Differential Equations
and
Dynamical Systems
Classnotes for Math 645
University of Massachusetts
v3: Fall 2008
Luc Rey-Bellet
Department of Mathematics and Statistics
University of Massachusetts
Amherst, MA 01003
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Study Guide for Differential Equations and Dynamical Systems | MATH 645, Study notes of Mathematics
Material Type: Notes; Class: Diff Eq&Dynmc Sys I; Subject: Mathematics; University: University of Massachusetts - Amherst; Term: Fall 2008;
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Differential Equations
and
Dynamical Systems
Classnotes for Math 645
University of Massachusetts
v3: Fall 2008
Luc Rey-Bellet
Department of Mathematics and Statistics
University of Massachusetts
Amherst, MA 01003
Contents
Chapter 1
Existence and Uniqueness
1.1 Introduction
An ordinary differential equation (ODE) is given by a relation of the form
F (t, x, x′, x′′, · · · , x(m)) = 0 , (1.1)
where t ∈ R, x, x′, · · · , x(m)^ ∈ Rn^ and the function F is defined on some open set of R × Rn^ × · · · × Rn. A function x : I → Rn, where I is an interval in R, is a solution of (1.1) if x(t) is of class Cm^ (i.e., m-times continuously differentiable) and if
F (t, x(t), x′(t), x′′(t), · · · , x(m)(t)) = 0 for all t ∈ I. (1.2)
We say that the ODE is of order m if the maximal order of the derivative occurring in (1.1) is m.
Example 1.1.1 Clairaut equation (1734) Let us consider the first order equation
x − tx′^ + f (x′) = 0 , (1.3)
where f is some given function. It is given, in implicit form, by a nonlinear equation in x′. It is easy to verify that the lines x(t) = Ct − f (C) are solutions of (1.3) for any C. Consider for example f (z) = z^2 + z, then one sees easily that given a point (t 0 , x 0 ) there exists either 0 or 2 such solutions passing by the point (t 0 , x 0 ) (see Figure 1.1).
As we see from this example, it is in general very difficult to obtain results on the uniqueness or existence of solutions for general equations of the form (1.1). We will therefore restrict ourselves to situations where (1.1) can be solved as a function of x(m),
x(m)^ = g(t, x, x′, · · · , x(m−1)). (1.4)
Figure 1.1: Some solutions for Clairault equation for f (z) = z^2 + z.
Such an equation is called an explicit ODE of order m. One can always reduce an ODE of order m to a first order ODE for a vector in a space of larger dimension. For example we can introduce the new variables
x 1 = x , x 2 = x′^ , x 3 = x′′^ · · · , xm = x(m−1)^ , (1.5)
and rewrite (1.4) as the system
x′ 1 = x 2 , x′ 2 = x 3 , ..
. (1.6) x′ m− 1 = xm , x′ m = g(t, x 1 , x 2 , · · · , xm).
This is an equation of order 1 for the supervector x = (x 1 , · · · , xm) ∈ Rnm^ (each xi is in Rn) and it has the form x′^ = f (t, x). Therefore, in general, it is sufficient to consider the case of first order equations (m=1). If f does not depend explicitly on t, i.e., f (t, x) = f (x), the ODE x′^ = f (x) is called autonomous. The function f : U → Rn, where U is an open set of Rn, defines a vector field. A solution of x′^ = f (x) is then a parametrized curve x(t) which is tangent to the vector field f (x) at any point, see figures 1.2 and (1.3). Note a non-autonomous ODE x′^ = f (t, x) with x ∈ Rn^ can be written as an autonomous ODE in Rn+1^ by setting
y =
( x t
) y′^ =
( x′ t′
)
( f (t, x) 1
) ≡ F (y). (1.7)
It can written as a first order system by setting y = x′
x′^ = y , y′^ = (1 − x^2 )y − x. (1.12)
It is a perturbation of the harmonic oscillator ( = 0) x′′^ + x = 0 whose solutions are the periodic solution x(t) = A cos(t − φ) and y(t) = x′(t) = −A sin(t − φ) (circles). When > 0 one observes that one periodic solution survives which is the deformation of a circle of radius 2 and all other solution are attracted to this periodic solution (limit cycle), see Figure 1.3.
Figure 1.3: The vector field for the van der Pol equation with = 0.1 as well as two solutions passing through the points (. 1 , .2) and (2, 3).
We will discuss these examples in more details later. For now we observe that, in both cases, the solutions curves never intersect. This means that there are never two solutions passing by the same point. Our first goal will be to find sufficient conditions for the problem x′^ = f (t, x) , x(t 0 ) = x 0 , (1.13)
to have a unique solution. We say that t 0 and x 0 are the initial values and the problem (1.13) is called a Cauchy Problem or an initial value problem (IVP).
1.2 Banach fixed point theorem
We will need some (simple) tools of functional analysis. Let E be a vector space with addition + and multiplication by scalar λ in R or C. A norm on E is a map ‖ · ‖ : E → R which satisfies the following three properties
- N1 ‖x‖ ≥ 0 and ‖x‖ = 0 if and only if x = 0 ,
- N2 ‖λx‖ = |λ|‖x‖ ,
- N3 ‖x + y‖ ≤ ‖x‖ + ‖y‖ (triangle inequality).
A vector space E equipped with a norm ‖ · ‖ is called a normed vector space. In a normed vector space E we can define the convergence of sequence {xn}. We say that the sequence {xn} converges to x ∈ E, if for any > 0, there exists N ≥ 1 such that, for all n ≥ N , we have ‖xn − x‖ ≤ . We say that {xn} is a Cauchy sequence if for any > 0, there exists N ≥ 1 such that, for all n, m ≥ N , we have ‖xn − xm‖ ≤ .
Definition 1.2.1 A normed vector space E is said to be complete if every Cauchy sequence in E converges to an element of E. A complete normed vector space E is called a Banach space.
Let ‖ · ‖ and ‖ · ‖? denote two norms on the vector space E. We say that the norms ‖ · ‖ and ‖ · ‖? are equivalent if there exist positive constants c and C such that
c‖x‖ ≤ ‖x‖? ≤ C‖x‖ for all x ∈ E.
It is easy to check that the equivalence of norm defines an equivalence relation. Further- more if a Cauchy sequence for a norm ‖ · ‖ is also a Cauchy sequence for an equivalent norm ‖ · ‖?.
Example 1.2.2 The vector space E = Rn^ or Cn^ with the euclidean norm ‖x‖ 2 = (
∑ i x^2 i )^1 /^2 is a Banach space.^ Other examples of norms are^ ‖x‖ 1 =^
∑ i |xi|^ or^ x∞ = supi |xi|. In any case Rn^ or Cn^ equipped with any norm is a Banach space, since all norm are equivalent in a finite-dimensional space (see exercises).
The previous example shows that for finite dimensional vector spaces the choice of a norm does not matter much. For infinite-dimensional vector spaces the situation is very different as the following example demonstrate.
Proposition 1.2.3 Let
C([0, 1]) = {f : [0, 1] → Rn^ ; f continuous}. (1.14)
With the norm ‖f ‖∞ = sup t∈[0,1]
|f (t)|. (1.15)
C([0, 1]) is a Banach space. With either of the norms
‖f ‖ 1 =
∫ (^1)
0
|f (t)| dt , or ‖f ‖ 2 =
(∫ (^1)
0
|f (t)|^2 dt
) 1 / 2 , (1.16)
C([0, 1]) is not complete.
- Let E and F be two Banach spaces and U ⊂ E. A function f : U → F is continuous at x 0 ∈ U if for all > 0, there exists δ > 0 such that x ∈ U and ‖x − x 0 ‖ < δ implies that ‖f (x) − f (x 0 )‖ < .
- The map x 7 → ‖x‖ is a continuous function of E to R, since |‖x‖ − ‖x 0 ‖| ≤ ‖x − x 0 ‖ by the triangle inequality.
Certain properties which are true in finite dimensional Banach spaces are not true in infinite dimensional Banach spaces such as the function spaces we have considered in Propositions 1.2.3 and 1.2.4. For example we show that
- The closed ball {x ∈ E ; ‖x‖ ≤ 1 } is not necessarily compact.
- Two norms on a Banach space are not always equivalent.
- The theorem of Bolzano-Weierstrass which says each bounded sequence has a convergent subsequence is not necessarily true.
- The equivalence of K compact and K closed and bounded is not necessarily true.
The proposition 1.2.3 shows that ‖ · ‖∞ and ‖ · ‖ 1 are not equivalent. For, if they were equivalent, any Cauchy sequence for ‖ · ‖ 1 would be a Cauchy sequence ‖ · ‖∞. But we have constructed explicitly a Cauchy sequence for ‖ · ‖ 1 which is not a Cauchy sequence for ‖ · ‖∞. Let us consider the Banach space B([0, 1]) and let fn(t) to be equal to 1 if 1 /(n + 1) < t ≤ 1 /n and 0 otherwise. We have ‖fn‖∞ = 1 for all n and ‖fn − fm‖∞ = 1 for any n, m. Therefore {fn} cannot have a convergent subsequence. This shows at the same time, that the unit ball is not compact, that Bolzano-Weierstrass fails, and that closed bounded sets are not necessarily compact. Let us suppose that we want to solve a nonlinear equation in a Banach space E. Let f be a function from E to E then we might want to solve
f (x) = x find a fixed point of f. (1.21)
The next theorem will provide a sufficient condition for the existence of a fixed point.
Theorem 1.2.5 (Banach Fixed Point Theorem (1922)) Let E be a Banach space, D ⊂ E closed and f : D → E a map which satisfies
- f (D) ⊂ D.
- f is a contraction on D, i.e., there exists α < 1 such that,
‖f (x) − f (y)‖ ≤ α‖x − y‖ , for all x, y ∈ D. (1.22)
Then f has a unique fixed point x in D, f (x) = x.
Proof: We first show uniqueness. Let us suppose that there are two fixed points x and y, i.e., f (x) = x and f (y) = y. Since f is a contraction we have
‖x − y‖ = ‖f (x) − f (y)‖ ≤ α‖x − y‖ (1.23)
with α < 1, this is possible only if x = y. To prove the existence we choose an arbitrary x 0 ∈ D and we consider the iteration x 1 = f (x 0 ), · · · , xn+1 = f (xn), · · ·. Since f (D) ⊂ D this implies that xn ∈ D for any n. Let us show that {xn} is a Cauchy sequence. We have ‖xn+1 − xn‖ = ‖f (xn) − f (xn− 1 )‖ ≤ α‖xn − xn− 1 ‖. Iterating this inequality we obtain
‖xn+1 − xn‖ ≤ αn‖x 1 − x 0 ‖. (1.24)
If m > n this implies that
‖xm − xn‖ ≤ ‖xm − xm− 1 ‖ + ‖xm− 1 − xm− 2 ‖ + · · · ‖xn+1 − xn‖ ≤
( αm−^1 + αm−^2 + · · · αn
) ‖x 1 − x 0 ‖
≤
αn 1 − α
‖x 1 − x 0 ‖. (1.25)
Therefore {xn} is a Cauchy sequence since αn^ → 0. Since E is a Banach space, this sequence converges to x ∈ E. The limit x is in D since D is closed. Since f is a contraction, it is continuous and we have
x = lim n→∞ xn+1 = f ( lim n→∞ xn) = f (x) , (1.26)
i.e., x is a fixed point of f.
The proof of the theorem is constructive and provides the following algorithm to construct a fixed point.
Method of successive approximations: To solve f (x) = x
- Choose an arbitrary x 0.
- Iterate: xn+1 = f (xn).
Even if the hypotheses of the theorem are difficult to check, one might apply this algorithm. If the algorithm converges this gives a fixed point, although not necessarily a unique one.
Example 1.2.6 The function f (x) = cos(x) has a fixed point on D = [0, 1]. By the mean value theorem there is ξ ∈ (x, y) such that cos(x) − cos(y) = sin(ξ)(y − x), thus | cos(x) − cos(y)| ≤ supt∈[0,1] | sin(t)||x − y| ≤ sin(1)|x − y|, and sin(1) < 1. One observes a quite rapid convergence to the solution 0. 7390 · · ·. For example we have x 0 = 0, x 1 = 1, x 2 = 0.5403, x 2 = 0.8575, x 3 = 0.6542, x 4 = 0.7934, x 5 = 0.7013, x 6 = 0.7639, · · ·.
The integral equation (1.32) can then be written as the fixed point equation
(T x)(t) = x(t) , (1.34)
i.e., we have transformed the differential equation (1.31) into a fixed point problem. The method of successive approximation for this problem is called
Picard-Lindel¨of iteration:
x 0 (t) = x 0 (or any other function) ,
xn+1(t) = x 0 +
∫ (^) t
t 0
f (s, xn(s)) ds. (1.35)
Example 1.3.1 Let us consider the Cauchy problem
x′^ = −x^2 , x(0) = 1. (1.36)
The solution is x(t) = (^) 1+^1 t. The Picard-Lindel¨of iteration gives x 0 = 1, x 1 = 1 − t, x 2 = 1 − t + t^2 − t^3 /3, and so on. One sees from Figure 1.4 that it converges in a suitable interval around 0 but diverges for larger values of t.
Figure 1.4: The first four iterations for the Picard Lindel¨of iteration scheme for the Cauchy problem x′^ = −x^2 , x(0) = 1.
The next result shows how to choose the interval I such that T maps a suitably chosen set D into itself. We have
Lemma 1.3.2 Let A = {(t, x) ; |t−t 0 | ≤ a , ‖x−x 0 ‖ ≤ b}, f : A → Rn^ be a continuous function with M = sup(t,x)∈A |f (t, x)|. We set α = min(a, b/M ). The map T given by (1.33) is well-defined on the set
B = {x : [t 0 − α, t 0 + α] → Rn^ , x continuous and ‖x(t) − x 0 ‖ ≤ b}. (1.37)
and it satisfies T (B) ⊂ B.
Proof: The lemma follows from the estimate
‖(T x)(t) − x 0 ‖ =
∥∥ ∥∥
∫ (^) t
t 0
f (s, x(s)) ds
∥∥ ∥∥ ≤ M |t − t 0 | ≤ M α ≤ b. (1.38)
We say that a function f : A → Rn^ (with A is in the previous lemma) satisfies a Lipschitz condition if
‖f (t, x) − f (t, y)‖ ≤ L‖x − y‖ for all (t, x), (t, y) ∈ A. (1.39)
The constant L is called the Lipschitz constant.
Remark 1.3.3 In order to illustrate the meaning of condition (1.39), let us suppose that f (t, x) = f (x) does not depend on t and that we have ‖f (x) − f (y)‖ ≤ L‖x − y‖ whenever x and y are in the closed ball Bb(x 0 ). This clearly implies that f is continuous in Bb(x 0 ) and f is called Lipschitz continuous. The opposite does not hold, for example
the function f (x) =
√ |x| is continuous but not Lipschitz continuous at 0. If f is of class C^1 , then f is Lipschitz continuous. To see this consider the line z(s) = x + s(y − x) which interpolates between x and y. We have
‖f (y) − f (x)‖ =
∥∥ ∥∥ ∥
∫ (^1)
0
d ds
f (z(s)) ds
∥∥ ∥∥ ∥ =
∥∥ ∥∥
∫ (^1)
0
f ′(z(s))(y − x) ds
∥∥ ∥∥
≤ sup z∈Bb(x 0 )
‖f ′(z)‖‖y − x‖ , (1.40)
and therefore f is Lipschitz continuous with L = sup{x , ‖x−x 0 ‖≤b} ‖f ′(x)‖. On the other hand Lipschitz continuity does not imply differentiability as the function f (x) = |x| demonstrates. The condition (1.39) requires that f (t, x) is Lipschitz continuous in x uniformly in t with |t − t 0 | ≤ a.
If f (t, x) satisfy a Lipschitz condition we have, for any t ∈ I = [t 0 − α, t 0 + α],
‖(T x)(t) − (T z)(t)‖ ≤
∫ (^) t
t 0
‖f (t, x(t)) − f (t, z(t))‖ dt
∫ (^) t
t 0
L‖x(t) − z(t)‖ dt
≤ αL sup t∈I
‖x(t) − z(t)‖ ≤ αL‖x − z‖∞. (1.41)
Taking the supremum over t on the left side shows that ‖T x − T z‖∞ ≤ αL‖x − z‖. If αL < 1 we can apply the Banach fixed point theorem to prove existence of a fixed
To show that x(t) is a solution of the Cauchy problem we take the limit n → ∞ in (1.35). The left side converges uniformly to x(t). Since f is continuous and A is compact f (t, xk(t)) converges uniformly to f (t, x(t)) on A. Thus one can exchange integral and the limit and x(t) is a solution of the integral equation (1.32). It remains to prove uniqueness of the solution. Let x(t) and z(t) be two solutions of (1.32). By recurrence we show that
‖x(t) − y(t)‖ ≤ 2 M Lk^
|t − t 0 |k+ (k + 1)!
We have x(t) − y(t) =
∫ (^) t t 0 (f^ (s, x(s))^ −^ f^ (s, y(s)))^ ds^ and therefore^ ‖x(t)^ −^ y(t)‖ ≤ 2 M |t − t 0 | which (1.45) for k = 0. If (1.45) holds for k − 1 we have
‖x(t) − y(t)‖ ≤
∫ (^) t
t 0
L‖x(s) − y(s)‖ ds ≤ 2 M Lk
∫ (^) t
t 0
|s − t 0 |k k!
dt
≤ 2 M Lk^
|t − t 0 |k+ (k + 1)!
and this proves (1.45). Since this holds for all k, this shows that x(t) = y(t).
1.4 Peano Theorem
In the previous section we established a local existence result by assuming a Lipschitz condition. Simple examples show that this condition is also necessary.
Example 1.4.1 Consider the ODE
x′^ = 2
√ |x|. (1.47)
We find that x(t) = (t − c)^2 for t > c and x(t) = −(c − t)^2 for t < c is a solution for any constant c. But x(t) ≡ 0 is also a solution. The Cauchy problem with, say, x(0) = 0 has infinitely many solutions. For t > 0, x(t) ≡ 0 is one solution, x = t^2 is another solution, and more generally x(t) = 0 for 0 ≤ t ≤ c and then x(t) = (t − c)^2 for
t ≥ c is also a solution for any c. This phenomenon occur because
√ |x| is not Lipschitz continuous at x = 0.
We are going to show that, without Lipschitz condition, we can still obtain existence of solutions, but not uniqueness. Instead of using the Picard-Lindel¨of iteration we are using another approximation scheme. It turns out to be the simplest algorithm used for numerical approximations of ODE’s.
Euler polygon (1736) Fix some h 6 = 0, the idea is to approximate the solution locally by x(t + h) ' x(t) + hf (t, x(t)). Let us consider now the sequence {tn, xn} given recursively by tn+1 = tn + h , xn+1 = xn + hf (tn, xn). (1.48)
We then denote by xh(t) the piecewise linear function which passes through the points (tn, xn). It is called the Euler polygon and is an approximation to the solution of the Cauchy problem.
0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.
1
Figure 1.5: Euler polygons for x′^ = −x^2 for h=0.5 and h = 0.25.
Lemma 1.4.2 Let A = {(t, x) ; |t − t 0 | ≤ a , ‖x − x 0 ‖ ≤ b}, and f : A → Rn^ be a continuous function with M = sup(t,x)∈A ‖f (t, x)‖. We set α = min(a, b/M ). If h = ±α/N , N an integer, the Euler Polygon satisfies (t, xh(t)) ∈ A for t ∈ [t 0 −α, t 0 +α] and we have the bound
‖xh(t) − xh(t′)‖ ≤ M |t − t′| for t, t′^ ∈ [t 0 − α, t 0 + α]. (1.49)
Proof: Let us consider first the interval [t 0 , t 0 + α] and choose h > 0. We show first, by induction that (tn, xn) ∈ A for n = 0, 1 , · · · , N. We have ‖xn − xn− 1 ‖ ≤ hM and so ‖xn − x 0 ‖ ≤ nhM ≤ αM ≤ b if n ≤ N. Since xh(t) is piecewise linear (t, xh(t)) ∈ A for any t ∈ [t 0 , t 0 + α]. The estimate (1.49) follows from the fact that the slope of xh(t) is nowhere bigger than M. On [t 0 − α, t 0 ] the argument is similar.
Definition 1.4.3 A family of functions fj : [a, b] → Rn, j = 1, 2 , · · ·, is equicontinuous if for any > 0 there exists δ > 0 such that for, for all j, |t − t′| < δ implies that ‖fj (t) − fj (t′)‖ ≤ .
Equicontinuity means that all the functions fj are uniformly continuous and that, moreover, δ can be chosen to depend only on , but not on j. The estimate (1.49) shows that the family xh(t), with h = α/N , N = 1, 2 , · · ·, is equicontinuous.
Theorem 1.4.6 (Peano 1890) Let A = {(t, x) ; |t − t 0 | ≤ a , ‖x − x 0 ‖ ≤ b}, f : A → Rn^ a continuous function with M = sup(t,x)∈A ‖f (t, x)‖. Set α = min(a, b/M ). The Cauchy problem (1.31) has a solution on [t 0 − α, t 0 + α].
Proof: Let us consider the Euler polygons with h = α/N , N = 1, 2 , · · ·. The sequence is bounded since ‖xh(t) − x 0 ‖ ≤ M |t − t 0 | ≤ M α and equicontinuous by Lemma 1.4.2. By Arzel`a-Ascoli Theorem, the family xh(t) has a subsequence which converges uniformly to a continuous function x(t) on [t 0 − α, t 0 + α]. It remains to show that x(t) is a solution. Let t ∈ [t 0 , t 0 + α] and let (tn, xn) the approximation obtained by Euler method for xh(t). If t ∈ [tl, tl+1] we have
xh(t) − x 0 = hf (t 0 , x 0 ) + hf (t 1 , x 1 ) + · · · + hf (tl− 1 , xl− 1 ) + (t − tl)f (tl, xl). (1.54)
Since f (t, x(t)) is a continuous function of t it is Riemann integrable and, using a Riemann sum with left-end points have ∫ (^) t
t 0
f (s, x(s)) ds = hf (t 0 , x(t 0 )) + hf (t 1 , x(t 1 )) + · · ·
· · · + hf (tl− 1 , x(tl− 1 )) + (t − tl)f (tl, x(tl)) + r(h) , (1.55)
where limh→ 0 ‖r(h)‖ = 0. By the uniform continuity of f on A and the uniform convergence of the subsequence of {xh(t)} to x(t) we have that ‖f (t, xh(t)−f (t, x(t))‖ ≤ if h is sufficiently small (and h is such that xh belongs to the convergent subsequence). Using that xh(tj ) = xj and subtracting (1.55) from (1.54) we find that
‖xh(t) − x 0 −
∫ (^) t
t 0
f (s, x(s)) ds‖ ≤ (l + 1)h + ‖r(h)‖ ≤ α + ‖r(h)‖ (1.56)
which converges to α as h → 0. Since is arbitrary x(t) is a solution of the Cauchy problem in integral form (1.32).
1.5 Continuation of solutions
So far we only considered local solutions, i.e., solutions which are defined in a neigh- borhood of (t 0 , x 0 ). Simple examples shows that the solution x(t) may not exist for all t, for example the equation x′^ = 1 + x^2 has solution x(t) = tan(t − c) and this solution does not exist beyond the interval (c − π/ 2 , c + π/2), and we have x(t) → ±∞ as t → c ± π/2. To extend the solution we solve the Cauchy problem locally, say from t 0 to t 0 + α and then we can try to continue the solution by solving the Cauchy problem x′^ = f (t, x) with new initial condition x(t 0 + α) and find a solution from t 0 + α to t 0 + α + α′, and so on... In order do this we should be able to solve it locally everywhere and we will therefore assume that f satisfy a local Lipschitz condition.
Definition 1.5.1 A function f : U → Rn^ (where U is an open set of R×Rn) satisfies a local Lipschitz condition if for any (t 0 , x 0 ) ∈ U there exist a neighborhood V ⊂ U such that f satisfies a Lipschitz condition on V , see (1.39).
Note that if the function f is of class C^1 in U , then it satisfies a local Lipschitz condition.
Lemma 1.5.2 Let U ⊂ R × Rn^ be an open set and let us assume that f : U → Rn is continuous and satisfies a local Lipschitz condition. Then for any (t 0 , x 0 ) ∈ U there exists an open interval Imax = (ω− , ω+) with −∞ ≤ ω− < t 0 < ω+ ≤ ∞ such that
- The Cauchy problem x′^ = f (t, x), x(t 0 ) = x 0 has a unique solution on Imax.
- If y : I → Rn^ is a solution of x′^ = f (t, x), y(t 0 ) = x 0 , then I ⊂ Imax and y = x|I.
Proof: a) Let x : I → Rn^ and z : J → Rn^ be two solutions of the Cauchy problem with t 0 ∈ I, J. Then x(t) = z(t) on I ∩ J. Suppose it is not true, there is point ¯t such that x(¯t) 6 = z(¯t). Consider the first point where the solutions separate. The local existence theorem 1.3.4 shows that it is impossible. b) Let us define the interval
Imax =
⋃ {I ; I open interval , t 0 ∈ I , there exists a solution on I}. (1.57)
This interval is open and we can define the solution on Imax as follows. If t ∈ Imax, then there exists I where the Cauchy problem has a solution and we can define x(t). The part (a) shows that x(t) is uniquely defined on Imax.
Theorem 1.5.3 Let U ⊂ R×Rn^ be an open set and let us assume that f : U → Rn^ is continuous and satisfies a local Lipschitz condition. Then every solution of x′^ = f (t, x) has a continuation up to the boundary of U. More precisely, if x : Imax → Rn^ is the solution passing through (t 0 , x 0 ) ∈ U , then for any compact K ⊂ U there exists t 1 , t 2 ∈ Imax with t 1 < t 0 < t 2 such that (t 1 , x(t 1 )) ∈/ K, (t 2 , x(t 2 )) ∈/ K.
Remark 1.5.4 If U = R × Rn, Theorem 1.5.3 means that either
- x(t) exists for all t,
- There exists t∗^ such that limt→t∗ ‖x(t)‖ = ∞,
The exists globally or the solution ”blows up” at a certain point in time.