
# 2.8: Theory of Existence and Uniqueness

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Recall the theorem that says that if a first order differential satisfies continuity conditions, then the initial value problem will have a unique solution in some neighborhood of the initial value. More precisely,

Theorem: A Result For Nonlinear First Order Differential Equations

Let

$y'=f(x,y) \;\;\; y(x_0)=y_0$

be a differential equation such that both partial derivatives

$f_x \;\;\; \text{and} \;\;\; f_y$

are continuous in some rectangle containing $$(x_0,y_0)$$/

Then there is a (possibly smaller) rectangle containing $$(x_0,y_0)$$ such that there is a unique solution $$f(x)$$ that satisfies it.

Although a rigorous proof of this theorem is outside the scope of the class, we will show how to construct a solution to the initial value problem. First by translating the origin we can change the initial value problem to

$y(0) = 0.$

Next we can change the question as follows. $$f(x)$$ is a solution to the initial value problem if and only if

$f'(x) = f(x,f(x)) \;\;\; \text{and} \;\;\; f(0) = 0.$

Now integrate both sides to get

$\phi (t) = \int _0^t f(s,\phi (s)) \, ds .$

Notice that if such a function exists, then it satisfies $$f(0) = 0$$.

The equation above is called the integral equation associated with the differential equation.

It is easier to prove that the integral equation has a unique solution, then it is to show that the original differential equation has a unique solution. The strategy to find a solution is the following. First guess at a solution and call the first guess $$f_0(t)$$. Then plug this solution into the integral to get a new function. If the new function is the same as the original guess, then we are done. Otherwise call the new function $$f_1(t)$$. Next plug in $$f_1(t)$$ into the integral to either get the same function or a new function $$f_2(t)$$. Continue this process to get a sequence of functions $$f_n(t)$$. Finally take the limit as $$n$$ approaches infinity. This limit will be the solution to the integral equation. In symbols, define recursively

$f_0(t) = 0$

$\phi_{n+1} (t) = \int _0^t f(s,\phi_n (s)) \, ds .$

Example $$\PageIndex{1}$$

Consider the differential equation

$y' = y + 2, \;\;\; y(0) = 0.$

We write the corresponding integral equation

$y(t) = \int_0^t \left(y(s)+2 \right) \, ds .$

We choose

$f_0(t) = 0$

and calculate

$\phi_1(t) = \int_0^t \left(0+2 \right) \, ds = 2t$

and

$\phi_2(t) = \int_0^t \left(2s+2 \right) \, ds = t^2 + 2t$

and

$\phi_3(t) = \int_0^t \left(s^2+2s+2 \right) \, ds = \frac{t^3}{3}+t^2 + 2t$

and

$\phi_4(t) = \int_0^t \left(\frac{s^3}{3}+s^2+2s+2 \right) \, ds = \frac{t^4}{3.4}+ \frac{t^3}{3}+t^2 + 2t.$

Multiplying and dividing by 2 and adding 1 gives

$\frac{f_4(t)}{2} + 1 = \frac{t^4}{4.3.2}+\frac{t^3}{3.2}+\frac{t^2}{2}+\frac{t}{1}+\frac{1}{1}.$

The pattern indicates that

$\frac{f_n(t)}{2} + 1 = \sum\frac{t^n}{n!}$

or

$\frac{f(t)}{2} + 1 = e^t.$

Solving we get

$f(t) = 2\left(e^t - 1\right).$

This may seem like a proof of the uniqueness and existence theorem, but we need to be sure of several details for a true proof.

1. Does $$f_n(t)$$ exist for all $$n$$. Although we know that $$f(t,y)$$ is continuous near the initial value, the integral could possible result in a value that lies outside this rectangle of continuity. This is why we may have to get a smaller rectangle.
2. Does the sequence $$f_n(t)$$ converge? The limit may not exist.
3. If the sequence $$f_n(t)$$ does converge, is the limit continuous?
4. Is $$f(t)$$ the only solution to the integral equation?

Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.