# Getting existence for $L^1$ data given existence for $L^\infty$ data and $L^1$ continuous dependence result

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.

• 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... – leo monsaingeon Jun 27 '14 at 7:31

Hy

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

• 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.. – riem Jul 19 '14 at 18:12
• ...but I would need data $u_0 \in L^{\frac 1m + 1}$ when $F(u) = u^{\frac 1m}$... – riem Jul 19 '14 at 18:14
• 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). – O.G. Jul 27 '14 at 20:17