Ordinary Differential Equations II
January 26
2017
Maximal solutions
(
x0 = f (t, x)
x(t0 ) = x0 ,
(1)
f continuous and Lipschitz in 2nd argument.
Suppose that x and y are solutions defined on I1 , I2 (open intervals). Let I =
I1 ∩ I2 := (T− , T+ ).
Proposition. x(t) = y(t) for all t ∈ I.
Proof. Suppose not. Then ∃ maximal open interval (t− , t+ ) on which x(t) = y(t),
with t+ < T+ , say, and x(t+ ) = y(t+ ) by continuity. Picard-Lindelöf ⇒ the
solutions agree in a neighbourhood of t+ . Contradiction!
Hence, ∃ maximal interval (T− , T+ ) to which the solution can be extended and the
extension is unique. This extension is called the maximal solution. From now on
we always consider the maximal solution when we discuss the IVP. If (T− , T+ ) = R,
the solution is called global.
Theorem. Let f ∈ C(U, Rn ), loc. Lip. in the 2nd argument, U ⊆ Rn+1 open. If
C ⊂ U is compact, then (t, x(t)) ∈
/ C for t sufficiently close to T± .
Proof. Suppose not. Then ∃{Pk }, Pk = (tk , x(tk )), s.t. Pk ∈ C ∀k and tk → T+ ,
say. Bolzano-Weierstrass ⇒ ∃ convergent subsequence
Pkj = (tkj , x(tkj )) → (T+ , y 0 ) ∈ C ⊂ U.
U open ⇒ ∃δ > 0 s.t.
V := [T+ − δ, T+ ] × Bδ (y 0 ) ⊂ U.
Let
M = max |f (t, x)|.
(t,x)∈V
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Then
Z t
|x(t) − x(s)| = f (τ, x(τ )) dτ ≤ M |t − s|,
s
if (τ, x(τ )) ∈ V , s ≤ τ ≤ t. But this implies that (t, x(t)) ∈ V and x(t) → y 0 as
t → T+ , since
|x(t) − y 0 | ≤ |x(t) − x(tkj )| + |x(tkj ) − y 0 | ≤ M |t − tkj | + |x(tkj ) − y| → 0
as t → T+ and j → ∞. Let y(t) be the solution of the ODE with y(T+ ) = y 0 and
set
(
x(t), t < T+ ,
z(t) =
y(t), t ≥ T+ .
Then z is an extension of the maximal solution, giving a contradiction.
Remark: z is (continuously) differentiable at T+ since e.g.
Z
z(T+ + h) − z(T+ )
1 T+ +h
=
f (s, x(s)) ds → f (T+ , y 0 )
h
h T+
as h ↑ 0 and z 0 (t) = f (t, x(t)) → f (T+ , y 0 ) as t ↑ T+ .
Suppose U = Rn+1 . If T+ = ∞ then the solution is global (forward in time). If
T+ < ∞, then limt→T+ |x(t)| → ∞. The solution blows up in finite time. If one
can show that the solution can’t blow up in finite time, then it has to be global.
Here’s an example of how this can be done.
Theorem. Suppose f as in previous theorem with U = Rn+1 and that ∀T > 0,
∃M (T ), L(T ) ∈ R s.t.
|f (t, x)| ≤ M (T ) + L(T )|x|,
(t, x) ∈ [−T, T ] × Rn .
Then all solutions of (1) are defined for all t ∈ R.
Proof. Take t0 = 0 w.l.o.g. Suppose that T+ < ∞. Then
Z t
|x(t)| ≤ |x0 | +
(M + L|x(s)|) ds,
t ∈ [0, T+ ),
0
Grönwall’s inequality ⇒
|x(t)| ≤ |x0 |eLT+ +
M LT+
(e
− 1),
L
t ∈ [0, T+ ),
so the solution doesn’t blow up at T+ . Contradiction! A similar argument works
for t ≤ 0.
Example. x0 = tx sin x + et . No explicit formula, but the solutions are defined for
all t since
|f (t, x)| ≤ eT + T |x|,
t ∈ [−T, T ],
so that we can take M (T ) = eT , L(T ) = T .
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4
Comparison theorems
Above we used Grönwall’s inequality to investigate the maximal existence time.
For scalar equations
x0 = f (t, x),
(2)
f : U → R, U ⊂ R2 open, one can also use the following comparison theorem.
Theorem (Combination of Lemma 1.2 and Theorem 1.3 in Teschl). Suppose f
is continuous and locally Lipschitz with respect to x. Let x(t) and y(t) be two
differentiable functions with
x0 (t) − f (t, x(t)) ≤ y 0 (t) − f (t, y(t)),
t ∈ [t0 , T ).
(i) If x(t0 ) ≤ y(t0 ) then x(t) ≤ y(t) for all t ∈ (t0 , T ).
(ii) If in addition x(t0 ) < y(t0 ) or x0 (t) − f (t, x(t)) < y 0 (t) − f (t, y(t)) for all
t ∈ [t0 , T ), then x(t) < y(t) for all t ∈ (t0 , T ).
Proof.
(i) Suppose that the claim is false. We can then find some interval [t1 , t1 +ε) such
that x(t1 ) = y(t1 ) and x(t) > y(t) for t ∈ [t1 , t1 + ε). Let ∆(t) = x(t) − y(t)
and note that
∆0 (t) = x0 (t) − y 0 (t) ≤ f (t, x(t)) − f (t, y(t)) ≤ L∆(t),
t ∈ [t1 , t1 + ε),
where L is a Lipschitz constant for f in a compact set containing the graphs
of x and y over the interval [t1 , t1 + ε]. Thus (∆(t)e−Lt )0 ≤ 0 and hence
∆(t) ≤ eL(t−t1 ) ∆(t1 ) = 0,
t ∈ [t1 , t1 + ε),
so that x(t) ≤ y(t) on [t1 , t1 + ε), contradicting the assumption.
(ii) The second part follows by setting ∆(t) = y(t) − x(t) ≥ 0 (by the first part).
Arguing as before one obtains ∆0 ≥ −L∆ and hence
∆(t) ≥ e−L(t−t0 ) ∆(t0 ) ≥ 0,
t ∈ (t0 , T ),
with strict inequality in the first step if x0 (t) − f (t, x(t)) < y 0 (t) − f (t, y(t))
and in the second one if x(t0 ) < y(t0 ). Hence, x(t) > y(t) in (t0 , T ).
This theorem is usually applied with either x0 (t) = f (t, x(t)) or y 0 (t) = f (t, y(t)).
If x0+ (t) ≥ f (t, x+ (t)) we call x+ a supersolution of (2) and if x0− (t) ≤ f (t, x− (t))
we call x− a subsolution of (2). This terminology differs slightly from Teschl, but
is more standard (he requires strict inequalities). Thus a supersolution will lie
above the solution with the same initial value for t > t0 , while a subsolution will
lie below the solution.
Let’s now use this to investigate the existence time.
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Example. Consider the IVP x0 = t2 + x2 , x(0) = 1. We have x0 (t) − x2 (t) ≥ 0 for
1
t ≥ 0, so x(t) is bounded from below by the solution 1−t
to the IVP x0 (t) = x2 (t),
0
2
2
x(0) = 1. On the other hand, x − x = t ≤ 1 there, so x is also bounded from
above by the solution tan(t + π4 ) to the IVP x0 = x2 + 1, x(0) = 1. Hence
1
π
≤ x(t) ≤ tan(t + ), 0 ≤ t < T+ ,
1−t
4
where T+ is the maximal existence time. It follows that π4 ≤ T+ ≤ 1.
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Peano’s theorem
Theorem (Peano’s theorem). Suppose that f ∈ C(U, Rn ), U ⊆ Rn+1 open,
(t0 , x0 ) ∈ U . There exists a solution of the IVP (1) on some interval containing t0 .
Take t ≥ t0 for simplicity. The proof is based on Euler’s method. Set
xh (tk+1 ) = xh (tk ) + f (tk , xh (tk ))h,
tk = t0 + kh,
where h > 0 is the step size. Use linear interpolation to define xh for t ∈ (tk , tk+1 ).
The idea is to show that for an appropriate interval [0, T0 ], there is a sequence
hm → 0 such that xhm converges to a solution.
For this, we first need a definition and a theorem.
Definition. Let I be a compact interval. A sequence {xm }∞
m=1 of functions I →
Rn is called uniformly equicontinuous if for every ε > 0 there is a δ > 0 s.t.
|xm (t) − xm (s)| < ε if |t − s| < δ,
m ≥ 1.
It is called uniformly bounded if there exists an M ≥ 0, s.t.
|xm (t)| ≤ M,
∀t ∈ I,
m ≥ 1.
Theorem (Arzelà-Ascoli). If the sequence {xm } is uniformly bounded and equicontinuous, then it contains a uniformly convergent subsequence {xmj }∞
j=1 .
Peano’s theorem follows by showing that {xhm } is uniformly bounded and equicontinuous for any sequence hm → 0 (if T0 > 0 is sufficiently small) and then showing
that the limit obtained from Arzelà-Ascoli satisfies the initial value problem. See
Chapter 2.7 in Teschl for details.
Examples.
• x0 = |x|α , x(0) = 0, 0 < α < 1, is an example where Peano’s theorem applies
but not Picard-Lindelöf. Solving the equation explicitly shows that there are
infinitely many solutions
• x0 = |x|α + |t|α , x(0) = 0, 0 < α < 1, is another example. No explicit
solutions in this case?
In the exercises there are some problems exploring uniqueness without Lipschitz
continuity in x.
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