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# Initial value problem

Module by: Aurèle Balavoine. E-mail the author

Summary: This module describes some conditions that guarantee the existence and uniqueness of the solution to an Ordinary Differential Equation with initial value condition.

Read more about ordinary differential equations, stability of equilibrium and Lyapunov direct method.

## Initial value problem

In this module, we are interested in the existence and uniqueness of solutions to Ordinary Differential Equations (ODE) that are defined as:

x ˙ ( t ) = f ( t , x ( t ) ) . x ˙ ( t ) = f ( t , x ( t ) ) .
(1)

The variable tt is a positive scalar and xx in RNRN is a vector of dimension NN. Let DD be a subset of R×RNR×RN, f:DRNf:DRN be a continuous function and x˙=dx/dtx˙=dx/dt.

A differential equation is generally satisfied by an infinite subset of solutions. For instance, the differential equation x˙(t)=2tx˙(t)=2t admits as a solution x(t)=t2+Cx(t)=t2+C, where CC is any constant in RR. In practice, we are interested in one specific solution, which is usually characterized by a constraint on its initial state. Finding a function that satisfies both the differential equation and the initial state condition is known as the initial value problem.

Given a point (t0,x0)D(t0,x0)D, we call solution to the initial value problem Equation 1 with initial condition (t0,x0)(t0,x0) a solution of Equation 1 on an interval II containing t0t0 satisfying x(t0)=x0x(t0)=x0, i.e.

x ˙ ( t ) = f ( t , x ( t ) ) x ( t 0 ) = x 0 . x ˙ ( t ) = f ( t , x ( t ) ) x ( t 0 ) = x 0 .
(2)

The solution φ(t,t0,x0)=x(t)φ(t,t0,x0)=x(t) is called the flow and the set of points (t,φ(t,t0,x0)):tI(t,φ(t,t0,x0)):tI is a trajectory.

## example

Consider the following differential equation for xR2xR2:

x ˙ 1 ( t ) = 2 x 1 ( t ) x ˙ 2 ( t ) = - 1 / 2 x 2 ( t ) . x ˙ 1 ( t ) = 2 x 1 ( t ) x ˙ 2 ( t ) = - 1 / 2 x 2 ( t ) .
(3)

The vector field associated with this differential equation maps the vector f(x1,x2)=(2x1,-1/2x2)f(x1,x2)=(2x1,-1/2x2) to each point (x1,x2)(x1,x2) in the plane. Since a solution has its first derivative defined by the ODE, it must be tangent to the vector field at every point. On Figure 1, the projection onto the plane of trajectory for the initial condition (t0,x0)=(0,(0.05,1))(t0,x0)=(0,(0.05,1)) is plotted in red on top of the phase portrait. The projection of the trajectory onto the phase space is often called path of the trajectory.

## Fundamental theorem of calculus

The following theorem is known as fundamental theorem of calculus and gives a different formulation of the initial value problem.

Theorem 3.1 If f(t,x(t))f(t,x(t)) is integrable on II, then the initial value problem Equation 2 is equivalent to

x ( t ) = x 0 + t 0 t f ( s , x ( s ) ) d s . x ( t ) = x 0 + t 0 t f ( s , x ( s ) ) d s .
(4)

Remark. From Equation 4, we can see that a solution to Equation 1 may exist even in the case where ff is not continuous on DD, but only integrable. In this case, one can use Carathéodory's existence theorem to show the existence of a solution to the initial value problem Equation 2, without requiring that ff be continuous, but this is beyond the scope of this module.

## Existence of solutions

The following theorem, known as Peano existence theorem, gives a condition for the initial value problem Equation 2 to admit a solution.

Theorem 4.1 If ff is continuous on DD, then for any (t0,x0)D(t0,x0)D, there exists at least one solution to the initial value problem Equation 1 with initial condition (t0,x0)(t0,x0).

## Uniqueness of solutions

Finally, the following theorem, known as Picard-Lindelöf theorem, provides conditions for solutions to the initial value problem Equation 2 to be unique.

Theorem 5.1 If ff is continuous on DD and locally Lipschitz continuous with respect to xx, i.e. for any closed bounded set UU in DD, there exists a constant k=kUk=kU such that for all (t,x)(t,x) and (t,y)(t,y) in UU:

f ( t , x ) - f ( t , y ) k x - y , f ( t , x ) - f ( t , y ) k x - y ,
(5)

then for any (t0,x0)D(t0,x0)D, there exists a unique solution of Equation 2.

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