P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Solutions to Exercises
Chapter 1 1.1
Let {x...
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P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Solutions to Exercises
Chapter 1 1.1
Let {x j } be a countable dense subset of X , and let {y j } be a countable dense subset of Y . Then the countable collection {(x j , yk )} is dense in X × Y , since for any (x, y) ∈ X × Y and any > 0 there exist x j and yk with x − x j X < /2
and
y − yk Y < /2,
and so (x j , yk ) − (x, y) X ×Y ≤ .
1.2
It follows that X × Y is separable, and by induction it follows that any finite product of separable spaces is separable. If M is a linear subspace of X then let {x j } be a countable subset of X such that for each x ∈ X there is an x j such that |x − x j | < . Now discard any element x j of this collection for which B(x j , ) does not intersect M. For each remaining x j , it follows that there exists an element m j ∈ M such that B(m j , 2) ⊃ B(x j , ). Thus this collection {m j } has the property that for each element m ∈ M there exists an m j such that |m − m j | < 2. Applying this construction for the sequence n = 2−n gives a countable dense subset of M, as required. Cover X with the collection of open balls
B(x, ).
x∈X
Since X is compact it follows that there exists a finite covering by such 1
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
2
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Char Count= 0
Solutions to Exercises balls: X⊂
N
B(x j , ).
j=1
1.3
It follows that for each x ∈ X there exists an x j with |x − x j | < as required. We first consider the case of bounded. If u ∈ Cc0 ( ) then clearly u = 0 on ∂ ; it follows that if u n ∈ Cc0 ( ) converges to u uniformly on then u = 0 on ∂ too. We now show that any function in C00 ( ) = {u ∈ C 0 ( ) : u = 0 on ∂ } can be arrived at in this way and hence that this space is the completion of Cc0 ( ) in the sup norm. Let θ be the continuous function x ≥ 1, x, θ (x) = 2x − 1, 1 > x > 12 , 0, x ≤ 12 , and define u (x) = θ(|u(x)|/)u(x). Clearly u is continuous on , and since u is uniformly continuous on
there exists a δ such that dist(x, ∂ ) < δ
⇒
|u(x)| < /2,
that is, such that u (x) = 0 when dist(x, ∂ ) < δ. It follows that u ∈ Cc0 ( ), and since, |u(x) − u (x)| ≤ , u converges uniformly to u on . It follows that C00 ( ) = Cc0 ( ) is the completion of Cc0 ( ) in the sup 0 norm, and Cc ( ) is therefore not complete. When = Rm the limit of any convergent sequence of functions in Cc0 (Rm ) must tend to zero as |x| → ∞. This is clear, since given > 0 there exists an N such that |u n − u| ≤ for all n ≥ N . In particular, u N is zero for all x > R N , say, and so |u| ≤ for all x > R N . The space of all such u, C00 (Rm ) = u ∈ Cb0 ( ) : u(x) → 0 as |x| → ∞ , is the appropriate completion of Cc0 (Rm ). For any u ∈ C00 (Rm ), we
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 1
1.4
3
can use the argument above to find an approximating sequence of u ∈ Cc0 (Rm ). If { f j } is Cauchy in the · c norm then it is Cauchy in each C n ( ) norm. Since each C n ( ) is complete, f j → f in each of these spaces, so that f ∈ C n ( ) for every n and thus f ∈ C ∞ ( ). It remains to show that in fact f c < ∞ and that f j − f c → 0 as j → ∞. Since { f j } is Cauchy it certainly follows that for j, k ≥ N we have l
cn f j − f k C n ( ) <
n=1
for each l < ∞, and taking the limit as k → ∞ gives l
cn f j − f C n ( ) < .
(S1.1)
n=1
Using the triangle inequality in each C n ( ), 0 ≤ n ≤ l, shows that l
cn f C n ( ) < +
n=1
l
cn f j C n ( ) ,
n=1
and so f c ≤ + f j c . Since (S1.1) holds for all l, we can let l → ∞ to show that ∞
cn f j − f C n ( ) < ,
n=1
1.5
and so f j → f in the · c norm. We show that C 0,γ ( ) is a Banach space; the case C r,γ then follows easily. If the sequence { f j } is Cauchy in C 0,γ ( ) then given > 0 there
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
4
10:52
Char Count= 0
Solutions to Exercises exists an N such that for j, k ≥ N we have |[ f j (x) − f k (x)] − [ f j (y) − f k (y)]| ≤ . |x − y|γ x,y∈
f j − f k ∞ + sup
Because C 0 ( ) is complete we know that f j converges to some f ∈ C 0 ( ). We just need to show that f is H¨older. However, since f k → f uniformly we have |[ f j (x) − f (x)] − [ f j (y) − f (y)]| ≤ |x − y|γ and so | f (x) − f (y)| ≤ | f j (x) − f j (y)| + |[ f j (x) − f (x)] + [ f j (y) − f (y)]| ≤ C j |x − y|γ + |x − y|γ ,
1.6
which shows that f ∈ C 0,γ ( ). If f ∈ C 1 ( ) then |D f (x)| is uniformly bounded on , by L, say. Since
is convex, given any two points x, y ∈ the line segment joining x and y lies entirely in . It follows that | f (x) − f (y)| =
0
1
D f (y + ξ(x − y)) · (x − y) dξ
≤ L|x − y|,
1.7
so f is Lipschitz. We have x−z y−z |u h (x) − u h (y)| = h −m ρ −ρ u(z) dz h h
x−z ≤ h −m ρ |u(z) − u(z + y − x)| dz h
≤ C|y − x|γ
1.8
by using (1.7) so that u h is also H¨older. We prove the result by induction, supposing that it is true for n = k. Then for n = k + 1 we take p such that k−1 1 1 + = 1, p p j j=1
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 1
5
to obtain | f 1 (x) · · · f k+1 (x)|d x ≤ f 1 L p1 · · · f k−1 L pk−1 f k f k+1 L p .
(S1.2)
Now, we use the standard H¨older inequality, noting that 1=
p p + ; pk pk+1
thus p/ pk
( f k f k+1 ) d x ≤ p
p fk k
dx
pk+1 f k+1
p/ pk+1 dx
,
and so f k f k+1 L p ≤ f k L pk f k+1 L pk+1 ,
1.9
which combined with (S1.2) gives (1.31) for n = k + 1. Since the standard H¨older inequality is (1.31) for n = 2 the result follows. Write p |u(x)| d x = |u(x)|q(r − p)/(r −q) |u(x)|r ( p−q)/(r −q) d x.
Now note that r−p p−q + = 1, r −q r −q and so using H¨older’s inequality we have
|u(x)| p d x ≤
(r − p)/(r −q) ( p−q)/(r −q) |u(x)|q d x |u(x)|r d x ,
which becomes q(r − p)/ p(r −q)
u L p ≤ u L q
r ( p−q)/ p(r −q)
u L r
as required. 1.10 If s ∈ S( ) then it is of the form of (1.10), s(x) =
n j=1
c j χ [I j ](x),
,
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
6
10:52
Char Count= 0
Solutions to Exercises where the I j are m-dimensional cuboids, each of the form I =
m [ak , bk ]. k=1
It clearly suffices to approximate χ [I ] to within in the L p norm using an element of Cc0 ( ). To do this, consider the function χη =
m
φη (xk ; ak , bk ),
k=1
where (x − a)/η, φη (x; b, a) = 1, (b − x)/η,
a ≤ x ≤ a + η, a + η < x < b − η, b − η ≤ x ≤ b.
Clearly χη ∈ Cc0 ( ) and converges to χ [I ] in L p ( ) as η → 0. 1.11 Since |g(x)| ≤ g∞ almost everywhere, it follows that | f (x)g(x)| ≤ | f (x)|g∞ almost everywhere, and so | f (x)g(x)| d x ≤ | f (x)|g∞ d x ≤ f L 1 g∞ ,
as claimed. 1.12 Since {x (n) } is Cauchy, given > 0 there exists an N such that (n) x − x (m) ∞ ≤ for all n, m ≥ N . l This implies that (n) x − x (m) ≤ j j
for all
n, m ≥ N .
(S1.3)
(n) In particular, we have x (n) j is Cauchy for each j. So x j → x j as n → ∞. ∞ It is then clear that x = {x j } ∈ l , and taking the limit m → ∞ in (S1.3) shows that (n) x − x j ≤ for all n ≥ N, for all j. j
It follows that x (n) → x in l ∞ , and so l ∞ is complete.
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 2
7
1.13 We know that the norm is positive definite, and so x + λy2 = (x + λy, x + λy) = x2 + 2λ(x, y) + λ2 y2 ≥ 0. In particular, the quadratic equation for λ, λ2 y2 + 2λ(x, y) + x2 = 0, can have only one distinct real root. Therefore the discriminant “b2 − 4ac” cannot be positive (which would give two real roots). In other words, 4(x, y)2 − 4y2 x2 ≤ 0 or |(x, y)| ≤ xy, which is the Cauchy–Schwarz inequality. We can now write x + y2 = x2 + 2(x, y) + y2 ≤ x2 + 2xy + y2 = (x + y)2 , giving the triangle inequality. 1.14 We simply expand the left-hand side, u + v2 + u − v2 = u2 + 2(u, v) + v2 + u2 − 2(u, v) + v2 = 2u2 + 2v2 , as required. 1.15 If {u j } is a dense subset of l 2 () then for each element γ ∈ there must exist a u j that is within of 1 at γ and within of 0 for all other elements of . Each such u j is distinct. It follows that if is uncountable then so are the {u j }, and so l 2 () cannot be separable.
Chapter 2 2.1
We can apply the contraction mapping theorem to h n to deduce that h n has a unique fixed point x ∗ , h n (x ∗ ) = x ∗ .
P1: FNT CB341/Robinson
CB341-Sol
8
February 5, 2001
10:52
Char Count= 0
Solutions to Exercises If we apply h to both sides then h(h n (x ∗ )) = h n+1 (x ∗ ) = h n (h(x ∗ ))
2.2
=
h(x ∗ ),
showing that h(x ∗ ) is also a fixed point of h n . Since the contraction mapping theorem guarantees that the fixed point of h n is unique, we must have h(x ∗ ) = h ∗ , and so h ∗ is also a fixed point of h. The interval [1, ∞) is closed but not compact, and the map h : [1, ∞) → [1, ∞) given by x → x + 1/x satisfies |h(x) − h(y)| = |x − y|(1 − (x y)−1 ) < |x − y| but clearly has no fixed point. However, if X is compact and h : X → X satisfies h(x) − h(y) < x − y,
(S2.1)
suppose that h has no fixed point. Then h(x) − x > 0
for all
x ∈ X,
and since h(x) − x is continuous from X into R it obtains its lower bound, so that h(x) − x ≥
for all
x ∈ X,
and there exists some y ∈ X such that h(y) − y = . However, if we take z = h(y) then from (S2.1) we have h(z) − z < ,
2.3
a contradiction. So h has at least one fixed point. Uniqueness follows as in the proof of the standard contraction mapping theorem. Take n = 2−n and apply the result of Exercise 1.2 so that there exists (k) −k finite set {x (k) j }, 1 ≤ j ≤ Mk , such that |x − x j | ≤ 2 . Set Nk = k j=1 M j , and let {x j } be the sequence (1) (2) x1(1) , . . . , x M , x1(2) , . . . , x M , x1(3) , . . . . 1 2
2.4
Suppose that there are solutions xn (t) of d x/dt = f (x)
with
x(0) = x0
(S2.2)
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 2
9
such that xn (τ ) → x ∗ . We need to show that there is a solution of (S2.2) with x(τ ) = x ∗ . Now, if f is bounded then the sequence xn (t) satisfies sup |xn (t)| ≤ |x0 | + τ f ∞
and
t∈[0,τ ]
2.5
|xn (t) − xn (s)| ≤ f ∞ |t − s|,
the conditions of the Arzel`a–Ascoli theorem (Theorem 2.5). It follows that there is a subsequence that converges uniformly on [0, τ ], and as in the proof of Theorem 2.6 the limit x(t) satisfies (S2.2). Since xn → x uniformly on [0, τ ], in particular we have x(τ ) = x ∗ as required. When |x| = 0 then it follows that d d |x|2 = 2|x| |x|, dt dt and (2.27) follows immediately. When |x(t0 )| = 0, since C(t) is continuous, for any > 0 we have 1 2
d |x|2 ≤ [C(t0 ) + ]|x| dt
for t − t0 small enough, and so it follows from Lemma 2.7 that 2 |x(t)|2 ≤ [C(t0 ) + ](t − t0 ) . Therefore |x(t + h)| ≤ [C(t0 ) + ]t, and so d |x| ≤ C(t0 ) + . dt + 2.6
Since this holds for any > 0 we have (2.27). If t y(t) = b(s)x(s) ds 0
then dy = b(t)x(t) ≤ a(t)b(t) + b(t)y(t), dt and so t t dy − b(t)y(t) exp − b(s) ds ≤ a(t)b(t) exp − b(s) ds . dt 0 0
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10
10:52
Char Count= 0
Solutions to Exercises If a(t) is increasing then we can replace a(t) on [0, T ] with a(T ), and so t
t d b(s) ds ≤ a(T )b(t) exp − b(s) ds . y(t) exp − dt 0 0 Integrating both sides between 0 and T gives us y(T ) exp −
T
T
≤ a(T )
b(s) ds
0
t b(t) exp − b(s) ds dt
0
0
and so
T
y(T ) ≤ a(T )
T
b(t) exp
b(s) ds dt.
0
t
We can integrate the right-hand side to obtain
y(T ) ≤ a(T ) exp
T
b(s) ds
−1 ,
0
and so, using (2.28), we have x(T ) ≤ a(T ) exp
T
b(s) ds 0
2.7
as claimed. As in the proof of Proposition 2.10 we consider the difference of two solutions, z(t) = x(t) − y(t), which satisfies dz = f (x) − g(y) dt = f (x) − f (y) + f (y) − g(y). We now use Lemma 2.9 to deduce that d |z| ≤ | f (x) − f (y)| + | f (y) − g(y)| dt + ≤ L|z| + f − g∞ . An application of Gronwall’s inequality [(2.21) in Lemma 2.8] now yields (2.29).
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 3
11
Chapter 3 3.1
We denote A1 = {smallest M such that AxY ≤ Mx X for all x ∈ X } and A2 = sup AxY . x X =1
First we take x = 0 and put y = x/x X ; then we have AyY ≤ A2
⇒
AxY ≤ A2 x X
for all x ∈ X , and so A1 ≤ A2 . Furthermore, it is clear that, for any M, AxY ≤ Mx X 3.2
x∈X
for all
⇒
A2 ≤ M,
and so A2 ≤ A1 . Thus A1 = A2 . I is clearly bounded from C 0 ([O, L]) into itself, since I ( f )∞ ≤ L f ∞ . For the L 2 bound, first observe, by using the Cauchy–Schwarz inequality, that I ( f )(x) is defined for all x if f ∈ L 2 . Then L |I ( f )|2 = |I ( f )(x)|2 d x 0
L
= 0
x
2 f (s) ds
0
L
=
dx
x
≤L
2
0 L
| f (s)| ds d x 2
ds 0
x
0
| f (s)|2 ds
0
≤ L 2 | f |2 . 3.3
Thus I is a bounded operator on both spaces. Suppose that A−1 y1 = x1 and that A−1 y2 = x2 . Then it is clear that A(x1 + x2 ) = y1 + y2 . Since the inverse is unique it follows that A−1 (y1 + y2 ) = A−1 y1 + A−1 y2 .
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
12 3.4
10:52
Char Count= 0
Solutions to Exercises For each x ∈ X, Pn x converges to x, and so it follows that the sequence {Pn x}∞ n=1 is bounded: sup Pn x X < ∞
n∈Z+
for each x ∈ X . From the principle of uniform boundedness (Theorem 3.7) we immediately obtain sup Pn op < ∞,
n∈Z+
3.5
as claimed. It is clear that φi (x)φ j (y) is an element of L 2 ( × ) and that [φi (x)φ j (y)][φk (x)φl (y)] d x d y = δik δ jl ,
×
and so they certainly form an orthonormal set. If k ∈ L 2 ( × ) then k(·, y) ∈ L 2 ( ), and we can write k(x, y) =
∞
u i (y)φi (x),
i=1
where
u i (y) =
Since
k(x, y)φi (x) d x.
2 |u i (y)| dy = k(x, y)φi (x) d x dy
≤ |k(x, y)|2 d x |φi (x)|2 d x dy
2
≤
×
|k(x, y)|2 d x d y,
we have u i ∈ L 2 ( ). So we can write ∞ u i (y) = u i (y)φ j (y) dy φ j (y), j=1
which yields the expression ∞ k(x, y) = k(x, y)φi (x)φ j (y) d x d y φi (x)φ j (y), i, j=1
as claimed.
×
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 3 3.6
13
We consider the approximations to A given by the truncated sums, An u =
n
λ j (u, w j )w j .
j=1
Using Lemma 3.12 we see that each operator An is compact. We now want to show that A − An op → 0, and it then follows from Theorem 3.10 that A is compact. However, this convergence is clear, since ∞ (A − An )u = λ j (u, w j )w j j=n+1 ∞ (u, w j )w j ≤ λn+1 ≤ λn+1
j=n+1 ∞
1/2 |(u, w j )|
2
j=n+1
≤ λn+1 u, and λn+1 → 0 as n → ∞. Thus A is compact. That A is symmetric follows by taking the inner product of Au with v to give (Au, v) =
∞
λ j (u, w j )(v, w j ) = (u, Av).
j=1
3.7
We know from Lemma 3.4 that A−1 exists iff Ker(A) = 0. So we show that if Ax = 0 then x = 0. Because A is bounded below we have 0 = AxY ≥ kx X , so that x X = 0. For y ∈ R(A) we can use the lower bound on A to deduce that A−1 y X ≤ so that A−1 is bounded.
1 1 A A−1 yY = yY , k k
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
14 3.8
10:52
Char Count= 0
Solutions to Exercises Since G is a solution of the homogeneous equation on both sides of x = y, we must have G(x, y) =
C1 (y)u 1 (x), a ≤ x < y, C2 (y)u 2 (x), y ≤ x ≤ b.
The conditions at y require C1 (y)u 1 (y) = C2 (y)u 2 (y), C1 (y)u 1 (y) + p(y)−1 = C2 (y)u 2 (y). Solving these simultaneous equations for C1 and C2 gives C1 (y) = u 2 (y)/W p (y)
and
C2 (y) = u 1 (y)/W p (y),
where W p (y) = p(y)[u 1 (y)u 2 (y) − u 2 (y)u 1 (y)]. Differentiating W p with respect to y and cancelling the pu 1 u 2 terms gives W p = p (u 1 u 2 − u 2 u 1 ) + p[u 1 u 2 − u 2 u 1 ].
3.9
If we use the differential equation L[u 1 ] = L[u 2 ] = 0 to substitute for the terms pu 1 and pu 2 we see that in fact W p = 0, so that W p is a constant. We therefore obtain (3.28), and G(x, y) is symmetric. Proposition 3.13 and Lemma 3.16 show that the integral operator K defined by [K u](x) = k(x, y)u(y) dy
is a compact symmetric mapping from L 2 ( ) into L 2 ( ). It follows from Theorem 3.18 that K has a set of eigenfunctions u n (x) with corresponding eigenvalues λn , so that K u n = λn u n : k(x, y)u n (y) dy = λn u n (x).
Since λ j = 0 for all j there is no nonzero u such that K u = 0. In this case KerK = {0}, and so we can expand any f ∈ L 2 ( ) in terms of the
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 3
15
eigenfunctions of K , ∞
f =
( f, u j )u j .
j=1
It is now easy to see that the solution of (3.29) is given by u(x) =
∞ ( f, u j )
λj
j=1
as claimed. 3.10 We have A−α w j =
1 (α)
Now,
∞
u j (x),
t α−1 e−λ j t dt w j .
(S3.1)
0
(x) =
∞
t x−1 e−t dt,
0
and so, substituting u = λ j t in (S3.1), we have ∞ du λ1−α u α−1 e−u = λ−α j j (α), λj 0 which gives A−α w j = λ−α j wj as required. Since A−α is characterised by its action on the eigenfunctions the two expressions are equivalent. 3.11 We have As u2 =
∞
2 λ2s j |c j |
j=1
=
∞
2(1−ϕ) λ2(s−α) |c j |2ϕ λ2α j j |c j |
j=1
≤
∞ j=1
ϕ 2(s−α)/ϕ λj |c j |2
∞
1−ϕ 2α/(1−ϕ) λj |c j |2
j=1
2ϕ 2(1−ϕ) ≤ A(s−α)/ϕ u Aα/(1−ϕ) u , which gives the result on setting ϕ = (k − s)/(k − l) and α = k(s − l)/(k − l).
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
16
10:52
Char Count= 0
Solutions to Exercises
3.12 Take x =
∞ j=1
x j w j and consider the series expansion
∞ −λ j h (e−Ah − I ) (e − 1) x + Ax = + λj xjwj. h h j=1
(S3.2)
Observe that
∞ −λ j h ∞ −λ j h (e (e − 1) − 1) + λj xjwj = + 1 λjxjwj. h λjh j=n+1 j=n+1 The mean-value theorem tells us that (e−z − 1)/z ≤ 1, and so 2
∞ −λ j h ∞ (e − 1) + λj xjwj ≤ 4 λ2j |x j |2 , h j=n+1
(S3.3)
j=n+1
which tends to zero as n → ∞. It follows that given > 0 we can choose an n such that the infinite sum in (S3.3) is bounded above by /2. It is then clear that the finite sum
n −λ j h (e − 1) + λj xjwj h j=1 converges to zero as h → 0, and so for small enough h the whole expression in (S3.2) is bounded by , as required. Chapter 4 4.1
4.2
Let P = {orthonormal subsets of H }, and define an order on P such that a ≤ b if a ⊆ b. If {Ci } is a chain (i ∈ I) then C = ∪i Ci is an upper bound. Zorn’s lemma implies that there is a maximal orthonormal set {ei }i∈I . The argument of the second part of Proposition 1.23 now shows that the {ei } form a basis. Take z ∈ / Y . Then if w is contained in the linear span of z and Y it has a unique decomposition of the form w = y + αz
with
y ∈ Y,
as in the proof of the Hahn–Banach theorem. We can therefore define a nonzero linear functional on the linear span of z and Y via f (y + αz) = α. The functional f is zero on Y , and we can extend it to a nonzero linear functional on X by using the Hahn–Banach theorem.
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 4 4.3
17
It is immediate from H¨older’s inequality that |L f (g)| ≤ f L ∞ g L 1 and so L f (L 1 )∗ ≤ f L ∞ .
(S4.1)
To show equality consider the sequence of functions g p (x) = | f (x)| p−2 f (x). Since f ∈ L ∞ ( ) and is bounded we have g p (x) ∈ L 1 ( ) for every p, with p−1
g p L 1 = f L p−1 . It follows from p
|L f (g p )| = f L p that p
L f (L 1 )∗ ≥
f L p p−1
f L p−1
.
Since f ∈ L ∞ we can use the result of Proposition 1.16, f L ∞ = lim f L p , p→∞
to deduce that L f (L 1 )∗ ≥ f L ∞ , 4.4
which combined with (S4.1) gives the required equality. Since M is a linear subspace of H it is also a Hilbert space. The Riesz theorem then shows that given a linear functional f on M there exists an m ∈ M such that f (x) = (m, x)
for all
x ∈ M.
Now define F on H by F(u) = (m, u); 4.5
it is clear that F is an extension of F and that F = f . If x ∈ / M then the argument of Solution 4.2 shows that there exists an element f ∈ X ∗ with f | M = 0 but f (x) = 0. So if f (x) = 0 for all
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
18
10:52
Char Count= 0
Solutions to Exercises such f then we must have x ∈ M. Now if xn x then for each f ∈ X ∗ with f | M = 0 we have f (x) = lim f (xn ) = 0, n→∞
and so it follows that x ∈ M. The linear span of the {xn } forms a linear subspace M of X , and clearly xn ∈ M for each n. It follows that x is contained in the linear span of the {xn } and so can be written in the form x=
∞
cjxj.
(S4.2)
j=1
4.6
[In fact x can be written as a convex combination of the {x j }, that is, (S4.2) with c j ≥ 0 and j c j = 1; see Yosida (1980, p. 120).] For any t ∈ [a, b], δt : x → x(t) is a bounded linear functional on C 0 ([a, b]). Since xn x, we have δt (xn ) → δt (x),
4.7
and so xn (t) → x(t) for each t ∈ [a, b]. Write xn − x2 = xn 2 + x2 − 2(x, xn ), and then take limits on the right-hand side, using norm convergence on xn 2 and weak convergence on (x, xn ), to show that lim xn − x2 = 0,
n→∞
which is xn → x.
Chapter 5 5.1
Simply write D α u, φn = (−1)|α| u, D α φn , and then using the definition of convergence in D( ) (Definition 5.2)
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 5
19
we have D α φn → D α φ in D( ), and so D α u, φn → (−1)|α| u, D α φ = D α u, φ, 5.2
so that D α u is indeed a distribution. If φn ∈ D( ) with φn → φ in D( ) then ψφn → ψφ in D( ). It follows that ψu, φn = u, ψφn → u, ψφn = ψu, φ, and so ψu ∈ D ( ). Given φ ∈ D( ) we have D(ψu), φ = −ψu, φ = −u, ψφ = −u, ψφ + φψ + u, φψ = Du, ψφ + u Dψ, φ = ψ Du + u Dψ, φ,
5.3
as claimed. Assume that | f n | ≤ M for every n. For every φ ∈ Cc∞ ( ) we know that fn φ d x (S5.1)
Since Cc∞ ( )
is a Cauchy sequence. is dense in L 2 ( ) (Corollary 1.14), 2 for each u ∈ L ( ) we can find a sequence of φn ∈ Cc∞ ( ) with φn → u in L 2 ( ). Then, given > 0, choose K such that |φk − u| ≤ /4M and then choose N such that ( f n − f m )φk d x ≤ /2
for all
for all
k≥K
n, m ≥ N .
It follows that for all n, m ≥ N ( f n − f m )u d x ≤ | f n − f m ||u − φk | + /2
≤ (2M)(/4M) + /2 = , and so (S5.1) is a Cauchy sequence for every u ∈ L 2 ( ), showing that f n f in L 2 ( ).
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
20 5.4
10:52
Char Count= 0
Solutions to Exercises Suppose that the result is true for k = n. We show that it holds for k = n + 1, which then gives a proof by induction since the statement of Proposition 5.8 gives (5.45) for k = 1. We know that u2H n+1 = u2H n + |D α u|2 , |α|=n+1
which along with the induction hypothesis becomes u2H n+1 ≤ C(n) |D α u|2 + |D α u|2 . |α|=n
(S5.2)
|α|=n+1
We therefore consider |D α u| for |α| = n. Since u ∈ H0n+1 ( ) we must have D α u ∈ H01 ( ), and so |D α u| ≤ C|D1 D α u| = C|D β u| with |β| = n + 1, by using (5.11) from the proof of Proposition 5.8. It follows from (S5.2) that u2H n+1 ≤ C(n + 1) |D α u|2 , |α|=n+1
5.5
which is the result for k = n + 1. Consider a sequence of u n ∈ C ∞ ( ) that approximates u in H k ( ). Then the derivatives of ψu n are given by the Leibniz formula (1.6) α α D (ψu n ) = D β ψ D α−β u n , β β≤α
and so |D α (ψu n )| ≤
α β≤α
≤
β
|D β ψ||D α−β u n |
α β≤α
β
|D β ψ| u n H k .
In this way the derivatives up to and including order k are bounded in L 2 by a constant (depending on ψ) times the H k norm of u n , and so ψu n H k ≤ C(ψ)u n H k . It follows that ψu n is Cauchy in H k ( ), and so in the limit as n → ∞ we have ψu ∈ H k ( ) with ψu H k ≤ C(ψ)u H k as required.
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 5 5.6
21
First we show that u ∈ L 2 (B(0, 1)):
B(0,1)
2 2π 1 1 log log 1+ dx dy = r log log(1+1/r ) dr dθ, |x| 0 0
which is finite since the integrand is bounded. Now, since ∂u 1 x = 2 ∂x log(1 + 1/|x|) |x| (1 + |x|) we have
|∇u(x)|2 d x d y B(0,1)
=
B(0,1) 2π
=
0
1 1 dx dy log(1 + 1/|x|)2 |x|2 (1 + |x|)2 1
1 1 dr. log(1 + 1/r )2 r (1 + r )2
0
(S5.3)
If we make the substitution u = 1/r this becomes ∞ 1 1 du. u + (1/u) log(1 + u)2 1 This integral is bounded by
∞
1
5.7
1 du, u(log u)2
and since the integrand is the derivative of −1/ log u it follows that the integral in (S5.3) is finite. Therefore u ∈ H 1 (B(0, 1)), even though it is unbounded. First integrate (5.46) with respect to x1 , so that
∞
−∞
|u(x)| d x1 ≤ 6 3
−∞
× ≤6
u D1 u dy1 1/2
∞
−∞
u D3 u dy3
−∞
−∞
1/2
∞
−∞
u D2 u dy2
d x1
u D1 u dy1 u D3 u d x1 dy3
1/2
∞
−∞
1/2
∞
−∞
∞
1/2
∞
×
1/2
∞
.
u D2 u d x1 dy2
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
22
10:52
Char Count= 0
Solutions to Exercises Now integrate with respect to x2 to obtain
∞
−∞
|u(x)|3 d x1 d x2 ≤ 6
×
−∞
u D2 u d x1 dy2
1/2
∞
−∞
1/2
∞
u D1 u dy1 d x2
1/2
∞
−∞
u D3 u d x1 d x2 dy3
.
Finally, integrating with respect to x3 gives |u(x)| d x ≤ 6 3
3 j=1
1/2 uDju dx
,
and so u3L 3 ≤ C|u|3/2 |Du|3/2 , which gives 1/2
u L 3 ≤ C|u|1/2 u H 1 , 5.8
as required. We simply apply the argument of Theorem 5.29 to the functions v = D α u for each α with |α| ≤ j. It follows that v ∈ H k− j ( ), and since k − j > m/2 we can use Theorem 5.29 to deduce that v ∈ C 0 ( ) with v∞ ≤ Cu H k− j ≤ Cu H k . Combining the estimates for each |α| ≤ j shows that u ∈ C j ( ) with uC j ( ) ≤ Cu H k ( )
5.9
as claimed. Suppose that the inequality does not hold. Then for each k ∈ Z+ there must exist u k ∈ V such that |u k | ≥ k|∇u k |.
(S5.4)
If we set vk = u k /|u k | so that |vk | = 1, (S5.4) becomes |∇vk | ≤ k −1 .
(S5.5)
It follows that vk is a bounded sequence in H 1 ( ), and so using Theorem 5.32 it has a subsequence that converges in L 2 ( ) to some
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 5
23
v ∈ V with
v(x) d x = 0
|v| = 1.
and
(S5.6)
However, we can also use (S5.5) along with the L 2 convergence of vk to v to show that for any φ ∈ D( ) and any j
v D α φ d x = lim
k→∞
vk D j φ d x = − lim
k→∞
D j vk φ d x = 0.
It follows that Dv = 0, and so, using the hint, v is constant almost everywhere. This contradicts (S5.6), and so we have the inequality (5.47). 5.10 Suppose that {u n } is a bounded sequence in L 2 ( ). Then, since L 2 is reflexive, there is a subsequence that converges weakly in L 2 ( ), i.e. for every φ ∈ L 2 ( ) we have (u n , φ) → (u, φ) for some u ∈ L 2 ( ). Now, suppose that u n does not converge to u in H −1 ( ), so that there exists an > 0 such that, for some subsequence {u n }, sup {φ∈H01 ( ):φ H 1 =1}
|(u n − u, φ)| ≥ .
0
Then there exist φn with φn H01 = 1 such that |(u n − u, φn )| ≥ /2. Since {φn } is a bounded sequence in H01 ( ) and H01 ( ) is compactly embedded in L 2 ( ), there exists a subsequence that is convergent in L 2 ( ) to some φ. It follows (on relabelling) that |(u n − u, φ)| ≥ /4 for n large enough. But this contradicts the weak convergence of u n to u in L 2 , and so we must have u n → u in H −1 ( ). 5.11 Since u=
k∈Zm
ck e2πik·x/L
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
24
10:52
Char Count= 0
Solutions to Exercises we have Du = It follows that |u|2 = L m |ck |2
2πik ck e2πik·x/L . L k∈Zm
|Du|2 = L m
and
k∈Zm
(4π/L)2 |k|2 |ck |2 ,
k∈Zm
and so
|u| ≤
L |Du| 2π
as claimed.
Chapter 6 6.1
Start with
∇u · ∇v d x =
f (x)v(x) d x,
and integrate the left-hand side by parts to give ( u − f )v d x = 0.
Since u ∈ C 2 ( ) and f ∈ C 0 ( ), we have ϕ ≡ u − f ∈ C 0 ( ). It therefore suffices to show that if ϕv d x = 0 for all
v ∈ Cc1 ( )
then ϕ = 0. Suppose that ϕ(x) = 0 for some x ∈ . Then since ϕ is continuous there is a neighbourhood N of x on which ϕ(x) is of constant sign. Taking a function v that is positive and has compact support within N implies that ϕv d x = ϕv d x = 0,
N
a contradiction. That u satisfies u|∂ = 0 follows from u ∈ H01 ( ) ∩ C 0 ( ), using Theorems 5.35 and 5.36.
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 6 6.2
25
Take the inner product of Lu = f with a v ∈ Cc1 ( ), m m ∂ ∂u ∂u − ai j (x) v(x)+ bi (x) v(x)+c(x)u(x)v(x) d x ∂x j ∂ xi
i, j=1 ∂ x i i=1 = f (x)v(x) d x,
and integrate the first term by parts, m
i, j=1
=
∂u ∂v ∂u + bi (x) v(x) + c(x)u(x)v(x) d x ∂ xi ∂ x j ∂ xi i=1 m
ai j (x)
f (x)v(x) d x.
We can now introduce a bilinear form a(u, v) =
m
i, j=1
∂u ∂v ∂u + bi (x) v(x) + c(x)u(x)v(x) d x, ∂ xi ∂ x j i=1 ∂ xi m
ai j (x)
and write the equation as a(u, v) = ( f, v)
for all
v ∈ Cc1 ( ).
As before we use the density of Cc1 ( ) in H01 ( ) to generalise to f ∈ H −1 ( ) and the weak form of the problem is thus to find u ∈ H01 ( ) such that a(u, v) = f, v 6.3
for all
v ∈ H01 ( ).
By definition a(u, u) =
m i, j=1
ai j (x)D j u Di u d x +
m i=1
bi (x)Di u u d x
+ c(x)u 2 d x
2 ∞ ≥θ |∇u| d x − max bi L |∇u| |u| d x i
− c L ∞ |u|2 d x.
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
26
10:52
Char Count= 0
Solutions to Exercises Now we use Young’s inequality with , ab ≤
a2 b2 + , 2 2
to split the second term, −1 a(u, u) ≥ 12 θ |∇u|2 d x − θ max bi L ∞ |u|2 d x i
− c L ∞ |u|2 d x,
and so a(u, u) ≥ Cu2H1 − λ|u|2 , 6.4
as required. Consider the bilinear form b(u, v) corresponding to the operator L + α. Then b(u, v) = a(u, v) + α (u, v) L2
is a continuous bilinear form on H01 : clearly (u, v) is, and a(u, v) is since |a(u, v)| ≤
m
|ai j ||D j u||Di v| d x +
i, j=1
+
|c||u||v| d x
m i=1
|bi ||Di u||v| d x
≤ Cu H 1 v H 1 . Furthermore, b satisfies the coercivity condition, since b(u, u) = a(u, u) + α(u, u) ≥ Cu2H1 − λ|u|2 + α|u|2 ≥ Cu2H 1 . 0
6.5
We can now apply the Lax–Milgram lemma to obtain the conclusion. First, it is easy to see that if (6.31) holds for all v ∈ H 1 ( ) then choosing v = 1 we have f (x) d x = 0.
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 6
27
We cannot immediately apply the Lax–Milgram lemma to the equation a(u, v) = ( f, v), since |a(u, u)| = |∇u|2 = u2H 1 − |u|2 , and so a(u, v) is not coercive. To deal with the L 2 part we need a Poincar´etype inequality. Note that if f (x) d x = 0, then ( f, v) =
f, v −
v(x) d x ,
since subtracting the constant from v does not make any difference, and similarly
a(u, v) = a u, v −
v(x) d x .
The weak form of the equation in this case [ equivalent to a(u, v) = ( f, v)
f (x) d x = 0] is therefore
for all
v ∈ V,
where V =
u ∈ H 1 ( ) :
u(x) d x = 0 .
It is was shown in Exercise 5.9 that in this space |u| ≤ C|∇u|, and so we have |a(u, u)| = |∇u|2 ≥
6.6
1 |u|2 + 12 |∇u|2 ≥ ku2H 1 . 2C
We can therefore apply the Lax–Milgram lemma to deduce the existence of a weak solution of the Neumann problem. Without the imposition of the condition Q u(x) d x = 0 Laplace’s equation on Q with periodic boundary conditions does not have a unique
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
28
10:52
Char Count= 0
Solutions to Exercises weak solution. In terms of the Lax–Milgram lemma this translates into the weak problem a(u, v) =
∇u · ∇v d x = f, v
with
f ∈ H −1 (Q),
Q
6.7
where we seek u ∈ L 2 (Q). But then a is not coercive on L 2 (Q), since a(c, c) = 0 for any constant c. First, u(x + hei )v(x + hei ) − u(x)v(x) Dih (uv)(x) = h
v(x + hei ) − v(x) u(x + hei ) − u(x) = u(x) + v(x + hei ) h h = u(x)Dih v(x) + v(x + hei )Dih u(x). Next, we write
u(x + hei ) − u(x) v(x) d x h
u(x + hei ) u(x) = v(x) d x − v(x) d x h
h
and change variables in the first integral, putting y = x + hei , to obtain
u(y) u(x) v(y − hei ) dy − v(x) d x h
h v(x − hei ) − v(x) = − u(x) dx −h
= − u(x)Di−h v(x) d x.
Finally, both expressions are equal to Di u(x + he j ) − Di u(x) . h 6.8
The inverse of xi =
is just the map y → x, given by
i = 1, . . . , m − 1, yi + z i , ym + ψ(y1 + z 1 , . . . , ym−1 + z m−1 ), i = m.
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 7
29
Therefore
D1 ψ D2 ψ D3 ψ . .. . 1
1 0 0 ... 0 1 0 ... ∇ψ = 0 0 1 . . . . . . . .. .. .. .. 0 0 0 ...
6.9
It follows immediately that det ∇! = 1, as required. The result of Lemma 3.26 shows that, for a general positive symmetric linear operator whose inverse is compact, |As u| ≤ C|Al u|(k−s)/(k−l) |Ak u|(s−l)/(k−l) for 0 ≤ l < s < k. A = − on with Dirichlet boundary conditions certainly satisfies these conditions. Taking u ∈ H0k ( ), Proposition 6.19 shows that u ∈ D(A j/2 ) for all j = 0, 1, . . . , k, and so u H j ( ) ≤ |A j/2 u| ≤ C j u H j ( ) . Therefore we have (k−s)/(k−l)
u H s ( ) ≤ Cu H l ( )
(s−l)/(k−l)
u H k ( )
(S6.1)
for all such u. Now take u ∈ H k ( ), and use Theorem 5.20 to extend u to a function Eu ∈ H0k ( ) for some ⊃ . Then (S6.1) holds for Eu, and since E j is bounded from H j ( ) into H0 ( ) for each 0 ≤ j ≤ k, we have (k−s)/(k−l)
u H s ( ) ≤ Eu H s ( ) ≤ CEu H l ( )
(k−s)/(k−l)
≤ Cu H l ( )
(s−l)/(k−l)
Eu H k ( ) (s−l)/(k−l)
u H k ( )
,
which is (6.32) for u ∈ H k ( ), as required.
Chapter 7 7.1
Define an element I ∈ X I, L = 0
T
∗∗
by
L , f (t) dt
for all
L ∈ X ∗.
(S7.1)
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
30
10:52
Char Count= 0
Solutions to Exercises This map I is clearly linear, and it is bounded since T |I, L| ≤ L X ∗ f (t) X dt 0
T
≤
f (t) X dt L X ∗ ,
0
and
T
f (t) X dt < ∞
0
from (7.31). Since X is reflexive, it follows that there exists an element y ∈ X such that I, L = L , y
for all
L ∈ X ∗.
Therefore, using (S7.1), we have (7.29). That the integral is well defined follows from Lemma 4.4, which shows that if L , y1 = L , y2 7.2
for all
L ∈ X∗
then y1 = y2 . Corollary 4.5 shows that there exists an element L ∈ X ∗ such that Lop = 1 and L y = y X . Then, using (7.29), we have T T ≤ f (t) dt |L , f (t)| dt 0
0
X
≤
T
f (t) X dt,
0
7.3
as required. An element v of L p (0, T ; V ) is the limit in the L p norm of a sequence of functions vn in C 0 ([0, T ]; V ). Since such functions are uniformly continuous on [0, T ], given > 0 we can find an integer N such that δ = T /N satisfies |t − s| < δ
⇒
vn (t) − vn (s)V ≤ /T 1/ p .
We can approximate vn to within in L p (0, T ; V ) by N j=1
vn ( jδ)χ [( jδ, ( j + 1)δ)],
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 7
7.4
31
an expression of the form (7.32). It follows that such elements are dense in L p (0, T ; V ). Since C 1 ([0, T ]) is dense in L p (0, T ) we could also use elements of the form of (7.32) with α j ∈ C 1 ([0, T ]); similarly, Cc∞ ( ) is dense in L p ( ), so we could take v j ∈ Cc∞ ( ). Taking the inner product of (7.33) with Ak u n yields 1 2
d k/2 2 k+1 2 k−1 k+1 |A u n | + A 2 u n ≤ A 2 f A 2 u n , dt
and so, using Young’s inequality, we obtain d k/2 2 k+1 2 k−1 2 |A u n | + A 2 u n ≤ A 2 f , dt which shows that
|Ak/2 u n (t)|2 +
t
2 k+1 A 2 u n (s)2 ds ≤ |Ak/2 u(0)|2 + A k−1 2 f ,
0
which yields (7.34), and then (7.35) follows from (7.33). Therefore, using Proposition 6.18, we get u n ∈ L ∞ (0, T ; H k ) ∩ L 2 (0, T ; H k+1 ) and du n /dt ∈ L 2 (0, T ; H k−1 ). Extracting a subsequence shows that the solution u satisfies u ∈ L 2 (0, T ; H k+1 ) 7.5
and
du/dt ∈ L 2 (0, T ; H k−1 ).
It follows from Corollary 7.3 that u ∈ C 0 ([0, T ]; H k ). Since the {w j } are orthogonal in H01 and orthonormal in L 2 the equation for u n becomes λ j un j = f j , where λ j ≡ w j 2 and f j = ( f, w j ). It follows that u n j = f j /λ j , independent of n. In particular we have un =
n fj wj, λ j=1 j
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
32
10:52
Char Count= 0
Solutions to Exercises and so, for m > n, m
u m − u n 2 ≤
j = n+1
f j2 λj
.
Since we have the Poincar´e inequality we must have λ j ≥ λ, for some λ, and so m 1 2 u m − u n ≤ f . λ j = n+1 j 2
Since f ∈ L 2 ( ) it follows that u n converges in H01 ( ) to u = ∞ j=1 f j w j /λ j . Now, we know that ((u n , v)) = (Pn f, v)
v ∈ Pn H01 ( ).
for all
Since ((u n , v)) = ((u n , Pn v))
and
(Pn f, v) = (Pn f, Pn v)
for all v ∈ H01 ( ), we in fact have ((u n , v)) = (Pn f, v)
v ∈ H01 ( ).
for all
Since u n → u in H01 ( ) we know that ((u n , v)) → ((u, v)), and since Pn f → f in L 2 ( ) we must have ((u, v)) = ( f, v)
for all
v ∈ H01 ( ),
and u is a weak solution of (7.36) as required.
Chapter 8 8.1
We show in general that if Z = X ∩ Y , with norm u Z = u X + uY ,
P1: FNT CB341/Robinson
CB341-Sol
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Char Count= 0
Chapter 8
33
then Z ∗ = X ∗ + Y ∗ . First, it is clear that if f = f 1 + f 2 , with f 1 ∈ X ∗ and f 2 ∈ Y ∗ , then for u ∈ X ∩ Y | f 1 + f 2 , u| ≤ | f 1 , u| + | f 2 , u| ≤ f 1 X ∗ u X + f 2 Y ∗ uY ≤ ( f 1 X ∗ + f 2 Y ∗ )u X ∩Y .
8.2
Thus X ∗ + Y ∗ ⊂ (X ∩ Y )∗ . Now, if f ∈ (X ∩ Y )∗ then, since it is a linear functional on a linear subspace of X , application of the Hahn–Banach theorem (Theorem 4.3) tells us it has an extension f 1 that is a linear functional on the whole of X (we could use Y rather than X here if we wished). Thus (X ∩ Y )∗ ⊂ X ∗ ⊂ X ∗ + Y ∗ , and so we have the required equality. Follow the argument of Theorem 7.2, except approximate u by a sequence u n ∈ C 1 ([0, T ]; H 1 ) such that un →
L 2 (0, T ; H 1 ) ∩ L p ( T )
in
and du n /dt → du/dt
in
L 2 (0, T ; H −1 ) + L q ( T ).
We will denote by X (t1 , t2 ) the space L 2 (t1 , t2 ; H 1 ) ∩ L p ( × (t1 , t2 )), and by X ∗ (t1 , t2 ) the space L 2 (t1 , t2 ; H −1 ) + L q ( × (t1 , t2 )). We now estimate T t 1 |u n (t)|2 d x = |u n (t)|2 dt d x + 2 u˙ n (s)u n (s) ds T 0
t∗ T 1 ≤ |u n (t)|2 dt d x + 2u˙ n X ∗ (t ∗ ,t) u n X (t ∗ ,t) T 0 T 1 ≤ |u n (t)|2 dt d x + 2u˙ n X ∗ (0,T ) u n X (0,T ) , T 0 showing once again that u n is also a Cauchy sequence in C 0 ([0, T ]; L 2 ) and hence that u ∈ C 0 ([0, T ]; L 2 ) as claimed.
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
34
Solutions to Exercises
8.3
Integrating by parts gives ∂ 2un − f (u n ) 2 d x ∂x j
j ∂u n 2 dx + = f (u n ) f (u n )∇u n · n d S. ∂x j
j ∂
We can estimate the extra term by f (u n )∇u n · n d S ≤ | f (0)| ∂
∂
|∇u n | d S
≤ | f (0)||∂ |1/2 ∇u n L 2 (∂ ) ≤ Cu n H 1 (∂ ) ≤ Cu n H 2 ( ) , using the trace theorem (Theorem 5.35). Since we have u H 2 ( ) ≤ C|Au| from Theorem 6.16, we can write 1 2
d u n 2 + |Au n |2 ≤ lu n 2 + C|Au n |. dt
Using Young’s inequality on the last term and rearranging finally gives d u n 2 + |Au n |2 ≤ 2lu n 2 + C, dt which integrates to give the bound T 2 2 u n (T ) + |Au n (s)| ds ≤ 2l 0
8.4
T
u n (t)2 dt + u 0 2 + C T.
0
Thus u n is uniformly bounded in L 2 (0, T ; D(A)) [and L ∞ (0, T ; V )], where (8.19) is used as before to guarantee that u n ∈ L 2 (0, T ; V ). In this case we can follow the proof of Proposition 8.6 until the line |F(u) − F(v)|2 ≤ C|u − v|2L 2 p 1 + |u|2L 2qγ + |v|2L 2qγ . We now have to be more careful with our use of the Sobolev embedding theorem, since the highest we can go is H 1 ⊂ L 6 . We therefore need 2p ≤ 6
and
2qγ ≤ 6,
where ( p, q) are conjugate.
The first conditions forces us to take p ≤ 3, and hence we must have
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 8
35
q ≥ 3/2, which shows that the largest possible value for γ is 2, as claimed. Provided that γ ≤ 2 we can write |F(u) − F(v)|2 ≤ Cu − v2H 1 (1 + u H 1 + v H 1 ), 8.5
as in Proposition 8.6. Since u, v ∈ L 2 (0, T ; D(A) we can use Corollary 7.3 to take the inner product of dw + Aw = F(u) − F(v) dt with Aw to obtain, using (8.31), 1 2
d w2 + |Aw|2 = (F(u) − F(v), Aw) dt ≤ C(1 + |Au| + |Av|)1/2 w1/2 |Aw|3/2 .
We now use Young’s inequality to split the right-hand side, 1 2
d 3 w2 + |Aw|2 ≤ |Aw|2 + C(1 + |Au| + |Av|)2 w2 , dt 4
and so 1 2
d w2 ≤ C(1 + |Au| + |Av|)2 w2 . dt
This yields w(t)2 ≤ w(0)2 exp
t
C(1 + |Au(s)|2 + |Av(s)|2 ) ds ,
0
8.6
which gives continuous dependence on initial conditions since we know that both u and v are elements of L 2 (0, T ; D(A)). Setting g(s) = e−A(t−s) u(s), we have ∂g du = Ae−A(t−s) u(s) + e−A(t−s) ∂s ds = Ae−A(t−s) u(s) + e−A(t−s) [−Au + f (u(s))] = e−A(t−s) f (u(s)), so that integrating with respect to s between 0 and t gives t g(t) − g(0) = e−A(t−s) f (u(s)) ds. 0
This rearranges to give (8.33).
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
36
10:52
Char Count= 0
Solutions to Exercises Chapter 9
9.1
Taking the divergence of the governing equation yields u = ∇ · f,
9.2
since all the other terms are divergence free. A solution of this equation in the periodic case when f ∈ L˙ 2 (Q) has been obtained as Equation (9.10). Note that if f ∈ H then this implies that p = 0 (or, equivalently, a constant). We have 1/2 1/2 |u|4L 4 = |u|4 d x ≤ |u|6 |u|2 Q
=
Q
Q
u3L 6 |u|
≤ ku3 |u|, 9.3
since H 1 (Q) ⊂ L 6 (Q) (see Theorem 5.31). Applying the Cauchy–Schwarz inequality first in the variable j and then in the variable i, we get m 1/2 m 1/2 m ai bi, j c j ≤ |ai bi, j |2 |c j |2 i, j=1
= ≤
i, j=1
i=1
m i=1
9.4
j=1
m 1/2 m m bi, j |c j |2 ai j=1
1/2 |ai |2
j=1
m
1/2
|bi, j |2
i, j=1
m j=1
as claimed. If m = 2, we have |b(u, v, w)| ≤ k|u|1/2 u1/2 |v|1/2 v1/2 w [using b(u, v, w) = −b(u, w, v)], so that B(u, u), w ≤ k|u|uw, and therefore B(u, u)V ∗ ≤ k|u|u.
1/2 |c j |2
,
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 9
37
If m = 3, we have |b(u, v, w)| ≤ k|u|1/4 u3/4 |v|1/4 v3/4 w [using b(u, v, w) = −b(u, w, v) again], and so B(u, u), w ≤ k|u|1/2 u3/2 w, giving B(u, u)V ∗ ≤ k|u|1/2 u3/2 . 9.5
Take ( p, q) = (2, 2) if m = 2 and ( p, q) = (4/3, 4) if m = 3. We know ∗ that Bn B in L p (0, T ; V ∗ ), where Bn = B(u n , u n ) and B = B(u, u). ∗ We need to show that Pn Bn B in the same sense. For ψ ∈ L q (0, T ; V ) we have
T
T
Pn Bn (t)− B, ψ dt =
0
Pn Bn − Bn , ψ dt +
0
T
Bn − B, ψ dt.
0 ∗
The second term converges since Bn B, so we have to treat only the first term. We rewrite this as
T
Bn (t), Q n ψ dt.
0
Since functions of the form ψ=
k
ψ j α j (t),
ψ j ∈ V, α j ∈ C 1 ([0, T ], R)
(S9.1)
j=1
are dense in L q (0, T ; V ) (see Exercise 7.3) we can consider 0
T
$ Bn ,
k
% Q n ψ j α j (t) dt.
j=1
Since Bn is uniformly bounded in L p (0, T ; V ∗ ) when m = 2, we can use the fact that Q n ψ j → ψ j in V to show the required convergence for all ψ of the form (S9.1). The density of such ψ in L q (0, T ; V ) then gives the full result.
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
38 9.6
10:52
Char Count= 0
Solutions to Exercises If u ∈ L 4 (0, T ; V ) then we can estimate b(w, u, w) in (9.41) differently, writing 1 2
d |w|2 + νw2 ≤ k|w|1/2 w3/2 u dt ν c ≤ w2 + 3 |w|2 u4 , 2 ν
which becomes, dropping the terms in w2 , d |w|2 ≤ C|w|2 u4 . dt Integrating gives |w(t)| ≤ |w(0)| exp 2
2
t
u(s) ds , 4
0
which implies uniqueness provided that u ∈ L 4 (0, T ; V ).
Chapter 10 10.1
If not, then there exist an > 0 and sequences δn → 0, xn ∈ K , yn ∈ H , such that |xn − yn | ≤ δn
and
| f (xn ) − f (yn )| > .
Since K is compact there is a subsequence of the {xn } (relabel this xn ) such that xn → x ∗ ∈ X . Now, |x ∗ − yn | ≤ |x ∗ − xn | + |xn − yn | → 0
as
n → ∞, (S10.1)
and | f (x ∗ )− f (yn )| ≥ | f (xn )− f (yn )|−| f (xn )− f (x ∗ )|≥/2
10.2
(S10.2)
if n is sufficiently large, since f is continuous at x ∗ . But then (S10.1) and (S10.2) say precisely that f is not continuous at x ∗ , which is a contradiction. The set in (10.23) is bounded since S(t)B, (S10.3) t≥t0 (B)
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 10
39
with t0 (B) from Definition 10.2, is a subset of B, and
S(t)B
(S10.4)
0≤t≤t0 (B)
10.3
is bounded since B is bounded and S(t) is continuous. Similarly, if B is compact then (S10.3) is a closed subset of B, and (S10.4) is the continuous image of the compact set B × [0, t0 (B)]: both parts are compact, and therefore so is (10.23). That (10.23) is positively invariant is clear by definition. In this example ω(0) = 0 and ω(x) = {|x| = 1} if x = 0. So $(B) = {(0, 0)} ∪ {|x| = 1} (which is clearly not connected). Since ω(x) = (1, 0) for all x with |x| = 1, $[$(B)] = {(0, 0), (1, 0)},
10.4
so that $[$(B)] = $(B) as claimed. We show that, for a bounded set X , ω1 (X ) = {y : S(tn )xn → y}, where tn → ∞ and xn ∈ X , is equal to ω2 (X ) =
&
S(s)X .
t≥0 s≥t
If y ∈ ω1 (X ) then clearly y∈
S(s)X
s≥t
for all t ≥ 0 and hence in y ∈ ω2 (X ). So ω1 (X ) ⊂ ω2 (X ). Conversely, if y ∈ ω2 (X ) then for any t ≥ 0 y∈
S(s)X ,
s≥t
and so there are sequences {τm(t) }, with τm(t) ≥ t, and {xm(t) } ∈ X with S(τm(t) )xm(t) → y. Now consider t = 1, 2, . . . and pick tn from τm(n) and
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
40
10:52
Char Count= 0
Solutions to Exercises xn from xm(n) such that (n) (n) S τ x − y ≤ 1/n. m
10.5
m
Then S(tn )xn → y with tn → ∞, since tn ≥ n, showing that y ∈ ω1 (X ). This gives ω2 (X ) ⊂ ω1 (X ), and so ω1 (X ) = ω2 (X ). If y ∈ S(t)B for all t ≥ 0 then for any tn there is an xn ∈ B with y = S(tn )xn , so clearly y ∈ ω2 (B) (as defined in the previous solution). Conversely, if y ∈ ω2 (B) then we must have y∈
S(s)B.
s≥t
Now, if τ ≥ t0 (B), then S(t)B ⊃ S(t + τ )B, and so then S(t)B ⊃
S(t + τ )B.
τ ≥t0 (B)
Since S(t)B is closed S(t)B ⊃
S(t + τ )B
y,
τ ≥t0 (B)
that is, y ∈ S(t)B for all t ≥ 0. Clearly we have & & S(t)B ⊂ S(nT )B. t≥0
n∈Z+
If u ∈ S(nT )B for all n ∈ Z+ then in particular u ∈ S(n 0 T )B, provided that n 0 is large enough that n 0 T ≥ t0 (B), where S(t)B ⊂ B
for all
t ≥ t0 (B).
Since u ∈ S(nT )B we have u = S(nT )y with y ∈ B, and it follows that for all t ≥ 0 S(t)u = S(t + nT )y = S(τ )y ∈ B, since τ ≥ t0 (B). Therefore u ∈ S(t)B for all t ≥ n 0 T . It follows that u ∈ ω2 (B), giving the required equality.
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 10 10.6
41
y ∈ ω(Y ) if S(tn )yn → y with tn → ∞ and yn ∈ Y . Then yn ∈ X also and so ω(X ) ⊃ ω(Y ). If Y absorbs X in a time t0 (assuming X to be bounded) and if S(tn )xn → x, then S(tn − t0 )[S(t0 )xn ] → x,
10.7
and since tn − t0 → ∞ and S(t0 )xn ∈ Y , ω(X ) ⊂ ω(Y ), so then ω(X ) = ω(Y ). First, the set ∞
Ki
(S10.5)
i= j
is clearly closed, and since all sets K i lie within 1/j of K j if i ≥ j it is also bounded, and hence compact. It follows that K ∞ , the intersection of a decreasing sequence of compact sets, is itself compact. Now, it is clear by a similar argument that dist(K ∞ , K j ) ≤ j −1 . Conversely, if u ∈ K j then dist(u, K i ) ≤ j −1 for all i ≥ j. So certainly ∞ dist u, K i ≤ j −1 . i= j
In particular, there exist points u i ∈ K i , i ≥ j, such that |u i − u| ≤ j −1 . Since each u i is contained in the compact set (S10.5) (with j = 1) then there exists a subsequence of the u i that converges to some u ∗ . It follows that u ∗ ∈ K ∞ , and by construction |u − u ∗ | ≤ j −1 . Therefore dist(K j , K ∞ ) ≤ j −1 , and so distH (K j , K ∞ ) ≤ j −1 : 10.8
K j converges to K ∞ in the Hausdorff metric. To show that the inverse is continuous, suppose not. Then there exist an > 0 and a sequence {xn } ∈ f (X ) with xn → y ∈ f (X ) but | f −1 (xn ) − f −1 (y)| ≥ . However, f −1 (xn ) ∈ X , and since X is
P1: FNT CB341/Robinson
CB341-Sol
42
10.9
February 5, 2001
10:52
Char Count= 0
Solutions to Exercises compact there exists a subsequence xn j such that f −1 (xn j ) → z. Since f is continuous, it follows that xn j → f (z). Since f is injective, it follows from f (z) = y that z = f −1 (y), which is a contradiction. So f −1 is continuous on f (X ). Proposition 10.14 says that, given 1 and T > 0, there exists a time τ1 such that, for all t ≥ τ1 , dist(u(t), A) ≤ δ(1 , T ). So we can track the trajectory u(t) within a distance 1 for a time T starting at any time t ≥ τ1 . We can replace T with 2T and apply the same argument for 2 = 1 /2, that is, there exists a time τ2 such that, for all t ≥ τ2 , dist(u(t), A) ≤ δ(2 , 2T ), and then the trajectory u(t) can be tracked for a time 2T starting at any time t ≥ τ2 . Thus u(t) can be followed from τ1 to τ2 by a distance 1 with a finite number of trajectories on A of time length T , and when we reach τ2 , we can start to track u(t) within a distance 2 with trajectories on A of time length 2T , until we reach a τ3 after which we can track within a distance 3 for a time length 3T , etc. The “jumps” are bounded by k + k+1 , since |vk+1 − S(tk+1 − tk )vk | ≤ |vk+1 − u(tk+1 )| + |u(tk + (tk+1 − tk )) − S(tk+1 − tk )vk | ≤ k+1 + k .
10.10 Take > 0. Then there is a T > 0 such that dist(S(t)B1 , A) + dist(S(t)B2 , A) <
for all
t ≥ T.
Also, by the uniform continuity of the semigroup, there is a δ > 0 such that dist(S(t)B1 , S(t)B2 ) ≤
for all
t ∈ [0, T ]
provided that dist(B1 , B2 ) ≤ δ. The argument is symmetric, which gives the result.
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 11
43
Chapter 11 11.1
Using Young’s inequality on (11.30) we can deduce that p |u| ≤ 2
2
|u| p d x +
p | |. p−2
So we can write (11.6) as 1 2
2α2 2 d |u|2 + u2 + |u| ≤ dt p
2 + k | |. p−2
Neglecting the u2 term we can write 1 2
11.2
d 2α2 2 |u|2 + |u| ≤ dt p
2 + k | |. p−2
We can now apply the Gronwall inequality to deduce an asymptotic bound on |u(t)|, as in Proposition 11.1. (The expression for the bound will be a more complicated expression than before.) Proceeding as advised, we obtain s s d g(τ ) dτ ≤ h(s) exp − g(τ ) dτ ≤ h(s), y(s) exp − ds t t and integrating both sides between s and t + r gives y(t + r ) ≤ y(s) exp +
t+r
g(τ ) dτ
s t+r
h(τ ) dτ
exp
s
t+r
g(τ ) dτ
s
≤ (y(s) + a2 ) exp(a1 ).
11.3
Integrating both sides for t ≤ s ≤ t + r gives the result as stated. Taking the inner product of du n + Au n = Pn f (u n ) dt with t 2 Au n we obtain
du n 2 , t Au n dt
+ t 2 |Au n |2 = (Pn f (u n ), t 2 Au n ),
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
44
10:52
Char Count= 0
Solutions to Exercises which, using the methods leading to (8.27) for the right-hand side, becomes 1 2
d tu n 2 − 2tu n 2 + t 2 |Au n |2 ≤ lt 2 u n 2 . dt
Integrating from 0 to T gives T T u n 2 + t 2 |Au n |2 dt ≤ 0
T
(2t + lt 2 )u n 2 dt.
0
Since we already know that u n ∈ L 2 (0, T ; V ), it follows that u n ∈ L 2 (t, T ; D(A)) for any t > 0. Since H 2 ( ) ⊂ C 0 ( ) if m ≤ 3 we also have Pn f (u n ) ∈ L 2 (t, T ; L 2 ), and so it follows that du n /dt ∈ L 2 (t, T ; H ). Taking limits shows that the solution u satisfies u ∈ L 2 (t, T ; D(A))
11.4
and
du/dt ∈ L 2 (t, T ; L 2 ).
Application of Corollary 7.3 then makes the “formal” calculations at the beginning of Section 11.1.2 rigorous. Observe that for s < 0 we have f (s)|s| ≥ α2 |s| p − k, and so in particular f (s) ≥ 0
for all
s < (k/α2 )1/ p .
(S11.1)
Now set M = (k/α2 )1/ p , multiply Equation (11.1) by (u(x) + M)− , and integrate to obtain 2 2 1 d (u(x) + M)− + |∇(u + M)− | = f (u)(u + M)− d x 2 dt
≤ 0, using (S11.1). It follows, using the Poincar´e inequality, that 2 1 d (u(x) + M) d x ≤ −C (u(x) + M)2− d x, − 2 dt
and so as in the last part of the argument given in Theorem 11.6, we must have (u(x) + M)2− d x = 0
for all u ∈ A.
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 12
45
Chapter 12 12.1
If u is smooth then Au = − u, and we have 2 2 b(u, u, A u) = u i (Di u j )Dk2 Dl2 u j d x i, j,k,l=1
2
=
i, j,k,l=1
Dk2 (u i (Di u j )) Dl2 u j d x
'
=
i, j,k,l
Dk2 u i (Di u j ) + 2(Dk u i )(Dk Di u j )
Dl2 u j = b(Au, u, Au) + 2 (Dk u i )(Dk Di u j ) Dl2 u j d x + u i Di Dk2 u j
(
i, j,k,l
+ b(u, Au, Au) = b(Au, u, Au) + 2
2
b(Dk u, Dk u, Au),
k=1
as claimed. The result follows for general u by taking limits. To obtain inequality (12.23), use (9.26) to give |b(u, u, A2 u)| ≤ k|A3/2 u||Au|u + 2
2
|b(D j u, Au, D j u)|
j=1
≤ k|A
3/2
u||Au|u + 2k
2
|D j u|D j uAu
j=1
≤ k|A3/2 u||Au|u 2 1/2 2 1/2 + 2k |D j u|2 D j u2 Au. j=1
j=1
Since u2 = a(u, u) = Au, u = (A1/2 u, A1/2 u) = |A1/2 u|2 , this becomes |b(u, u, A2 u)| ≤ 3k|A3/2 u||Au|u ≤ as required.
ν 3/2 2 9k 2 |A u| + u2 |Au|2 , 4 ν
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
46 12.2
10:52
Char Count= 0
Solutions to Exercises Take the inner product of du/dt + ν Au + B(u, u) = f with A2 u to obtain 1 2
d |Au|2 + ν|A3/2 u|2 = −b(u, u, A2 u) + ( f, A2 u), dt
and use the estimate (12.23) from the previous exercise to write 1 2
d |Au|2 + ν|A3/2 u|2 dt ≤ |b(u, u, A2 u)| + f |A3/2 u| C f 2 ν ν ≤ |A3/2 u|2 + u2 |Au|2 + + |A3/2 u|2 , 4 ν ν 4
so that d C 2 f 2 |Au|2 + ν|A3/2 u|2 ≤ + u2 |Au|2 . dt ν ν Using a similar trick as we did for the absorbing set in V , we integrate this equation between s and t, with t < s < t + 1, so that 2M C t+1 |Au(t + 1)|2 ≤ |Au(s)|2 + u(s)2 |Au(s)|2 ds, + ν ν t where we have used (12.24). Integrating again with respect to s between t and t + 1 gives t+1 2M C t+1 |Au(t + 1)|2 ≤ |Au(s)|2 ds + u(s)2 |Au(s)|2 ds. + ν ν t t (S12.1) Now, if t ≥ t1 (|u 0 |) then we know that t+1 u(s) ≤ ρV and |Au(s)|2 ds ≤ I A , t
and so if it follows that then |Au(t + 1)|2 ≤ ρ A ≡ I A +
12.3
C 2M + ρV2 I A , ν ν
an absorbing set in D(A). Suppose that u n ∈ V with u n ≤ k and that u n → u in H . Then there exists a subsequence u n j such that u n j v in V , so that v ≤ k. Since V ⊂⊂ H , it follows that u n j → v in H , and so in particular we must have u = v, which implies that u ≤ k.
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 12 12.4
If u ∈ D(A) with u=
47
u k e2πik·x/L
k∈Z2
then we can estimate u∞ by u∞ ≤
|u k |.
k∈Z2
Split the sum into two parts, u∞ ≤
|u k | +
|k|≤κ
|u k |.
|k|>κ
We now use the Cauchy–Schwarz inequality on each piece, u∞ ≤ (|u k | × 1) + (|u k ||k|2 × |k|−2 ) |k|≤κ
≤
|k|>κ
1/2 |u k |
2
|k|≤κ
+
1/2 1
|k|≤κ
1/2 |u k | |k| 2
4
|k|>κ
Since
1 ≤ Cκ 2
|k|≤κ
1/2 |k|
−4
.
|k|>κ
and
|k|4 ≤ Cκ −2 ,
|k|>κ
this becomes u∞ ≤ C(κ|u| + κ −1 |Au|). To make both terms on the right-hand side the same, we choose κ = |Au|1/2 |u|−1/2 , obtaining u∞ ≤ C|u|1/2 |Au|1/2 . 12.5
We have already derived in (12.20) the inequality d 2| f |2 u2 + ν|Au|2 ≤ + Cu6 , dt ν and since we have a uniform bound on u for t large enough, we obtain a uniform bound on the integral of |Au(s)|2 , t0 +1 |Au(s)|2 ds ≤ C1 . (S12.2) t0
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
48
10:52
Char Count= 0
Solutions to Exercises Following the analysis in Proposition 12.4, we estimate |u t | ≤ ν|Au| + |B(u, u)| + | f |, and using (12.25) this becomes |u t | ≤ ν|Au| + ku3/2 |Au|1/2 + | f |. An application of Young’s inequality yields |u t | ≤ c|Au| + Cu2 + | f |, and so for t large enough, |u t | ≤ c|Au| + CρV2 + | f |. The bound in (S12.2) therefore implies a bound on t0 +1 |u t (s)|2 ds ≤ C2 .
|u t |2 , (S12.3)
t0
Now differentiate u t + ν Au + B(u, u) = f with respect to t to obtain u tt + ν Au t + B(u t , u) + B(u, u t ) = 0 and take the inner product with u t so that 1 2
d |u t |2 + νu t 2 ≤ |b(u t , u, u t )| dt ≤ ku|u t |1/2 u t 3/2 ≤
3ν k 4 u4 |u t |2 . u t 2 + 4 4ν 3
Using once again the asymptotic bound on u, we have for t ≥ t0 that d |u t |2 ≤ C3 |u t |2 . dt We use the usual trick, integrating between s and t + 1, with t < s < t + 1, t+1 2 2 |u t (t + 1)| ≤ |u t (s)| + C4 |u t (s)|2 ds, t
P1: FNT CB341/Robinson
CB341-Sol
February 5, 2001
10:52
Char Count= 0
Chapter 13
49
and then between t and t + 1 (with respect to s) so that t+1 |u t (t + 1)|2 ≤ (1 + C4 ) |u t |2 ds t
≤ (1 + C4 )C3 ,
(S12.4)
by (S12.3). To end, we show that |u t | bounds |Au|. From the equation we have ν|Au| ≤ |u t | + |B(u, u)| + | f |, or with (12.25) ν|Au| ≤ |u t | + k|Au|1/2 u3/2 + | f |, and so after using Young’s inequality and rearranging we have |Au| ≤ C(|u t | + u3 + | f |). Together with (S12.4) we obtain |Au(t)| ≤ ρ D for all t ≥ 1 + t0 (u 0 ). So we have an absorbing set in D(A) and hence a global attractor for the 3D equations.
Chapter 13 13.1
Let G(X, ) be the number of boxes in a fixed cubic lattice, with sides , that are necessary to cover X . Since each cube with side sits inside a ball of radius , N (X, ) ≤ G(X, ), and so d f (X ) ≤ dbox (X ). Also, since any ball with side is contained within at most 2m different boxes in the grid, we have G(X, ) ≤ 2m N (X, ). Therefore dbox (X ) = lim sup →0
log G(X, ) − log
≤ lim sup
m log 2 + log N (X, ) − log
= lim sup
log N (X, ) − log
→0
→0
= d f (X ),
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50
13.2
10:52
Char Count= 0
Solutions to Exercises giving equality between box-counting dimension and fractal dimension in Rm . If n+1 ≤ < n then we have log N (X, ) log N (X, n+1 ) ≤ − log − log n ≤
log N (X, n+1 ) − log n+1 + log(n+1 /n )
≤
log N (X, n+1 ) , − log n+1 + log α
and so lim sup →0
13.3
log N (X, ) log N (X, n ) . ≤ lim sup − log − log n n→∞
That this inequality holds in the opposite sense is straightforward, and hence we obtain the desired equality. √ The sequence m = ( 2 log m)−1 , m ≥ 2, satisfies m+1 log m log 2 = ≥ , m log(m + 1) log 3 and so we can use the result of the previous exercise. Note that we have en ek 2 1 1 2 log n − log k = (log n)2 + (log k)2 ≤ (log n)2 for n > k, and so the first m − 1 elements from Hlog will belong to distinct balls of radius m . It follows that N (Hlog ) ≥ m − 1, and so d f (Hlog ) ≥ lim sup m→∞
≥ lim sup m→∞
13.4
log N (Hlog , m ) log m log(m − 1) √ = ∞, log( 2 log m)
which implies that d f (Hlog ) = ∞, as claimed. At the jth stage of construction the middle-α set Cα consists of 2 j intervals of length β j , where β = (1 − α)/2. It follows that N (Cα , β j ) = 2 j .
P1: FNT CB341/Robinson
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Chapter 13
51
Therefore, using the result of Exercise 13.2 we can calculate d f (Cα ) = lim sup j→∞
13.5
log 2 log 2 j = . j log β log β
Clearly µ
∞
X k , d,
≤
k=1
∞
µ(X k , d, ).
k=1
Since µ(X k , d, ) is nondecreasing in we have µ(X k , d, ) ≤ Hd (X k ) for each k, and so for every > 0 we have ∞ ∞ µ X k , d, ≤ Hd (X k ). k=1
k=1
We can now take the limit as → 0 on the left-hand side to obtain ∞ ∞ d H Xk ≤ Hd (X k ) k=1
13.6
k=1
as claimed. The map L taking e(i) into v (i) (1 ≤ i ≤ n) is given by L=
n
v (k) (e(k) )T ,
k=1 ( j)
and since ei(k) = δik , the components of L are L i j = vi : ( j)
(L T L)i j = vk(i) vk = v (i) · v ( j) = Mi j . 13.7
M is real and symmetric since Mi j = δx (i) · δx ( j) . It follows that its eigenvalues λ j are real, and one can find an orthonormal set of eigenvectors e(k) with Me(k) = λk e(k) .
P1: FNT CB341/Robinson
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Char Count= 0
Solutions to Exercises To show that λk > 0, consider (k) 2 λk = e(k) Me(k) = ei(k) δxs(i) δxs( j) e(k) j = |v | ≥ 0, T
where v (k) is the vector given by its components vs(k) = ei(k) δxs(i) . If v (k) = 0 then the two different initial conditions δx(0) = 0
δx(0) =
and
n
ei(k) δx (i)
i=1
13.8
have the same solution at time t, contradicting uniqueness. So all the eigenvalues are strictly positive. Writing M as M=
n
λ j e j e Tj ,
j=1
we have log M =
n
log λ j e j e Tj .
j=1
Clearly Tr[log M] =
n
log λ j ,
j=1
and since detM =
n
λj
j=1
the required result follows immediately. Since Tr[log M] is given by (S13.1), we have n ˙ d λi . Tr[log M] = dt λ i=1 i
(S13.1)
P1: FNT CB341/Robinson
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Chapter 13
53
The right-hand side of (13.34) is n −1 d M ei , M ei dt i=1 * ) n n n T = λ˙ k ek ekT + λk e˙ k ekT + λk ek e˙ kT ei ei , λ−1 j ejej i=1
=
n i=1
=
n
j=1
)
λi−1 eiT
)
λi−1
k=1
n
+
λk e˙ k ekT
+
λk ek e˙ kT
k=1
λ˙ i eiT
+
λi e˙ iT
i=1
=
λ˙ k ek ekT +
n
* ei
* λk (ei , e˙ k )ekT
ei
k=1
n ( ' −1 λi λ˙ i + (˙ei , ei ) + (ei , e˙ i ) i=1
=
n
λi−1 λ˙ i ,
i=1
13.9
since dtd (ei , ei ) = 0. Since the eigenvalues are proportional to the sums of squares of m integers, we will have reached the eigenvalue mk 2 once we have taken k m combinations of integers. Thus λk m = Cmk 2 , and so if k m < n < (k + 1)m we obtain Cmk 2 ≤ λn ≤ Cm(k + 1)m . We now have k < n 1/m < (k + 1) and so 1 1/m n 2
< k < k + 1 < 2n 1/m .
This gives cn 2/m ≤ λn ≤ Cn 2/m , as required.
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Solutions to Exercises
13.10 Taking the inner product of (13.27) with U we obtain 1 2
d |U |2 + νU 2 = −b(U, u, U ), dt
and so 1 2
d |U |2 + νU 2 ≤ k|U |U |u|. dt
Using Young’s inequality and rearranging we get d |U |2 + νU 2 ≤ C|U |2 . dt
(S13.2)
That bounded sets in L 2 are mapped into bounded sets in L 2 follows by neglecting the term in U 2 and applying Gronwall’s inequality (Lemma 2.8), |U (t)|2 ≤ eCt |U (0)|2 = eCt |ξ |2 .
(S13.3)
To show that we in fact obtain a bounded set in H 1 , we first return to (S13.2) and integrate between t/2 and t to obtain t t 2 ν U (s) ds ≤ C |U (s)|2 ds + |U (t/2)|2 ≤ C(t)|U (t/2)|2 , t/2
t/2
(S13.4) using (S13.3). Now we take the inner product of (13.27) with AU , which gives 1 2
d U 2 + ν|AU |2 = −b(u, U, AU ) − b(U, u, AU ). dt
Using (9.27) we obtain 1 2
d U 2 + ν|AU |2 ≤ k |u|1/2 u1/2 U 1/2 |AU |3/2 dt + |U |1/2 U 1/2 u1/2 |Au|1/2 |AU | ,
and after using Young’s inequality and rearranging we have d U 2 + ν|AU |2 ≤ CU 2 . dt Expression (S13.4) allows us to use the “uniform Gronwall” trick and find a bound on U valid for all t > 0. Thus $(t; u 0 ) is compact for all t > 0.
P1: FNT CB341/Robinson
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Char Count= 0
Chapter 14
55
13.11 If we integrate (12.6) between 0 and T we obtain T T | f |2 ν |Au(s)|2 ds ≤ + u(0)2 . ν 0 Dividing by T and taking the limit as T → ∞ yields 1 T | f |2 lim sup |Au(s)|2 ds ≤ 2 , ν T →∞ T 0 since there is an absorbing set in V . Therefore χ≤
| f |2 = ν 3 L −6 G 2 . L 2ν
The only length that can be formed from χ and ν is 3 1/6 ν Lχ = , χ and this implies (13.35).
Chapter 14 14.1
Since A is compact, it is bounded and certainly contained in B(0, r ) for some r > 0. So Nr (A) = 1. We consider S(B(0, r ) ∩ A), which by our assumption can be covered by K 0 balls, centred in A, and of radius r/2. So N (A, r/2) = K 0 . Now consider each one of the balls in this covering, and apply our assumption again to show that S(B(ai , r/2) ∩ A) can be covered by K 0 balls of radius r/4, so that N (A, r/4) = K 02 . Iterating this argument, we can see that N (A, 2−k r ) = K 0k .
P1: FNT CB341/Robinson
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Char Count= 0
Solutions to Exercises So therefore, using the result of Exercise 13.2, we have log(N (A, 2−k r )) k→∞ log(2k ) k log K 0 ≤ k log 2 log α ≤ n0 , log 2
d f (A) = lim
14.2
precisely (14.33). We have, for any u ∈ D(A1/2 ), u2 = a(u, u) = (A1/2 u, A1/2 u) = |A1/2 u|. Expanding p in terms of the eigenfunctions of A gives p=
n
( p, w j )w j ,
j=1
and so p2 =
n
λ j |( p, w j )|2 ≤ λn | p|2 .
j=1
Similarly, q2 =
n
λ j |(q, w j )|2 ≥ λn+1 |q|2 .
j=1
14.3
14.4
The other two inequalities in the exercise follow easily from these. Differentiating gives
d da λb db λa = exp(λa/C(a +b)) 1− + 1+ . dt dt C(a+b) dt C(a+b) Since we have (14.34), the coefficient of da/dt is negative, whereas the coefficient of db/dt is positive. It follows that we can substitute in the inequalities for da/dt and db/dt, which gives d /dt ≤ 0. Write |u(x) − u(y)| ≤ |e2πik·x/L − e2πik·y/L ||ck |, k∈Z2
P1: FNT CB341/Robinson
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Chapter 14
57
and use (14.35) to deduce that |u(x) − u(y)| ≤ C|x − y|1/2 ≤ C|x − y|
1/2
|ck ||k|1/2
k∈Z2
1/2
(1 + |k| )|ck | 4
2
k∈Z2
≤ Cu H 2 |x − y|
14.5
1/2
k∈Z2
|k| (1 + |k|4 )
1/2
.
[ k∈Z2 |k|/(1 + |k|4 ) is finite.] Since (6.14) shows that u H 2 = C|Au| for u ∈ D(A), we can use the result of the previous exercise to deduce that |u(x) − u(y)| ≤ c|Au||x − y|1/2 .
14.6
Expression (14.36) follows immediately from this and the definitions of d(N ) and η(u). Choose > 0. Then there exists a T such that b(t) ≤ /2 for all t ≥ T . Hence for t ≥ T , dX + a X ≤ /2. dt By Gronwall’s inequality (Lemma 2.8), X (T + t) ≤ X (T )e−at + /2, and so choosing τ large enough that ke−aτ < /2, we have X (t) ≤
14.7
for all
t ≥ T + τ,
so that X (t) → 0. Using the bound on b given in (9.25), we can write 1 2
d w2 + ν|Aw|2 ≤ w∞ w|Au| dt ≤ [η(w) + cd(N )1/2 |Aw|]w|Au| −1/2
≤ η(w)w|Au| + cd(N )1/2 λ1
|Aw|2 |Au|,
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58
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Char Count= 0
Solutions to Exercises using (14.36), and therefore 1 2
' ( d −1/2 w2 + ν − cλ1 d(N )1/2 |Au| λ1 w2 ≤ η(w)w|Au|. dt
Now, we know that A is bounded in V and D(A), so that 1 2
d w2 + [ν − cλ1/2 ρ A d(N )1/2 ]λ1 w2 ≤ 2ρV ρ A η(w). dt
Now, choose δ such that 1/2
µ = ν − cλ1 ρ A δ 1/2 > 0. Then we have, for d(N ) < δ, 1 2
14.8
d w2 + µw2 ≤ 2ρV ρ A η(w). dt
(S14.1)
By assumption, we know that η(w) → 0, and since the attractor is bounded in V we have w(t)2 ≤ 4ρV2 . The result of the previous exercise applied to (S14.1) now shows (14.37). (i) Take the inner product of (14.38) with qn = Q n u to obtain 1 2
d |qn |2 + (Au, qn ) = (F(u), qn ). dt
Now, notice that |(Au, qn )| = |(Aqn , qn )| ≥ λn+1 |qn |2 , and so 1 2
d |qn |2 + λn+1 |qn |2 ≤ C0 |qn |, dt
from which, using the result of Exercise 2.5, we see that d |qn | ≤ −λn+1 |qn | + C0 , dt + which gives |Q n u(t)| ≤
C0 + |Q n u(0)|, λn+1
using the Gronwall lemma (Lemma 2.8).
(S14.2)
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Char Count= 0
Chapter 15
59
(ii) Writing p(t) = Pn u(t) and q(t) = Q n u(t), p solves the equation d p/dt + Ap = Pn F( p + q). Thus the equation for w = p − pn is dw/dt + Aw = Pn F( pn ) − Pn F( p + q). Taking the inner product with w and using the Lipschitz property of F gives 1 2
d |w|2 + w2 ≤ C1 |w|2 + C1 |q||w|. dt
Hence d |w| ≤ C1 |w| + C1 |q|, dt + and so, using the bound in (S14.2) and the Gronwall lemma as above we obtain
C0 −1 |Pn u(t) − pn (t)| ≤ C1 + |Q n u(0)| eC1 t . λn+1 Combining this with (S14.2) yields
C0 |u(t) − pn (t)| ≤ C1−1 + |Q n u(0)| (C1 + eC1 t ), λn+1 and since we know that λn+1 → ∞ and |Q n u(0)| → 0 as n → ∞, it follows that pn (t) converges to u(t) as claimed.
Chapter 15 15.1
For any point v ∈ H , dist(v, M)2 = inf
p∈P H
|Pv − p|2 + |Qv − φ( p)|2
and |Qv − φ(Pv)|2 = |Qv − φ( p) + φ( p) − φ(Pv)|2 ≤ 2|Qv − φ( p)|2 + 2|φ( p) − φ(Pv)|2 ≤ 2|Qv − φ( p)|2 + 2l 2 |Pv − p|2 ≤ c2 |Qv − φ( p)|2 + |Pv − p|2
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Char Count= 0
Solutions to Exercises for all p ∈ P H , where c2 = 2 max(l 2 , 1). Therefore |Qv − φ(Pv)| ≤ c dist(v, M).
15.2
The other implication is obvious. Using Proposition 15.3 we see that the attractor lies in the graph of some Lipschitz function : Pn H → Q n H . We can therefore project the dynamics on A onto Pn H by writing d p/dt + Ap = Pn F( p +
( p)).
(S15.1)
It is easy to show that (S15.1) is a Lipschitz ODE on Pn H , since |Pn F( p+ ( p))−Pn F( p+ ( p))| ≤ |F( p +
( p))−F( p +
( p))|
≤ C| p + ( p) − p − ( p)| ≤ C | p − p| + | ( p) − ( p)| ≤ 2C| p − p|.
15.3
We know that if u(t) is a solution in A then p(t) = Pn u(t) is a solution of (S15.1) lying in Pn A. Since (S15.1) is Lipschitz its solutions are unique, and so in particular Pn A is an invariant set. Thus (S15.1) is a finite-dimensional system that reproduces the dynamics on A. [The advantage of the inertial form over (S15.1) is that Pn A is the attractor of the finite-dimensional system, not just an invariant set.] Since F = 0 outside B(0, ρ), 't0 ⊂ S(t0 ){u : u ∈ Pn H : ρ ≤ |u| ≤ ρeλn+1 t0 }. The cone invariance part of the strong squeezing property then shows that for any two points u 1 and u 2 in 't0 we must have |Q n (u 1 − u 2 )| ≤ |Pn (u 1 − u 2 )|. If we write
'=
S(t)
0≤t
