# Understanding "well-posedness"

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.

• Are you familiar with weak formulations ? Oct 13, 2021 at 13:09
• I am hearing the term "weak formulations" for the first time. Oct 13, 2021 at 13:14

• $$V_1=D(L)=\{ f \in C^1([a,b] : f(0)=u_0 \}$$ for some $$a<0. $$\| 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. $$\| f \|_2 = \| f \|_\infty$$. (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.

• What is the Definition of D()? And how do you get to your result Oct 13, 2021 at 19:00
• @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).
– Ian
Oct 13, 2021 at 19:38
• Thank you very much Oct 13, 2021 at 19:49
• 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. Oct 14, 2021 at 12:33
• @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.
– Ian
Oct 14, 2021 at 12:57