[Math] Stability of pde in some $L^p$ norm and stability of a numerical scheme for it equivalence.

finite differencesnumerical methodspartial differential equations

I would like to get some light on how to proceed and my confusion. I consider some IBVP of the form $$u_t+L(t,x)u=0, x\in D, t\in [0,T]$$ with some BC and initial data. And I use some numerical method to solve it. First, I start with a question of well-posedness of IBVP. Assume I proved existence and uniqueness in some general space, say only bounded measurable functions, so in fact I showed existence and uniqueness only of a weak solution. Now, I still have to show stability and I can see it perfectly overlaps with a question of stability of a numerical scheme. And this is where I am stuck. So, I can pick a space first, and as it is done in many books, I choose $L^2$ and try to show stability, that is to prove that $$||u(t,x)||_2 \leq C ||u(0,x)||_2$$ for some $C$ using energy methods, for example. I can also use discrete energy methods and discrete energy norms to show the same for a numerical method that $$||u_h(t_n,x)||_2 \leq C_h ||u_h(0,x)||_2.$$ I am dealing with a problem where I have these stability results for any $t,t_n \in [0,T-\epsilon]$, with $C=C(\epsilon)$, but not on $[0,T]$. If I picked $L^{\infty}$ instead of $L^2$, I can show it for $t\in [0,T]$ and $C$ is a fixed constant. So I have the following questions:

  1. So far, looks like a problem is well posed only in $L^{\infty}$ on $[0,T]$. However, I happen to note that my numerical method does exhibit $L^2$ convergence, so it might be a coincidence or there is some limiting result I can show in $L^2$ in this case?

  2. What is the intuition behind choosing $L^2$ or $L^{\infty}$(I can also pick some Sobolev space but I know nothing about them, so I restrict myself to $L^p$ spaces). Is one better than the other? Is it stronger result in one space? I know that $L^{\infty} \in L^2 \in L^1$ but there is no inequality between those norms though, so I can't say that stability in one norm imply stability in another, can I?

  3. Related to 2. So, If I have a stable IBVP in $L^p$ my numerical scheme has to be stable at least in that space? Say I have $L^2$ well posed problem, can I try to prove $L^1$ stability of a numerical scheme or it doesn't make sense since the initial problem is not $L^1$ stable?

Best Answer

  1. Stability in $L^\infty$ should imply $L^2$-stability (if your domain is bounded) since you have the bound $\|\cdot\|_{L^2}\leq C\|\cdot\|_{L^\infty}$.

  2. You have the continuous embedding $L^p(U)\subset L^q(U)$ for $p>q$, provided $U$ is bounded. This means that $\|u\|_{L^q}\leq C\|u\|_{L^p}$ for $u\in L^p(U)$.

  3. Generally, a good numerical scheme ought to be stable in the same spaces that come from continuous theory. On the other hand, instability of a continuous problem means that you cannot solve it numerically in any straightforward manner. If you really need to "solve" it you have to regularize the problem first, but what it is is you replace your original problem by a different problem that behaves better at continuous level.

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