Basic concepts · Definition

Existence and uniqueness theorem

differential equationsexistence and uniqueness

Before you start you need to know

Before looking for the solution of an initial value problem it helps to know whether one exists and whether it is the only one; this theorem answers both questions without solving the equation.

Definition

The existence and uniqueness theorem guarantees that an initial value problem has a unique solution near the initial point, if the equation and its partial derivative with respect to the unknown are continuous there.

\[f,\ \frac{\partial f}{\partial y}\ \text{continuous near }(x_0,y_0)\ \Longrightarrow\ \text{a unique solution}\]

The idea is simple: if \(f(x,y)\) is smooth on a region, the initial point \((x_0,y_0)\) fixes a direction at each instant and the solution curve is determined step by step. The precise conditions are that \(f\) and \(\partial f/\partial y\) are continuous on a rectangle containing the point. The result is local: the solution exists on an interval around \(x_0\), which may be small.

To see that «local» is real, consider \(y'=y^2\) with \(y(0)=1\). Both \(y^2\) and \(2y\) are continuous on the whole plane, and the solution is \(y=1/(1-x)\), which blows up at \(x=1\). The equation is perfectly smooth, and yet the solution does not live on the whole line.

When the hypothesis on \(\partial f/\partial y\) fails, uniqueness can be lost. The classic case is \(y'=\sqrt{y}\) with \(y(0)=0\): both \(y=0\) and \(y=x^2/4\) (for \(x\ge 0\)) solve the problem, because \(\partial f/\partial y=1/(2\sqrt{y})\) is not continuous at \(y=0\). A common mistake is to read this backwards: if the hypotheses fail, the theorem does not say there are several solutions, only that it concludes nothing.

A useful geometric consequence is that, under the hypotheses, two distinct solution curves never cross or touch. This allows, for example, arguing that a solution starting between two constant solutions stays trapped between them. The next concept, the direction field, draws exactly those curves without solving the equation.

Think about it

Why does \(y'=\sqrt{y}\), \(y(0)=1\) have a unique solution although \(y'=\sqrt{y}\), \(y(0)=0\) does not?

Because near \((0,1)\) the function \(\sqrt{y}\) and its derivative \(1/(2\sqrt{y})\) are continuous; the problem appears only on \(y=0\).

Exercises 10

References

  • William E. Boyce, Richard C. DiPrima, Douglas B. Meade, Elementary Differential Equations and Boundary Value Problems, 12th edition, Wiley, 2021. §2.4
  • Dennis G. Zill, A First Course in Differential Equations with Modeling Applications, 11th edition, Cengage, 2018. §1.2
  • Morris W. Hirsch, Stephen Smale, Robert L. Devaney, Differential Equations, Dynamical Systems, and an Introduction to Chaos, 3rd edition, Academic Press, 2012. Cap. 8