Let $F:\mathbb{R} \to \mathbb{R}$ be locally Lipschitz, monotone and continuous. For the sake of concreteness only let us suppose it is of porous medium type (eg. $F(r) = r^{\frac 1m}$.)

Let $\Omega \subset \mathbb{R}^n$ be a bounded domain and let $Q=(0,T)\times\Omega$.

Given $u_0 \in L^\infty(\Omega)$ and $f \in L^\infty(Q)$, I have $F(u) \in L^\infty(Q) \cap L^2(0,T;H^{-1})$ with $u \in L^\infty(Q) \cap L^2(0,T;H^1)$ such that $$\int_0^T \langle (F(u))_t, \varphi \rangle + \int_0^T\int \nabla u \cdot \nabla \varphi = \int_0^T \int f\varphi$$ for all test functions $\varphi$. I also have the continuous dependence result for two solutions corresponding two two data: $$\lVert F(u_1) - F(u_2) \rVert_{L^1(0,T;L^1)} \leq C\left(\lVert u_{01} - u_{02}\rVert_{L^1} + \lVert f_1 - f_2 \rVert_{L^1(0,T;L^1)}\right).$$ Now I wish to extend my existence result to $L^1$ data satisfying a weaker formulation $$-\int_0^T \int F(u) \varphi_t - \int_0^T\int u \Delta \varphi = \int_0^T \int f\varphi$$ for smooth $\varphi$. Are there any standard tricks to do this using this continuous dependence result?

Of course we can approximate the $L^1$ data by $L^\infty$ data and using the above estimate (we obtain a Cauchy sequence and so) we find $F(u_n) \to F$ in $L^1(L^1)$ for some $F$ where $F(u_n)$ is the solution with data that approximates the $L^1$ data. From this I can obtain $u_n \to u$ for some $u$ pointwise a.e., but this is not enough pass to the limit.

Edit: I am aware of the book on PME by J Vazquez. If I recall correctly he handles $L^1$ data rather differently and I would like to know whether the above approach can work.

  • $\begingroup$ By the strong $L^1(Q)$ convergence $F(u_n)\to v$ you get $F(u_n)(t,x)\to v(t,x)$ a.e. $t,x$. By continuity of $F^{-1}(z)=z^m$ you see that $u_n(t,x)\to u(t,x)=F^{-1}(v)(t,x)$ a.e., so all you need now is prove that $u_n\to u$ in $L^1$. By dominated convergence it should be enough to prove some uniform $L^1(Q)$ bounds. Have you tried taking $\varphi=u$ as a test function in your strong formulation? I guess it should give you an $L^{\infty}(0,T;L^{1+1/m})$ estimate since formally $u\partial_t F(u)= C\partial_t(u^{1+1/m})$. But that's just a suggestions... $\endgroup$ Jun 27, 2014 at 7:31

1 Answer 1



First I have two remarks. The dependence continuous should be modified : it depends on $F(u_{01})-F(u_{02})$ and not on $u_{01}-u_{02}$. As far as the "weaker formulation" you do not precise the space where $\varphi$ is lying and there is no initial data. A formulation with $\varphi\in C_c([0,T)\times \Omega)$ can include the initial data term.

For your question, you need to have some compactness result in $L^1_{loc}$ (at least) for $u_\varepsilon$. Your approach can work

1) if $|F(r)|>C|r|$ for large value of $r$. But it is too restrictive

2) if $F(r)$ verifies a growth assumption $|F(r)|>C |r|^{\alpha}$, using the Boccardo-Gallouët estimates technique you can obtain a lower bound of $\alpha$ depending upon the dimension which insures and $L^1$ bound. But it is too restrictive.

The two previous methods do not use strongly the linear character of the operator :

3) use the regularity properties of $-\Delta$ operator and integrate with respect to time, like in the paper you cite in your question A question about PDE argument involving monotone convergence theorem and Sobolev space

  • $\begingroup$ Yes you are right that it should depend on $F(u_0i)$ not $u_{0i}$. For the weaker formulation we just take $\varphi$ to be as smooth as necessary for the integrals to make sense given the conditions on the data: The initial data $F(u_0)$ must be in $L^1$ (so for $F(u) = u^{\frac 1m}$, $u_0 \in L^{\frac 1m}$, and the RHS data $f \in L^1(0,T;L^1)$ too. Hmm, I think the arguments you cited in 3) are not amenable to the case when there is time-dependence in the operators. So I was looking for maybe a different approach. Leo's comment, if I am correct, almost works.. $\endgroup$
    – riem
    Jul 19, 2014 at 18:12
  • $\begingroup$ ...but I would need data $u_0 \in L^{\frac 1m + 1}$ when $F(u) = u^{\frac 1m}$... $\endgroup$
    – riem
    Jul 19, 2014 at 18:14
  • $\begingroup$ For the initial data I mean do not forget to give a sense to the initial data condition. Of course the argument 3) uses the linear character of the operator and cannot be used in a general case. But you take here the $-\Delta$ operator and you use strongly its properties in the weak formulation. Moreover for time dependent and nonlinear operator the $L^1$ contraction could be not true (or not known). So the approach can depend on the problem. At last using $u$ as a test function for an $L^1$ data problem is not allowed in general (it is possible in the approximate process but gives nothing). $\endgroup$
    – O.G.
    Jul 27, 2014 at 20:17

Your Answer

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

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