# Initial value problem

In mathematics, in the field of differential equations, an initial value problem (also called the Cauchy problem by some authors) is an ordinary differential equation together with a specified value, called the initial condition, of the unknown function at a given point in the domain of the solution. In physics or other sciences, modeling a system frequently amounts to solving an initial value problem; in this context, the differential equation is an evolution equation specifying how, given initial conditions, the system will evolve with time.

## Definition

An initial value problem is a differential equation

$y'(t) = f(t, y(t)) \,$ with $f: \Omega \subset \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}^n$ where $\Omega \,$ is an open set of $\mathbb{R} \times \mathbb{R}^n$,

together with a point in the domain of ƒ

$(t_0, y_0) \in \Omega,$

called the initial condition.

A solution to an initial value problem is a function y that is a solution to the differential equation and satisfies

$y(t_0) = y_0. \,$

In higher dimensions, the differential equation is replaced with a family of equations $y_i'(t)=f_i(t, y_1(t), y_2(t), ...)$, and $y(t)$ is viewed as the vector $(y_1(t), ..., y_n(t))$. More generally, the unknown function y can take values on infinite dimensional spaces, such as Banach spaces or spaces of distributions.

Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. $y''(t)=f(t,y(t),y'(t))$.

## Existence and uniqueness of solutions

For a large class of initial value problems, the existence and uniqueness of a solution can be illustrated through the use of a calculator.

The Picard–Lindelöf theorem guarantees a unique solution on some interval containing t0 if ƒ is continuous on a region containing t0 and y0 and satisfies the Lipschitz condition on the variable y. The proof of this theorem proceeds by reformulating the problem as an equivalent integral equation. The integral can be considered an operator which maps one function into another, such that the solution is a fixed point of the operator. The Banach fixed point theorem is then invoked to show that there exists a unique fixed point, which is the solution of the initial value problem.

An older proof of the Picard–Lindelöf theorem constructs a sequence of functions which converge to the solution of the integral equation, and thus, the solution of the initial value problem. Such a construction is sometimes called "Picard's method" or "the method of successive approximations". This version is essentially a special case of the Banach fixed point theorem.

Hiroshi Okamura obtained a necessary and sufficient condition for the solution of an initial value problem to be unique. This condition has to do with the existence of a Lyapunov function for the system.

In some situations, the function ƒ is not of class C1, or even Lipschitz, so the usual result guaranteeing the local existence of a unique solution does not apply. The Peano existence theorem however proves that even for ƒ merely continuous, solutions are guaranteed to exist locally in time; the problem is that there is no guarantee of uniqueness. The result may be found in Coddington & Levinson (1955, Theorem 1.3) or Robinson (2001, Theorem 2.6). An even more general result is the Carathéodory existence theorem, which proves existence for some discontinuous functions ƒ.

## Examples

A simple example is to solve $y' = 0.85 y$ and $y(0) = 19$. We are trying to find a formula for $y(t)$ that satisfies these two equations.

Start by noting that $y' = \frac{dy}{dt}$, so

$\frac{dy}{dt} = 0.85 y$

Now rearrange the equation so that $y$ is on the left and $t$ on the right

$\frac{dy}{y} = 0.85dt$

Now integrate both sides (this introduces an unknown constant $B$).

$\ln | y | = 0.85t + B$

Eliminate the $\ln$

$| y | = e^Be^{0.85t}$

Let $C$ be a new unknown constant, $C = \pm e^B$, so

$y = Ce^{0.85t}$

Now we need to find a value for $C$. Use $y(0) = 19$ as given at the start and substitute 0 for $t$ and 19 for $y$

$19 = C e^{0.85 * 0}$
$C = 19$

this gives the final solution of $y(t) = 19e^{0.85t}$.

Second example

The solution of

$y'+3y=6t+5,\qquad y(0)=3$

can be found to be

$y(t)=2e^{-3t}+2t+1. \,$

Indeed,

\begin{align} y'+3y &= \tfrac{d}{dt} (2e^{-3t}+2t+1)+3(2e^{-3t}+2t+1) \\ &= (-6e^{-3t}+2)+(6e^{-3t}+6t+3) \\ &= 6t+5. \end{align}