Consider the following problem
$-(x^2+1)u''-2xu'=\sin(2\pi x), \;x \in (0,1)$
$u(0)=0$, $u(1)=0$.
I want to show that there exists a unique solution $u \in H^1_0(0,1)$ and find an estimate for
$\|u'\|_{L^2(0,1)}$
Here is the solution:
I have a few questions: What is $a_{ij}$ ? Is it $-(x^2+1)$? and if so how does that form a matrix? Is it possible to prove uniqueness by using Lax-Milgram?
I don't understand the integrating by parts. What is our $u$ and $dv$?
Why are we inner producting with u? Shouldn't we use a test function but then again it is stated that u is in $H^1_0(0,1)$. I'd still like an answer on this, if possible
Finally the lines after the Cauchy Schwarz inequality are a bit mysterious. Please shed some light on it.
And it looks like the problem was written incorrectly. Is should be find an estimate $||u'||_{H^1_0(0,1)}$ but I could be wrong
If you have a different solution that you would like to share, please feel free to do so
Suppose $u_1$ and $u_2$ on $[0, 1]$ both satisfy $$0 = (x^2 + 1)u''(x) + 2xu'(x) + \sin(2\pi x)$$ on $(0, 1)$ with the boundary conditions $u(0) = u(1) = 0$. Then $w := u_1 - u_2$ satisfies \begin{align} 0 &= (x^2 + 1)w''(x) + 2xw'(x) \\ &= ((x^2 + 1)w'(x))' \end{align} with $w(0) = w(1) = 0$. This shows $(x^2 + 1) w'(x)$ is a constant, say $C$, and that $$ w(1) = w(0) + C \int_0^1 (x^2 + 1)^{-1} dx.$$ You know that $w(0) = 0$, that the integrand is positive on $(0, 1)$, so the only possible way you'll get $w(1) = 0$ is if $C = 0$.
You might consider more generally the ODE $u''(x) + b(x)u'(x) = f(x)$, where $b$ and $f$ are continuous on $[0, 1]$, same boundary conditions. With $u_1$ and $u_2$ also satisfying these, again set $w := u_1 - u_2$. Using the integrating factor $I(x) := \exp(\int_0^x b(\xi)d\xi)$ you can write \begin{align} 0 &= w''(x) + b(x)w'(x) \\ &= (I(x)w'(x))'/I(x) \end{align} and use the same trick to show $w = 0$.
All that being said, I understand it doesn't really answer the question about weak solutions, in this case in $H^1_0(0, 1)$, being unique.
Your original ODE can be written in the form $au'' + a'u' = f$, and for any $v \in H^1_0(0,1)$ you have $(au'' + a'u')v = fv$. Integrating over $(0, 1)$ you get \begin{align} \int_0^1 fv &= \int_0^1 (au'' + a'u')v \\ &= \int_0^1 (au')'v \\ &= \int_0^1 (au'v)' - \int_0^1 au'v' \\ &= -\int_0^1 au'v'. \end{align} which provides a weak formulation. Doing the $w := u_1 - u_2$ trick again you find $w$ has to satisfy $$ 0 = \int_0^1 a w'v'.$$ The specific choice $v = w$ shows $\int_0^1 a (w')^2 = 0$.