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((anl v \ PM dose wo'f (bu/(3:692 4"? I wH—L YQ/QPe/cl” 10 “HQ O’HWLN mm W Exercise 8.5 Let ’H = L2(T3;R3) be the Hilbert space of 21rperiodic, square
integrable, vector—valued functions u : T3 —+ R3, with the inner product (u, v) = / u(x)  v(x) dx.
T3
We deﬁne subspaces V and W of ’H by V = {veC°°(T3;Ra)Vv=0},
W = {WEC°°(T3;R3)w=V<pforsorne<pzT3—+R}. Show that ’H = M @N’ is the orthogonal direct sum of M = V and N = W. Let P be the orthogonal projection onto M. The velocity v(x, t) e R3 and
pressure p(x,t) e R of an incompressible, viscous ﬂuid satisfy the NotiﬁerStokes
equations Vc+v~Vv+Vp=uAv,
Vv==0. Show that the velocity v satisﬁes the nonlocal equation
v; + P[v  Vv] = VAv. We ﬁrst show that V and W are orthogonal. Let v E V and w E W. Then, integrating by
parts, the inner product (v,w)= f13u(x)v(x)dx = (p(x)v(x)T3 .. fT3(V.v)V¢dx = 0,
since V is 2m periodic (so the surface term vanishes) and V  v = 0 (so the integral also vanishes). Since the kernel of an orthogonal projection closed, we have N = W C ker PV and M = \7 C ker PW. Next, we show that W J. V. To see this note that for any
limit point v’ of V there exists a sequence {vn} E V such that vn —> v', and v' is orthogonal to W since (v’,w) =<lir_gv,,,w> = lim(vn,w) = 0 for all w E W. Futhermore W J. V since for any limit point w’ of W there exists a sequence {wn} E W such that wn ——> v', and w' is orthogonal to Vsince (w’,v) = (lim wn,v> = lim(wn,v) = 0 for all v E V. Now ifx E W H V—Lthen since x E W, xJ. l7, and sincex E \7, xi W, so it must
hold thatx=0, thus W O V = {0}. F036 9' It remains to be shown that W U \7 spans H. Since H = L2(T3, R3), it follows that has a
Fourier basis {exp[in.(x+y+z)]i + exp[inj(x+y+z)]j + exp[ink(x+y+z)]k}. So expanding
about this basis, we see that if v E V, then v = Eunieim'm’mi + E anje""'(x+y+2)}t + Eameimmymk and then "k n, "h "1'
I _ . in (x+y+z).‘ . in(x+y+z)'t . ' (x+ + )d .
V v—Emianie ’ 1+Eznjanje ’ 1+2mkankem‘ y zk. This can only be
n.
I zero if a,” = ani = a,“ = 0, when ni,nj,nk are nonzero. So v =anii + a,” j + ankk , where the coefﬁcients are any real number.
On the other hand if w E W, then _ I _', . '( )1 r a
w = Eame'w‘+y ")1 + Zanje'" “y H _] + Zane'mmymk =V(p for some realvalued
,,. "t J "" function (p. But this implies that f Eanlein,(x+y+z)? + EanjeinJ(nyu); + Bankein.(x+y+z)k dx = fTS dexe T3 :1; nj nk ' eiru: But fem‘dx = —.— , so the left hand side of the above equation only gives a real value if
T T a =anj =anh =0 whenni=nj=nk=0. "I m Therefore any vector x E H can be expressed as x = v + w, where the v E Vpart contains
the Fourier components for ni = nj = nk = 0, and that w E W part contains the Fourier
components when ni,nj,nk are nonzero. So we can conclude that H =W (43 V = M @N. Given the equations v, + vVv +Vp = vAv and V  v: 0, we project rewrite the ﬁrst
equation as vt + v Vv +Vp  vAv = 0 so that P(vt + v  Vv + Vp  vAv) = 0. Now the
time derivative commutes with the divergence, so vt E V and P(vt)= vt. Since the
pressure p is a scalar function, we have Vp E W which implies that P(Vp) = 0. Next,
VvAv=(Vv)vAv+(VvA) v= (Vv)vAv+vA(Vv)=0, so VvAv e v and
P(V  vAv) = V  vAv. So by the above relations, we have P(vt + v Vv + Vp — vAv) = vt + P(v  Vv) — vAv = 0,
and we can conclude that vt + P(v  Vv + Vp) = vAv. Exercise 8.9 Let A C H be such that
M = {2: E H  a: is a ﬁnite linear combination of elements in A} is a dense linear subspace of H. Prove that any bounded linear functional on H
is uniquely determined by its values on A. If {no} is an orthonormal basis, find a.
necessary and sufﬁcient condition on a family of complex numbers ca for there to
be a bounded linear functional <p such that (Mun) = ca. ch that cp(x) = cp’(x) for all x E A.
Now let y E AC. Then since M is dense in H there exists a sequence {xn} E M such that xn >y. Then since both (p and cp’ are linear we have
W) = wqig} x.) = gig; w.) = gig;an akz.» = gig: we...)
k k = gig; We...) = <p’qi3; x.) = My)
1: where Eakznk is a ﬁnite linear combination for xn
k E M. Thus, we see that q)(y) = CP'O’)
for all y E AC , showing that cp is uniquely determined by its values on A. Since the span of an orthonormal basis for H is dense in H, it follows that any bounded linear function is uniquely determined by its values on the basis. Let (p(x) = <Ec u x>. (111’ Then it is clear that <p(ua) = ca. Since {ua} is a basis for H, there exists y = Ecaua E H and (p(y)=<2caua,2caua> = Elcalz. In order for Elcml2 to converge, only a (1 em. But in order for cp(x) to be bounded for every x
= Zbaua, it follows that only a ﬁnite number of cacan be nonzero. The reason is because for every set {ca} containing inﬁnitely (but countable) nonzero terms such that series Elca] < 00, we can an x = Ebaua, where bu = l/c—a and then zc—aba does not
a a converge. Therefore in order for to be bounded it is necessary for {Cu} to have only
ﬁnitely many nonzero terms. On the other hand if {ca} does have only ﬁnitely many
nonzero terms then llinrnlrp(x)=llilr_r11<2cmum,x>=c=c”x“ where c = max{ca}. So it (1 follows that cp is bounded. So a necessary and sufﬁcient condition for there to be a bounded linear operator cp such that <p(ua) = Cu is that the family of complex numbers
ca} has only ﬁnitely many nonzero terms. Exercise 8.11 Prove that if A : H —r H is a linear map and dim’H < 00, then dimkerA+dimranA = dim'H. Prove that, if dim'H < 00, then dimkerA = dim ker A‘. In particular, kerA = {0}
if and only if ker/1“ == If dim H < 00, then there exists some n E N such that dim H = n, so then the linear map A can be expressed as an n x 11 matrix. But then the dim ran A = rank of A and dim ker A
= nullity of A. So by the ranknullity theorem, dim ker A ' ' detail, since dim H < 00, then is ﬁnite and A is a bounded ' W 9 Theorem 8.17, ran A = (ker A*)*. Furthermore H = ker A* 69 ran A. Now dim ran A = dim ran A and dim ker A* = dim ker A (as shown below) so dim H = dim ker A + dim
ran A. Since the kernel of a matrix is invariant under transposition and solutions to linear equations are invariant under complex conjugation, it follows that the nullity of a A is
equal to the nullity of A*, when dim H < 00. Thus, dim ker A = dim ker A“. In particular, dim ker A = 0 iff ker A = {0}, and likewise for A*, so it follows that ker A =
{O} iffker A* = {0}. Exercise 8.12 Suppose that A : H —+ H is a bounded, selfadioint linear operator
such that there is a constant c > 0 with alkali 5 “A2” for all a: E H. Prove that there is a unique solution a: of the equation Ax = y for every y E H. By Proposition 5.30, A has a closed range and the only solution to the equation Axu= 0 is x = 0. Since A is selfadjoint, a similar statemﬂtiiolds for A*. So it follows that kerA* = {0}. It follows from Theorem 8.17 that ran A = (ker A*)1 = {0}* = H . Now, by Theorem 8.18, since A has a closed range, the equation Ax = y has a solution if y is orthogonal to ker A*. But ker A“ = {0} so every y E H is orthogonal to ker A", and thus
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 Spring '10
 Thomases
 Linear Algebra, dim ker, dim ker A*, W J., m1 Xlekerp

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