First Order Ordinary Differential Equations

First Order Ordinary Differential Equations

We have already looked extensively into solving first order ordinary differential equations. We will now begin to look at more of the theory regarding such differential equations. We begin by giving a more formal definition of a first order ODE.

 Definition: Let $D \subseteq \mathbb{R}^2$ be a domain and $f \in C(D, \mathbb{R})$. A First Order Ordinary Differential Equation is of the form $x' = f(t, x)$.

When we say that "$D \subseteq \mathbb{R}^2$ is a domain" we mean that $D$ is an nonempty, open, and connected subset of $\mathbb{R}^2$. The notation $f \in C(D, \mathbb{R})$ will mean that the function $f : D \to \mathbb{R}$ is continuous on $D$. Also, $x$ is a function of $t$ and $\displaystyle{x' = \frac{dx}{dt}}$.

Some examples of first order ordinary differential equations are:

(1)
\begin{align} \quad x' = t + 2x \end{align}
(2)
\begin{align} \quad x' = \cos (2t) x^2 \end{align}

In the first example $f(t, x) = t + 2x$, and in the second example, $f(t, x) = \cos (2t) x^2$.

 Definition: A Solution to the first order ordinary differential equation $x' = f(t, x)$ on an open interval $J = (a, b)$ is a continuously differentiable function $\phi \in C^1(J, \mathbb{R})$ such that for all $t \in J$ we have that $(t, \phi(t)) \in D$ and $\phi'(t) = f(t, \phi(t))$.

The notation that "$\phi \in C^1 (J, \mathbb{R})$" means that the function $\phi : J \to \mathbb{R}$ is real-valued continuous and differentiable function on $J$. We require that for all $t \in J$ that $(t, \phi(t)) \in D$ so that $f(t, \phi(t))$ is well-defined. Furthermore, the requirement that for all $t \in J$, $\phi'(t) = f(t, \phi(t))$ means that $\phi$ satisfies the differential equation $x' = f(t, x)$ on $J$.

 Definition: An Initial Value Problem is a first order ordinary differential equation $x' = f(t,x)$ with an Initial Condition $x(\tau) = \xi$ where $(\tau, \xi) \in D$. A Solution to the initial value problem $x' = f(t, x)$ with $x(\tau) = \xi$ on the open interval $J = (a, b)$ with $\tau \in J$ is a solution $\phi$ to the differential equation $x' = f(t, x)$ with $\phi(\tau) = \xi$.

We require that $(\tau, \xi) \in D$ so that $f(\tau, x(\tau)) = f(\tau, \xi)$ is defined for the initial value problem $x' = f(t, x)$ with $x(\tau) = \xi$.

In the following theorem we show that every first order ordinary differential equation initial value problem is equivalent to an integral equation. This will be very useful later on.

 Theorem 1: The initial value problem $x' = f(t, x)$ with $x(\tau) = \xi$ is equivalent to an integral equation of the form $\displaystyle{\phi(t) = \xi + \int_{\tau}^t f(s, \phi(s)) \: dx}$.
• Proof: $\Rightarrow$ Let $\phi$ be a solution on $J = (a, b)$ to the initial value problem:
(3)
\begin{align} \quad x' &= f(t, x) \\ \quad x(\tau) &= \xi \end{align}
• Then for all $t \in J$ we have that:
(4)
\begin{align} \quad \phi'(t) &= f(t, \phi(t)) \\ \quad \phi(\tau) &= \xi \end{align}
• We integrate both sides of the first equation above from $\tau$ to $t$, use the Fundamental theorem of Calculus, and use the second equation above to get:
(5)
\begin{align} \quad \int_{\tau}^{t} \phi'(s) \: ds &= \int_{\tau}^{t} f(s, \phi(s)) \: ds \\ \quad \phi(t) - \phi(\tau) &= \int_{\tau}^{t} f(s, \phi(s)) \: ds \\ \quad \phi(t) - \xi &= \int_{\tau}^{t} f(s, \phi(s)) \: ds \\ \quad \phi(t) &= \xi + \int_{\tau}^{t} f(s, \phi(s)) \: ds \end{align}
• $\Leftarrow$ Now let $\displaystyle{\phi(t) = \xi + \int_{\tau}^{t} f(s, \phi(s)) \: ds}$. We differentiate both sides of this equation with respect to $t$ and use the Fundamental Theorem of Calculus again to get:
(6)
\begin{align} \quad \frac{d}{dt} \left ( \phi(t) \right ) &= \frac{d}{dt} \left ( \xi + \int_{\tau}^{t} f(s, \phi(s)) \: ds \right ) \\ \quad \phi'(t) &= f(t, \phi(t)) \end{align}
• Furthermore, plugging in $t = \tau$ to $\phi(t)$ gives us:
(7)
\begin{align} \quad \phi(\tau) &= \xi + \int_{\tau}^{\tau} f(s, \phi(s)) \: ds \\ &= \xi \end{align}
• Therefore $\phi$ is a solution to the initial value problem $x' = f(t, x)$ with $x(\tau) = \xi$. $\blacksquare$