {u'v'} dx$$