2.8: Theory of Existence and Uniqueness
A Sketch of the Proof of the Existence and Uniqueness Theorem
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
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.

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.

Does the sequence \(f_n(t)\) converge? The limit may not exist.

If the sequence \(f_n(t)\) does converge, is the limit continuous?

Is \(f(t)\) the only solution to the integral equation?
Contributors
 Larry Green (Lake Tahoe Community College)