ec = 2

is sometimes called the conformal energy of the solution z(t, y).

It is also sometimes convenient to set

n’1

ed = rp0pr + zp0

2

so that

•dil ≡ (ed + te)dy (mod {dt}),

and also

•inv ≡ (ted + 1 (t2 + |y|2 )e ’ n’1 2

(mod {dt}).

4 z )dy

2

The fact that the integrals of these quantities are constant in t yields results

about growth of solutions.

In fact, in the analysis of wave equations that are perturbations of the con-

formally invariant equation (3.68), the most e¬ective estimates pertain to the

quantities appearing in these conservation laws. One can think of the conser-

vation laws as holding for our “¬‚at” non-linear wave equation (3.68), and then

the estimates are their analogs in the “curved” setting.

3.5.3 Energy in Three Space Dimensions

We conclude this section by discussing a few more properties that involve the

energy in three space dimensions.

The fact that the energy E(t) is constant implies in particular that Rn zt dy

t

is bounded with respect to t. This allows us to consider the evolution of the

spatial L2 -norm of a solution to z = z 3 as follows:

d2

z 2 dy 2

= 2 (z ztt dy + zt )dy

dt2 R3 R3

t t

(z∆z ’ z 4 + zt )dy

2

= 2

R3

t

2

+ z 4)dy + 2 2

’2

= (| y z| zt dy

R3 Rn

t t

¤ 4E,

where the second equality follows from the di¬erential equation and the third

from Green™s theorem (integration by parts). The conclusion is that ||z||2 2(Rn )

L t

grows at most quadratically,

z 2dy ¤ 2Et2 + C1 t + C2 , (3.77)

R3

t

128 CHAPTER 3. CONFORMALLY INVARIANT SYSTEMS

z 2 dy = O(t2 ).

and in particular

The energy plays another interesting role in the equation

z = ’z 3 (3.78)

Here, it is possible for the energy to be negative:

z4

12

+ || z||2) ’

E= (z dy. (3.79)

2t 4

R3

t

We will prove that a solution to (3.78), with compactly supported initial data

z(0, y), zt (0, y) satisfying E < 0, must blow up in ¬nite time ([Lev74]). No-

tice that any non-trivial compactly supported initial data may be scaled up to

achieve E < 0, and may be scaled down to achieve E > 0.

The idea is to show that the quantity

def 12

I(t) = z dy

2

R3

t

becomes unbounded as t T for some ¬nite time T > 0. We start by computing

its derivatives

I (t) = zzt dy,

2

I (t) = zt dy + zztt dy

2

| z|2dy + z 4 dy.

zt dy ’

=

The last step uses Green™s theorem, requiring the solution to have compact y-

support for each t ≥ 0. To dispose of the z 4 term, we add 4E to each side

using (3.79):

2

| z|2dy.

I (t) + 4E = 3 zt dy +

We can discard from the right-hand side the positive gradient term, and from

the left-hand side the negative energy term, to obtain

2

I (t) > 3 zt dy.

To obtain a second-order di¬erential inequality for I, we multiply the last in-

equality by I(t) to obtain

3

z 2dy 2

I(t)I (t) > zt dy

2

(t)2 .

3

≥ I

2

3.5. CONSERVATION LAWS FOR WAVE EQUATIONS 129

The last step follows from the Cauchy-Schwarz inequality, and says that I(t)’1/2

has negative second derivative. We would like to use this to conclude that I ’1/2

vanishes for some T > 0 (which would imply that I blows up), but for this we

would need to know that (I ’1/2 ) (0) < 0, or equivalently I (0) > 0, which may

not hold.

To rectify this, we shift I to

J(t) = I(t) ’ 1 E(t + „ )2 ,

2

with „ > 0 chosen so that J (0) > 0. We now mimic the previous reasoning to

show that (J ’1/2) (t) < 0. We have

zzt dy ’ E(t + „ ),

J (t) =

2

zztt dy ’ E

J (t) = zt dy +

2

|| z||2dy ’ 5E

=3 zt dy +

2

zt dy ’ E .

>3

From this we obtain

2

J(t)J (t) ’ 3 J (t)2 > 3

z 2 ’ E(t + „ )2 2

zt ’ E ’ zzt ’ E(t + „ ) ,

2 2

which is positive, again by the Cauchy-Schwarz inequality. This means that

(J ’1/2 ) (t) < 0. Along with J ’1/2(0) > 0 and (J ’1/2 ) (0) < 0, this implies

that for some T > 0, J ’1/2(T ) = 0, so J(t) blows up.

We conclude by noting that the qualitative behavior of solutions of z =

f(z) depends quite sensitively on the choice of non-linear term f(z). In contrast

to the results for z = ±z 3 described above, we have for the equation

z = ’z 2 (n = 3),

that every solution must blow up in ¬nite time ([Joh79]). We will outline the

proof in case the initial data are compactly supported and satisfy

u(0, t)dy > 0, ut(0, t)dy > 0.

Note that replacing z by ’z gives the equation z = z 2 , which therefore has

the same behavior.

This proof is fairly similar to the previous one; we will derive di¬erential

inequalities for

def

J(t) = z dy

Rn

t

130 CHAPTER 3. CONFORMALLY INVARIANT SYSTEMS

which imply that this quantity blows up. We use integration by parts to obtain

(∆z + z 2)dy = z2,

J (t) = ztt dy =

and using H¨lder™s inequality on Supp z ‚ {|y| ¤ R0 + t} in the form ||z||L1 ¤

o

||z||L2 ||1||L2, this gives

2

(R0 + t)’3 ≥ C(1 + t)’3 J(t)2 .

J (t) ≥ C z dy (3.80)

This is the ¬rst ingredient.

Next, we use the fact that if z0 (y, t) is the free solution to the homogeneous

wave equation z0 = 0, with the same initial data as our z, then

z(y, t) ≥ z0 (y, t)

for t ≥ 0; this follows from a certain explicit integral expression for the solution.

Note that if we set J0(t) = Rn z0 , then it follows from the equation alone that

t

J0 (t) = 0, and by the hypotheses on our initial data we have J0 (t) = C0 + C1 t

with C0, C1 > 0. Another property of the free solution is that its support at

time t lies in the annulus At = {t ’ R0 ¤ |y| ¤ t + R0}. Now

C0 + C 1 t ¤ z dy

At

¤ ||z||L1(A)

¤ ||z||L2(A) ||1||L2(A)

2

2

¤ C(1 + t) z dy .

This gives

1/2

C0 + C 1 t

2

z dy ≥

J (t) = ,

C(1 + t)

and in particular, J > 0. With the assumptions on the initial data, this gives

J (t) >0

≥ C(1 + t)2.

J(t)

We can use (3.80, 3.81, 3.81) to conclude that J must blow up at some ¬nite

time. This follows by writing

C(1 + t)’3 J 3/2J 1/2

≥

J (t)

C(1 + t)’2 J 3/2.

≥

Multiply by J and integrate to obtain

J (t) ≥ C(1 + t)’1 J(t)5/4 .

3.5. CONSERVATION LAWS FOR WAVE EQUATIONS 131

Integrating again, we have

’4

J(t) ≥ J(0)’1/4 ’ 1 C ln(1 + t) ,

4

and because J(0) > 0 and C > 0, J(t) must blow up in ¬nite time.

132 CHAPTER 3. CONFORMALLY INVARIANT SYSTEMS

Chapter 4

Additional Topics

4.1 The Second Variation

In this section, we will discuss the second variation of the Lagrangian func-