My title - Departamento de Matemática da Universidade do Minho

6 FLOWS 43
Counter-example.
Both the curves x(t) = 0 and x(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(x) = 3x 2/3 , although
continuous, is not differentiable and not even Lipschitz at the origin. You may notice that the
solution starting, for example, at x 0 = 1 reaches (or better comes from) the singular point x − = 0
in finite time, since
∫ 0
t(x − ) − t(x 0 ) =
1
1 3 y−2/3 dy
= −1 .
One-dimensional Newtonian motion in a time independent force field. The onedimensional
motion of a particle of mass m subject to a force F (x) that does not depend on time
is described by the Newton equation
mẍ = −U ′ (x) ,
where the potential U(x) = − ∫ F (x)dx is some primitive of the force. The total energy
E (x, ẋ) = 1 2 mẋ2 + U(x)
(which of course is defined up to an arbitrary additive constant) of the system is a constant of
the motion, i.e. is constant along solutions of the Newton equation. In particular, once a value E
of the energy is given (depending on the initial conditions), the motion takes place in the region
where U(x) ≤ E, since the kinetic energy 1 2 mẋ2 is non-negative. Conservation of energy allows to
reduce the problem to the first order ODE
ẋ 2 = 2 (E − U(x)) ,
m
which has the unpleasant feature to be quadratic in the velocity ẋ. Meanwhile, if we are interested
in a one-way trajectory going from some x 0 to x, say with x > x 0 , we may solve for ẋ and find the
first order autonomous ODE
√
2
ẋ = (E − U(x)) .
m
There follows that the time needed to go from x 0 to x is
t(x) =
∫ x
x 0
dy
√
.
2
m (E − U(y))
The inverse function of the above t(x)
√
will give the trajectory x(t) with initial position x(0) = x 0
2
and initial positive velocity ẋ(0) =
m (E − U(x 0)), at least for sufficiently small times t.
6.3 Exponential
The exponential. The exponential function, according to Walter Rudin “the most important
function in mathematics” ([Ru87], 1st line of page 1), is the unique solution of the (autonomous)
differential equation
ẋ = x
with initial condition x(0) = 1. Actually, it is convenient to complexify time, i.e. take z = t+iθ ∈ C
with t, θ ∈ R, and define the exponential as the power series
exp(z) = 1 + x + x2
2 + x3
6 + x4
24 + · · · = ∑ z n
n!
n≥0
Since lim sup n→∞ (1/n!) 1/n = 0, the radius of convergence is ∞, hence the power series defines an
entire function, i.e. a holomorphic function exp : C → C. Deriving each term of the series, we