Functional Solutions for a Plane Problem in Magnetohydrodynamics
GIOVANNICIMATTI
ABSTRACT - We study the flow in a channel of a fluid obeying the equations of magnetohydrodynamics under the hypotheses that viscosity, resistivity and thermal conductivity depend on the temperature. The special class of solutions for which two functional dependences between temperature and magnetic field and velocity and magnetic field exist is considered.
1. Introduction
We study in this paper a class of steady, incompressible and recti- linear flows of viscous, electrically and thermally conducting fluids along cylindrical channels of arbitrary cross-section in the framework of the equations of magnetohydrodynamics. The open and bounded subset of R2, representing the cross-section of the pipe is referred to the orthogonal frame Ox1x2x3.Ox3 is the axis of the channel with unit vector i3. The magnetic fieldH and the velocityvare assu med to be of the form
Hh(x1;x2)i3
(1:1)
vv(x1;x2)i3: (1:2)
Moreover, we suppose
pp(x1;x2):
(1:3)
The electrical resistivityr, the viscositynand the thermal conductivitykare assumed to be given functions of the temperature uand ofv andh. The
(*) Indirizzo dell'A.: Department of Mathematics, University of Pisa, Largo Bruno Pontecorvo 5, 56127, Pisa, Italy.
E-mail: cimatti@dm.unipi.it
domainVis multiply connected with its boundaryGconsisting of two dis- joint closed curvesG1 andG2 of classC1. From the general equations of magnetohydrodynamics and from the energy equation [6] we obtain, in view of (1.1) and (1.2) the following boundary value problem (P) [4]. For more details we refer to the Appendix.
r (r(u;v;h)rh)0 inV (1:4)
r (n(u;v;h)rv)0 inV (1:5)
r
k(u;v;h)run(u;v;h)vrvr(u;v;h)hrh
0 inV (1:6)
h0 onG1; hh onG2
v0 onG1; vvonG2 u0 onG1; uu onG2;
whereh,vanduare given positive constants. We prove in Section 2 that if the functionsk,nandrhave the special form (2.9) below, problem (P) has one and only one solution. Section 3 deals with the functional solutions of problem (P) i.e. with solutions for which there exist functional de- pendences betweenuandhand betweenvandh. We prove the existence of a one-to-one correspondence between the functional solutions and the solutions of a two-point problem for a system of first order differential equations depending on two parameters. First order ordinary differ- ential equations depending on parameters whose solutions must satisfy two boundary conditions have been considered, among others, in [2], [3], [7] and [8]. In Section 4 a theorem of existence and uniqueness is proved when resistivity, viscosity and thermal conductivity depend only on the temperature.
2. A result of existence and uniqueness
To prove the next theorem we need the following elementary LEMMA1. Letk(u)^ 2C1([0;1)); ^n(v)2C1([0;v]); 2C1([0;h])
^
k(u)k0 >0 (2:1)
and
^n(u)n0>0:
(2:2)
The transformation
uK(u)M(v)h2 (2:3) 2
cN(v);
(2:4)
defined for u2[0;1), v2[0;h]and depending on the parameter h2[0;h], where
K(u) Zu
0
^
k(t)dt; M(v) Zv
0
t^n(t)dt; N(v) Zv
0
^ n(t)dt
is globally invertible for every h2[0;h].
PROOF. By (2.2), (2.4) has an inverse vN 1(c) (2:5)
defined in [0;N(v)]. Substituting (2.5) into (2.3) we obtain uK(u)M(N 1(c))h2
2 : (2:6)
Since by (2.1)K(u) is strictly increasing and
u!1lim K(u) 1 (2:7)
(2.6) has, with respect tou, a unique solution uK 1
uM(N 1(c))h2 2
: (2:8)
The inverse of (2.3), (2.4) is therefore (2.8), (2.5). p THEOREM1. Ifkandnare given by
k(u;v;h)r(u;v;h)^k(u); n(u;v;h)r(u;v;h)^n(v);
(2:9)
wherer(u;v;h),^k(u)and^n(v)are C1functions such that r(u;v;h)r0>0; ^k(u)k0>0; ^n(v)n0>0 (2:10)
when u0,vv0,hh0, then problem(P)has one and only one solution.
PROOF. By the maximum principle the solutions of problem (P) satisfy u(x)0; vv(x)0; hh(x)0 inV; x(x1;x2):
DefineK(u),M(v) andN(v) as in Lemma 1. The transformation uK(u)M(v)h2
2 cN(v) by Lemma 1 is globally invertible with inverse
uU(u;c;h)K 1
uM(N 1(c))h2 2 (2:11)
vV(c)N 1(c):
(2:12) Define
a(u;c;h)r(U(u;c;h);V(c);h):
(2:13)
Let (u(x);v(x);h(x)) be any classical solution of problem (P) and set u(x)K(u(x))M(v(x))h(x)2
2 c(x)N(v(x)):
Problem (P) can be reformulated in terms ofu,candhas problem (P) r (a(u;c;h)ru)0 inV
(2:14)
r (a(u;c;h)rc)0 inV (2:15)
r (a(u;c;h)rh)0 inV (2:16)
u0 onG1; uuK(u)M(v)h2 2 onG2 c0 onG1; cc N(v) on G2
h0 onG1; hh onG2:
We claim that problem (P) has one and only one solution. For, let (u(x);c(x);h(x)) be any solution of (P). Define
z(x)u(x) u hh(x) j(x)c(x) c
hh(x):
We havez0 andj0 onG. On the other hand, by (2.14) and (2.16) r (a(u;c;h)rz)r (a(u;c;h)ru) u
hr (a(u;c;h)rh)0 inV (2:17)
and, by (2.15) and (2.16),
r(a(u;c;h)rj) r(a(u;c;h)rc) c
hr(a(u;c;h)rh)0 inV:
(2:18)
Multiplying (2.17) byzand integrating by parts overVwe have Z
V
a(u;c;h)jrzj2dx1dx20:
Similarly from (2.18) we obtain Z
V
a(u;c;h)jrjj2dx1dx20:
Hence
z(x)0; j(x)0 inV:
Thusu(x),c(x) andh(x) are related by the functional relations u(x)u
hh(x) (2:19)
c(x)c hh(x):
(2:20)
Hence the equation (2.16) can be rewritten r a
u hh;c
hh;h
rh
!
0 (2:21)
h0 onG1; hhonG2: (2:22)
This problem has one and only one solution. For, let w F(h)
Zh
0
a u
ht;c ht;t
dt:
Equation (2.21) under this transformation becomes
Dw0 inV; w0 onG1; w F(h) onG2:
Hence
h(x) F 1(w(x)); u(x)u
hF 1(w(x)); c(x)c
hF 1(w(x)):
By (2.11) and (2.12) we obtain as the only solution of problem (P) u(x)K 1
u
hF 1(w(x))M
N 1 c
hF 1(w(x))
1 2
F 1(w(x)) 2
v(x)N 1 c
hF 1(w(x))
h(x) F 1(w(x)):
p
3. The functional solutions
The assumptions (2.9) are quite special. To treat more general situa- tions we use a different definition of solutions suggested by the proof of Theorem 1 and in particular by the functional relations (2.19) and (2.20) betweenu,candh.
DEFINITION. We say that a classical solution (u(x);v(x);h(x)) of problem (P) is a functional solution if two functions U(h)2C2([0;h]), V(h)2C2([0;h]) exist such that
u(x) U(h(x)); v(x) V(h(x)):
THEOREM2. If
r(u;v;h)r0>0 (3:1)
then there exists a one-to-one correspondence between the setfFSgof the functional solutions of problem (P) and the set fTPPg of the solutions (U(h);V(h);g;m)of the two-point problem(TPP)
n(U;V;h)dV
dh gr(U;V;h) (3:2)
k(U;V;h)dU
dhn(U;V;h)VdV
dhhr(U;V;h)mr(U;V;h) (3:3)
U(0)0; U(h)u (3:4)
V(0)0; V(h)v:
(3:5)
PROOF. We define the map I :fTPPg ! fPg as follows. Suppose (U(h);V(h);g;m)2 fTPPgand consider the nonlinear boundary value pro- blem
r (r(U(h);V(h);h)rh)0 inV (3:6)
h0 onG1; hhonG2: (3:7)
It is easily seen that (3.6), (3.7) has one and only one solution. For, let wR(h)
Zh
0
r(U(t);V(t);t)dt:
(3:8)
By (3.1) R maps one-to-one [0;h] onto [0; R(h)]. Moreover, under (3.8) problem (3.6), (3.7) becomes
Dw0 inV (3:9)
w0 onG1; wR(h) on G2: (3:10)
Problem (3.9), (3.10) has one and only one solution. Therefore, the only solution to problem (3.6), (3.7) is given byh(x)R 1(w(x)). We claim that
(u(x);v(x);h(x))(U(h(x));V(h(x));h(x))2 fPg:
(3:11)
We have by (3.2) and (3.6)
r (n(u(x);v(x);h(x))rv) r (n(U(h(x));V(h(x));h(x))dV dhrh) (3:12)
gr (r(U(h(x));V(h(x));h(x))rh)0 inV:
Moreover, by (3.3) and (3.6) r
k(u;v;h)ruvn(u;v;h)rvhr(u;v;h)rh
(3:13)
r
(k(U(h);V(h);h)dU
dh V(h)n(U(h);V(h);h)dV
dhhr(U(h);V(h);h)rh
mr (r(U(h);V(h);h)rh)0 inV:
All the boundary conditions of problem (P) are also satisfied in view of (3.4) and (3.5). Hence I is well-defined. On the other hand,I is one-to-one. For, let (U(h);V(h);g;m)2 fTPPg; (U(h);~ V(h);~ ~g;m)~ 2 fTPPg and (u(x);v(x);h(x)) I(U(h);V(h);g;m), (~u(x);~v(x);~h(x)) I(U(h);~ V(h);~ ~g;m). If~ I(U(h);V(h);g;m) I(U~(h);V(h);~ ~g;~m) we have u(x)u(x),~ v(x)~v(x) and h(x)~h(x), but u(x) U(h(x)), u(x)~ U(h(x)) and~ v(x) V(h(x)), ~v(x)V(h(x)). Hence~
U(h(x))U(~~h(x))U(h(x)) and~ V(h(x))V(~h(x))~ V(h(x)). Thus~ U(h)U~(h); V(h)V(h):~
(3:14)
This implies also g~g and m~m. Finally I is surjective. For, let (u(x);v(x);h(x))(U(h(x));V(h(x));h(x))2 fFSg. Define
u Zh
0
n(U(t);V(t);t)dV dt(t)dt (3:15)
c Zh
0
k(U(t);V(t);t)dU
dt(t) V(t)n(U(t);V(t);t)dV
dt(t)tr(U(t);V(t);t)
dt
z Zh
0
r(U(t);V(t);t)dt (3:16)
and
U(x)u(h(x)); C(x)c(h(x)); Z(x)z(h(x)):
(3:17)
By (1.4), (1.5) and (1.6) we have
DU0; DC0; DZ0 inV with
U0 onG1; UAonG2 (3:18)
C0 onG1; CBonG2
(3:19)
Z0 onG1; ZConG2 where
A Zh
0
n(U(t);V(t);t)dV dt(t)dt
B Zh
0
k(U(t);V(t);t)dU
dt (t) V(t)n(U(t);V(t);t)dV
dt (t)tr(U(t);V(t);t)
dt
C Zh
0
r(U(t);V(t);t)dt
andC60 by (3.1). Letz(x) be given by the solution of the problem Dz0 inV; z0 onG1; z1 onG2:
(3:20) We have
U(x)Az(x); C(x)Bz(x); Z(x)Cz(x):
(3:21) Setting
gA
C; mB (3:22) C
we obtain
U(x)gZ(x); C(x)mZ(x) (3:23)
and by (3.15), (3.16), and (3.17) we haveugzandcmzi.e.
Zh
0
n(U(t);V(t);t)dV
dt(t)dtg Zh
0
r(U(t);V(t);t)dt (3:24)
Zh
0
k(U(t);V(t);t)dU
dt(t)V(t)n(U(t);V(t);t)dV
dt(t)tr(U(t);V(t);t)
dt (3:25)
m Zh
0
r(U(t);V(t);t)dt:
Taking the derivative with respect to h in (3.24) and (3.25) we obtain (3.2) and (3.3). Hence, with g and m given by (3.22) we have
I(U(h);V(h);g;m)(u(x);v(x);h(x)). p
As an immediate consequence of Theorem 2 problem (P) has one and only one functional solution if the corresponding problem (TPP) has one and only one solution. Moreover, the search of the functional solution of problem (P) is broken into two steps: (i) to find the solutions (U(h);V(h);g;m) of problem (TPP) and (ii) to solve problem (3.20). Step (ii) contains, so-to-speak, the geometry of problem (P) and step (i) the nonlinear part.
To give a theorem of existence for the two-point problem (TPP) we note that (3.2), (3.3), (3.4) and (3.5) can be rewritten as follows
dV
dh g r(U;V;h) n(U;V;h) (3:26)
dU
dh ( gV hm)r(U;V;h) k(U;V;h) (3:27)
U(0)0; U(h)u (3:28)
V(0)0; V(h)v:
(3:29)
THEOREM3. If
k(u;v;h)>0; r(u;v;h)>0; n(u;v;h)>0 (3:30)
K2r(u;v;h)
k(u;v;h))K1>0 (3:31)
and
N2r(u;v;h)
n(u;v;h))N1>0 (3:32)
the two-point problem(3.26)-(3.29)has at least one solution.
PROOF. We transform problem (3.26)-(3.29) in a system of two Vol- terra-Fredholm integral equations. From (3.26) and (3.29) we have
g~g v Rh
0
r
n]dt (3:33)
where
r n
r(U(t);V(t);h(t)) n(U(t);V(t);h(t)): Moreover, from (3.27) and (3.28) we obtain
mm~ 1 Rh
0
r k
dt
u~g Zh
0
V(t) r
k
dt Zh
0
t r
k
dt (3:34)
where
r k
r(U(t);V(t);h(t)) k(U(t);V(t);h(t)):
Substituting (3.33) and (3.34) in (3.26) and (3.27) and integrating with re-
spect tohwe have the system of integral equations U(h)~m
Zh
0
r k
dt ~g Zh
0
V(t) r
k
dt Zh
0
t r
k
dt (3:35)
V(h)~g Zh
0
r n
dt:
(3:36)
From (3.33) we have
0~gG2: v N1h (3:37)
and from (3.36) and (3.26)
0 V(h)V2:N2
N1vh2 (3:38)
dV
dh(h)N2
N1vh:
(3:39)
Moreover, by (3.34) and (3.37) 0~mM2: 1
K1h
uvh3N2K2 N12 h2K2
2
: (3:40)
Hence, by (3.35) and (3.27)
jU(h)j U2:M2K2hv2h4K2N2
N21 K2h2 (3:41)
dU dh
D2:(V2G2hM2)K2: (3:42)
Using (3.38), (3.39), (3.41) and (3.42) we can apply the Schauder fixed point theorem. For, define
B f(U(h);V(h))2C0([0;h]) C0([0;h]); jU(h)j U2; jV(h)j V2g and letT :B!C0([0;h])C0([0;h]),T (T1;T2)
T1(U;V)~g Zh
0
r n
dt:
T2(U;V)~m Zh
0
r k
dt ~g Zh
0
V(t) r
k
dt Zh
0
t r
k
dt:
We haveT(B)B. Moreover, proceeding as in the proof of (3.39) and (3.42) we find thatT(B) is bounded inC1([0;h])C1([0;h]) and thus compact in C0([0;h])C0([0;h]) by Arzela's theorem. We conclude that the two-point problem (3.26)-(3.29) has at least one solution. p As a consequence of the theorem just proved and of Theorem 1 problem (P) has at least one functional solution if (3.1), (3.30), (3.31) and (3.32) are satisfied.
4. A result of uniqueness
In general [5] the two-point problem (TPP) has more than one solution.
It is therefore interesting to give conditions under which the solution is unique. For example, if condition (2.9) holds we have as corresponding two- point problem
^n(V)dV dhg
^k(U)dU
dh V^n(V)dV
dhh^r(h)m U(0)0; U(h)u
V(0)0; V(h)v
and it is easily seen that in this case the solution is unique. In the next theorem a result of uniqueness is given whenk,nandrdepend only on the temperatureu. In practical cases the main dependence ofk,nandr is precisely from the temperature. In the proof we use the following Lemma [1] .
LEMMA2. Ifb(u)2C1([0;1)),b(u)0andb0(u)0then the problem d2u
dh21g2b(u); u(0)0; u(h)u>0 (4:1)
has, for everyg2R1, one and only one solution.
PROOF. Simply note thatu(h)0 by the maximum principle and that d
du(1g2b(u))g2b0(u)0. p
THEOREM4. Ifk(u),n(u)andr(u)2C1([0;1))and satisfy k(u)>0; r(u)>0; n(u)>0
(4:2)
Z1
0
k(t) r(t)dt 1 (4:3)
r0(u)n(u)r(u)n0(u) (4:4)
and
r(u)
n(u)r0>0 (4:5)
then there exists one and only one functional solution of problem(P).
PROOF. According to Theorem 2 the two-point problem corresponding to problem (P) is, in this case:
n(U)dV
dhgr(U) (4:6)
k(U) r(U)
dU
dhgV hm (4:7)
U(0)0; U(h)u (4:8)
V(0)0; V(h) v:
(4:9)
We note that
g>0 (4:10)
since g0 is incompatible with (4.2) and (4.8). By (4.3) the function uF(U)
F(U) ZU
0
k(t) r(t)dt (4:11)
maps diffeomorphically [0;1) onto [0;1). The system (4.6)-(4.9) becomes, in terms ofuandV,
dV
dh gb(u); V(0)0; V(h)v (4:12)
du
dhgV hm; u(0)0; u(h)uF(u);
(4:13)
where
b(u)r(F 1(u)) n(F 1(u)): (4:14)
From (4.13) we have
du2 dh2gdV
dh10:
(4:15)
Substituting (4.12) into (4.15) we obtain du2
dh21g2b(u) (4:16)
u(0)0; u(h) u:
(4:17)
If (4.4) holds we haveb0(u)0. Thus, by Lemma 2 problem (4.16), (4.17) has for everygone and only one solutionu(h;g) and
U(h;g)F 1(u(h;g)):
(4:18)
Moreover, by (4.12)
V(h)g Zh
0
b(u(t;g))dt:
(4:19)
The parameterg, still at our disposal, is now used to satisfy the boundary conditionV(h) vwhich, by (4.19), reads
vg Zh
0
b(u(t;g))dt:
(4:20)
To prove that (4.20), as an equation ing, has one and only one solution we define
G(g)g Zh
0
b(u(t;g))dt:
We haveG(0)0 andG(g)>for allg>0. Moreover,
G0(g) Zh
0
b(u(t;g))dtg Zh
0
b0(u(t;g))@u
@g(t;g)dt:
(4:21)
We claim that @u
@g(t;g)0. By the continuous differentiability with re- spect to the parameterg and by (4.13) we have
dug
dh V(h) (4:22)
ug(0)0:
(4:23)
On the other hand, by (4.12) and (4.10) we haveV(h;g)>0. Thus, from (4.22) and (4.23) we obtain
ug(h;g)50:
(4:24)
Hence, by (4.21) and (4.5)
G0(g) Zh
0
b(u(t;g;g))dtr0h>0:
Therefore (4.20) has, for everyv, one and only one solution~g. We conclude that the only solution of problem (4.6)-(4.9) is given by
U(h)F 1(u(h;~g)) (4:25)
V(h)~g Zh
0
b(u(t;~g))dt:
(4:26)
By Theorem 2 there is only one functional solution of problem (P) which we obtain in the usual way: we solve the problem
r (r(U(h))rh)0 inV; h0 onG1; hhonG2 (4:27)
defining
w H(h) Zh
0
r(U(t))dt:
By (4.2)Hmaps ono-to-one [0;h] onto [0;w] where w H(h). In term ofw problem (4.27) reads
Dw0 inV; w0 onG1; ww onG2: (4:28)
Letw(x) be the solution of (4.28). We have h(x) H 1(w(x)) (4:29)
and
u(x) U(h(x)) U(H 1(w(x))) (4:30)
v(x) V(h(x)) V(H 1(w(x))):
(4:31)
The only functional solution of problem (P) is given by (4.29), (4.30) and
(4.31). p
We stress the difference between Theorem 1, where we found un- iqueness in the general class of classical solutionsfCSg, and Theorem 4 where uniqueness is proved in the smaller class of functional solutions fFSg. It would be interesting to show that the solution, under the hy- potheses of Theorem 4, is unique also infCSg.
5. Appendix
We derive here the equations (1.4), (1.5) and (1.6) entering in pro- blem (P). We assume the magnetic permeability m0 to be a constant.
Moreover, the pressure p, the magnetic field H and the velocity v are supposed to have the special form (1.1), (1.2) and (1.3). Using the Maxwell equation
J c 4pr H (5:1)
the magnetohydrodynamic body forces f m0
c JH by (1.1) take the form
f m0 2ph @h
@x1i1 m0 2ph @h
@x2i2: (5:2)
Thus, by (5.2), (1.2) and (1.3) the Navier-Stokes equations reduce to r (nrv)0
(5:3)
@p
@x1h @h
@x10 (5:4)
@p
@x2h @h
@x20:
(5:5)
In view of (1.1) and (1.2), the Ohm's lawrJEvm0Hgives rJE:
(5:6)
Moreover, sinceEis irrotational, we have, taking thecurlof (5.1), r (rr H)0
(5:7) and by (1.1)
r (rrh)0:
(5:8)
The energy equation reads
r (kru)EJnX3
i;k1
@vi
@xk@vk
@xi
@vi
@xk
(5:9)
where in the right hand side the first term represents the Joule heating and the second the viscous heating. Specializing (5.9) in our case we obtain, in view of (1.1), (1.2) and (5.6),
r (kru)rjrhj2njrvj2: (5:10)
This equation, by (5.3) and (5.8), can also be written as follows r
krurhrhvnrv]0:
Once h(x1;x2) is known as part of the solution of problem (P), we can computep(x1;x2) from (5.4) and (5.5).
REFERENCES
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Manoscritto pervenuto in redazione il 25 gennaio 2011.