Consider an ordinary vector-valued differential equation of the form $$ \begin{align*} \dot y(t) &= f(t,y(t)), \\ y(0) &= y_0 \in \mathbb{R}^n. \end{align*} $$ It is well known that if $f$ is continuous and furthermore Lipschitz continuous in an appropriate sense, then there exists a unique solution $y \in C^1([0,T], \mathbb{R}^n)$ satisfying the differential equation at every point of the time interval $(0,T)$ (Picard–Lindelöf).

The question: Which general assumptions on $f$ ensure solvability in a (Sobolev) weak sense? In particular, I want the solution $y$ to be in $H^1 = W^{1,2}$, i.e. $y$ should be $L^2$, $y$ should be weakly differentiable, $y'$ should be $L^2$ and the weak derivative of $y$ should coincide with $f(t,y(t))$ almost everywhere.

This should be an extremely basic question, but I do not know relevant literature.

[I seem to have an existence proof in the case $f(t,y) = A(t) y$, where $A \in L^2([0,T], \mathbb{R}^{n \times n})$, by mimicking the usual proof of the Picard–Lindelöf theorem and simply considering the usual integral operator as an operator $L^2([0,T], \mathbb{R}^n) \to L^2([0,T], \mathbb{R}^n)$. This case is particularly important to me. Note that here, $f(t,y)$ is not continuous in $t$.]

  • 1
    $\begingroup$ You may see this: math.stackexchange.com/a/460809/254733 $\endgroup$
    – Svetoslav
    Commented Mar 10, 2016 at 14:27
  • $\begingroup$ Also, a weak formulation for this first order equation would involve $L^2$ as a trial and test space, so maybe it is more or less the same as your second method with the integral operator from $L^2\to L^2$ $\endgroup$
    – Svetoslav
    Commented Mar 10, 2016 at 14:29
  • $\begingroup$ @Svetoslav: Thanks! I'll check that. (Sorry I didn't reply sooner -- didn't get a notification of your comment, even though of the bounty.) Meanwhile, I stumbled on Carathéodory's existence theorem (which was curiously absent from by education) and which might also provide an answer. It is formulated for absolutely continuous functions instead of Sobolev functions. $\endgroup$ Commented Mar 15, 2016 at 23:29
  • $\begingroup$ @IngoBlechschmidt By the way, you may know, but just to mention that in $1D$ all Sobolev functions are absolutely continuous. In dimension $\ge 2$ they are absolutely continuous on almost all lines parallel to the coordinate axes (the ACL characterization). $\endgroup$
    – Svetoslav
    Commented Mar 16, 2016 at 7:22

1 Answer 1


This is mostly for anyone who stumbles upon this question and wants a more detailed answer. There are also partly answers in the question itself and in the comments. I try to extend it a bit with references and give a typical application for this formalism which arises in the study of evolutionary PDEs.

We want to investigate the ODE $$ \begin{cases} \dot y(t)=f(t,y) \\ y(0)=y_0. \tag{$\star$} \end{cases}$$ Your aim is to equip $f$ with conditions such that the ODE has a solution $y \in H^1(0,T;\mathbb{R}^d)$ almost everywhere.

The key-word was already given: Carathéodory's theorem

It is the extension of Peano's existence theorem to ODE with discontinuous right hand side. It gives the existence of a solution in a Sobolev space. A short version can be found in "Nonlinear Partial Differential Equations with Applications" by Tomas Roubicek.

Theorem (Carathéodory). Let $T$ be fixed and $f: (0,T) \times \mathbb{R}^k \to \mathbb{R}^k$ such that $f(t,\cdot) \in C(\mathbb{R}^k)$, $f(\cdot,r) \in L^1(0,T)$ and $|f(t,r)| \leq \gamma(t)+C|r|$ for some $\gamma \in L^1(I)$. Then ($\star$) has almost everywhere a solution $y \in W^{1,1}(0,T;\mathbb{R}^k)$. Moreover, if $f(t,\cdot)$ is Lipschitz continuous, then the solution is unique.

  • Here $f(\cdot,r)$ indicates the mapping $t \mapsto f(t,r)$ for fixed $r$ and analogously for $f(t,\cdot)$
  • Other sources are:

    • "Finite Element Methods for Incompressible Flow Problems" by Volker John
    • "Ordinary Differential Equations" by Wolfgang Walter
    • "Ordinary Differential Equations" by Jack Hale
  • We come back to a comment who mentions the ACL characterization. Since $y \in W^{1,1}(0,T;\mathbb{R}^k) \hookrightarrow C([0,T];\mathbb{R}^k)$ and $y' \in L^1(0,T,\mathbb{R}^k)$ we have $y \in AC([0,T];\mathbb{R}^k)$

  • Now you want a solution in $H^1$. This can be easily achieved if we additionally assume that $f(\cdot,r) \in L^2(0,T)$.

  • Probably the main application for this theorem is in the existence of weak solutions of evolutionary PDEs when doing a Galerkin approximation. There we reduce the infinite-dimensional problem to a finite-dimensional one and reduce the PDE to an ODE. Mostly the right hand side is in a space such as $L^2(0,T;H^{-1})$ and classical theorems like Peano are not applicable.

    • Example: We consider the parabolic PDE $y'+Ay=f$ on $\Omega \times (0,T)$ with homogeneous Dirichlet boundary and data $f\in L^2(0,T;H^{-1})$, $y_0 \in L^2$. $A$ is typically a second order partial differential operator in divergence form. The weak form reads $\langle y',v \rangle_{H^{-1}} + a(y,v)= \langle f,v \rangle_{H^{-1}}$ for all $v\in H_0^1$ where the bilinear form $a$ corresponds to $A$. The finite dimensional problems reads $\langle y_k',v \rangle_{H^{-1}} + a(y_k,v)= \langle f,v \rangle_{H^{-1}}$ for all $v \in V_k$ where $V_k$ is just that $\cup_k V_k$ is dense in $H_0^1$. With the Galerkin ansatz this problem reduces to an ODE with the vector-valued right hand side $F(t)=(\langle f(t),v_j \rangle_{H^{-1}})_j \in L^2(0,T;\mathbb{R}^k)$. Thus Carathéodory's theorem gives a solution $y_k \in H^1(0,T;V_k)$ to the Galerkin problem. In the end one derives energy estimates and get that a subsequence of $y_k$ converges to a $y$ which is a weak solution to the original PDE. For a reference see for example "Optimization with PDE Constraints" by Hinze, Pinnau, Ulbrich.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .