My title - Departamento de Matemática da Universidade do Minho

6 FLOWS 46
• Verifique que
A =
( ρ 1
0 ρ
• Verifique que
( ) 0 ω
A =
−ω 0
( ) ρ ω
A =
−ω ρ
)
⇒ e tA = e ρt ( 1 t
0 1
( )
⇒ e tA cos(ωt) sin(ωt)
=
− sin(ωt) cos(ωt)
( )
⇒ e tA = e ρt cos(ωt) sin(ωt)
− sin(ωt) cos(ωt)
A origem é dita foco estável se ρ < 0, foco instável se ρ > 0.
• Considere o sistema linear
ẋ = x − y
ẏ = x + y
Determine a solução com condições iniciais x(0) = 1 e y(0) = 0.
6.5 Existence and uniqueness theorems
Here we consider a generic first order ODE of the form
ẋ = v(x, t)
where the velocity field v is a (continuous) function defined in some extended phase space X × R.
The phase space X may be some interval of the real line, an open subset of some Euclidean R n ,
or a differentiable manifold.
The problem we address is the existence and uniqueness of solutions of the initial value (or
Cauchy) problem. A local solution passing through the point (x 0 , t 0 ) ∈ X×R is a solution t ↦→ ϕ(t),
defined in some neighborhood I of t 0 , such that ϕ(t 0 ) = x 0 . Eventually, we’ll be interested also in
the possibility of extending such local solutions to larger intervals of times.
The basic existence theorem is 13
Peano’s theorem. Let v(x, t) be a continuous velocity field in some domain A of the extended
phase space R 2 . Then for any point (x 0 , t 0 ) ∈ A passes at least one integral curve of the differential
equation ẋ = v(x, t).
Idea of the proof. Natural guesses for the solutions are Euler lines starting through (x 0 , t 0 ). If
we restrict to a sufficiently small neighborhood of (x 0 , t 0 ), we can assume that the velocity field is
bounded, say |v(x, t)| ≤ K, and that all such Euler lines lies in the “papillon” made of two triangles
touching at (x 0 , t 0 ) with slopes ±K. Construct a family of Euler lines, graphs of ϕ n (t), such that
the maximal step ε n of the n-th line goes to 0 as n → ∞. One easily sees that the family (ϕ n )
is bounded and equicontinuous. By the Ascoli-Arzelá theorem it admits a (uniformly) convergent
subsequence. Finally, we claim that the sublimit ϕ ni → ϕ solves the differential equation. q.e.d.
Both existence and uniqueness may fail.
(ẋ) 2 − xt + 1 = 0
)
The Hamilton-Jacobi equation
cannot have solutions satisfying the initial condition x(0) = 0, for otherwise we would have a
negative “kinetic energy” (ẋ) 2 = −1 at that point!
Some regularity of the functions involved in a differential equation is also needed to ensure the
uniqueness of solutions. For example, both curves t ↦→ 0 and t ↦→ t 3 solve the equation
ẋ = 3x 2/3
with initial condition x(0) = 0. The problem here is that the velocity field v(t, x) = 3x 2/3 , although
continuous, is not differentiable and not even Lipschitz at the origin.
13 G. Peano, Sull’integrabilità delle equazioni differenziali del primo ordine, Atti Accad. Sci. Torino 21 (1886),
677-685. G. Peano, Demonstration de l’intégrabilité des équations différentielles ordinaires, Mathematische Annalen
37 (1890) 182-228.