My title - Departamento de Matemática da Universidade do Minho

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from zero. Integrating, from x 0 to x ∈ J on the left and from t 0 to t on the right, we obtain a
differentiable function x ↦→ t(x) defined as
t(x) = t 0 +
∫ x
x 0
dy
v(y)
for any x ∈ J. Now, observe that the derivative dt/dx is equal to 1/v. Since, by continuity, 1/v
does not change its sign in J, our t(x) is a strictly monotone continuously differentiable function.
We can invoke the inverse function theorem and conclude that the function t(x) is invertible. This
prove that the above relation defines actually a continuously differentiable function t ↦→ x(t) in
some interval I = t(J) of times around t 0 . Finally, you may want to check that the function
t ↦→ x(t) solves the Cauchy problem: just compute the derivative (using the inverse function
theorem),
( ) dt
ẋ(t) = 1/
dx (x(t))
= v(x) ,
and check the initial condition. Observe that the function t(x) − t 0 has then the interpretation of
the “time needed to go from x 0 to x”.
At the end of the story, if you are lucky enough and know how to invert the function t(x), you’ll
get an explicit solution as
x(t) = F −1 (t − t 0 + F (x 0 )) ,
where F is any primitive of 1/v. Close inspection of the above reasoning shows that the local
solution you’ve found is indeed the unique one. Namely, we have the following
Existence and uniqueness theorem for autonomous ODEs near a non-singular point.
Let v(x) be a continuous velocity field and let x 0 be a non-singular point of v. Then there exist one
and only one solution of the Cauchy problem ẋ = v(x) with initial condition x(t 0 ) = x 0 in some
sufficiently small interval I around t 0 . Moreover, the solution x(t) is the inverse function of
t(x) = t 0 +
defined in some small interval J around x 0 .
∫ x
x 0
dy
v(y) ,
Proof.
Here we give the pedantic proof. Let J be as above. Define a function H : R × J → R as
H(t, x) = t − t 0 −
∫ x
x 0
dy
v(y) .
If t ↦→ ϕ(t) is a solution of the Cauchy problem, then computation shows that d dtH (t, ϕ(t)) = 0
for any time t. There follows that H is constant along the solutions of the Cauchy problem.
Since H(t 0 , x 0 ) = 0, we conclude that the graph of any solution belongs to the level set Σ =
{(t, x) ∈ R × J s.t. H(t, x) = 0}. Now observe that H is continuously differentiable and that its
differential dH = dt + dx/v(x) is never zero. Actually, both partial derivatives ∂H/∂t and ∂H/∂x
are always different from zero. Hence we can apply the implicit function theorem and conclude
that the level set Σ is, in some neighborhood I × J of (t 0 , x 0 ), the graph of a unique differentiable
function x ↦→ t(x), as well as the graph of a unique differentiable function t ↦→ x(t), the inverse of
t, which as we have already seen solves the Cauchy problem.
On the failure of uniqueness near singular points. The interval I = t(J) where the
solution is defined need not be the entire real line: solutions may reach the boundary of J, i.e. one
of the singular points x ± of the velocity field, in finite time. Since singular points are themselves
equilibrium solutions, this imply that solutions of the Cauchy problem at singular points may not
be unique, under such mild conditions (continuity) for the velocity field. Later we’ll see Picard’s
theorem, which prescribes stronger regularity conditions on the velocity field v under which the
Cauchy problem admits unique solutions for any initial condition in the extended phase space.