EXISTENCE AND UNIQUENESS THEOREM: PROOF, EXAMPLES AND EXERCISES - MATHS - 2024

The existence and uniqueness theorem establishes the necessary and sufficient conditions for a first-order differential equation, with a given initial condition, to have a solution and for this solution to be the only one.

However, the theorem does not give any technique or indication of how to find such a solution. The existence and uniqueness theorem is also extended to higher-order differential equations with initial conditions, which is known as the Cauchy problem.

Figure 1. A differential equation with initial condition and its solution is shown. The Existence and Uniqueness Theorem guarantees that it is the only possible solution.

The formal statement of the existence and uniqueness theorem is as follows:

“For a differential equation y '(x) = f (x, y) with initial condition y (a) = b, there exists at least one solution in a rectangular region of the XY plane that contains the point (a, b), if f (x, y) is continuous in that region. And if the partial derivative of f with respect to y: g = ∂f / ∂y is continuous in that same rectangular region, then the solution is unique in a neighborhood of the point (a, b) contained in the region of continuity of fy g. "

The usefulness of this theorem lies first in knowing which are the regions of the XY plane in which a solution can exist and also, knowing if the solution found is the only possible one or if there are others.

Note that in case the uniqueness condition is not satisfied, the theorem cannot predict how many solutions in total the Cauchy problem has: perhaps it is one, two, or more.

Proof of the existence and uniqueness theorem

Figure 2. Charles Émile Picard (1856-1941) is credited with one of the first proofs of the Existence and Uniqueness Theorem. Source: Wikimedia Commons.

For this theorem, two possible proofs are known, one of them is the proof of Charles Émile Picard (1856-1941) and the other is due to Giuseppe Peano (1858-1932) based on the works of Augustin Louis Cauchy (1789-1857).

It is noteworthy that the most brilliant mathematical minds of the nineteenth century participated in the proof of this theorem, so it can be intuited that neither of the two is simple.

To formally prove the theorem, it is necessary to first establish a series of more advanced mathematical concepts, such as Lipschitz-type functions, Banach spaces, Carathéodory's existence theorem and several others, which are beyond the scope of the article.

A large part of the differential equations that are handled in physics deal with continuous functions in the regions of interest, therefore we will limit ourselves to showing how the theorem is applied in simple equations.

Examples

- Example 1

Let's consider the following differential equation with an initial condition:

y '(x) = - y; with y (1) = 3

Is there a solution for this problem? Is it the only possible solution?

Answers

In the first place, the existence of the solution of the differential equation is evaluated and that it also fulfills the initial condition.

In this example f (x, y) = - and the existence condition requires knowing if f (x, y) is continuous in a region of the XY plane that contains the point of coordinates x = 1, y = 3.

But f (x, y) = - y is the affine function, which is continuous in the domain of real numbers and exists throughout the range of real numbers.

Therefore it is concluded that f (x, y) is continuous in R ², so the theorem guarantees the existence of at least one solution.

Knowing this, it is necessary to evaluate if the solution is unique or if, on the contrary, there is more than one. For this, it is necessary to calculate the partial derivative of f with respect to the variable y:

Then g (x, y) = -1 which is a constant function, which is also defined for all R ² and is also continuous there. It follows that the existence and uniqueness theorem guarantees that this initial-value problem does have a unique solution, although it does not tell us what it is.

- Example 2

Consider the following first-order ordinary differential equation with initial condition:

y '(x) = 2√y; and (0) = 0.

Is there a solution y (x) to this problem? If so, determine if there is one or more than one.

Reply

We consider the function f (x, y) = 2√y. The function f is defined only for y≥0, since we know that a negative number lacks a real root. Furthermore f (x, y) is continuous in the upper half plane of R ² including the X axis, so the existence and uniqueness theorem guarantees at least one solution in said region.

Now the initial condition x = 0, y = 0 is on the edge of the solution region. Then we take the partial derivative of f (x, y) with respect to y:

∂f / ∂y = 1 / √y

In this case the function is not defined for y = 0, precisely where the initial condition is.

What does the theorem tell us? It tells us that although we know that there is at least one solution in the upper half-plane of the X axis including the X axis, as the uniqueness condition is not met, there is no guarantee that there will be a unique solution.

This means that there could be one or more than one solution in the region of continuity of f (x, y). And as always, the theorem doesn't tell us what they could be.

Solved exercises

- Exercise 1

Solve the Cauchy problem in Example 1:

y '(x) = - y; with y (1) = 3.

Find the function y (x) that satisfies the differential equation and the initial condition.

Solution

In Example 1 it was determined that this problem has a solution and is also unique. To find the solution, the first thing to note is that it is a first degree differential equation of separable variables, which is written as follows:

Dividing between and in both members to separate the variables we have:

The indefinite integral is applied in both members:

Solving the indefinite integrals we have:

where C is a constant of integration that is determined by the initial condition:

Substituting the value of C and rearranging it remains:

Applying the following property of logarithms:

The above expression can be rewritten like this:

The exponential function with base e in both members is applied to obtain:

y / 3 = e ^{(1 - x)}

Which is equivalent to:

y = 3e e ^-x

This is the unique solution of the equation y '= -y with y (1) = 3. The graph of this solution is shown in Figure 1.

- Exercise 2

Find two solutions for the problem posed in Example 2:

y '(x) = 2√ (y); and (0) = 0.

Solution

It is also an equation of separable variables, which, written in differential form, looks like this:

dy / √ (y) = 2 dx

Taking the indefinite integral in both members remains:

2 √ (y) = 2 x + C

Since we know that y≥0 in the solution region we have:

y = (x + C) ²

But since the initial condition x = 0, y = 0 must be fulfilled, then the constant C is zero and the following solution remains:

y (x) = x ².

But this solution is not unique, the function y (x) = 0 is also a solution to the problem posed. The existence and uniqueness theorem applied to this problem in Example 2 had already predicted that there could be more than one solution.

References

1. Coddington, Earl A.; Levinson, Norman (1955), Theory of Ordinary Differential Equations, New York: McGraw-Hill.
2. Encyclopedia of Mathematics. Cauchy-Lipschitz theorem. Recovered from: encyclopediaofmath.org
3. Lindelöf, Sur l'application de la méthode des approximations successives aux équations différentielles ordinaires du premier ordre; Comptes rendus hebdomadaires des séances de l'Académie des sciences. Vol. 116, 1894, pp. 454–457. Recovered from: gallica.bnf.fr.
4. Wikipedia. Picard's successive approximations method. Recovered from: es.wikipedia.com
5. Wikipedia. Picard-Lindelöf theorem. Recovered from: es.wikipedia.com.
6. Zill, D. 1986. Elementary Differential Equations with Applications. Prentice Hall.