Ordinary Differential Equations#

Definition: Ordinary Differential Equation

An ordinary differential equation (ODE) of order \(n\) is a function

\[F: \mathcal{D}_F \subseteq \mathbb{R} \times \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}\]

which is dependent on its last argument.

Notation

Most commonly, an ODE is written directly as follows:

\[F(x, y, y', y'', \dotsc, y^{(n)}) = 0\]

Definition: Explicit ODE

We say that \(F\) is explicit if there exists a function

\[f: \mathcal{D}_f \subseteq \mathbb{R} \times \mathbb{R} \times \mathbb{R}^{n-1} \to \mathbb{R}\]

with

\[F(x, y, y', y'', \dotsc, y^{(n)}) = y^{(n)} - f(x, y, y', \dotsc, y^{(n-1)})\]

for all \((x, y, y', y'', \dotsc, y^{(n)}) \in \mathcal{D}_F\).

Definition: Solution

We say that a function \(\phi: \mathcal{D}_{\phi} \subseteq \mathbb{R} \to \mathbb{R}\) is a solution of \(F\) on some subset \(S \subseteq \mathcal{D}_{\phi}\) if \(\phi\) is \(n\)-times differentiable on \(S\) with

\[F\left(x, \phi(x), \phi'(x), \phi''(x), \dotsc, \phi^{(n)}(x)\right) = 0\]

for all \(x \in S\).

Example: \(y'(x) = a y(x)\)

Consider the ordinary differential equation

\[y'(x) = a y(x)\]

for some \(a \in \mathbb{R}\). It is a first-order ODE with the following canonical form \(F: \mathbb{R} \times \mathbb{R} \times \mathbb{R} \to \mathbb{R}\):

\[F(x, y, y') = y' - a y\]

Its solutions on \(\mathbb{R}\) have an exponential form

\[y(x) = c \mathrm{e}^{ax}\]

for some \(c \in \mathbb{R}\). We easily see that any \(y\) of this form is a solution on \(\mathbb{R}\) by differentiating:

\[y'(x) = (c \mathrm{e}^{ax})' = ca \mathrm{e}^{ax} = a (c\mathrm{e}^{ax}) = ay(x)\]

To show that all solutions on \(\mathbb{R}\) have such a form, let \(v\) be any arbitrary solution on \(\mathbb{R}\) and let \(q(x) = v(x) \mathrm{e}^{-ax}\). By differentiating \(q\), we get:

\[q'(x) = v'(x)\mathrm{e}^{-ax} - av(x)\mathrm{e}^{-ax} = \mathrm{e}^{-ax}(v'(x) - a v(x))\]

Since \(v\) is a solution, we have \(v'(x) = a v(x)\) and so \(q'(x) = 0\). This means that \(q\) must be constant on \(\mathbb{R}\), i.e. \(q(x) = c\) for some \(c \in \mathbb{R}\). Therefore, \(c = v(x)\mathrm{e}^{-ax}\) and so \(v(x) = c\mathrm{e}^{ax}\).

Example: \(y''(t) + \omega^2 y(t) = 0\)

Consider the ordinary differential equation

\[y''(t) + \omega^2 y(t) = 0\]

for some \(\omega \in \mathbb{R}_{\gt 0}\). It is very common in physics. It is a second-order ODE with the following canonical form:

\[F: \mathbb{R} \times \mathbb{R} \times \mathbb{R}^{2} \to \mathbb{R} \qquad F(t, y, y', y'') = y'' + \omega^2 y\]

It can be shown that its solutions on \(\mathbb{R}\) have the form

\[y(t) = A \cos (\omega t) + B \sin (\omega t)\]

for some \(A, B \in \mathbb{R}\).

Theorem: Reduction of an ODE to a First-Order System

Every explicit ODE

\[y^{(n)} = f(t, y, y', \dotsc, y^{(n-1)})\]

of order \(n\) is equivalent to an explicit system of ODEs

\[\boldsymbol{u}' = \boldsymbol{f}(t, \boldsymbol{u}) \qquad \boldsymbol{f}(t, \boldsymbol{u}) = \begin{bmatrix} u_2 \\ u_3 \\ \vdots \\ u_n \\ f(t, u_1, u_2, \dotsc, u_n) \end{bmatrix}\]

with \(n\) equations of order \(1\). Specifically, a function \(\phi: \mathcal{D}_{\phi} \subseteq \mathbb{R} \to \mathbb{R}\) is a solution to the ODE if and only if the function \(\boldsymbol{\phi}: \mathcal{D}_{\phi} \subseteq \mathbb{R} \to \mathbb{R}^n\) defined as

\[\boldsymbol{\phi}(t) = \begin{bmatrix} \phi(t) \\ \phi'(t) \\ \phi''(t) \\ \vdots \\ \phi^{(n-1)}(t) \end{bmatrix}\]

is a solution to the system of ODEs.

Proof

TODO