I have a variable coefficient pde of the form $$u_t=c(t,x)u_{xx}, t\in [0,T], x\in [0,1]$$ with initial data $u_0=u(0,x)$ and $c(x,t)\in C([0,T]\times[0,1])$. I use three point discretization for the diffusion part. I would like to have en error estimate by proving stability and consistency. Doing Taylor expansion, I can show consistency. However, I am confused how to choose the norm? I have seen in the book the author chose discrete $L^2$ norm and by using discrete energy methods the stability followed.

First question for the discretized equation: If I pick another discrete norm, which one should I pick (I mean one of the $L^p$ norms), which one is easier or harder? So far, I have obtained the estimate that $||\vec{e}||_2=O(h^2)$, where $\vec{e}:=\vec{u}_h(T)-\vec{u}(T)$. But the other similar estimates would follow in an analogous way once I prove the stability in those norms. Please provide some hints how to choose another norm and what are the "tools" to prove stability for such an equation in other discrete $L^p$ norms.

And second question: is the choice affected by the stability estimate of the original pde? That is, once I establish continuous dependence on initial data, (it follows from maximum principle that I have stability in $L^{\infty}$) and assume I have managed to show it is well-posed in $L^2([0,T]\times [0,1])$(by the same energy methods in continuous setting), does imply it is well posed in $L^1$ or $L^{\infty}$? How is that connected to the fact $L^{\infty}\in L^{2} \in L^1$?

  • $\begingroup$ An excellent question! $\endgroup$
    – Paul
    Commented Jan 12, 2013 at 22:32

2 Answers 2


For your case, the $L^2$ norm is natural. If you had an equation of the kind $$ \rho(x) u_t - (a(x) u_x)_x = f $$ then the natural norms would be $\|u\|_{L^2_\text{weighted}} = \|\sqrt{\rho} u\|_{L^2}$ as the analog of the $L^2$ norm, and $|u|_{H^1_\text{weighted}} = \|\sqrt{a} u_x\|_{L^2}$ as the analog of the energy ($H^1$) norm. You get these naturally if you multiply the equation by $u$ or $u_x$, respectively, and integrate over the domain. If the two coefficients are bounded pointwise from below and above, these norms are of course equivalent to the $L^2$ and $H^1$ norms, respectively. Thus, if you have a result in the weighted norms, then you immediately have the same result in the original norms.

Using this technique, show convergence in $L^2_\text{weighted}$, then you immediately have convergence in $L^2$ as well, and consequently in $L^1$.

I don't seem to see much sense in showing convergence in any other of the $L^p$ norms for $p$ not either $1,2,\infty$. Nobody seems to care about other values of $p$ and I can not see a reason to care either.

  • $\begingroup$ ok, so there are two issues: First, we pick a norm in order to show $||u^{n+1}||_p^2\leq C||u^{n}||_p^2$. Thus, doesn't the $L^{\infty}$ such estimate follow from the maximum principle? If it does, then I have $||e||_{\infty}=O(h^2+k^2)$ and other convergences would follow from embeddings. why not just show in $L^{\infty}$ instead? Second, I am sure I can construct such a $c(x,t)$, not very smooth so that I don't have $L^2$ well-posedness and have only say the fact that it is well posed in $L^{\infty}$. Does it affect my analysis of the numerical method, or I just choose $L^2$ by default? $\endgroup$
    – Kamil
    Commented Jan 13, 2013 at 21:00
  • $\begingroup$ I am trying to get this connections between the theoretical side of the pde and analysis of the numerical algorithm used to solve it. I start by looking at the problem first, prove it is existence, uniqueness and some energy estimates and only then start explore the method. So, how are they related, does results from theory imply how I choose the norm for the numerical method? $\endgroup$
    – Kamil
    Commented Jan 13, 2013 at 21:04
  • $\begingroup$ I think you miss the point that the $L^\infty$ convergence is typically very difficult to show. Sure, if you can show it, then most everything usually follows. But it's hard to get it (see, for example, how difficult it is to show $L^\infty$ convergence for the Laplace equation) and so people work in other norms for which convergence is simpler to show. $\endgroup$ Commented Jan 13, 2013 at 21:31
  • $\begingroup$ ok, this is what I have been looking for, so it get's harder to prove for "larger" $p$. However, convergence follows directly from stability, so it should be harder to show $||u^{n+1}||\leq C||u^n||$ in $L^{\infty}$, but doesn't it follow from the maximum principle? $\endgroup$
    – Kamil
    Commented Jan 14, 2013 at 0:58
  • $\begingroup$ Well, the maximum principle gives you stability, but does that give you convergence? I'm not enough of a finite difference person to really know the answer in arbitrary norms. But yes, in general, it is harder to prove convergence the larger the $p$ gets. $\endgroup$ Commented Jan 14, 2013 at 13:46

In general, well-posedness (with combines existence, uniqueness, and stability) holds in the space for which you showed it only. There can be no solutions in a smaller space, and multiple solutions in a larger space. The continuous dependence of the solution on the data also depends on the norms (and thus on the normed space) in which you take data and solution.

As Wolfgang Bangerth already pointed out, there is a natural norm (the energy norm) in which stability holds for the continuous problem and thus for any consistent approximation. If(!) your data is sufficiently smooth, you can of course improve this to get stability in higher norms.

If you focus on a specific discretization, you are correct that equivalence of norms in finite dimensions gives you stability in any norm as long as you have stability in one norm. However (and this is the crucial point), the equivalence constants depends on your discretization, and usually blow up as the discretization parameter $h$ tends to 0. Needless to say, this makes the equivalence argument useless for proving error estimates as $h$ tends to 0...

I should also point out that $L^\infty$ in time error estimates are pointwise estimates for $u_h(t)$ at any $t\in[0,T]$, not just for the time steps $u_n = u(t_n)$. This makes these estimates much harder to obtain than (say) $L^2$ estimates.

Finally, I think you have the question the wrong way around: It's not "Let's show convergence, what norm should I pick?" but "I need convergence in that norm, how do I show it?" (If any norm will do, use the energy norm.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.