lecture 13 ordinary differential · pdf fileoutlinelinear equations with constant coe cients....

Outline Linear equations with constant coefficients. Variation of constants.

Lecture 13Ordinary differential equations.

Shlomo Sternberg

Shlomo Sternberg



1 Linear equations with constant coefficients.

2 Variation of constants.The parametrix expansion.

Shlomo Sternberg



Today I want to discuss some facts about ordinary differentialequations. A course on dynamical systems given 40 years agowould consist almost entirely in the study of ordinary differentialequations. We are only going to devote six or so lectures to thissubject.

Today’s lecture will be some mix of slides and demonstrationsusing pplane.

Shlomo Sternberg



We have proved the local existence theorem for equations of theform

x ′(t) = F (t, x), x(0) = x0

under Lipschitz assumptions on F via the contraction fix pointtheorem. I first want to concentrate on equations where F doesnot depend on t.

Shlomo Sternberg



Linear homogenous equations with constant coefficients.

These are equations of the form

x ′ = Ax

where A is a constant bounded linear operator on a Banach space,X . You may as well think of X as a finite dimensional vectorspace. We must also specify the initial conditions, of course.

In case X = R, so A = a is a scalar, we know that the generalsolution to the above equation is x(t) = ceat where the constant cis determined by the initial condition, c = x(0) and

eat = 1 + at +1

2a2t2 +

1

3!a3t3 + · · · .

Exactly the same method works in general!Shlomo Sternberg



The exponential series for an operator.

Define

etA = I + tA +1

2t2A2 +

1

3!t3A3 + · · · .

Here I is the identity operator so e0A = I is the identity. The exactsame proof of the convergence of the exponential series in onevariable shows that this series converges for all t and that

e(s+t)A = esAetA.

It is not true in general that etAetB = et(A+B). This lack ofequality stems from the fact that A and B may not commute, andthis fact is manifested in the physical world by quantummechanics. But if A and B do commute then etAetB = et(A+B).

Shlomo Sternberg



The derivative of the exponential.

We may differentiate the exponential series with respect to t termby term and we find that

d

dtetA = AetA = etAA.

In particular, if we setx(t) = etAx0

thenx ′(t) = Ax(t) and x(0) = x0.

So the study of linear homogeneous differential equations withconstant coefficients reduces to the analysis of etA.

Shlomo Sternberg



etPAP−1

= PetAP−1.

Suppose that P is an invertible operator and

B = PAP−1.

then B2 = PA2P−1, B3 = PA3P−1 etc. so

etB = PetAP−1.

We know from linear algebra that (in finite dimensions) every B isof the form B = PAP−1 where A has a “nice” normal form. So ourstudy is reduced to understanding etA for normal forms (and alsounderstanding the effect of conjugating by P).

Here is a complete analysis in two (real) dimensions:

Shlomo Sternberg



B has two real distinct eigenvalues.

Then B can be diagonalized, i.e. B = PAP−1 where

A =

(a 00 b

)and

etA =

(eta 00 etb

).

The x and y axes are invariant. If a < 0 and b < 0 then etA fort > 0 contracts all points at different rates toward the origin. Theeffect of the conjugation by P is to replace the x and y axes byother lines through the origin.

Shlomo Sternberg



If a > 0 and b < 0 then etA is hyperbolic for t 6= 0, for t > 0expanding in the x-direction and contracting in the y -direction. Ifa > 0 and b > 0, then etA expands for positive t. If a = 0 andb < 0 then points on the x-axis are stationary and points allconverge toward the x-axis. Similar analysis for a > 0 and b = 0.

Shlomo Sternberg



Two equal real eigenvalues.

Here there are two cases: if

A =

(a 00 a

)then

etA = etaI .

So uniform expansion if a > 0, uniform contraction if a < 0 andetA ≡ I if A = 0. Since an A of this form commutes with all othermatrices, if B = PAP−1 then B = A.

Shlomo Sternberg



The second case is

A =

(a 10 a

).

This is of the form A = aI + C where

C =

(0 10 0

)so

etA = etaetC .

To compute etC observe that C 2 = 0 so

etC =

(1 t0 1

).

Shlomo Sternberg



Non-real eigenvalues.

If the eigenvalues are a± ib with b 6= 0 then a normal form is

A =

(a b−b 0

)= aI + b

(0 1−1 0

).

So if

C =

(0 1−1 0

)then

etA = etaetbC

so we must compute etC .

Shlomo Sternberg



et

0 1−1 0

=

(cos t sin t− sin t cos t

).

This follows from

C 2 = −I so C 3 = −C , C 4 = I .

Thus

etC =

(1− 1

2 t2 + 14! t

4 + · · · t − 13! t

3 + · · ·−t + 1

3! t3 − · · · 1− 1

2 t2 + 14! t

4 + · · ·

).

Shlomo Sternberg



Thus if a = 0, the trajectories of etA are circles with velocity ofrotation b. If a < 0 the trajectories are circular spirals heading intothe origin as t →∞ , and if a > 0 the trajectories are circularspirals spiraling out.

The effect of conjugating by P is to replace the circles by ellipses.

Shlomo Sternberg



Hyperbolicity.

If no eigenvalue has real part zero, we have the “infinitesimalversion” of hyperbolicity, and an old differential equation theoremof mine says that if there is a zero of the vector field (say at 0) andthe matrix of partial derivatives at the zero is hyperbolic in theabove sense, then the behavior of the non-linear system nearenough to the fixed point is equivalent, via a homoemorphism, tothe linearized equation.

Shlomo Sternberg



Bifurcations.

But, just as we studied in the behavior of maps in one dimensionnear fixed points with f ′(p) = ±1 there is a similar study ofbifurcations of non-hyperbolic zeros of vector fields. A famousexample is the Hopf bifurcation. Here is an illustration of thisphenomenon, without going into the technical details. Suppose oursystem of differential equations is

x ′ = ax + y − x(x2 + y2)

y ′ = ay − x − y(x2 + y2).

Shlomo Sternberg



If a < 0, all trajectories spiral into the origin. If a > 0 and small,then the origin has become an unstable fixed point, and alltrajectories starting near the origin spiral outward towards aperiodic trajectory while points outside the periodic trajectoryspiral in towards it.In other words, as a passes from negative to positive, an attractivefixed point has become repulsive and a nearby attractive periodicorbit has appeared.

Shlomo Sternberg



So far we have been studying the homogeneous equation

x ′(t) = Ax(t), x(0) = x0.

Suppose we want to solve the “inhomogeneous” equation

d

dtx(t) = Hx(t) + f (t)

where f is given, and with the initial condition x(0) = x0. Thesolution is given by Lagrange’s variation of constants formula

x(t) = etHx0 +

∫ t

0e(t−s)H f (s)ds

as can be checked by differentiating the right hand side. Thisformula is also known as Duhamel’s formula.

Shlomo Sternberg



The operator version.

If F is an operator valued function of t then the operator versionof the above says that

X (t) = etH +

∫ t

0e(t−s)HF (s)ds (1)

is the solution to the differential equation

d

dtX = HX + F

with the initial conditions

X (0) = I .

Shlomo Sternberg



For example, suppose that H = H0 + H1 and we takeF (t) = H1e

tH . We want to find a solution to

d

dtY (t) = HY (t) = H0Y (t) + H1Y (t), Y (0) = I .

The variation of constants formula tell us that

Y (t) = etH0 +

∫ t

0e(t−s)H0H1e

sHds. (2)

If we substitute this formula into itself (i.e. use this formula for theesH occurring in the integral on the right) we get

etH = etH0+

∫ t

0e(t−s)H0H1e

sH0dt+

∫ t

0

∫ s

0e(t−s)H0H1e

(s−τ)H0H1eτHdτds.

Shlomo Sternberg



The Born series.

Clearly we can keep going. The usefulness of this scheme is asfollows. Suppose that H1 is small, for example suppose that wecan ignore all terms involving three products of H1. Then in theabove expression, replacing eτH by eτH0 on the right, we get anapproximate expression for etH in terms of integrals involving etH0

and products of H1. The corresponding series (and approximation)is known as the Volterra series (or approximations) tomathematicians, and is known as the Born series (orapproximations) to physcists.

Shlomo Sternberg



The parametrix expansion.

Suppose we want to find etH and we only found an approximatesolution - we have found an operator valued function K (t) suchthat

dK (t)

dt= HK (t) + R(t), K (0) = I .

The variation of constants formula tells us that

K (t) = etH +

∫ t

0e(t−s)HR(s)ds

which we shall write as

etH = K (t)−∫ t

0e(t−s)HR(s)ds.

Substitute this back into itself to obtain

etH = K (t)−∫ t

0K (t−s)R(s)ds+

∫ t

0

∫ t−s

0e(t−s−τ)HR(τ)R(s)dτds.

Keep going. This suggests the following:Shlomo Sternberg




Let ∆k denote the k-simplex

∆k = {(t1, . . . , tk |0 ≤ t1 ≤ t2 ≤ · · · ≤ tk ≤ 1}.

If all the ti are unequal, there are k! ways of of reordering them.Since we may ignore possible equalities in computing volume, thisshows that the Euclidean volume of ∆k is 1/k!.So the volume of t∆k is tk/k!. Define the operators Q(k, t) by

Q(k , t) :=

∫t∆k

K (t − tk)R(tk − tk−1) · · ·R(t)dt1 · · · dtk .

So this integral is over 0 ≤ t1 ≤ t2 · · · ≤ tk ≤ t. To shorten theformulas, I will drop the dt1 · · · dtk .

Shlomo Sternberg




If K and R are uniformly bounded, say by C , in the interval [0,T ],then the Q(k , t) are bounded by C ktk/k! and so the series

∞∑k=0

(−1)kQ(k , t)

converges uniformly. R(k , s) by

R(k, s) =

∫s∆k−1

R(s − tk−1)rR(tk−1 − tk−2) · · ·R(t2 − t1)R(t1)

so that

Q(k , t) =

∫ t

0K (t − s)R(k , s)ds.

Shlomo Sternberg




Q(k , t) =

∫ t

0K (t − s)R(k , s)ds.

If we apply(

ddt − H

)to Q(k , t) we get

R(k, s) + R(k + 1, s)

so the sum in [d

dt− H

]( ∞∑k=0

(−1)kQ(k , t)

)telescopes to zero. Hence

∞∑k=0

(−1)kQ(k , t)

is a solution to our search for etH starting with an approximatesolution. This method is known as the parametrix expansion.

Shlomo Sternberg


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