Dirichlet's principle on $-u''+u=f$ on $I=(0,1)$,

Consider the ODE $-u''+u=f$ on the interval $I=(0,1)$ with boundary conditions $u(0)=u(1)=0$, and assume $f\in L^2(I)$. I wish to find a weak solution to this problem (following Brezis's book on Functional Analysis, Chapter 8.4).

Now, what he does, is he moves to the space $H_0^1(I)$, and notes that a weak solution $u$ is precisely one such that $$\langle u,v\rangle_{H_0^1(I)}=\int_I u'v'+\int_I u v = \int_I vf$$ for all $v\in H_0^1(I)$, and since $\left(v\mapsto\int_I vf\right)\in H_0^1(I)'$ then we know that by Riesz-representation, there exists a unique solution in $H_0^1(I)$.

My confusion is with his restatement of this ODE as a variational problem, where he states that $u$ is obtained by searching for the minimizer of $$\min_{v\in H_0^1}\left(\frac12\int_I(v'^2+v^2)-\int_I fv\right)$$ however, I don't see where this comes from. I don't see why the solution should minimize this value. Have I misunderstood what is meant by this?

• I am confused. Do you know how to obtain the variational form given an ODE? May 1, 2017 at 15:55
• No, I unfortunately do not. May 1, 2017 at 15:55
• Multiply by a test function and integrate by parts. I'm sure that it's in that book. May 1, 2017 at 15:58
• That was precisely I did in order to use Riesz representation. If I'm missing some important tool, I don't think it's that one. May 1, 2017 at 16:02

The equivalent variational problem was likely found from Euler-Lagrange equations $$\frac{d}{dt} \frac{\partial F}{\partial x'} - \frac{\partial F}{\partial x} = 0.$$ Where $$F$$ is the functional defined in your minimization problem. For details see https://en.wikipedia.org/wiki/Euler–Lagrange_equation
Note that $$E[\phi]:=\frac12\int [(\phi')^2+ \phi^2]-\int f\phi=:\frac12a[\phi,\phi]-\langle f,\phi\rangle$$ is a quadratic+linear functional on $$\operatorname H^1_0(0,1)$$ and the form $$a$$ is symmetric. As such it has a unique minimum (already proven) $$u$$ which satisfies for any $$\phi$$ $$$$\langle \operatorname DE[u],\phi\rangle=0.$$$$ Formally since $$E[v]$$ is quadratic in $$v$$, it's derivative $$\langle \operatorname DE[u],\phi\rangle$$ is an affine in $$u$$ and of course linear in the increment $$\phi$$ and given by $$$$\frac12(a[u,\phi]+a[\phi,u])-\langle f,\phi\rangle,$$$$ from which you deduce the weak form.