数理解析研究所講究録
1495 巻 2006 年 1-7

1Weak Solutions to theEquations-Poisson$\mathrm{N}\mathrm{a}\mathrm{v}\mathrm{i}\mathrm{e}\mathrm{r}-\mathrm{S}\mathrm{t}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{s}$Dedicated to Professor Tai-Ping Liu on his 60th birthday
佐賀大理工 小林 孝行
Takayuki KOBAYASHI
Departmcnt of Mathematios

Faculty of Science and Engineering, Saga University
大阪大基礎工鈴木貴

Takashi SUZUKI
Division of Mathematical ScienceGraduate School of Engineering Science, Osaka University

We study the global existence of weak solutions to the Navier-StokesPoisson equation$\rho_{t}+\nabla$. (pu) $=0$$(\rho u)_{t}+\nabla\cdot(pu\otimes u)+\rho\nabla\Phi+a\nabla\rho^{\gamma}=\mu\Delta u+(\lambda+\mu)\nabla(\nabla\cdot u)$$\Delta\Phi=4\pi g(\rho-\frac{1}{|\Omega|}\int_{\mathrm{f}l}\rho)$in(1)$\Omega \mathrm{x}(0, T)$with the initial-boundary condition$u=0$ ,$\frac{\partial\Phi}{\partial\nu}=0$$\rho|_{t=0}=\rho_{0}(x)$,on$\partial\Omega \mathrm{x}(0, T)$$(\rho u)|_{t=0}=q_{0}(x)$in$\Omega$,(2)

where Stis a bounded domain withboundary ast $(0<\theta<1),$
the outer normal vector, $\rho=\rho(x, t)$ the density,$C^{2,\theta}$$\subset \mathrm{R}^{3}$

$\nu$

’

$u=u(x, t)=(u^{1}(x, t),$ $u^{2}(x, t),$ $u”(x, t))$the velocity, $\Phi=\Phi(x, t)$ the Newtonian gravitational potential, $\gamma>1$ the,adiabatic constant, $\mu>0$ and the viscosity constants satisfying$a=e^{S}$ the constant determined by the entropy, and $g>0$ the gravitational constant. Physically, this system describes the motion of compressible$\lambda+\frac{2}{3}\mu\geq 0$$\lambda$$S$

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viscous isentropic gas flow under the self-gravitational force. Such a fluid
may be formulated as the Euler-Poisson equation, and several mathematicalstudies have been done for this Euler-Poisson equation [16, 17, 2, 10], where,
the viscosity is neglected, the equation is considered in the whole space
and the solution admits to have the compact support. Then, the contact
angle between fluid and vaccumc is not zcro in the equilibrium state, andestablishing the existence of the solution in an appropriate function space
including the equilibrium state is not easy because of this, even locally intime with spherically symmetry in space.
In a series of papers [20, 18, 19], T. Makino and his co–workers studied
such a problem of vaccume for spherically symmetric Navier-Stokes-Poisson
equation (of a specified pair of the viscosity constants) with solid core. Here,
we study the Navier-Stokes-Poisson equation on the fixed domain without
radial symmetry or solid core, and show the existence of the weak solution$\mathrm{R}^{3}$$\Omega$,although the equilibrium state is everywhere positive in this problem.
Similarly to the above mentioned equations, equation (1) is provided withand the decrease
the properties of the conservation of total massof total energy ;$M= \int_{\Omega}\rho$$E$$E= \int_{\Omega}(\frac{\rho}{2}|u|^{2}+\frac{P}{\gamma-1})+\frac{g}{2}\int\int_{\Omega}$

xrt$G(x, y)\rho(x)\rho(y)dxdy$$= \frac{a}{\gamma-1}||\rho||_{\gamma}^{\gamma}+\frac{1}{2}||\sqrt{\rho}?\mathit{4}||_{2}^{2}-\frac{1}{8\pi g}||\nabla\Phi||_{2}^{2}$,

and $G=G(x, y)$ denote the pressure and the Green’sand here,
fimction of the Poisson part, respectively, so that $\Phi(x)=g\int_{\Omega}G(x, y)\rho(y)dy$
if and only if$P=a\rho^{\gamma}$$\triangle\Phi=4\pi g(\rho-\frac{1}{|\Omega|}\int_{\Omega}\rho)$

In this Poisson equation,. More precisely,

in$\Omega$,$\rho\in L^{\gamma}(\Omega)$in$\frac{\partial\Phi}{\partial\nu}=0$

implies$\Phi_{x}$.

$\partial\Omega$,$\int_{\Omega}\Phi=0$. (3)$\in L^{2}(\Omega)(i=1,2,3)$

for$\gamma\geq\frac{6}{5}$$||\nabla\Phi||_{2}\leq gK||\rho||_{\S}$(4)elliptic estimate and Sobolev’s inequality, where $K$ is a constant deterby
mined by . In accordance with the energy $E$ given above, therefore,$L^{\frac{6}{6}}$$\Omega$$\gamma>\frac{6}{5}$3andare the subcritical and critical exponents of the equilibrium, respectively. In more detail, the equilibrium state is realized by $u=0$ , andhence it holds that (3) and$\gamma=\frac{6}{5}$$\Phi+\frac{a\gamma}{\gamma-1}\rho^{\gamma-1}=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$in$\Omega$.

Then, this problem has a variational solution in the case of, while itdoes not admit a solution ifis star-shaped [21].
andOur result on the non-equilibrium state, on the other hand, is regarded
as the generalization of Feireisl, Nevotn\’y, and Petzeltov\’a [12] concerning theNavier-Stokes equation without the Poisson term; more precisely,$\gamma>\frac{6}{5}$$\Omega\subset \mathrm{R}^{3}$$1< \gamma<\frac{6}{5}$Theorem 1 Let $T>0$ and. Then, given
andwithand $q_{0}^{i}(x)=0$ for of
we have a finite
energy weak solution
to (1) satisfying the following.
$\rho_{0}=\rho_{0}(x)\geq 0$$\rho,$

1.

$u,$$|q_{0}^{i}|^{2}/\rho_{0}\in$$\rho_{0}\in L^{\gamma}(\Omega)$$\gamma>\frac{3}{2}$$L^{1}(\Omega)$$x$$\rho_{0}(x)=0_{f}$$\Phi$$u^{i}\in L^{2}(0, T;H_{0}^{1}(\Omega))$$\rho=\rho(x, t)\geq 0_{f}\rho\in L^{\infty}(0, T;L^{\gamma}(\Omega)),$.

2. $E=E(t)\in L_{loc}^{1}(0, T)$ .

3.

in$\frac{dE}{dt}+\mu||\nabla u||_{2}^{2}+(\lambda+\mu)||\nabla\cdot u||_{2}^{2}\leq 0$

4. The first two equations of (1) hold in
5.

$\Phi(\cdot, t)=g\int_{\Omega}G(\cdot, y)\rho(y, t)dy$for a. .

is taken outside$\Omega$to$\rho,$$u$.

$\mathcal{D}’(\Omega \mathrm{x}(0, T))$$t\in(\mathrm{O}, T)$$e$

6. The first equation of (1) holds in 7)’$D’(\mathrm{O}, T)$.

$(\mathrm{R}^{3}\cross(0, T))$.

.

if the zero extension

7. The first equation of (1) is satisfied in the sense of the renormalized
solution, $i.e.$ ,$\frac{d}{df,}b(\rho)+\nabla\cdot(b(\rho)u)+(b’(\rho)\rho-b(\rho))\nabla\cdot u=0$

in $D’(\Omega\cross((), T))$ for any$b\in C^{1}(\mathrm{R})$(5)such that $b’(z)=0if|z|$ is large.

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Similarly to the Navier-Stokes equation without the Poisson term, any
finite energy weak solution satisfies$\rho\in C([0, T];L_{weak}^{\gamma}(\Omega))$(6)$\rho u^{i}\in C([0, T];L^{\frac{2\prime}{w\mathrm{e}\gamma+1}}’ ak(\Omega))$by the first two equations of (1) and consequently, the initial conditions makesense in the above weak solution.
by Sobolev’s inequalIn fact, we have $\rho u^{i}\in L^{2}(0,T;L^{p}(\Omega))$ for
ity, and therefore,$\frac{1}{\mathrm{p}}=\frac{5}{6}-\frac{1}{\gamma}$$| \frac{d}{dt}\int_{\Omega}\rho\phi|=|\int_{\Omega}\rho u\cdot\nabla\phi|\leq C||\nabla\phi||_{p}$for$\phi\in C_{0}^{\infty}(\Omega)$, whereby$\frac{1}{p},$$+ \frac{1}{p}=1$,. This implies the continuity of$t\in[0, T]-\rangle$, and hence, where. The second relation of (6) is shown similarly, because,by (4) andthe energy inequality guarantees.
follows fromand therefore,
B. Ducomet, E. Feireisl, H. Petzeltov\’a, and I. Straskrab\’a [7] prove theabove Theorem 1 in the case that the solution of the Poisson equation
is expressed by the corresponding Green function provided isextended to be zero outside .
We shall follow the scheme $[12, 7]$ to construct our weak solution. In moredetail, taking large $\beta>0$ , we construct an approximate solution
to$\int_{\Omega}\rho\phi$for$\phi\in L^{\gamma’}(\Omega)$$\rho\in L^{\infty}(\mathrm{O}, T;L^{\gamma}(\Omega))$$+ \frac{1}{\gamma}=1$$\frac{1}{\gamma},$$\rho\in C([0, T];L_{w\mathrm{e}ak}^{\gamma}(\Omega))$$\sqrt{\rho}u\in L^{\infty}(\mathrm{O}, T;L^{2}(\Omega))$$\rho n\in L^{\infty}(\mathrm{O}, T;L^{\frac{2\gamma}{\gamma+1}}(\Omega))$$\gamma\geq\frac{6}{5}$$\rho\in L^{\infty}(\mathrm{O}, T;L^{\gamma}(\Omega))$$\Phi$$\Delta\Phi=4\pi g\rho$$\rho$$\Omega$$u=u_{\delta,\epsilon}(x, t)$.

$\rho_{t}+\nabla\cdot(\rho u)=\in\Delta\rho$$(pu^{i})_{t}+\nabla\cdot(\rho u^{i}u)+a(\rho^{\gamma})_{x:}+\rho\Phi_{x_{i}}+\delta(\rho^{\beta})_{x}$$=\mu\Delta u^{i}+(\lambda+\mu)(\nabla\cdot u)_{x}$.

$\Delta\Phi=4\pi g(\rho-\frac{1}{|\Omega|}\int_{\Omega}\rho)$

in$+\epsilon\nabla u^{i}\cdot\nabla\rho$(7)$\Omega\cross(0, T)$with (2) by the Faedo-Galerkin method. Then, we obtain the actual solution,
by vanishing the artificial viscosity and then the artificial pressure 6. The
is necessary to take the limit of the Faedo-Galerkin approxcondition-curl lemma. Thisimation and construct the solution to (7), using the$\epsilon$$\gamma>\frac{3}{2}$$\mathrm{d}\mathrm{i}\mathrm{v}$5restriction is the same as that of [12], because the total mass conservation$||\rho(\cdot, t)||_{1}=M$ guarantees$||\rho||_{6}\leq M^{1-\theta}||\rho||_{\gamma}^{\theta}\mathrm{E}$, and therefore, iffor, thenand the contribution of
the Poisson termof the total energy is absorbed into that ofby (4). We note that this condition
the pressure termis actually
weaker than.
Also, we use the following lemma on
convergence used in taking limitsof the approximate solution.
$2\theta<\gamma$$\gamma>\frac{4}{3}$$\frac{1-\theta}{1}+\frac{\theta}{\gamma}=\frac{6}{6}$$E$$- \frac{1}{8\pi g}||\nabla\Phi||_{2}^{2}$$/_{\Omega}P$$\gamma>\frac{4}{3}$$\gamma>\frac{3}{2}$$L^{1}$

Lemma 0.1 Ifconvex function,$\Phi$:

is a proper, lower semi-continuous,
is a domain with bounded measure, and$\mathrm{R}arrow(-\infty, +\infty]$$D\subset \mathrm{R}^{m}$(8)$\mathrm{v}\mathrm{u}_{k}\mathrm{p}||v_{k}||_{p}<+\infty$$v_{k}arrow v$$\Phi(v_{k})arrow\overline{\Phi(v)}$weakly in $L^{1}(D)$weakly in $L^{1}(D)$$\int_{D}\Phi(v)=\int_{D}\overline{\Phi(v)}$

(9)
(10)

(11)

with $p>1$ , then it holds that$v_{k}arrow v$

strongly in $L^{1}(D)$ .

(12)

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$L^{p}$