What is an ODE

Describing change through derivatives; existence and uniqueness of solutions.

about 8 min

Start from a problem

Problem

A cup of hot water cools on a table. We do not know the temperature function $T(t)$ ; we only know the law of its change: it cools at a rate proportional to how far it is above room temperature.

Many laws of nature come this way: not the quantity itself, but a relation between its rate of change and its value. We need a mathematical object to express such laws, and a way to recover the function from the law.

Observe

A cup of hot water sits on a table. It does not cool at a constant rate: fast while hot, slowly as it nears room temperature. We do not know the temperature as a function of time, $T(t)$ , but we do know the law of its change:

$T'(t)=-k\,\bigl(T(t)-T_{\text{room}}\bigr).$

In the experiment below choose "Exponential growth", make $k$ negative and vary the initial value. What you see are solutions of this kind of equation.

Interactive experimentNumerical ODE explorer

Equation

y' = k\,y

Time series

k1.00

Initial value y(0)1.00

step h0.250 · 12 steps

Method Euler RK4 exactEuler error at end: 5.5336RK4 error at end: 0.0016

Conjecture

Many natural laws do not hand us the quantity itself but a relation between its rate of change and its value. Such a relation is an equation containing an unknown function and its derivatives. If we can recover the function from the equation, we can predict the future.

Definitions

Definition 1.1Ordinary differential equation

An ordinary differential equation (ODE) of order $n$ is an equation $F\bigl(t,\,y,\,y',\,y'',\dots,y^{(n)}\bigr)=0$ for an unknown function $y=y(t)$ of one variable. The order is the highest derivative that appears. If it can be solved for that derivative, $y^{(n)}=f(t,y,\dots,y^{(n-1)})$ , the equation is explicit.

Definition 1.2Solution and initial value problem

A function $\varphi$ on an interval $I$ is a solution if substituting $y=\varphi(t)$ makes the equation hold for every $t\in I$ . An initial value problem adds $y(t_0)=y_0$ (for order $n$ , the values of $y,y',\dots,y^{(n-1)}$ at $t_0$ ).

Definition 1.3Linear equation

If the equation can be written $a_n(t)y^{(n)}+\cdots+a_1(t)y'+a_0(t)y=g(t)$ , so that $y$ and its derivatives appear only to the first power and never multiplied together, it is linear; if $g\equiv0$ it is homogeneous.

$y'''+y\,y'=\sin t$ is third order and nonlinear (because of $y\,y'$ ). Order and linearity are independent attributes.

Derivation: the simplest equation

Proposition 1.1

The solutions of $y'=ky$ are exactly $y=Ce^{kt}$ , $C\in\mathbb R$ .

Proof

$Ce^{kt}$ is a solution by direct differentiation. Conversely let $y$ be any solution and set $u(t)=y(t)e^{-kt}$ . Then $u'=y'e^{-kt}-ky\,e^{-kt}=(y'-ky)e^{-kt}=0,$ so $u$ is a constant $C$ and $y=Ce^{kt}$ .

We used the central fact $(e^{kt})'=ke^{kt}$ from The derivative, and that a function with zero derivative is constant (a corollary of the mean value theorem). The initial condition $y(0)=y_0$ fixes $C=y_0$ .

Back to the cooling cup: with $u=T-T_{\text{room}}$ , $u'=-ku$ , so $u=u_0e^{-kt}$ and $T(t)=T_{\text{room}}+(T_0-T_{\text{room}})e^{-kt}.$

Existence and uniqueness

An equation only states a law. Must it have a solution? Is the solution unique? These questions decide whether "predicting the future with a differential equation" is legitimate.

Theorem 1.2Picard–Lindelöf

Let $f(t,y)$ be continuous on the rectangle $R=\{|t-t_0|\le a,\ |y-y_0|\le b\}$ and Lipschitz in $y$ : there is $L$ with $|f(t,y_1)-f(t,y_2)|\le L|y_1-y_2|$ . Then the initial value problem $y'=f(t,y),\ y(t_0)=y_0$ has a unique solution on some neighbourhood of $t_0$ .

Proof

We prove uniqueness, which shows what the Lipschitz condition is for. Let $\varphi,\psi$ be solutions and $w(t)=|\varphi(t)-\psi(t)|$ . Integrating the equation (see The integral) gives $\varphi(t)=y_0+\int_{t_0}^tf(s,\varphi(s))\,ds$ and likewise for $\psi$ , so for $t\ge t_0$ $w(t)\le\int_{t_0}^t|f(s,\varphi)-f(s,\psi)|\,ds\le L\int_{t_0}^tw(s)\,ds.$ Let $W(t)=\int_{t_0}^tw$ . Then $W'\le LW$ and $W(t_0)=0$ , so $(We^{-Lt})'=(W'-LW)e^{-Lt}\le0$ : $We^{-Lt}$ is non-increasing, starts at $0$ and is non-negative, hence $W\equiv0$ and $w\equiv0$ .

Existence is obtained by Picard iteration $\varphi_{n+1}(t)=y_0+\int_{t_0}^tf(s,\varphi_n(s))\,ds$ and a uniform convergence argument; this is one of the roots of Numerical solutions.

Example 1.1Uniqueness fails without Lipschitz

$y'=2\sqrt{|y|},\ y(0)=0$ . Both $y\equiv0$ and $y=t^2$ (for $t\ge0$ ) are solutions. $f(y)=2\sqrt{|y|}$ is not Lipschitz near $y=0$ .

Common mistake

The general solution of $y'=ky$ is $Ce^{kt}$ , not $e^{kt}+C$ . Where the constant goes is dictated by the derivation: here $C$ comes from " $ye^{-kt}$ is constant". Also, the existence–uniqueness theorem guarantees only a local solution; the solution of $y'=y^2$ blows up at $t=1$ .

Application

ApplicationSolutions need not live forever

$y'=y^2,\ y(0)=1$ satisfies the theorem and has the unique local solution $y=\dfrac1{1-t}$ , which blows up at $t=1$ . The theorem guarantees a local solution. This is the first exercise of First-order equations.

Remark

Three levels of working with differential equations: modelling (writing the law), solving (finding the function), analysis (knowing things without solving). Existence and uniqueness belongs to the third level: under broad conditions, the present state completely determines the future. This is the mathematical form of Newtonian determinism.

Exercises

Solve

y'=2y,\ y(0)=3

and give

y(1)

to two decimals.

What is the order of

y'''+y\,y'=\sin t