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?

  • $\begingroup$ I am confused. Do you know how to obtain the variational form given an ODE? $\endgroup$ May 1, 2017 at 15:55
  • $\begingroup$ No, I unfortunately do not. $\endgroup$ May 1, 2017 at 15:55
  • $\begingroup$ Multiply by a test function and integrate by parts. I'm sure that it's in that book. $\endgroup$ May 1, 2017 at 15:58
  • $\begingroup$ 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. $\endgroup$ May 1, 2017 at 16:02

2 Answers 2


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 $\phi$ \begin{equation} \langle \operatorname DE[u],\phi\rangle=0. \end{equation} 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 \begin{equation} \frac12(a[u,\phi]+a[\phi,u])-\langle f,\phi\rangle, \end{equation} from which you deduce the weak form.


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