C2 FUNDAMENTAL THEORY OF DYNAMICAL SYSTEMS
HANDOUT 1
UNIQUENESS OF SOLUTIONS OF DIFFERENTIAL
EQUATIONS
.
In lectures, I stated that under reasonably mild conditions on F, the equation x = F(x) had a unique
solution for each initial condition x(0). Normally, uniqueness follows from the proof of existence.
Such a proof however is rather technical (see Handout 3). Here therefore, I give a direct proof of the
fact that if solutions exist, then they must be unique, and explore some of its consequences. In
particular, this answers most of Q8 and Q9 on Example Sheet 1.
Consider the differential equation:
dx(t )
dt
=
F(x(t))
(1)
Fix some x0 and a δ > 0 and suppose that there exists a constant K such that for all x and h such that
x-x0 < δ and (x+h)-x0 < δ we have
F(x+h) - F(x)
≤
K h
Using Taylor’s theorem we can seen that this condition will for instance be satisfied if F is smooth
(differentiable with a continuous derivative).
Now suppose that we have two trajectories x(t) and x'(t) such thatx(0) = x'(0). Let
h(t)
=
x(t) - x'(t)
Note that h(t) → 0 as t → 0. Then
dh(t )
dt
=
d
(x(t) - x'(t))
dt
=
F(x(t)) - F (x'(t))
=
d
d
h(t)†.h(t) + h(t)†. h(t)
dt
dt
=
2h(t)†.[F(x(t)) - F (x'(t))]
Thus
d h(t )
2
=
dt
=
d
[h(t)†.h(t)]
dt
d
2 h(t)†. h(t)
dt
For some t0 > 0, we have that for all -t 0 < t < t0, the points x(t) and x'(t) are within δ of x0 and hence
F(x(t)) - F(x'(t)) ≤ (x(t) - x'(t)) = K h(t) . Thus
d h(t )
2
≤
dt
2
2K h(t )
2
Let ε(t) = h(t ) and take any t1, t2 with 0 < t1 < t2 < t0 . Then
Unique Solutions
∫
ε (t 2 )
ε (t 1 )
dε (t )
ε (t )
≤
2K
∫
t2
2
dt
t1
and so
log ε(t2) - log ε(t1)
≤
≤
2K (t2 - t1)
2Kt0
Thus
ε(t2)
ε(t1) e 2Kt
≤
0
2
Now let t1 → 0. Then ε(t1) → 0, and e 2Kt is fixed. Since ε(t2) = h(t ) ≥ 0, we must have ε(t2) = 0! But we
are free to choose t2 arbitrarily in (0,t0) and hence we conclude that ε(t) = 0 for all 0 ≤ t < t 0. A similar
argument works for t0 < t ≤ 0.
0
This proof shows that for sufficiently short times, solutions must be unique. It can easily be
extended to a global argument. Thus, for instance, suppose that F is smooth. Then, by above, for
any initial condition x there is an interval of time (-T,T) such that any two solutions x(t) and x'(t)
with x(0) = x'(0) = x are identical for t∈ (-T,T). Suppose that x(t) and x'(t) were not identical for all t >
0. Then, let T0 be the maximal t such that x(t) and x'(t) agree on [0,T]. Apply the above argument at
x(T0) = x'(T0) to see that there is some τ > 0 such that x(t) = x'(t) for t∈[T0,T0+τ). This contradicts the
maximality of T0, and hence we must have x(t) = x'(t) for t ≥ 0. The proof for negative t is similar.
CONSEQUENCES: 1
Denote the unique solution with initial condition x at t = 0 as ϕ(x,t). What is the solution with initial
condition ϕ(x,s) at t = 0 ? Obviously, it must be ϕ(ϕ(x,s),t). On the other hand, let z(t) = ϕ(x,t+s). Then
dz(t )
dt
dϕ(x ,t + s )
dt
=
=
F(ϕ(x,t+s))
=
F(z(t))
Hence z(t) is a solution of (1), and has initial condition z(0) = ϕ(x,s). Hence
ϕ(ϕ(x,s),t)
=
z(t)
=
ϕ(x,t+s)
(2)
CONSEQUENCES: 2
Now consider
dx(t )
dt
=
- F(x(t))
(3)
Let the unique solution of this with initial condition x at t = 0 be denoted ψ(x,t). Observe that
dϕ(x , −t )
dt
- F(ϕ(x,-t))
=
and ϕ(x,0) = x. Hence ϕ(x,-t) is a solution of (3) with initial condition x at t = 0, and hence
ϕ(x,-t)
ψ(x,t)
=
Then by (2) we have
ϕ(ψ(x,t),t)
=
ϕ(ϕ(x,-t),t)
=
ϕ(x,t-t)
=
ϕ(x,0)
=
x