1 for ImAo ^ 0. This contradicts (5.1.14). Consequently, sinh(Im Ao a) — 0, i.e., ImAo = 0. Thus, only real eigenvalues may have algebraic multiplicity greater than one. 2) Suppose that Ao is an eigenvalue satisfying the equalities d_(Ao) = d'_(Ao) = d" (Ao) = 0. As we have shown above, Ao is real. Furthermore, by the equality df!_(Xo) = —X^1 a2 sin Xoa, we have sin Xoa = 0. Since sin Aoa = Ao sin a, this yields sin a = 0. This contradicts the equality d'_(Xo) = 0. Analogously, the equalities d+(Ao) = d'+(Xo) — d+(Xo) — 0 lead to a contradiction. • LEMMA 5.1.2. If Ao is a multiple root of one of the equations cL(A) = 0 or d+(Ao) = 0, then a generalized eigenvector to (5.1.11) and (5.1.12), respectively, can be obtained by the formula
(5.1.15)
o?(*o,¥>) \
d
( u?(\0,) ^(Ao,^) / \ p^(X0,ip)
\=\{ 0 Proof: Let c\ = ci(A), C3 = cs(X) be the functions defined in (5.1.10). It can be easily verified that v?\\,a/2)\
„ (A,a/2) J
+C3(A)
fu?\\,a/2)\ _ 3)
( cos(l + A)a/2
Xd {X)
U (A,«/2)J ~ -
l- S in(l + A)a/2
for all A G C. Using the fact that Ur , uKr } are even functions of the variable ip and Um , Uin are odd functions, we get U
^ V^1}(A,-a/2);
3K }
\u$\\,-a/2)J
Since (ui- \u^\p^) and (wf \u\p\pW) last two equalities yield
V}
Vsin(l + A)a/2 y
are solutions of the system (5.1.4), the
x ~/,x / r \ x , ,,N fr, ( cos(l + A)a/2 \ /cos(l + A ) a / 2 \ \ 21(A) u~ = A d _ ( A ) [ 0 , ( .\ , * ' ) , ( . ,\ ^ /„ ) • v M *. I V V-sin(l + A)a/2/ Vsin(l + X)a/2J J
v(5.1.16) }
Let A = Ao be a multiple root of the equation cL(A) = 0. Differentiating (5.1.16) with respect to A and setting A = Ao, we get / u-(\0,(ft)x ft£(fi) such that (A + 2) ^
where F G ( Wlityx p = 0 and u = t^°).
0)
— (ul , uL ) G
]
+ V^ • u^ = #. Then we have
hk(ty)* is given by the left-hand sides of (5.2.2)-(5.2.4) with
We show that there exists a vector function (u^\p) such that
(,,,0,
EW^ity*
baity
x
^ity
A 1, #ien —1 — Ao is also an eigenvalue and the vector functions /
.,(*).
,(*)>
(;;;H-»> K> Q +
(
.,(*)«
+ 4°) = 0.
If we set vr = 0, q = 0 in (5.4.23), we get (5.4.25)
Q(u£\vu)
- J (uW-Vu
+ 2(Vwu) • vu + p ( 1 ) V w • t; w ) dw = 0.
Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms
5.4. T H E E I G E N V A L U E S A = l A N D A = - 2
167
Furthermore, the equality f ((Vc.4 1 )) • V^vr + 2 (V w • i^1*) vr ~ 3p (1) iJr + ( 3 4 0 ) + p ( 0 ) ) U r ) dw = 0 n holds, if one sets vu — 0, q = 0 in (5.4.23). Prom the last equality together with the equalities p^ = 2ur * — c and (5.4.24) it follows that J ((VctzW) • Vjur + (3^°) - 3p (1) - c) vr) da; = 0 n Obviously, the equations (5.4.24)-(5.4.25) are equivalent to (5.4.23). By (5.4.24), (5.4.25), we have (5.4.26)
(5.4.27)
Q{v£\v„)
- j
(u^
• vu + p™ V w • Uw + ( V . • u ^ ) do;
for all ^ eWl{n), q G L 2 (fi), i.e., (^ 1 } ,P ( 1 ) ) (5.4.4) with / = 2deUr , g — 2{sm0)~1d^ur 5.4.1, this problem is solvable if and only if j From (5.4.26) it follows that Sur = Sur (5.4.28)
is a solution of the auxiliary problem \ ijj — ur . According to Lemma dw = 0. Q ur — 3 p ^ — c or, what is the same,
SuW + 6uW = 3^°) - 3p - c
with p = p^ — 2ur . Furthermore, (5.4.27) is equivalent to Q(u£\v«>) ~ J (u^Vu
+ p V „ -iJw + (V w • uW)q) du =
ft
ju^qdu. n
This means that (uL ,p) is a solution of the auxiliary problem (5.4.4) with / = g = 0, ip — Ur • Under our assumption on ur , this solution exists. We show that problem (5.4.28) is solvable if and only if c ^ 0 and ur * is orthogonal to all w GW'^(fi) such that 1 such that
(5.5.6)
^(A*) = 0
forj = k,k + 1,..., k +
v-1,
where \ij is as in (5.5.5). The corresponding eigenspace is spanned by the functions 1
(5.5.7)
A * + 2 V " U" ,0') un •
We derive a transcendental equation for the eigenvalues of the pencil C from the following representation of solutions (U1P) of the homogeneous system (5.0.1) in terms of three harmonic functions \I>, ©, A, which is the analogue of the Boussinesq representation [23] of solutions of the homogeneous system of linear three-dimensional elastostatics (see Section 3.3): U P
(5.6.3)
= =
aVtf + 2 6 V x ( e e 3 ) + c(V(x 3 A)-2Ae 3 ), 2cdX3A,
where V is the gradient in Cartesian coordinates, e% denotes the unit vector in x% direction and a, 6, c are arbitrary constants. In spherical coordinates this representation becomes =
a
+
1
(rsinfl)- ^*
2br
eos0(dr(rA)-2AJ + c
00(Acos0) + 2Asin0
cot Od^A
cot 9 dpQ •smOdr(rQ)-de{ecosO) \
J
and
P=
2c(cos6drA-—deA
respectively. Based on (5.6.1) and (5.6.3), we select the potentials (5.6.4)
#m
=
vA+1 P ^ J (cos 0) cos{m(f),
Om
=
rA+1 P~^ (cos 0) sin(ras0))).
(5.6.6)
The homogeneous Dirichlet condition for the velocity components ur,uo,u^ at 6 = a gives the following condition for the existence of nontrivial solutions of (5.6.1): det(M^ m ) (a)) = 0
(5.6.7)
which is obtained from (5.6.1), (5.6.5), (5.6.6) and from the eigenvector (a,6,c)Gker(M^ m ) (a)). Condition (5.6.7) is the desired transcendental equation for the eigenvalues A of the pencil C which was solved numerically. Figure 11 below shows the eigenvalues Ai(fia) and A2(fia) obtained for m = 1 and 0, respectively, in the case n/2 < a < IT. As we remarked earlier (see Theorem 5.4.1), A = 1 is an eigenvalue of C for all a £ (0,7r], corresponding to zero velocity U and constant pressure P . It is obtained in (5.6.3) for # = O = 0 and A = x3.
i
alpha
FIGURE 13. Eigenvalues Afc(fia) e [0,1) for ir/2 < a < n. Solid line: simple eigenvalue corresponding to axisymmetric eigenstates, dashed line: twofold eigenvalue corresponding to the first spherical harmonic.
Furthermore, it is interesting to observe that the double eigenvalue Ai(fi a ) corresponding t o r n = 1 (with the eigenvectors given by (5.6.1) and (5.6.2)) lies
Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms
178
5. T H E D I R I C H L E T P R O B L E M F O R T H E S T O K E S S Y S T E M
below the eigenvalue \2(Qa) corresponding to m = 0. The latter eigenvalue and A = 1 arise from two coalescing simple eigenvalues of the pencil corresponding to the Dirichlet problem of linear elastostatics as the Poisson ratio tends to 1/2 from below, i.e. in the incompressible limit. 5.6.2. The case a = 27r/3. By Theorem 5.4.3, there correspond exactly one eigenvector (0,0,0,1) and the generalized eigenvector (5.4.20) to the eigenvalue ^2(^271-/3) = 1- Figure 11 indicates that Ai(0 27r /3) is either equal or very close to 1/2. To determine Ai(f£27r/3), w e evaluate
detM«(|) = l(i 7 (P 3 7 2 (-i)) 2 P 1 7 2 (-i) +22F 3 - /2 (-i)P 1 7 2 (-i)-80(P 3 7 2 (-i)) 3 + (P172(-i))3). Using
it was found by Kozlov, Maz'ya and Schwab [141], with accurate series expansions for the hypergeometric function, that 0 ^ detM^A
= -0.0010545...
with all digits shown correct, i.e., Ai(f227r/3) ¥" \- A bisection method showed the inclusion 0.666496TT < a* < 0.666507TT
for the critical angle a* for which Ai(Q a *) = 1/2. Hence a* < 27r/3. Furthermore, by means of a bisection method, the inclusion 0.49967 < Ai(fi 27r / 3 ) < 0.49968 was obtained in [141]. 5.7. The Dirichlet problem for the Stokes system in a dihedron Now we consider the special case when the cone K coincides with the dihedral angle /C = /Ca x R 1 , where a G (0,2ir], JCa = {(xux2)
G l 2 : 0 < p < 00, , (p) are the polar coordinates in the (#i, #2)-plane. We construct all solutions to the homogeneous problem (5.0.1), (5.0.2) which have the form (5.0.3). This means that we get a description of all eigenvalues, eigenvectors and generalized eigenvectors of the operator pencil C = Ca defined on the subdomain Qa = K D S 2 of the unit sphere. The construction of such solutions is done in several steps. First we describe all solutions of the form (5.0.3) with Re A > —1/2 which are independent of X3. These solutions can be directly expressed by the eigenvectors and generalized eigenvectors of the pencil generated by the plane Stokes problem. Next we give a description of all other solutions of the form (5.0.3) with Re A > —1/2. The solutions of this form with Re A < —1/2 are constructed by means of the relation between the eigenvectors and generalized eigenvectors corresponding to the eigenvalues A and — 1 — A which was given in Theorem 5.2.2. Since this relation is not one-to-one in the case A = 1, we consider the eigenvalue A = — 2 separately in the last subsection.
Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms
5.7. T H E D I R I C H L E T P R O B L E M F O R T H E S T O K E S S Y S T E M IN A D I H E D R O N
179
5.7.1. Special solutions of the homogeneous Stokes system which are independent of # 3 . First we give an explicit description of all solutions of the problem (5.0.1), (5.0.2) which have the form (5.0.3) and are independent of £3. Obviously, ^-independent solutions U = U(x\,X2), P — P(x\,x) s - f e H ( , c ) M = (Vry^UiVx), (s-k)\
j =
0,l,...,s,
for the functions u^k\ Since the coefficients determinant of this system is nonzero (multiplying this determinant by 1! • • • s\, one obtains the Vandermonde determinant), the functions u^ in (5.0.3) can be represented in terms of the functions U(2jx), 1 < \x\ < 2, j = 0 , 1 , . . . ,s. Hence, from our assumption on U it follows that v,W eWli^a)3 LEMMA
for fc = 0 , 1 , . . . s. Analogously, we obtain pW G L2(^a)-
•
5.7.3. If (U,P) is a solution of problem (5.0.1), (5.0.2) of the form
(5.0.3); where A G C, vW G ^ W 3 , P^ G L 2 (O a ) ; then the vector functions (di3U, d^3P) have the same properties for an arbitrary positive integer £.
Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms
5.7. T H E D I R I C H L E T P R O B L E M F O R T H E S T O K E S S Y S T E M IN A D I H E D R O N
183
Proof: Let £, r\ be an arbitrary functions from CQ°(JC\{0}) such that (rj = £. We show that C3r3tf G W^(/C)3 and C ^ P G L2(/C). Prom (5.0.1), (5.0.2) it follows that the functions V = (,11 and Q — (P satisfy the equations (5.7.12)
-AV
+ VQ = F,
V-V = G
in/C,
where F = -2(VC • V) U - U A( + P VC and G = U • VC For an arbitrary function in /C let £ denote the mollification of $ in £3 with radius e: $ e ( x i , x 2 , - ) = *(xi,ar 2 ,-) * ^ > where h e (x 3 ) = £ h(x3/e), h(t) = c exp ( - 1/(1 - t 2 )) for —1 < t < 1, h(t) = 0 for \t\ > 1 and c is such that J_ x h(t) dt = 1. Then, by (5.7.12), we have _1
-AV;
+ vg £ = F £ , v . K = G£
which implies (5.7.13)
-AdX3V£
+ VdX3Q£
= dX3F£,
V • dX3V£ = dX3GE .
Multiplying the first equation of (5.7.13) by dX3V£ and integrating over /C, we arrive at
l i v a ^ m ^ s + (vdX3Q£,dX3v£)lc = {dxaF£,dxav£)K.
Since (VdX3Q£,dX3V£)K
= -(dX3Q£,V
-dX3V£)K
= -(dX3Q£
,dX3G£)K,
this implies
(5.7.14) 1 1 0 * ^ 1 1 ^ , < l|fl-.^ll^-1(JC), \WM\^K)>
+ \(dX3Qs,dX3G£)K\.
O
There exists a vector function W GVF2(/C)3 with compact support such that V- W = dX3G£ and \\W\\^{ic)3u3(y>) + s i n 6»Mp()) ^ r^ (sinOY ( — sm.6uz(tp) + cos6up() logp + g(y?))
(5.7.31)
be a non-trivial solution to the Dirichlet problem for the system (5.0.1). This means that fi = Xj for some j > 0, (u,p) is the corresponding eigenvector of the operator pencil £, and (v, q) is its generalized eigenvector. The solution (5.7.31) takes the following form in the spherical coordinates I rM (ur log p+ vr ) \ r^ (ue logp+5^ ) r^ ( Uy log p+ v^ )
/ Ur \
V PJ
\ r^- 1 (p iogp+5) y
where
/ °
\
(5.7.32)
/
V P )
v
fir
/
"if
\
o
(5.7.33)
ve o
V
vv q
J
(sin0)^ (cos#i£3 + sm6up) ^ (sin 6Y (— sin 6 us + cos 0 u p ) (sin0)^i^ (sintf)^ 1 ? /
(sin 0y (cos 0v3+ sin 0 v p ) \ (sin QY ( - sin 0 u3 + cos 6 vp) {wa6Yvv (sin0) M - i , \ /
According to Theorem 5.2.2, the vector (Vr, Ve, V^, Q), where K
=
r " 1 " ^ ((/i + 2) (2 S r log r - ur log p - vr ) - ur J,
Ve
=
r ~ 1 - / i m - /i) (2 u^ l o g r - ue \ogp-
K.
0 =
r-\-n
,-2-M
ve)+ue),
h^ _ ^ (2x1^ l o g r - u^ \ogp- v^ ) + u^ J, ( ( l - / x ) (2P l o g r - P logp-