Part I: One-dimensional systems: flows on the line - Chapter 2

Existence, uniqueness and lifetime of solutions

If we are given a system $\dot{x}=f(x)$ and an initial state $x_0$, does there exist a solution $x(t)$ to this equation such that $x(0)=x_0$? It turns out that if $f$ is a continuous function, then existence is guaranteed. The example

$$ f(x)= \begin{cases} 1 & \text{if}\quad x< 0\\ -1 & \text{if}\quad x\geq 0 \end{cases} $$

shows that when $f$ is discontinuous, things can go wrong: there is no solution that satisfies $x(0)=0$.

The next natural question is about uniqueness. What can go wrong? A good example is the leaky bucket.

Leaky bucket versus radioactive decay

Consider a bucket with a hole in the bottom. If at a given time you see the bucket empty, can you figure out when (if ever) it was full? You cannot, obviously! Let’s look at an oversimplified model to see what happens in terms of differential equations. If we let $x(t)$ be the height of the water remaining in the bucket at time $t$, the equation is

$$ \dot{x}=-C\sqrt{x} $$

where $C=\sqrt{2g}\times s/S$ is a positive constant determined by the acceleration of gravity $g$, the cross-section of the hole $s$ and the cross-section of the bucket $S$. This is known as Torricelli’s law. We shall study the differential equation for $0\leq x\leq 1$, where $x=0$ corresponds to an empty bucket and $x=1$ to a full bucket.

The solution with $x(0)=1$ is easily found by separation of variables:

$$ x(t)= \begin{cases} \frac{C^2}{4}\ (t-t_e)^2 & \text{for}\quad 0\leq t\leq t_e\\ 0 & \text{for}\quad t>t_e \end{cases} $$

where $t_e=2/C$ is the time required for the bucket to go from full to empty. This solution for a full bucket initial condition $x(0)=1$ uniquely determines the height $x(t)$ of the water at any time $t\geq 0$. The problem of non-uniqueness arises when we look backwards from an empty bucket initial condition. For example, an empty bucket initial condition $x(t_e)=0$ could have resulted from a full bucket at any $t\leq 0$!

Contrast the leaky bucket equation $\dot{x}=-\sqrt{x}$ with the deceptively similar equation $\dot{x}=-x$. The latter equation was our very first example used to describe a population of bacteria. We interpret it here as describing the disintegration of a radioactive substance over time. We have

$$ x(t)=x(t_0)\thinspace e^{-(t-t_0)}. $$

As we have seen, uniqueness holds for these solutions and working backwards from any initial condition is no problem. This is used in carbon dating, $x(t)$ being the ratio of carbon-14 to carbon-12 at time $t$.

So, from current data the radioactive decay equation can tell you exactly what happened a given number of years ago, but the bucket equation cannot! We thus need a mathematical criterion to tell why these differential equations behave in such different ways. We have the following theorem:

Existence and Uniqueness

Fundamental theorem. Consider the initial value problem

$$ \dot{x}=f(x),\quad x(0)=x_0. $$

Suppose that $f$ and $f’$ are continuous on an open interval $I$ of the $x$-axis, and suppose that $x_0\in I$. Then there exists a time interval $(-\tau,\tau)$ about $t=0$, with $\tau>0$, such that the equation has a unique solution $x(t)$ satisfying $x(0)=x_0$.


If we take an open interval $J$ that is included in $(-\tau,\tau)$, then there is a unique solution $x’(t)$ defined on $J$ such that $x’(0)=x_0$. By the theorem, this solution has to match in $J$ with the solution $x(t)$ defined on $(-\tau,\tau)$. Conversely, we can look for the largest time interval in which one can define the solution that is then called the ‘maximal solution’. The term ‘solution’ will be always understood as ‘maximal solution’.

The solution traces out a segment that is called a trajectory. (Mathematically speaking, this is the image of the interval $(-\tau,\tau)$ by the function $t\mapsto x(t)$.)
In the leaky bucket example, the function $f(x)=\sqrt{x}$ is not differentiable at $x=0$.

The theorem says that if $f$ is smooth enough, then solutions exist and are unique, but there’s no guarantee that solution exist forever, as shown by the next example.

Life-time and blow-up

Consider $\dot{x}=x^2$ with an initial condition $x_0>0$. By separation of variables we get

$$ x(t)=\frac{1}{\frac{1}{x_0}-t}. $$

Thus the solution exists in the time interval $(-\infty,\frac{1}{x_0})$. Moreover, the system reaches infinity in finite time ! This phenomenon is called blow-up. It is of physical relevance in models of combustion and other runaway processes.

From now on, we will not worry about issues of existence and uniqueness because we will deal with smooth enough $f$’s.

Shifting solutions

One may wonder why we restricted ourselves to look for solutions about the time $t=0$. There is in fact no loss of generality in doing so. This is intuitively obvious if we come back to the interpretation of the equation $\dot{x}=f(x)$ as describing a particle moving on the real line whose velocity at point $x$ is $f(x)$. If a particle is at time $0$ at position $x_0$, then the above theorem tells us that it is at time $t$ at a well-defined position $x(t)$. Now, if we drop another particle at time $s>0$ at position $x_0$, it will have the same evolution as the first particle, but shifted in time: if the first particle is at time $t$ at position $x(t)$, the second one will be at the same position at time $t+s$. Obviously, the two particles have the same trajectory. Mathematically, what we’ve just said is a consequence of the fact that $f$ does not depend explicitly on time.

Indeed, given an equation $\dot{x}=f(x)$, let $x(t)$ be the solution, defined in some interval $(-\tau,\tau)$, such that $x(0)=x_0$. For a given $t_0\in\mathbb{R}$, let $y(t)=x(t-t_0)$. It is clear that $y(t)$ is also a solution of the equation. By the above theorem, $y(t)$ is the (unique) solution such that $y(t_0)=x_0$ and it is defined in the interval $(-\tau+t_0,\tau+t_0)$.