Chapter 5
Existence, Uniqueness and
Dependence on Parameters
5.1
Existence
A system of n first order equations in the explicit form is given by:
x01 = f1 (t, x1 , x2 , · · · , xn ),
x02 = f2 (t, x1 , x2 , · · · , xn ),
······
x0n = fn (t, x1 , x2 , · · · , xn ),
or in matrix form:
x0 = f (t, x),
(5.1.1)
where x is the vector with components xi , i = 1, 2, · · · , n, and f is the vector with components fi , i = 1, 2, · · · , n.
Definition 5.1.1 Suppose that F = {f } is an infinite set of functions on the closed and
bounded interval α ≤ t ≤ β. If there exists a constant M0 > 0 such that
|f (t)| < M0 ,
∀f ∈ F, ∀t ∈ [α, β],
then we say that the family of functions F is uniformly bounded.
Definition 5.1.2 (equicontinuous) Suppose that F = {f } is an infinite set of functions on
the closed and bounded interval α ≤ t ≤ β. If for given > 0, there exists a δ(), which
depends only on , such that for any f ∈ F
|f (t0 ) − f (t00 )| < ,
∀t0 , t00 ∈ [α, β], |t0 − t00 | < δ,
then F is called equicontinuous on [α, β].
Theorem 5.1.1 (Ascoli-Arzela) Suppose that F = {f } is an infinite set of functions on
the closed and bounded interval α ≤ t ≤ β, F is uniformly bounded and equicontinuous on
53
54
Differential Equations
[α, β]. Then there is a sequence fn of elements of F such that fn converges uniformly on
[α, β].
Proof. Let {xi }i=1,2,3,... denote all rational numbers in [α, β]. Since F is uniformly
bounded, the set of points {f (x1 )}f ∈F is bounded. From the Bolzano-Weierstrass theorem,
there is a subsequence {fn1 } of distinct functions in F such that {fn1 (x1 )} converges.
Similarly, the sequence of points {fn1 (x2 )} is bounded, and there is a subsequence {fn2 }
of {fn1 } such that {fn2 (x2 )} converges. Continue in this way, we have
{fn1 } ⊃ {fn2 } ⊃ {fn3 } ⊃ · · ·
and for each k = 1, 2, 3, · · · ,
be the diagonal subsequence
subsequence {fnm }. Thus, fm
> 0 and rational number xk
the subsequence {fnk } converges at x1 , · · · , xk . Let {fm }
such that the mth term fm is the mth term in the mth
converges at every rational number of [α, β], i.e., for given
∈ [α, β], there is an integer Nk such that
|fn (xk ) − fm (xk )| < ,
3
n, m ≥ Nk .
Since F is equicontinuous, then for this fixed and for every x ∈ [α, β], there is an open
interval Ix containing x such that
|f (t0 ) − f (t00 )| < ,
3
∀f ∈ F, ∀t0 , t00 ∈ [α, β] ∩ Ix .
The collection of intervals Ix , x ∈ [α, β], forms an open cover of [α, β]. Since [α, β] is closed
and bounded, there is a finite subcover I1 , · · · , IJ . Thus, for any t ∈ [α, β], there are j
and k so that t and xk belong to the same interval Ij , j = 1, 2, · · · , J. For this J, let
N = max{N1 , N2 , · · · , NJ }, then
|fn (t) − fm (t)| ≤ |fn (t) − fn (xk )| + |fn (xk ) − fm (xk )| + |fm (xk ) − fm (t)|
<
+ + = , ∀n, m > N.
3 3 3
By Cauchy convergence criterion, the sequence {fn } converges uniformly on [α, β].
Lemma 5.1.2 Suppose that f (t, x) is continuous on the domain R : |t−t0 | ≤ a, |x−x0 | ≤
b. Let M = max(t,x)∈R |f (t, x)| and h = min{a, b/M }. On [t0 , t0 + h], define the function
ϕ(t) by
ϕ(t0 ) = x0 ,
ϕ(t) = ϕ(tk−1 ) + f (tk−1 , ϕ(tk−1 ))(t − tk−1 ),
tk−1 < t ≤ tk , k = 1, 2, · · · , n.
(5.1.2)
Similarly, ϕ(t) can be defined on [t0 − h, t0 ]. Then for given > 0, there is partition P of
[t0 − h, t0 + h], such that
dϕ(t)
− f (t, ϕ(t)) ≤ ,
dt
where
dϕ(t)
dt
is understood as the left or right derivative at the points tk , k = 0, 1, · · · , n.
Proof. It is easy to see that ϕ(t) is well defined for t ∈ [t0 − h, t0 + h], and
|ϕ(t) − ϕ(t̃)| ≤ M |t − t̃|,
∀t, t̃ ∈ [t0 − h, t0 + h].
Chapter 5: Existence, Uniqueness and Dependence on Parameters
55
Since f (t, x) is uniformly continuous on R, there exists a δ > 0 such that
|f (t, x) − f (t̃, x̃)| ≤ ,
∀(t, x) ∈ R, (t̃, x̃) ∈ R, |t − t̃| ≤ δ, |x − x̃| ≤ δ.
Let max1≤k≤n |tk − tk−1 | ≤ min{δ, δ/M }, then if tk−1 < t ≤ tk , we have
|ϕ(t) − ϕ(tk−1 )| ≤ δ,
and
dϕ(t)
− f (t, ϕ(t)) = f (tk−1 , ϕ(tk−1 )) − f (t, ϕ(t)) ≤ .
dt
Similarly, the above is true for the interval [t0 − h, t0 ].
Theorem 5.1.3 (Cauchy–Peano) Suppose that f (t, x) is continuous on the domain
R : |t − t0 | ≤ a, |x − x0 | ≤ b. Let M = max(t,x)∈R |f (t, x)|. Then the initial value problem
x0 = f (t, x),
x(t0 ) = x0
(5.1.3)
has a solution x = ϕ(t) on |t − t0 | ≤ h, where h = min{a, b/M }.
Proof. Let {m } be any sequence such that m > 0, m = 1, 2, · · · , and m → 0 as m → ∞.
For each m , let ϕm (t) be the function defined in Lemma 5.1.2, then
|ϕm (t) − ϕm (t̃)| ≤ M |t − t̃|,
∀|t − t0 | ≤ h, |t̃ − t0 | ≤ h,
which means that {ϕm (t)} is equicontinuous on [t0 − h, t0 + h]. Let t̃ = t0 we get
|ϕm (t)| ≤ |x0 | + M h,
∀|t − t0 | ≤ h,
which means that {ϕm (t)} is uniformly bounded on [t0 − h, t0 + h]. From Ascoli-Arzela
Theorem, {ϕm (t)} has a subsequence {ϕmk (t)} which converges uniformly on [t0 −h, t0 +h].
Denote the limit by {ϕ(t)}, i.e.,
lim ϕmk (t) = ϕ(t).
k→∞
Let
dϕm (t)
= f (t, ϕm (t)) + δ m (t),
dt
∀t ∈ [t0 − h, t0 + h].
Then from Lemma we have
dϕ (t)
|δ m (t)| = m
− f (t, ϕm (t)) ≤ m ,
dt
∀t ∈ [t0 − h, t0 + h].
Integrating (5.1.4) on both sides from t0 to t we get
Z t
Z t
ϕm (t) = x0 +
f (τ, ϕm (τ ))dτ +
δ m (t)dτ,
t0
∀t ∈ [t0 − h, t0 + h].
t0
Especially, we have
Z
ϕmk (t) = x0 +
t
t0
Z
f (τ, ϕmk (τ ))dτ +
t
δ mk (t)dτ,
t0
∀t ∈ [t0 − h, t0 + h].
(5.1.4)
56
Differential Equations
Since f (t, x) is uniformly continuous on R, and {ϕmk (t)} converges uniformly on [t0 −
h, t0 + h], we have
Z t
ϕ(t) = x0 +
f (τ, ϕ(τ ))dτ, ∀t ∈ [t0 − h, t0 + h].
t0
Since ϕ(t) and f (t, x) are both continuous, then ϕ(t) is differentiable. Taking the derivative
on the above equation, we get the result.
Corollary 5.1.4 Suppose that f (t, x) is continuous on an open domain G. Then the initial
value problem (5.1.3) has a solution x = ϕ(t) on some neighborhood of t0 .
Theorem 5.1.5 Suppose that f (t, x) is continuous on the domain R : |t − t0 | ≤ a, |x −
x0 | ≤ b, and satisfies the Lipschitz condition
|f (t, x1 ) − f (t, x2 )| ≤ L|x1 − x2 |,
∀(t, x1 ), (t, x2 ) ∈ R
(5.1.5)
where L is a constant. Let M = max(t,x)∈R |f (t, x)|. Then the initial value problem (5.1.3)
has a solution x = ϕ(t) on |t − t0 | ≤ h, where h = min{a, b/M }.
Proof. It is easy to see that the initial value problem is equivalent to the following integral
equation:
Z
t
ϕ(t) = x0 +
f (τ, ϕ(τ ))dτ.
t0
Construct a sequence of vectors:
ϕ0 (t) = x0 ,
Z
t
ϕk (t) = x0 +
f (τ, ϕk−1 (τ ))dτ,
k = 1, 2, · · · .
(5.1.6)
t0
It is easy to see by induction that ϕk (t), k = 0, 1, 2, · · · , is well defined, continuous, and
satisfies the inequality
|ϕk (t) − x0 | ≤ M h ≤ b,
∀t ∈ [t0 − h, t0 + h].
Also by induction, we can prove that
|ϕk (t) − ϕk−1 (t)| ≤ M Lk−1
Thus, the function series
x0 +
∞
X
|t − t0 |k
,
k!
k = 1, 2, · · · .
(ϕk (t) − ϕk−1 (t))
k=1
converges uniformly on [t0 − h, t0 + h], i.e., the sequence {ϕk (t)} converges uniformly to
some function ϕ(t), and ϕ(t) is continuous on [t0 − h, t0 + h]. Taking the limit in (5.1.6),
we get
Z
t
ϕ(t) = x0 +
f (τ, ϕ(τ ))dτ,
t0
i.e., ϕ(t) is the solution of the initial value problem (5.1.3).
Chapter 5: Existence, Uniqueness and Dependence on Parameters
5.2
57
Continuation of Solutions
In the last section, we showed that the solution of the initial value problem (5.1.3)
exists on on |t − t0 | ≤ h. Since h could be very small, the interval |t − t0 | ≤ h could be
very small. Thus, the theorems given in the last section are usually called local existence
theorems. Suppose that f (t, x) is continuous on the open domain G, and x = ϕ(t) is
a solution of (5.1.3) on the interval α ≤ t ≤ β, we extend the solution in the following
way. Since (α, ϕ(α)) ∈ G, then from Corollary 5.1.4 the initial value problem (5.1.3) has a
solution x = ϕ1 (t) on some interval [α−h1 , α+h1 ]. Similarly, there is a solution x = ϕ2 (t)
on some interval [β − h2 , β + h2 ].
Lemma 5.2.1 Suppose that f (t, x) is continuous on an open domain G, and x = ϕ(t) is
a solution of (5.1.3) on the interval α ≤ t ≤ β, then the solution x = ϕ(t) can be extended
to a bigger interval [α − h1 , β + h2 ] for some numbers h1 and h2 .
Proof. Let x = ϕ1 (t) and x = ϕ2 (t) be defined as above, and define

 ϕ1 (t),
ϕ(t),
ϕ̃(t) =

ϕ2 (t),
α − h1 ≤ t < α,
α ≤ t ≤ β,
β < t ≤ β + h2 .
Clearly, ϕ̃(t) satisfies the initial condition and the differential equation except at the two
points α and β. Since x = ϕ1 (t) is a solution of (5.1.3) on [α − h1 , α], then the left
derivative of ϕ̃(t) at α is
ϕ̃0− (α) = ϕ01− (α) = f (α, ϕ1 (α)) = f (α, ϕ(α)) = f (α, ϕ̃(α)).
Since x = ϕ(t) is a solution of (5.1.3) on [α, β], then the right derivative of ϕ̃(t) at α is
ϕ̃0+ (α) = ϕ0+ (α) = f (α, ϕ(α)) = f (α, ϕ̃(α)).
Thus, ϕ̃(t) satisfies the differential equation at t = α. Similarly, ϕ̃(t) satisfies the differential equation at t = β.
Following the above procedure, the solution x = ϕ(t) of (5.1.3) can be extended step
by step until the largest possible interval is reached. This interval must be an open
interval, since otherwise we can extend again. Let ρ(t) be the distance between the solution
x = ϕ(t) to the boundary of G, defined by
ρ(t) =
inf
(t0 ,x0 )∈∂G
[|t − t0 |2 + |ϕ(t) − x0 |2 ]1/2 ,
where ∂G is the boundary of G.
Theorem 5.2.2 Suppose that f (t, x) is continuous on an bounded open domain G, and let
x = ϕ(t) be a solution of (5.1.3) on the largest possible interval a < t < b, then we have
lim ρ(t) = 0,
t→a+
lim ρ(t) = 0.
t→b−
(5.2.1)
58
Differential Equations
Proof. Suppose that (5.2.1) is not right, for example,
lim ρ(t) 6= 0.
t→b−
Then there exists a ρ0 > 0 such that for any 1/k > 0, there is a tk ∈ (b − 1/k, b), and
ρ(tk ) ≥ ρ0 ,
k = 1, 2, · · · .
Since G is bounded, {ϕ(tk )} is bounded. Then {ϕ(tk )} has a converged subsequence
{ϕ(tki )}. Let
lim ϕ(tki ) = y0 .
i→∞
Clearly, (b, y0 ) ∈ G. Take δ > 0 such that the domain
R:
|t − b| ≤ δ, |x − y0 | ≤ δ
is in G. Since f (t, x) is continuous on R, f (t, x) is bounded on R. Let
M = max |f (t, x)|.
(t,x)∈R
Then for any t1 , t2 ∈ (b − δ, b), we have
Z
t1
ϕ(t1 ) = x0 +
f (τ, ϕ(τ ))dτ,
t0
Z t2
ϕ(t2 ) = x0 +
f (τ, ϕ(τ ))dτ.
t0
Subtracting these two equations we get
Z t2
|ϕ(t2 ) − ϕ(t1 )| = f (τ, ϕ(τ ))dτ t1
Z t2
≤
|f (τ, ϕ(τ ))|dτ
t1
≤ M |t2 − t1 |.
From Cauchy Convergence Criterion,
lim ϕ(t)
t→b−
exists. Thus, the solution can be extended to the right of b again, which is a contraction.
Theorem 5.2.3 Suppose that f (t, x) is continuous on the domain G : t0 < t < t1 , |x| <
∞, and let x = ϕ(t) be a solution of (5.1.3) on the largest possible interval a < t < b. If
there is a constant C such that |ϕ(t)| ≤ C, then the solution x = ϕ(t) exists on the whole
interval (t0 , t1 ).
Chapter 5: Existence, Uniqueness and Dependence on Parameters
5.3
59
Uniqueness
Lemma 5.3.1 (Gronwall inequality) Suppose that f (t) and g(t) are continuous nonnegative functions on the interval [α, β], which satisfy the inequality
Z
t
f (t) ≤ K +
f (s)g(s)ds,
∀t ∈ [α, β],
α
where K ≥ 0 is a constant. Then we have
Rt
f (t) ≤ Ke
α
g(s)ds
Proof. Let
∀t ∈ [α, β].
,
t
Z
h(t) = K +
f (s)g(s)ds,
α
then h0 (t) = f (t)g(t). Since f (t) ≤ h(t), we have h0 (t) ≤ h(t)g(t). Then we have
h0 (t)e−
Rt
α
g(s)ds
≤ h(t)g(t)e−
Rt
α
g(s)ds
,
i.e.,
h0 (t)e−
or
Rt
α
g(s)ds
− h(t)g(t)e−
Rt
α
g(s)ds
≤ 0,
i
Rt
dh
h(t)e− α g(s)ds ≤ 0.
dt
Integrating from α to t we get
h(t)e−
Rt
α
g(s)ds
− h(α) ≤ 0.
Thus, we have
f (t) ≤ h(t) ≤ h(α)e
Rt
α
g(s)ds
Rt
= Ke
α
g(s)ds
.
Theorem 5.3.2 Suppose that f (t, x) is continuous on the domain R : |t − t0 | ≤ a, |x −
x0 | ≤ b, and satisfies the Lipschitz condition
|f (t, x1 ) − f (t, x2 )| ≤ L|x1 − x2 |,
∀(t, x1 ), (t, x2 ) ∈ R
(5.3.1)
where L is a constant. Let M = max(t,x)∈R |f (t, x)|. Then the initial value problem (5.1.3)
has a unique solution x = ϕ(t) on |t − t0 | ≤ h, where h = min{a, b/M }.
Proof. Let ϕ1 (t) and ϕ2 (t) be two solutions of the initial value problem (5.1.3) on
|t − t0 | ≤ h. Then we have
Z
t
ϕ1 (t) = x0 +
f (τ, ϕ1 (τ ))dτ,
t0
Z t
ϕ2 (t) = x0 +
f (τ, ϕ2 (τ ))dτ.
t0
60
Differential Equations
Subtracting these two equations we get
Z t
|ϕ2 (t) − ϕ1 (t)| = [f (τ, ϕ2 (τ )) − f (τ, ϕ1 (τ ))]dτ t0
Z t
|ϕ2 (τ ) − ϕ1 (τ )|dτ ,
≤ L
∀t ∈ [t0 − h, t0 + h].
t0
If t ≥ t0 , then the Gronwall inequality implies that |ϕ2 (t) − ϕ1 (t)| ≤ 0, i.e., ϕ2 (t) =
ϕ1 (t), ∀t ∈ [t0 , t0 + h]. Similarly, if t ≤ t0 , we have ϕ2 (t) = ϕ1 (t), ∀t ∈ [t0 − h, t0 ].
For the uniqueness of the global solution, we have
Theorem 5.3.3 Suppose that f (t, x) is continuous on the open domain G. If for any
(t0 , x0 ) ∈ G, there is a closed domain
R : |t − t0 | ≤ a, |x − x0 | ≤ b,
such that R ⊂ G and the initial value problem (5.1.3) has a unique solution on R, then
if x = ϕ(t) is the local solution of the initial value problem (5.1.3), and x = ϕ̃(t) is the
continuation of x = ϕ(t) on the largest possible interval, then x = ϕ̃(t) is unique.
Proof. Let ϕ1 (t) and ϕ2 (t) are two continuations of ϕ(t) on the largest possible domains,
and ϕ1 (t1 ) 6= ϕ2 (t1 ) for some t1 > t0 . Then there exists a t̃0 ≥ t0 such that
ϕ1 (t̃0 ) = ϕ2 (t̃0 ),
ϕ1 (t) 6= ϕ2 (t),
∀t ∈ (t̃0 , t1 ].
(5.3.2)
Let x̃0 = ϕ1 (t̃0 ) = ϕ2 (t̃0 ). Since (t̃0 , x̃0 ) ∈ G, there is a closed domain |t − t̃0 | ≤ a, |x −
x̃0 | ≤ b, such that R ⊂ G and the initial value problem (5.1.3) has a unique solution on
R, which contradicts with (5.3.2).
5.4
Dependence on initial data and Parameters
Suppose that f (t, x) is continuous on the domain R : |t − t0 | ≤ a, |x − x0 | ≤ b, and
satisfies the Lipschitz condition on R. Then from Theorem 5.3.2 the initial value problem
(5.1.3) has a unique solution x = ϕ(t) in a neighborhood of t0 . Clearly, the solution ϕ(t)
is also dependent on the initial value (t0 , x0 ). Thus, we can also write the solution in the
form x = ϕ(t, t0 , x0 ), to emphasis the dependence on (t0 , x0 ).
Theorem 5.4.1 Suppose that f (t, x) is bounded and continuous on the open domain G,
and satisfies the Lipschitz condition
|f (t, x1 ) − f (t, x2 )| ≤ L|x1 − x2 |,
∀(t, x1 ), (t, x2 ) ∈ G,
where L is a constant. Let ϕ(t, t0 , x0 ) be the solution of
x0 = f (t, x)
(5.4.1)
and satisfy the initial condition ϕ(t0 ) = x0 , and let ψ(t, t̃0 , x̃0 ) be the solution of (5.4.1)
and satisfy the initial condition ψ(t̃0 ) = x̃0 . Suppose that ϕ(t, t0 , x0 ) and ψ(t, t̃0 , x̃0 ) both
Chapter 5: Existence, Uniqueness and Dependence on Parameters
61
exist on some interval α < t < β. Then for given > 0, there exists δ > 0 such that for
α < t < β, α < t̃ < β, we have
|ϕ(t, t0 , x0 ) − ψ(t̃, t̃0 , x̃0 )| < ,
∀|t − t̃| < δ, |t0 − t̃0 | < δ, |x0 − x̃0 | < δ.
Proof. Since ϕ(t, t0 , x0 ) and ψ(t, t̃0 , x̃0 ) are solutions of (5.4.1), we have
Z t
ϕ(t, t0 , x0 ) = x0 +
f (τ, ϕ(τ, t0 , x0 ))dτ,
t0
t
Z
f (τ, ψ(τ, t̃0 , x̃0 ))dτ.
ψ(t, t̃0 , x̃0 ) = x̃0 +
t̃0
Subtracting these two equations we get
t
Z
Z
t
f (τ, ϕ(τ, t0 , x0 ))dτ −
ϕ(t, t0 , x0 ) − ψ(t, t̃0 , x̃0 ) = x0 − x̃0 +
= x0 − x̃0 +
f (τ, ψ(τ, t̃0 , x̃0 ))dτ
t̃0
t0
Z t
[f (τ, ϕ(τ, t0 , x0 ))dτ − f (τ, ψ(τ, t̃0 , x̃0 ))]dτ
t̃0
t̃0
Z
+
f (τ, ϕ(τ, t0 , x0 ))dτ.
t0
Then we have
Z t
|ϕ(t, t0 , x0 ) − ψ(t, t̃0 , x̃0 )| = |x0 − x̃0 | + L
|ϕ(τ, t0 , x0 )) − ψ(τ, t̃0 , x̃0 ))|dτ + M |t̃0 − t0 |,
t̃0
where |f (t, x)| ≤ M, ∀(t, x) ∈ G. If |t0 − t̃0 | < δ and |x0 − x̃0 | < δ, we get
Z t
|ϕ(t, t0 , x0 ) − ψ(t, t̃0 , x̃0 )| = (M + 1)δ + L
|ϕ(τ, t0 , x0 )) − ψ(τ, t̃0 , x̃0 ))|dτ .
t̃0
Using Gronwall inequality we obtain
|ϕ(t, t0 , x0 ) − ψ(t, t̃0 , x̃0 )| ≤ (M + 1)δeL|t−t̃0 |
≤ (M + 1)δeL(β−α) .
Then if |t − t̃| < δ, we have
|ϕ(t, t0 , x0 ) − ψ(t̃, t̃0 , x̃0 )| ≤ |ϕ(t, t0 , x0 ) − ψ(t, t̃0 , x̃0 )| + |ψ(t, t̃0 , x̃0 ) − ψ(t̃, t̃0 , x̃0 )|
Z t
L(β−α)
≤ (M + 1)δe
+
|f (τ, ψ(τ, t̃0 , x̃0 ))|dτ t̃
L(β−α)
≤ (M + 1)δe
Then for given > 0, let δ <
,
M +(M +1)eL(β−α)
+ M δ.
we have the proof.
Theorem 5.4.2 Suppose that f (t, x) and g(t, x) are bounded and continuous on the open
domain G, and satisfies the Lipschitz condition
|f (t, x1 ) − f (t, x2 )| ≤ L|x1 − x2 |,
∀(t, x1 ), (t, x2 ) ∈ G,
(5.4.2)
|g(t, x1 ) − g(t, x2 )| ≤ L|x1 − x2 |,
∀(t, x1 ), (t, x2 ) ∈ G,
(5.4.3)
62
Differential Equations
where L is a constant. Let ϕ(t) be the solution of the initial value problem
x0 = f (t, x),
x(t0 ) = x0 ,
(5.4.4)
and ψ(t) the solution of the initial value problem
x0 = g(t, x),
x(t0 ) = x̃0
(5.4.5)
existing on the same interval α < t < β. Suppose that |f (t, x) − g(t, x)| < , ∀(t, x) ∈ G.
Then the solution satisfies the inequality
|ϕ(t) − ψ(t)| ≤ |x0 − x̃0 |eL(β−α) + (β − α)eL(β−α) .
Proof. Since ϕ(t) and ψ(t) are solutions of (5.4.4) and (5.4.5), we have
Z t
f (τ, ϕ(τ ))dτ,
ϕ(t) = x0 +
t0
t
Z
ψ(t) = x̃0 +
g(τ, ψ(τ ))dτ.
t0
Subtracting these two equations we get
t
Z
ϕ(t) − ψ(t) = x0 − x̃0 +
[f (τ, ϕ(τ ))dτ − g(τ, ψ(τ ))]dτ
t0
Z t
= x0 − x̃0 +
[f (τ, ϕ(τ ))dτ − f (τ, ψ(τ ))]dτ
t0
Z
t
[f (τ, ψ(τ ))dτ − g(τ, ψ(τ ))]dτ
+
t0
Then we have
Z t
|ϕ(t) − ψ(t)| ≤ |x0 − x̃0 | + L
|ϕ(τ ) − ψ(τ )|dτ + |t − t0 |
t0
Z t
≤ |x0 − x̃0 | + (β − α) + L
|ϕ(τ ) − ψ(τ )|dτ .
t0
Using Gronwall inequality we obtain
|ϕ(t) − ψ(t)| ≤ |x0 − x̃0 |eL(β−α) + (β − α)eL(β−α) .
5.5
Exercise
1. Consider the initial value problem:
x0 =
t+x
,
t−x
x(t0 ) = x0 ,
x0 6= t0 .
Find the domain R such that the initial-value problem has a solution in R.
Hint. Use Theorem 5.1.3. For any t0 6= x0 , there exist a and b, such that f (t, x) =
continuous on R : |t − t0 | ≤ a, |x − x0 | ≤ b.
t+x
t−x
is
Chapter 5: Existence, Uniqueness and Dependence on Parameters
63
2. Consider the initial value problem:
x0 = t + sin x,
x(t0 ) = x0 .
Find the domain R such that the initial-value problem has a unique solution in R.
3. Find the largest possible interval for the solution x(t) of the differential equation
x0 = (x2 − 1)/2 passing through the points (0, 0) and (ln 2, −3). Find the limit of
x(t) as t tends to the endpoints of the intervals. Hint. Find the solution.
4. Consider the initial value problem:
dx
= 3x2/3 ,
dt
x(t0 ) = x0 ,
where −∞ < t0 < +∞, −∞ < x0 < +∞. Discuss the existence and uniqueness of
the solution x(t).
5. Prove that the initial value problem
dx
= 3x + 15 cos x,
dt
x(0) = 1,
has a solution x(t) defined for all t ∈ (−∞, +∞).
Hint. Consider the domain R : |t − t0 | ≤ a, |x − x0 | ≤ b. Select suitable a and b, such that the
solution exists in a neighborhood of t0 which does not depend on a and b. Then, the solution can
be extended to (−∞, +∞).
6. Let t0 ∈ (−∞, +∞) and x0 ∈ (0, 1). Consider the initial value problem:
x2 − x
dx
= 2
,
dt
t + x2 + 1
x(t0 ) = x0 .
Show that the largest possible interval for the existence of the solution x = φ(t) of
the above initial value problem is (−∞, +∞).
Hint. Use Theorem 5.2.3. For any t0 , t1 , show that the solution x = ϕ(t) is bounded.
7. Consider the initial value problem:
(t − 1)y 0 =
y2
1
,
+ t3/2 + 1
y(2) = 2.
Discuss the existence and uniqueness of the solution, and find the largest possible
interval for the existence of the solution.
8. Consider the initial value problem:
dy
2y 3 − 8ty 2 + 8(t2 + 1)y − 4t
=
,
dt
y 2 − 4ty + 4t2 + 3
y(2) = 4.
(a) Discuss the existence and uniqueness of the solution.
(b) If a solution exists, find the largest possible interval for the existence of the
solution.
64
Differential Equations
9. Discuss the existence and uniqueness of the solution for the initial value problem:
p
y 00 − y 0 + sin y = sin t,
y(0) = 1, y 0 (0) = 1,
and if a solution exists, find the largest possible interval for the solution and verify
your answer.
Hint. Change to system.
10. Consider the initial value problem:
dx
x
=
,
dt
(t − 3)2
x(t0 ) = x0 ,
x < 3.
(a) Discuss the existence and uniqueness. (b) Find the largest possible interval for
the solution passing through (0, 0) and (5, 0).
11. Consider the initial value problem:
dx
x2
= 2 2
,
dt
x t + x2 + 1
x(0) = 1,
on the domain: R : −∞ < t < +∞, −3 < x < 3. (a) Show that the initial value
problem has a unique local solution. (b) Show that the largest possible interval for
the solution is (−∞, +∞).
12. Consider the initial value problem:
dx
x−1
=
,
dt
(t − 1)2
x(t0 ) = x0 ,
x < 5.
(a) Discuss the existence and uniqueness of the solution. (b) Find the largest possible
interval for the solution passing through (t0 , x0 ) = (2, 0).
13. Consider the initial value problem:
dy
2
=
,
dt
ty − y − t + 1
y(t0 ) = y0 .
(a) Discuss the existence and uniqueness of the solution for any (t0 , y0 ) ∈ R2 . (b) If
(t0 , y0 ) = (0, 0), find the largest possible interval for the existence of the solution.
14. Consider the initial value problem:
x+1
dx
=
,
dt
(t − 1)2
x(t0 ) = x0 ,
x < 1.
(a) Discuss the existence and uniqueness. (b) Find the largest possible interval for
the solution passing through (0, 0) and (5, 0).
Chapter 6: The Laplace Transform
65
15. Suppose that f1 (t, x) and f2 (t, x) are continuous on the domain G and satisfy the
condition
f1 (t, x) < f2 (t, x), ∀(t, x) ∈ G.
Consider the initial value problems
dx
= f1 (t, x),
dt
x(t0 ) = x0 ,
(5.5.1)
dx
= f2 (t, x),
dt
x(t0 ) = x0 ,
(5.5.2)
and
where (t0 , x0 ) ∈ G. Let x1 (t) and x2 (t) be the solutions of (5.5.1) and (5.5.2)
respectively. If both x1 (t) and x2 (t) exist on the same interval (a, b) (a < t0 < b),
prove that x1 (t) < x2 (t), ∀ t0 < t < b.
66
Differential Equations