Until now I have learned that well-posedness means:

  • the problem has a solution
  • the solution is unique
  • the solution changes continuously with the initial conditions

But now I have a "proper definition", which I don't understand.

Let $L: D(L) \subseteq V_1 \rightarrow V_2$ be a linear function and $(V_1,||.||_1)$ and $(V_2,||.||_2)$ be normed vector spaces.

The problem, "For a given $f \in V_2$ find a $u \in D(L)$ such that $L(u)=f$ holds", is called well-posed if $L$ is bijective and $L^{-1}$ is continuous. This means the problem must have a unique solution and a constant C such that $||u||_1 \leq C ||f||_2$

My questions: I don't see what $D$ is in the first place.

I will try to show how I understood it:

Considering the ODE:

$u'(t)=k(\overline{u}-u(t))$ where $k$ and $\overline{u}$ are some constants.

Some basic knowledge in ODE is enough to understand that there exists a solution and with the boundary value, the solution is unique.

The solution is $u(t)=\overline{u}+ u_0e^{-kt} -\overline{u}e^{-kt}$ From this solution one can see that the estimation $|u(t)| \leq |\overline{u}|+|u_0|$ holds for t greater or equal 0.

Let $u_1(t)$ be a solution with the conditions $\overline{u}+ a$, and $u_0+b$, then $u_1(t)-u(t)$ is the solution of the ODE with the conditions $a,b$.

Considering now the estimation above, we get $|u_1(t)-u(t)|\leq |a|+|b|$

This should actually be enough given my intuitive understanding (the 3 things listed at the beginning). I don't really see that $V_1,V_2,f,u$ and $L$ are in my given example.

I would be thankful if someone could explain it to me.

  • $\begingroup$ Are you familiar with weak formulations ? $\endgroup$
    – nicomezi
    Oct 13, 2021 at 13:09
  • $\begingroup$ I am hearing the term "weak formulations" for the first time. $\endgroup$
    – John.W
    Oct 13, 2021 at 13:14

1 Answer 1


In just your example:

  • $V_1=D(L)=\{ f \in C^1([a,b] : f(0)=u_0 \}$ for some $a<0<b$. $\| f \|_1 = \| f \|_\infty + \| f' \|_\infty$. (This is the standard norm of $C^1$ on a compact interval.)
  • $V_2$ is just the image of $V_1$ under $L$, which is a subset of $C^0([a,b])$ for the same $a<0<b$. $\| f \|_2 = \| f \|_\infty$. (This is again the standard norm of $C^0$ on a compact interval). Notice that less regularity is demanded of the forcing than is demanded of the solution.
  • $Lu=u'-ku$
  • $f=k\overline{u}$.

Note that in the general situation this topic is a massive rabbit hole that consumes the entire careers of many analysts, so you should not expect a clean general purpose answer for an arbitrary situation.

  • $\begingroup$ What is the Definition of D()? And how do you get to your result $\endgroup$
    – John.W
    Oct 13, 2021 at 19:00
  • 1
    $\begingroup$ @John.W $D(L)$ means "the domain of $L$". The point of this formalism is simply to have all the derivatives that you need to use be defined and to have $L$ itself be continuous (as a map from $V_1$ to $V_2$), and not impose any other restrictions (besides the need to restrict to $[a,b]$ which is a technical assumption needed to define the norms). $\endgroup$
    – Ian
    Oct 13, 2021 at 19:38
  • $\begingroup$ Thank you very much $\endgroup$
    – John.W
    Oct 13, 2021 at 19:49
  • $\begingroup$ Applying this leads to the a non unique solution of $Lu=u$ if I have fully understood your answer. The solutions form an affine space of dimesnion $1$ though. But this ODE problem is not well-posed (if my interpretation of being unique is correct) unless we set an initial condition, but in this case $L$ is no longer defined as you did. Do not hesitate to correct me. Actually I have wrapped my head around this question yesterday, I came to the same "solution" but I found it non-satisfactory because of what I have mentionned here. $\endgroup$
    – nicomezi
    Oct 14, 2021 at 12:33
  • $\begingroup$ @nicomezi That's a good point (although for $Lu=u$ that's not correct, you need a derivative in there somewhere). In the setting of ODE you can force it through a domain restriction on $L$ (which I will edit in now) but in the setting of PDE this does not always work. $\endgroup$
    – Ian
    Oct 14, 2021 at 12:57

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