Ordinary Differential Equations II
January 22
2016
Initial value problems
(
x0 = f (t, x)
x(t0 ) = x0 ,
(1)
f ∈ C(U, Rn ), U ⊂ Rn+1 open.
Theorem (Picard-Lindelöf). Suppose f ∈ C(U, Rn ) and (t0 , x0 ) ∈ U . If f is loc.
Lipschitz in the 2nd argument, then ∃ a unique local solution x ∈ C 1 (I, Rn ) of (1)
in some interval I containing t0 .
Proof. Assume for simplicity that t0 = 0.
(
Z t
x0 = f (t, x)
⇔ x(t) = x0 +
f (s, x(s)) ds.
x(0) = x0 ,
0
Fixed point problem x = K(x), where
Z
K(x)(t) = x0 +
t
f (s, x(s)) ds,
0
and x belongs to some function space X. Need to identify X and C to apply the
contraction principle. For simplicity consider the case t ≥ 0. Set
V = [0, T ] × Bδ (x0 ),
Bδ (x0 ) = {x ∈ Rn : |x − x0 | < δ},
and choose T and δ so small that V ⊂ U . Let
M = max |f (t, x)| (f continuous, V compact).
(t,x)∈V
Suppose that (s, x(s)) ∈ V for all s ∈ [0, t]. Then
Z t
(2)
|K(x)(t) − x0 | ≤
|f (s, x(s))| ds ≤ tM ≤ δ
0
if
δ
0 ≤ t ≤ T0 := min T,
.
M
1
x
|x
x0 +
x0 | = M t
V
x0
x0
T
T0
t
Take
X = C([0, T0 ], Rn ),
C = {x ∈ X : kx − x0 k ≤ δ}.
Then K : C → C. It remains to show that K is a contraction. Let L be a Lipschitz
constant for f on [0, T0 ] × Bδ (x0 ).
Z t
|K(x)(t) − K(y)(t)| ≤
|f (s, x(s)) − f (s, y(s))| ds
0
Z t
≤
L|x(s) − y(s)| ds
0
≤ LT0 kx − yk,
0 ≤ t ≤ T0 . Hence kK(x) − K(y)k ≤ LT0 kx − yk. By possibly decreasing T0 we
can assume that θ := LT0 < 1, so that K is a contraction. It follows that K has
a unique fixed point x ∈ C. The integral eq. shows that x ∈ C 1 ([0, T0 ], Rn ). It
remains to prove that x is the unique sol. of the integral eq. in X. Let x ∈ X be
any solution. From (2) it follows that
|x(t) − x0 | ≤ tM < δ
as long as x(t) remains in Bδ (x0 ) and t < T0 . By continuity it follows that it stays
there for t close to 0. If it leaves (for the first time) at time t = T? < T0 we get
the contradiction
δ ≤ T? M < δ.
Hence, x ∈ C and it follows that x = x.
The proof shows that the solution exists at least on the interval
δ 1
|t − t0 | < min T, ,
.
M L
This can be improved to
δ
|t − t0 | ≤ min T,
M
2
by using Weissinger’s theorem. We prove that
|K n (x)(t) − K n (y)(t)| ≤
(Lt)n
kx − yk,
n!
n = 0, 1, 2, 3, . . . .
by induction. It’s clearly true for n = 0 and if it’s true for some n ≥ 1, then
Z t
n+1
n+1
|f (s, K n (x)(s)) − f (s, K n (y)(s))| ds
|K (x)(t) − K (y)(t)| ≤
Z0 t
L|K n (x)(s) − K n (y)(s)| ds
≤
Z0 t n+1 n
L s
≤
kx − yk ds
n!
0
Ln+1 tn+1
≤
kx − yk.
(n + 1)!
Taking θn =
point.
(LT0 )n
n!
in Weissinger’s theorem, we find that K has a unique fixed
Proposition. Suppose that [t0 , T ]×Rn ⊂ U and that f satisfies the global Lipschitz
condition
|f (t, x) − f (t, y)| ≤ L|x − y|,
x, y ∈ Rn ,
t ∈ [t0 , T ]
for some 0 ≤ L < ∞. Then the solution x is defined for all t ∈ [t0 , T ].
Proof. Repeat the proof of Picard-Lindelöf, but with C = X.
This applies in particular to linear systems
x0 = A(t)x + b(t),
where A(t) is a matrix and b(t) a vector, both with continuous entries. If A and
b are defined on some interval I, then the solution is also defined on I.
Continuous dependence
Lemma (Grönwall’s inequality — simple version). Assume that ψ ∈ C[0, T ] satisfies
Z t
ψ(t) ≤ α +
(βψ(s) + γ) ds, t ∈ [0, T ],
0
where α, γ ∈ R and β ≥ 0. Then
ψ(t) ≤ αeβt +
γ βt
(e − 1),
β
3
t ∈ [0, T ].
Proof. Let
Z
φ(t) = α +
t
(βψ(s) + γ) ds.
0
Then
φ0 (t) = βψ(t) + γ ≤ βφ(t) + γ,
t ∈ [0, T ].
This implies that
φ0 (t) − βφ(t) ≤ γ ⇒ (e−βt φ(t))0 ≤ γe−βt
γ
⇒ e−βt φ(t) − φ(0) ≤ (1 − e−βt )
|{z} β
=α
⇒ ψ(t) ≤ φ(t) ≤ αeβt +
γ βt
(e − 1),
β
t ∈ [0, T ].
Theorem. Suppose that f , g ∈ C(U, Rn ) and that f is loc. Lipschitz in the second
argument. If x(t), y(t) solve
(
(
x0 = f (t, x)
y 0 = g(t, y)
and
x(t0 ) = x0
y(t0 ) = y 0 .
Then
|x(t) − y(t)| ≤ |x0 − y 0 |eL|t−t0 | +
M L|t−t0 |
(e
− 1)
L
where
L=
sup
(t,x),(t,y)∈C
x6=y
|f (t, x) − f (t, y)|
,
|x − y|
M = sup |f (t, x) − g(t, x)|
(t,x)∈C
and C ⊂ U is a compact set containing the graphs of x and y.
Proof. Assume w.l.o.g. that t0 = 0 and t ≥ 0. Then
Z t
|x(t) − y(t)| ≤ |x0 − y 0 | +
|f (s, x(s)) − g(s, y(s))| ds,
0
where
|f (s, x(s)) − g(s, y(s))| ≤ |f (s, x(s)) − f (s, y(s))| + |f (s, y(s)) − g(s, y(s))|
≤ L|x(s) − y(s)| + M.
Apply Grönwall’s ineq. with α = |x0 − y 0 |, β = L and γ = M .
If f = g we obtain
|x(t) − y(t)| ≤ |x0 − y 0 |eL|t−t0 | ,
The solution depends continuously on the initial value, but the bound grows exponentially!
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Example. If f (t, x) = ax, we have L = |a|, so |x(t) − y(t)| ≤ |x0 − y0 |e|a|t , t ≥ 0.
If a > 0, we have equality:
x(t) = x0 eat ,
y(t) = y0 eat
|x(t) − y(t)| = |x0 − y0 |eat ,
⇒
so we can’t hope to do better in general.
x
y0
x0
t
Note: one can take M = supt∈I |f (t, y(t)) − g(t, y(t))| (see proof).
Example (Simple pendulum).
x00 + sin x = 0.
Assume x ≈ 0 ⇒ sin x ≈ x. Approximate model
x00 + x = 0.
2nd order ⇒ 2 initial conditions. How big is the difference between the solutions
if
x(0) = 0, x0 (0) = 1?
First order systems:
(
y10 = y2 ,
y20 = −y1 .
(
x01 = x2 ,
x02 = − sin x1 ,
Explicit solution
y1 (t) = sin t,
y2 (t) = cos t.
|f (t, y(t)) − g(t, y(t))| = |(y2 (t), − sin y1 (t)) − (y2 (t), −y1 (t))| = | sin y1 (t) − y1 (t)|
|y2 (t)|3
|t|3
1
1
≤
≤
≤ 3
= ,
3!
3!
2 · 3!
48
if |t| ≤ 1/2, so M = 1/48. Can take L = 1:
|f (t, x)−f (t, y)|2 = (x2 −y2 )2 +(sin x1 − sin y1 )2 ≤ (x2 −y2 )2 +(x1 −y1 )2 = |x−y|2 .
|
{z
}
≤(x1 −y1 )2
The difference at t = 1/2 can be estimated by
1 t
1
(e − 1) = (e0.5 − 1) ≈ 0.01.
48
48
If f is C k one can show that φ(·, x0 ) ∈ C k+1 (Lemma 2.3) and φ ∈ C k (Thm 2.10),
where φ(t, x0 ) is the solution with initial value x0 .
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