Radon measures as solutions of the Cauchy problem for evolution equations

Abstract

We consider a standard Navier–Stokes system on the n-D torus ${\mathbb {T}}^n={\mathbb {R}}^n/{\mathbb {Z}}^n$, valid when the flow is not very compressible and the temperature does not vary too much. We construct a sequence of approximate solutions that tend to satisfy the equations in a weak sense for arbitrary physical initial conditions. By weak compactness, we obtain Radon measure solutions in density and momentum when the velocity of the flow is finite, as numerically observed in all tests, and the absence of void regions in the flow. We notice that the method also applies without viscosity and complements results on other systems of evolution equations such as the systems of isothermal and isentropic gas flows considered in Colombeau (Z Angew Math Phys 66(5):2575–2599, 2015).

Appendix: Proof of Theorem 1

We first establish a priori inequalities to prove existence of a global solution to (14–18). For fixed $\epsilon >0$ and for some $\delta (\epsilon )>0$, we assume existence of a solution to the ODEs (14–18) on the interval $[0,\delta (\epsilon )[$

$$\begin{aligned} {[}0,\delta (\epsilon )[\longmapsto & {} (L^\infty ({\mathbb {T}}))^2\nonumber \\ t\longmapsto & {} [x\mapsto (\rho (x,t,\epsilon ),(\rho u)(x,t,\epsilon ))] \end{aligned}$$

(30)

continuously differentiable on $[0,\delta (\epsilon )[$ having the following properties a.e. $\forall \delta '<\delta (\epsilon )$

$$\begin{aligned}&\exists m=m(\epsilon )>0 \ / \ \rho (x,t,\epsilon )\ge m \ \forall x\in {\mathbb {T}} \ \forall t\in [0,\delta '], \end{aligned}$$

(31)

$$\begin{aligned}&\exists M=M(\epsilon )>0 \ / \ \Vert u(\cdot ,t,\epsilon )\Vert _\infty \le M, \ \ \Vert \rho (\cdot ,t,\epsilon )\Vert _\infty \le M \ \forall t\in [0,\delta ']. \end{aligned}$$

(32)

Note that m and M depend on $\epsilon$ and $\delta '$ and that, from the Lipschitz property of the ODEs (14–18) in the Banach space $L^\infty ({\mathbb {T}})^2$ or ${\mathcal {C}}({\mathbb {T}})^2$, for each $\epsilon >0$ there exists such a local solution, provided the initial conditions $\rho _{0,\epsilon }(x)=\rho (x,0,\epsilon ), u_{0,\epsilon }(x)=u(x,0,\epsilon )$ satisfy bounds (31, 32) for some values $m_0$, $M_0$ at time $t=0$, $m_0>m,M_0<M$. One will prove that the solution is global in positive time by obtaining uniform a priori estimates on this solution in the interval $[0,\delta (\epsilon )[$ open on the right. Note that $m=m(\epsilon )$ tends to 0 when $\epsilon \rightarrow 0$ in the case of the presence of a void region in the initial condition or in the solution; this does not cause trouble because the proof is done for fixed $\epsilon$. The quantities m and M are used in the reasoning to obtain a priori inequalities, and they disappear in the results (33–37).

Proof of Proposition 2

(a priori inequalities) For fixed $\epsilon >0$ as soon as (31–33) hold for some $\delta >0$, one has

$$\begin{aligned} \forall t\in [0,\delta (\epsilon )[ \int \limits _0^1\rho (x,t,\epsilon )\mathrm{d}x = \int \limits _0^1\rho _{0,\epsilon }(x)\mathrm{d}x+\epsilon ^\beta t, \end{aligned}$$

(33)

and $\exists \ C>0$, depending only on $\Vert \rho _0\Vert _{L^1({\mathbb {T}})}$ and $\delta$, not on m, M and $\epsilon >0$ small, such that

$$\begin{aligned}&\forall t\in [0,\delta (\epsilon )[ \Vert \partial _x\Phi (\cdot ,t,\epsilon )\Vert _{L^\infty ({\mathbb {T}})} \le \frac{C}{\epsilon ^{3\alpha }}, \end{aligned}$$

(34)

$$\begin{aligned}&\forall t\in [0,\delta (\epsilon )[ \Vert u(\cdot ,t,\epsilon )\Vert _{L^\infty ({\mathbb {T}})} \le \Vert u_{0,\epsilon }\Vert _{L^\infty ({\mathbb {T}})} +\frac{C}{\epsilon ^{3\alpha }} t. \end{aligned}$$

(35)

Set

$$\begin{aligned} k(\epsilon )=\Vert u_{0,\epsilon }\Vert _{L^\infty ({\mathbb {T}})}+ \frac{C \ \delta (\epsilon )}{\epsilon ^{3\alpha }}. \end{aligned}$$

(36)

Then,

$$\begin{aligned} \forall t\in [0,\delta (\epsilon )[, \forall x\in {\mathbb {R}}, \ \rho _{0,\epsilon }(x) \mathrm{exp}\left( \frac{-k(\epsilon )t}{\epsilon }\right) \le \rho (x,t,\epsilon ) \le 2\Vert \rho _{0,\epsilon }\Vert _{L^\infty ({\mathbb {T}})} \mathrm{exp}\left( \frac{2k(\epsilon )t}{\epsilon }\right) . \end{aligned}$$

(37)

Proof of Proposition 2

From (14–18), formulas (33) and (34) are proved in [10] where it is noticed that the constant C in (34) depends only on $\Vert \rho _0\Vert _{L^1({\mathbb {T}})}$ and $\delta$ for fixed N. Due to the presence of viscosity, the proof in [10] has to be modified to prove (35). This is done as follows.

One introduces artificially $\frac{\mu }{\epsilon }$ in formula (14) by stating it in the following form since the terms $\frac{\mu }{\epsilon }$ simplify

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\rho (x,t,\epsilon )=\frac{1}{\epsilon }\left[ \left\{ (\rho {\tilde{u}}^+)(x-\epsilon ,t,\epsilon )\right. \right. \nonumber \\&\left. \left. \quad +\,\frac{\mu }{\epsilon }\right\} -\left\{ (\rho ({\tilde{u}}^++{\tilde{u}}^-))(x,t,\epsilon )+2\frac{\mu }{\epsilon }\right\} +\left\{ (\rho {\tilde{u}}^-)(x+\epsilon ,t,\epsilon )+\frac{\mu }{\epsilon }\right\} \right] +\epsilon ^\beta . \end{aligned}$$

(38)

Taylor’s formula in time gives

$$\begin{aligned} \rho (x,t+\mathrm{d}t,\epsilon )= & {} \rho (x,t,\epsilon )+ \frac{\mathrm{d}t}{\epsilon }\left[ \left\{ (\rho {\tilde{u}}^+)(x-\epsilon ,t,\epsilon )+\frac{\mu }{\epsilon }\right\} \right. \nonumber \\&\left. \quad -\,\left\{ (\rho ({\tilde{u}}^++{\tilde{u}}^-) )(x,t,\epsilon )+2\frac{\mu }{\epsilon }\right\} \right. \nonumber \\&\left. \quad +\left\{ (\rho {\tilde{u}}^-)(x+\epsilon ,t,\epsilon )+\frac{\mu }{\epsilon }\right\} \right] +\mathrm{d}t r(x,t,\epsilon ,\mathrm{d}t)+\epsilon ^\beta \mathrm{d}t\nonumber \\= & {} \rho (x,t,\epsilon )\left( 1-\frac{\mathrm{d}t}{\epsilon }\left\{ ({\tilde{u}}^++{\tilde{u}}^-)(x,t,\epsilon )+2\frac{\mu }{\epsilon \rho (x,t,\epsilon )}\right\} \right) \nonumber \\&\quad +\,\frac{\mathrm{d}t}{\epsilon }\left\{ (\rho {\tilde{u}}^+)(x-\epsilon ,t,\epsilon )+\frac{\mu }{\epsilon }\right\} \nonumber \\&\quad +\,\frac{\mathrm{d}t}{\epsilon }\left\{ (\rho {\tilde{u}}^-)(x+\epsilon ,t,\epsilon )+\frac{\mu }{\epsilon }\right\} +\mathrm{d}t \ r(x,t,\epsilon ,\mathrm{d}t)+\epsilon ^\beta \mathrm{d}t \end{aligned}$$

(39)

where the remainder r is such that $\Vert r\Vert _{L^\infty ({\mathbb {T}})}\rightarrow 0$ when $\mathrm{d}t\rightarrow 0$ for fixed $\epsilon$. For fixed $\epsilon$, for $\mathrm{d}t>0$ small enough depending on $\epsilon$, the single term

$$\begin{aligned} \rho (x,t,\epsilon )\left( 1-\frac{\mathrm{d}t}{\epsilon }\left\{ ({\tilde{u}}^++{\tilde{u}}^-)(x,t,\epsilon )+2\frac{\mu }{\epsilon \rho (x,t,\epsilon )}\right\} \right) \end{aligned}$$

is positive and dominates the term $\mathrm{d}t \ r(x,t,\epsilon )(\mathrm{d}t)$ uniformly in $t\in [0,\delta '] \forall \delta '<\delta$; note that the other terms in right-hand side of (39) are positive since $\rho {\tilde{u}}^\pm \ge 0$. Therefore, one can invert (39)

$$\begin{aligned}&\frac{1}{\rho (x,t+\mathrm{d}t,\epsilon )}= \left[ \frac{\mathrm{d}t}{\epsilon }\left\{ (\rho {\tilde{u}}^+)(x-\epsilon ,t,\epsilon )\right. \right. \\&\left. \left. \quad +\,\frac{\mu }{\epsilon }\right\} +\rho (x,t,\epsilon )\left( 1-\frac{\mathrm{d}t}{\epsilon }\left\{ ({\tilde{u}}^++{\tilde{u}}^-)(x,t,\epsilon )+2\frac{\mu }{\epsilon \rho (x,t,\epsilon )}\right\} \right) \right. \\&\left. \quad +\,\frac{\mathrm{d}t}{\epsilon }\left\{ (\rho {\tilde{u}}^-)(x+\epsilon ,t,\epsilon )+\frac{\mu }{\epsilon }\right\} +\epsilon ^\beta \mathrm{d}t\right] ^{-1}+\mathrm{d}t \ r(x,t,\epsilon ,\mathrm{d}t) \end{aligned}$$

where the new remainder r has still the property that $\Vert r(\cdot ,t,\epsilon ,\mathrm{d}t)\Vert _{L^\infty ({\mathbb {T}})} \rightarrow 0$ when $\mathrm{d}t\rightarrow 0$ uniformly for $t\in [0,\delta ']$ if $\delta '<\delta$, but, as always in this proof, without any uniformness in $\epsilon$.

From (15), one obtains the following analog of (39) for the Euler equation, where now the terms involving $\frac{\mu }{\epsilon }$ originate from the last term in (15)

$$\begin{aligned} (\rho u)(x,t+\mathrm{d}t,\epsilon )= & {} \frac{\mathrm{d}t}{\epsilon }u(x-\epsilon ,t,\epsilon )\left\{ (\rho {\tilde{u}}^+)(x-\epsilon ,t,\epsilon )+\frac{\mu }{\epsilon }\right\} \nonumber \\&\quad +\,u(x,t,\epsilon )\left[ \rho (x,t,\epsilon )-\frac{\mathrm{d}t}{\epsilon }\left\{ (\rho ({\tilde{u}}^++{\tilde{u}}^-))(x,t,\epsilon )+2\frac{\mu }{\epsilon }\right\} \right] \nonumber \\&\quad +\,\frac{\mathrm{d}t}{\epsilon }u(x+\epsilon ,t,\epsilon )\left\{ (\rho {\tilde{u}}^-)(x+\epsilon ,t,\epsilon )+\frac{\mu }{\epsilon }\right\} \nonumber \\&\quad -\,\mathrm{d}t\rho (x,t,\epsilon )\partial _x\Phi (x,t,\epsilon )+\mathrm{d}t \ r(x,t,\epsilon ,\mathrm{d}t) \end{aligned}$$

(40)

where r is another remainder such that $\Vert r(\cdot ,t,\epsilon ,\mathrm{d}t)\Vert _{L^\infty ({\mathbb {T}})}\rightarrow 0$ when $\mathrm{d}t\rightarrow 0$.

One obtains

$$\begin{aligned} u(x,t+\mathrm{d}t,\epsilon )= & {} \frac{(\rho u)(x,t+\mathrm{d}t,\epsilon )}{\rho (x,t+\mathrm{d}t,\epsilon )}\nonumber \\= & {} \frac{u(x-\epsilon )\frac{\mathrm{d}t}{\epsilon }\left\{ (\rho {\tilde{u}}^+)(x-\epsilon )+\frac{\mu }{\epsilon }\right\} +u(x)[\rho (x)-\frac{\mathrm{d}t}{\epsilon }\left\{ \rho ({\tilde{u}}^++{\tilde{u}}^-)(x)+2\frac{\mu }{\epsilon }\right\} ]+u(x+\epsilon )\frac{\mathrm{d}t}{\epsilon }\left\{ (\rho {\tilde{u}}^-)(x+\epsilon )+\frac{\mu }{\epsilon }\right\} }{\nonumber }\\&\quad \frac{\mathrm{d}t}{\epsilon }\left\{ (\rho {\tilde{u}}^+)(x-\epsilon )+\frac{\mu }{\epsilon }\right\} +\left[ \rho (x)-\frac{\mathrm{d}t}{\epsilon }\left\{ \rho ({\tilde{u}}^++{\tilde{u}}^-)(x)+2\frac{\mu }{\epsilon }\right\} \right] \nonumber \\&\quad +\,\frac{\mathrm{d}t}{\epsilon }\left\{ (\rho {\tilde{u}}^-)(x+\epsilon )+\frac{\mu }{\epsilon }\right\} +\epsilon ^\beta \mathrm{d}t\nonumber \\&\quad -\,\mathrm{d}t\frac{\rho (x,t,\epsilon )}{\rho (x,t+\mathrm{d}t,\epsilon )}\partial _x\Phi (x,t,\epsilon )+ \mathrm{d}t \ r(x,t,\epsilon ,\mathrm{d}t) \end{aligned}$$

(41)

where the new r has the same property as in (39) for fixed $\epsilon$ and where in the large fraction we have omitted to note t and $\epsilon$ inside $\rho ,u$ and ${\tilde{u}}^\pm$.

For $\mathrm{d}t>0$ small enough, the large fraction is a barycentric combination with positive coefficients of $u(x-\epsilon ,t,\epsilon ), u(x,t,\epsilon )$ and $u(x+\epsilon ,t,\epsilon )$. For fixed $\epsilon$$\Vert \frac{\rho (x,t+\mathrm{d}t,\epsilon )}{\rho (x,t,\epsilon )}\Vert _{L^\infty ({\mathbb {T}})}$ tends to 1 when $\mathrm{d}t\rightarrow 0$ uniformly for $t\in [0,\delta ']$ using (31, 32).

It follows from (34) and (41)

$$\begin{aligned} \Vert u(\cdot ,t+\mathrm{d}t,\epsilon )\Vert _{L^\infty ({\mathbb {T}})}\le \Vert u(\cdot ,t,\epsilon )\Vert _{L^\infty ({\mathbb {T}})}+\mathrm{d}t\frac{\mathrm{const}}{\epsilon ^{3\alpha }}+\mathrm{d}t \Vert r(\cdot ,t,\epsilon ,\mathrm{d}t)\Vert _{L^\infty ({\mathbb {T}})} \end{aligned}$$

(42)

where, when $\mathrm{d}t \rightarrow 0$, $\mathrm{const}\rightarrow C$ uniformly if $t\in [0,\delta ']$ and where, for fixed $\epsilon$, $\Vert r\Vert _{L^\infty ({\mathbb {T}})}$ converges to 0 when $\mathrm{d}t\rightarrow 0$ uniformly if $t\in [0,\delta ']$. One divides the interval [0, t] into n intervals $[\frac{it}{n},\frac{(i+1)t}{n}], \ 0\le i \le n-1$.

Application of (42) in each subinterval gives

$$\begin{aligned} \Vert u\left( \cdot ,(i+1)\frac{t}{n},\epsilon \right) \Vert _{L^\infty ({\mathbb {T}})} \le \Vert u\left( \cdot ,i\frac{t}{n},\epsilon \right) \Vert _{L^\infty ({\mathbb {T}})}+\frac{t}{n}\frac{\mathrm{const}}{\epsilon ^{3\alpha }}+\frac{t}{n}{\overline{r}}\left( \frac{t}{n}\right) , \end{aligned}$$

where $\mathrm{const}\rightarrow C$ uniformly if $t\in [0,\delta ']$ when $n\rightarrow \infty$ and ${\overline{r}}$ is a function of the increment $\mathrm{d}t$, depending on $\epsilon$, such that ${\overline{r}}(\mathrm{d}t)= \mathrm{sup}_{x,t}|r(x,t,\epsilon ,\mathrm{d}t)|$.

Then, one sums on i and uses that ${\overline{r}}(\frac{t}{n})\rightarrow 0$ when $n\rightarrow \infty$. One notices that when $n\rightarrow +\,\infty$, then $\mathrm{const}$ can be chosen so as to tend to C and that ${\overline{r}}$ disappears. Finally, one obtains (35) in which the auxiliary value $\delta '$ has disappeared.

The ODEs (14, 15) for fixed $\epsilon \ (X=\rho _\epsilon ,Y=\rho _\epsilon u_\epsilon )$ have the standard form

$$\begin{aligned} X'(t)= & {} F(X(t),Y(t)), \ Y'(t)=G(X(t),Y(t)),\nonumber \\ F(X,Y)= & {} x\longmapsto \frac{1}{\epsilon }\left[ X\left( \tilde{\frac{Y}{X}}\right) ^+(x-\epsilon )-X\left( \tilde{\frac{Y}{X}}\right) ^+(x)-X\left( \tilde{\frac{Y}{X}}\right) ^-(x)\right. \nonumber \\&\left. \quad +\,X\left( \tilde{\frac{Y}{X}}\right) ^-(x+\epsilon )\right] +\epsilon ^\beta ,\nonumber \\ G(X,Y)= & {} x\longmapsto \frac{1}{\epsilon }\left[ Y\cdot \left( \tilde{\frac{Y}{X}}\right) ^+(x-\epsilon )-Y\cdot \left( \tilde{\frac{Y}{X}}\right) ^+(x)-Y\left( \tilde{\frac{Y}{X}}\right) ^-(x)+Y\left( \tilde{\frac{Y}{X}}\right) ^-(x+\epsilon )\right] \nonumber \\&\quad -\,K X(x) \left[ \mathrm{ln}(X+\epsilon ^N)*(\phi _{\epsilon ^\alpha })'\right] (x)+\frac{\mu }{\epsilon ^2}\left[ \frac{Y}{X}(x-\epsilon )-2\frac{Y}{X}(x)+\frac{Y}{X}(x+\epsilon )\right] . \end{aligned}$$

(43)

The existence of a unique global solution to (14–18) for fixed $\epsilon$ is obtained from the a priori $L^\infty$ estimates in the variables $\rho ,\rho u$ (35, 37) valid on $[0,\delta [$ open on the right in Proposition 1 since F and G are uniformly Lipschitz continuous in the sets defined by (31, 32).

It remains to prove that the solution of system (14, 18) provides the limits (4–6) for system (9–11) when $\epsilon \rightarrow 0$. The proof is identical to the one in [10] since the viscous term in (15) converges weakly to $\mu \partial _{xx} u$ from the bound (35) in u if $3\alpha <2$. Indeed,

$$\begin{aligned}&\int \left\{ u(x,t,\epsilon )\psi ''(x)-\frac{1}{\epsilon ^2}[u(x-\epsilon )-2u(x)+u(x+\epsilon )]\psi (x)\right\} \mathrm{d}x=\int u(x,t,\epsilon )\left[ \psi ''(x)\right. \\&\left. \quad - \frac{\psi (x+\epsilon )-2\psi (x)+\psi (x+\epsilon )}{\epsilon ^2}\right] \mathrm{d}x=O(\epsilon ^{2-3\alpha }). \end{aligned}$$

$\square$

In n-D, we denote the velocity vector by $(u_1,\dots ,u_n)$. If $x=(x_1,\dots ,x_n)$, we set ${\hat{x}}_i =(x_1,\dots ,x_n)$ with $x_i$ omitted. Then, the n-D ODEs are stated as

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\rho (x,t,\epsilon )\nonumber \\&\quad =\sum _{i=1}^n\frac{1}{\epsilon }\left[ (\rho {\tilde{u}}_i^+)({\hat{x}}_i,x_i-\epsilon ,t,\epsilon )-(\rho ({\tilde{u}}_i^++{\tilde{u}}_i^-))(x,t,\epsilon )+(\rho {\tilde{u}}_i^-)({\hat{x}}_i,x_i+\epsilon ,t,\epsilon )\right] + \epsilon ^\beta , \end{aligned}$$

(44)

$$\begin{aligned}&\hbox { for } j=1,\dots ,n \ \ \frac{\mathrm{d}}{\mathrm{d}t}(\rho u_j)(x,t,\epsilon )=\sum _{i=1}^n \frac{1}{\epsilon }\left[ (\rho u_j {\tilde{u}}_i^+)({\hat{x}}_i,x_i-\epsilon ,t,\epsilon )\right. \nonumber \\&\left. \quad -\,(\rho u_j ({\tilde{u}}_i^++{\tilde{u}}_i^-))(x,t,\epsilon )+(\rho u_j {\tilde{u}}_i^-)({\hat{x}}_i,x_i+\epsilon ,t,\epsilon )\right] - \rho (x,t,\epsilon )\partial _{x_j}\Phi (x,t,\epsilon )\nonumber \\&\quad +\,\frac{\mu }{\epsilon ^2}\sum _{i=1}^n\left[ u_j({\hat{x}}_i,x_i-\epsilon ,t,\epsilon ) -2u_j(x,t,\epsilon )+u_j({\hat{x}}_i,x_i+\epsilon ,t,\epsilon )\right] . \end{aligned}$$

(45)

The n-D proof extends exactly the 1-D proof. Besides the n-D notation, the difference with the 1-D case lies in the formula $\phi _\epsilon (x)=\frac{1}{\epsilon ^n}\phi (\frac{x}{\epsilon })$ for convolution in n-D. This implies some minor changes within the proof limited to this occurrence of power n in bounds such as in Proposition 1.