Fractional porous media equations: existence and uniqueness of weak solutions with measure data

Abstract

We prove existence and uniqueness of solutions to a class of porous media equations driven by the fractional Laplacian when the initial data are positive finite Radon measures on the Euclidean space ${\mathbb R}^d$. For given solutions without a prescribed initial condition, the problem of existence and uniqueness of the initial trace is also addressed. By the same methods we can also treat weighted fractional porous media equations, with a weight that can be singular at the origin, and must have a sufficiently slow decay at infinity (power-like). In particular, we show that the Barenblatt-type solutions exist and are unique. Such a result has a crucial role in Grillo et al. (Discret Contin Dyn Syst 35:5927–5962, 2015), where the asymptotic behavior of solutions is investigated. Our uniqueness result solves a problem left open, even in the non-weighted case, in Vázquez (J Eur Math Soc 16:769–803, 2014).

Appendices

Appendix A

We recall here some basic properties of the fractional Laplacian (and of a similar nonlocal, nonlinear operator) of functions in $\mathcal {D}(\mathbb {R}^d)$. We omit the proofs of the first two lemmas, since they follow by exploiting the same strategy of [5, Lemma 2.1].

Lemma 6.1

The s-Laplacian $(-\Delta )^s(\phi )(x)$ of any $\phi \in \mathcal {D}(\mathbb {R}^d)$ is a regular function which decays (together with its derivatives) at least like $|x|^{-d-2s}$ as $|x| \rightarrow \infty$.

Lemma 6.2

For any $\phi \in \mathcal {D}(\mathbb {R}^d)$, the function

$$\begin{aligned} l_s(\phi )(x) := \int _{\mathbb {R}^d} \frac{(\phi (x)-\phi (y))^2}{|x-y|^{d+2s}} \, \mathrm {d}y \quad \forall x \in \mathbb {R}^d \end{aligned}$$

is regular and decays (together with its derivatives) at least like $|x|^{-d-2s}$ as $|x| \rightarrow \infty$.

Lemma 6.3

For any $R>0$, let $\xi _R$ be the cut-off function

$$\begin{aligned} \xi _R(x):=\xi \left( \frac{x}{R}\right) \quad \forall x \in \mathbb {R}^d , \end{aligned}$$

where $\xi (x)$ is a positive, regular function such that $\Vert \xi \Vert _\infty \le 1$, $\xi \equiv 1$ in $B_1$ and $\xi \equiv 0$ in $B_2^c$. Then, $(-\Delta )^s(\xi _R)$ and $l_s(\xi _R)$ enjoy the following property:

$$\begin{aligned} (-\Delta )^s(\xi _R)(x)=\frac{1}{R^{2s}}(-\Delta )^s(\xi )\left( \frac{x}{R} \right) , \quad l_s(\xi _R)(x)=\frac{1}{R^{2s}}l_s(\xi )\left( \frac{x}{R} \right) \quad \forall x \in \mathbb {R}^d . \end{aligned}$$

Proof

We only prove the result for $l_s(\xi _R)$, since the proof for $(-\Delta )^s(\xi _R)$ is identical. Letting $\widetilde{y}=y/R$, one has:

$$\begin{aligned} l_s(\xi _R)(x)= \int _{\mathbb {R}^d} \frac{(\xi _R(x)-\xi _R(y))^2}{|x-y|^{d+2s}} \, \mathrm {d}y = \frac{1}{R^{2s}} \int _{\mathbb {R}^d} \frac{(\xi ( {x}/{R} )-\xi (\widetilde{y}) )^2}{|x/R-\widetilde{y}|^{d+2s}} \, \mathrm {d}\widetilde{y} = \frac{1}{R^{2s}}l_s(\xi )\left( \frac{x}{R} \right) . \end{aligned}$$

$\square$

The next lemmas contain technical ingredients concerning fractional Sobolev spaces and Riesz potentials, which we need in the proofs of our existence and uniqueness results.

Lemma 6.4

Let $d>2s$ and assume that $\rho$ satisfies (1.2) for some $\gamma \in (0,d+2s]$. Consider a function $v \in L^2_\mathrm{loc}((0,\infty );\dot{H}^s(\mathbb {R}^d))$ such that, for all $t_2>t_1>0$,

$$\begin{aligned} \int _{t_1}^{t_2} \int _{\mathbb {R}^d} \left| v(x,t) \right| ^2 \rho (x) \mathrm {d}x \mathrm {d}t&\le C_0 ,\end{aligned}$$

(6.1)

$$\begin{aligned} \int _{t_1}^{t_2} \int _{\mathbb {R}^d} | (-\Delta )^{\frac{s}{2}} \left( v \right) (x,t) |^2 \mathrm {d}x \mathrm {d}t&\le C_0 \end{aligned}$$

(6.2)

and

$$\begin{aligned} \int _{t_1}^{t_2} \int _{\mathbb {R}^d} \left| v_t(x,t) \right| ^2 \rho (x) \mathrm {d}x \mathrm {d}t \le C_0 , \end{aligned}$$

(6.3)

where $C_0$ is a positive constant depending only on $t_1$ and $t_2$. Take any cut-off functions $\xi _1 \in C^\infty _c(\mathbb {R}^d)$, $\xi _2 \in C^\infty _c((0,\infty ))$ and define $v_c: \mathbb {R}^d \rightarrow \mathbb {R}$ as follows:

$$\begin{aligned} v_c(x,t):= \xi _1(x) \xi _2(t) v(x,t) \quad \forall (x,t) \in \mathbb {R}^d \times \mathbb {R} , \end{aligned}$$

where we implicitly assume $\xi _2$ and v to be zero for $t<0$. Then

$$\begin{aligned} \Vert v_c \Vert _{{H^s} (\mathbb {R}^{d+1})}^2 = \Vert v_c \Vert _{L^2(\mathbb {R}^{d+1})}^2 + \Vert v_c \Vert _{\dot{H}^s(\mathbb {R}^{d+1})}^2 \le C^\prime \end{aligned}$$

(6.4)

for a positive constant $C^\prime$ that depends only on $\rho$, $\xi _1$ and $\xi _2$ (also through $C_0$).

Proof

The validity of

$$\begin{aligned} \Vert v_c \Vert _{L^2(\mathbb {R}^{d+1})}^2 \le C^\prime \end{aligned}$$

(6.5)

is an immediate consequence of (6.1) and of the fact that $\rho$ is bounded away from zero on compact sets (from now on $C^\prime$ will be a constant as in the statement that may change from line to line). Moreover, since $(v_c)_t = \xi _1 \xi _2^\prime v + \xi _1 \xi _2 v_t$, by (6.1), (6.3) and again the fact that $\rho$ is bounded away from zero on compact sets we deduce that

$$\begin{aligned} \left\| (v_c)_t \right\| _{L^2(\mathbb {R}^{d+1})}^2 \le C^\prime . \end{aligned}$$

(6.6)

Now we have to handle the spatial regularity of $v_c$. Straightforward computations show that

$$\begin{aligned} \left\| v_c(t) \right\| _{\dot{H}^s(\mathbb {R}^d)}^2&= \frac{C_{d,s}}{2} \, \xi _2^2(t) \int _{\mathbb {R}^d} \xi _1^2(x) \left( \int _{\mathbb {R}^d} \frac{\left( v(x,t) - v(y,t) \right) ^2}{|x-y|^{d+2s}} \, \mathrm {d}y \right) \mathrm {d}x \nonumber \\&\quad + \frac{C_{d,s}}{2} \, \xi _2^2(t) \int _{\mathbb {R}^d} \left| v(y,t)\right| ^2 \left( \int _{\mathbb {R}^d} \frac{\left( \xi _1(x) - \xi _1(y) \right) ^2}{|x-y|^{d+2s}} \, \mathrm {d}x \right) \mathrm {d}y \nonumber \\&\quad + C_{d,s} \, \xi _2^2(t) \int _{\mathbb {R}^d} \int _{\mathbb {R}^d} \xi _1(x) v(y,t) \frac{\left( v(x,t) - v(y,t) \right) \left( \xi _1(x) - \xi _1(y) \right) }{|x-y|^{d+2s}} \, \mathrm {d}x \mathrm {d}y . \end{aligned}$$

(6.7)

The Cauchy–Schwarz inequality allows us to bound the third integral on the r.h.s. of (6.7) by the first two integrals. As concerns the first one, we have:

$$\begin{aligned}&\frac{C_{d,s}}{2} \, \xi _2^2(t) \int _{\mathbb {R}^d} \xi _1^2(x) \left( \int _{\mathbb {R}^d} \frac{\left( v(x,t) - v(y,t) \right) ^2}{|x-y|^{d+2s}} \, \mathrm {d}y \right) \mathrm {d}x\nonumber \\&\quad \le \chi _{{\text {supp}}{\xi _2} }(t) \left\| \xi _2 \right\| _\infty ^2 \left\| \xi _1 \right\| _\infty ^2 \left\| v(t) \right\| _{\dot{H}^s(\mathbb {R}^d)}^2 . \end{aligned}$$

(6.8)

In order to bound the second integral, it is important to recall that the function $l_s(\xi _1)(y)$ is regular and decays at least like $|y|^{-d-2s}$ as $|y| \rightarrow \infty$ (for the definition and properties of $l_s$ see Lemmas 6.2, 6.3). Hence, thanks to the assumptions on $\rho$ and $\gamma$, we infer that

$$\begin{aligned}&\xi _2^2(t) \int _{\mathbb {R}^d} \left| v(y,t)\right| ^2 \left( \int _{\mathbb {R}^d} \frac{\left( \xi _1(x) - \xi _1(y) \right) ^2}{|x-y|^{d+2s}} \, \mathrm {d}x \right) \mathrm {d}y\nonumber \\&\quad \le C^\prime \chi _{{\text {supp}}\xi _2 }(t) \left\| \xi _2 \right\| _\infty ^2 \int _{\mathbb {R}^d} \left| v(y,t) \right| ^2 \rho (y) \mathrm {d}y . \end{aligned}$$

(6.9)

Integrating in time (6.7), using (6.8), (6.9), (6.1), (6.2) and recalling the validity of the identity $\Vert (-\Delta )^{\frac{s}{2}} (v_c)(t) \Vert _{L^2(\mathbb {R}^d)}^2 = \left\| v_c(t) \right\| _{\dot{H}^s(\mathbb {R}^d)}^2$, we then get

$$\begin{aligned} \Vert (-\Delta )^{\frac{s}{2}} ( v_c ) \Vert _{L^2(\mathbb {R}^{d+1})}^2 \le C^\prime . \end{aligned}$$

(6.10)

By exploiting (6.5), (6.6) and (6.10) one deduces (6.4), e.g. by using Fourier transform methods. $\square$

Lemma 6.5

Let $d>2s$ and $\phi :\mathbb {R}^d \rightarrow \mathbb {R}$ be a continuous function which belongs to $L^1(\mathbb {R}^d)$ and decays at least like $|x|^{-d}$ as $|x| \rightarrow \infty$. Then, the convolution $I_{2s} *\phi$ (namely, the Riesz potential of $\phi$) is also a continuous function, decaying at least like $|x|^{-d+2s}$ as $|x| \rightarrow \infty$.

Proof

The idea of the proof is to split the convolution $(I_{2s} *\phi )(x)$ in the three regions $B^c_{2|x|}(0)$, $B_{{|x|}/{2}}(x)$, $B_{2|x|}(0) \setminus B_{|x|/2}(x)$ and use there the decay and integrability properties of $\phi$ and $I_{2s}$. We omit the details. $\square$

Lemma 6.6

Let $d > 2s$ and assume that $\rho$ satisfies (1.2) for some $\gamma \in (0,2s)$. Let $v \in L^1_{\rho }(\mathbb {R}^d) \cap L^\infty (\mathbb {R}^d)$ and $U^{v}_{\rho }$ be the Riesz potential of $\rho v$. Then $U^{v}_\rho$ belongs to $C(\mathbb {R}^d) \cap L^p(\mathbb {R}^d)$ for all p such that

$$\begin{aligned} p \in \left( \frac{d}{d-2s} , \infty \right] . \end{aligned}$$

(6.11)

Proof

In order to prove that $U^v_\rho$ belongs to $C(\mathbb {R}^d) \cap L^p(\mathbb {R}^d)$ for all p satisfying (6.11), we proceed as follows:

$$\begin{aligned} U^v_\rho (x) = \underbrace{ \int _{B_1(0)} \rho (y)\,v(y) \, I_{2s}(x-y) \, \mathrm {d}y}_{U^v_{\rho ,1}(x)} + \underbrace{ \int _{\mathbb {R}^d} \chi _{B_1^c(0)}(y) \, \rho (y)\,v(y) \, I_{2s}(x-y) \, \mathrm {d}y }_{U^v_{\rho ,2}(x)} . \end{aligned}$$

Exploiting the fact that $v \in L^\infty (\mathbb {R}^d)$ and $\gamma <2s$ (so that $|y|^{-d+2s}\,\rho (y)$ is locally integrable), it is easily seen that $U^v_{\rho ,1}(x)$ is a continuous function which decays at least like $|x|^{-d+2s}$ as $|x| \rightarrow \infty$. In particular, it belongs to $L^p(\mathbb {R}^d)$ for all p satisfying (6.11). As concerns $U^v_{\rho ,2}(x)$, notice that since $v \in L^1_{\rho }(\mathbb {R}^d) \cap L^\infty (\mathbb {R}^d)$ we have that the function $\chi _{B_1^c(0)} \rho v$ belongs to $L^1(\mathbb {R}^d) \cap L^\infty (\mathbb {R}^d)$. Hence $U^v_{\rho ,2}(x)$ is continuous too. To prove that it belongs to $L^p(\mathbb {R}^d)$ for all p satisfying (6.11), we write:

$$\begin{aligned} U^v_{\rho ,2} = ( \chi _{B_1(0)} \, I_{2s} ) *( \chi _{B_1^c(0)} \rho v ) + ( \chi _{B_1^c(0)} \, I_{2s} ) *(\chi _{B_1^c(0)}\rho v ) ; \end{aligned}$$

(6.12)

since $\chi _{B_1(0)} \, I_{2s} \in L^1(\mathbb {R}^d)$ and $\chi _{B_1^c(0)} \rho v \in L^1(\mathbb {R}^d) \cap L^\infty (\mathbb {R}^d)$, the first convolution in (6.12) belongs to $L^1(\mathbb {R}^d) \cap L^\infty (\mathbb {R}^d)$. Using the fact that $\chi _{B_1^c(0)} \, I_{2s} \in L^p(\mathbb {R}^d)$ for all p as in (6.11) and $\chi _{B_1^c(0)} \rho v \in L^1(\mathbb {R}^d)$, we infer that the second convolution in (6.12) belongs to $L^p(\mathbb {R}^d)$ for all such p . The latter property is then inherited by $U^v_{\rho ,2}$. $\square$

Appendix B

This section is devoted to give a sketch of the proofs of Theorem 3.7 and of the forthcoming Proposition 7.1.

Sketch of proof of Theorem 3.7

We start from the validity of the fractional “integration by parts” formula

$$\begin{aligned} \frac{C_{d,s}}{2} \int _{\mathbb {R}^d} \int _{\mathbb {R}^d} \frac{(\phi (x)-\phi (y))(\psi (x)-\psi (y))}{|x-y|^{d+2s}} \, \mathrm {d}x \mathrm {d}y = \int _{\mathbb {R}^d} \phi (x) (-\Delta )^s(\psi )(x) \, \mathrm {d}x \end{aligned}$$

(7.1)

for all $\phi ,\psi \in \mathcal {D}(\mathbb {R}^d)$, and our aim is to extend it to all functions of $X_{s,\rho }$. In order to do it, the first step consists in showing that $C^\infty (\mathbb {R}^d) \cap X_{s,\rho }$ is dense in $X_{s,\rho }$. This can be done by mollification arguments, which however are slightly more complicated than the standard ones, since we work with the weighted spaces $L^2_{\rho }(\mathbb {R}^d)$ and $L^2_{\rho ^{-1}}(\mathbb {R}^d)$ instead of $L^2(\mathbb {R}^d)$. Hence, given $v,w \in C^\infty (\mathbb {R}^d) \cap X_{s,\rho }$, one plugs the cut-off functions $\phi := \xi _R v$ and $\psi := \xi _R w$ into (7.1) and lets $R \rightarrow \infty$. The problem is that on the r.h.s. there appear terms involving $\Vert \xi _R w \Vert _{\dot{H}^s}$, and a priori we do not know whether $C^\infty (\mathbb {R}^d) \cap X_{s,\rho }$ is continuously embedded in $\dot{H}^s(\mathbb {R}^d)$. But this turns out to be true: the inequality

$$\begin{aligned} \frac{C_{d,s}}{2} \int _{\mathbb {R}^d} \int _{\mathbb {R}^d} \frac{(w(x)-w(y))^2}{|x-y|^{d+2s}} \, \mathrm {d}x \mathrm {d}y \le \int _{\mathbb {R}^d} w(x) (-\Delta )^s(w)(x) \, \mathrm {d}x \quad \forall w \in C^\infty (\mathbb {R}^d) \cap X_{s,\rho } \end{aligned}$$

(7.2)

can be proved just by repeating the above scheme with $\phi =\psi = \xi _R w$. In fact, on the r.h.s. of (7.1) we still have terms involving $\Vert \xi _R w \Vert _{\dot{H}^s}$, but the latter are small and can be absorbed into the l.h.s.; passing to the limit as $R \rightarrow \infty$ yields (7.2). Therefore, we can now let $R \rightarrow \infty$ safely in (7.1) (with $\phi = \xi _R v$ and $\psi =\xi _R w$) and obtain that

$$\begin{aligned} \frac{C_{d,s}}{2} \int _{\mathbb {R}^d} \int _{\mathbb {R}^d} \frac{(v(x)-v(y))(w(x)-w(y))}{|x-y|^{d+2s}} \, \mathrm {d}x \mathrm {d}y = \int _{\mathbb {R}^d} v(x) (-\Delta )^s(w)(x) \, \mathrm {d}x \end{aligned}$$

(7.3)

for all $v,w \in C^\infty (\mathbb {R}^d) \cap X_{s,\rho }$, which in particular shows that (7.2) is actually an equality. Notice that in all these approximation procedures using cut-off functions, to prove that “remainder” terms go to zero we deeply exploit the results provided by Lemmas 6.1–6.3. It is in fact here that the condition $\gamma < 2s$ plays a fundamental role: in particular, it ensures that both $\Vert \rho ^{-1} (-\Delta )^s(\xi _R) \Vert _\infty$ and $\Vert \rho ^{-1} l_s(\xi _R) \Vert _\infty$ vanish as $R \rightarrow \infty$. As already mentioned, we refer the reader to the note [31] for the details. However, for similar computations involving $(-\Delta )^s(\xi _R)$ and $l_s(\xi _R)$, see also the proofs of Proposition 4.1, Lemmas 4.3 and 5.5.

By the claimed density of $C^\infty (\mathbb {R}^d) \cap X_{s,\rho }$, we are allowed to extend (7.3) to the whole of $X_{s,\rho }$. Clearly, the r.h.s. of (7.3) can be rewritten as

$$\begin{aligned} \int _{\mathbb {R}^d} v(x) \, A(w)(x) \, \rho (x) \mathrm {d}x , \end{aligned}$$

and letting $v=w$ we obtain that the operator A is positive. The fact that it is densely defined is trivial since, for instance, $\mathcal {D}(\mathbb {R}^d) \subset X_{s,\rho }$. Because in (7.3) one can interchange the role of v and w, we also have that A is symmetric. In order to prove that it is self-adjoint we need to show that $D(A^*) \subset D(A)$, namely that any function of $D(A^*)$ also belongs to $X_{s,\rho }$. It is indeed straightforward to check this fact, and we leave it to the reader.

We finally deal with the quadratic form Q associated to A. Thanks to (7.3), we have that

$$\begin{aligned} Q(v,v)= \frac{C_{d,s}}{2} \int _{\mathbb {R}^d} \int _{\mathbb {R}^d} \frac{(v(x)-v(y))^2}{|x-y|^{d+2s}} \, \mathrm {d}x \mathrm {d}y \quad \forall v \in D(A) . \end{aligned}$$

(7.4)

As it is well known (see e.g. [13]), the domain D(Q) of Q is just the closure of D(A) w.r.t. the norm

$$\begin{aligned} \left\| v \right\| _{Q}^2 := \left\| v \right\| _{2,\rho ^{-1}}^2 + Q(v,v) = \left\| v \right\| _{2,\rho ^{-1}}^2 + \left\| v \right\| _{\dot{H}^s}^2 . \end{aligned}$$

It is then easy to see that such a closure is nothing but $L^2_{\rho }(\mathbb {R}^d) \cap \dot{H}^s(\mathbb {R}^d)$ and the quadratic form on $D(Q)=L^2_{\rho }(\mathbb {R}^d) \cap \dot{H}^s(\mathbb {R}^d)$ is still represented by (7.4).

By classical results (we refer again to [13]), proving that A generates a Markov semigroup is equivalent to proving that if v belongs to D(Q) then both $v \vee 0$ and $v \wedge 1$ belong to D(Q) and satisfy

$$\begin{aligned} Q(v \vee 0,v \vee 0) \le Q(v,v) , \quad Q(v \wedge 1,v \wedge 1) \le Q(v,v) . \end{aligned}$$

But the latter properties are straightforward consequences of the characterization of Q given above.

The last assertions follow from the general theory of symmetric Markov semigroups (cf. [13, Section 1.4]) and from their known analiticity properties (cf. [13, Theorem 1.4.2]). See also the discussion in the proof of Lemma 5.3. $\square$

The next proposition extends the symmetry property of the operator $A=\rho ^{-1}\,(-\Delta )^s$ to functions which belong to other suitable $L^p_{\rho }$ spaces. This is essential in proving our uniqueness Theorem 3.4 for certain values of $\gamma$ and s in low dimensions $d \le 3$, more precisely whenever $(d-\gamma )/(d-2s) > 2$.

Proposition 7.1

Let $d>2s$ and assume that $\rho$ satisfies (1.2) for some $\gamma \in [0,2s) \cap [0,d-2s]$ and $\gamma _0 \in [0,\gamma ]$. Let $p \in [2,2(d-\gamma )/(d-2s))$ and $p^\prime ={p}/(p-1)$ be its conjugate exponent. Suppose that $v,w \in L^p_{\rho }(\mathbb {R}^d)$ are such that $A(v),A(w) \in L^{p^\prime }_{\rho }(\mathbb {R}^d)$. Then $v,w \in \dot{H}^s(\mathbb {R}^d)$ and the following formula holds:

$$\begin{aligned} \int _{\mathbb {R}^d} v(x) (-\Delta )^s(w)(x) \, \mathrm {d}x&= \int _{\mathbb {R}^d} (-\Delta )^s(v)(x) \, w(x) \, \mathrm {d}x \\&= \frac{C_{d,s}}{2} \int _{\mathbb {R}^d} \int _{\mathbb {R}^d} \frac{(v(x)-v(y))(w(x)-w(y))}{|x-y|^{d+2s}} \, \mathrm {d}x \mathrm {d}y . \end{aligned}$$

Sketch of proof

The method of proof proceeds along the lines of the one of Theorem 3.7. The main difference here lies in the fact that, when using the approximation procedure by cut-off functions mentioned above, if p is strictly larger than 2 in order to prove that “remainder” terms go to zero one cannot exploit the fact that $\rho ^{-1} (-\Delta )^s(\xi _R)$ and $\rho ^{-1} l_s(\xi _R)$ vanish in $L^\infty (\mathbb {R}^d)$ as $R \rightarrow \infty$. In fact, such remainder terms are of the form

$$\begin{aligned} \int _{\mathbb {R}^d} v^2(x) (-\Delta )^s(\xi _R)(x) \, \mathrm {d}x \quad \text {or} \quad \int _{\mathbb {R}^d} v^2(x) \, l_s(\xi _R)(x) \, \mathrm {d}x . \end{aligned}$$

(7.5)

Thanks to Lemmas 6.1–6.3, it is direct to see that $\Vert \rho ^{-1} (-\Delta )^s(\xi _R) \Vert _{q,-\gamma }$ and $\Vert \rho ^{-1} l_s(\xi _R) \Vert _{q,-\gamma }$ vanish as $R\rightarrow \infty$ provided $q > (d-\gamma )/(2s-\gamma )$, whence the condition $p \in [2,2(d-\gamma )/(d-2s))$ to ensure that also the integrals in (7.5) go to zero as $R\rightarrow \infty$. $\square$