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,
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)) and f(0) = 0 Now integrate both sides to get
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
Example Consider the differential equation y' = y + 2 y(0) = 0 We write the corresponding integral equation
We choose f_{0}(t) = 0 and calculate
and
and
and
Multiplying and dividing by 2 and adding 1 gives f_{4}(t)
t^{4}
t^{3} t^{2}
t 1 The pattern indicates that f_{n}(t)
t^{n} or f(t) Solving we get f(t) = 2(e^{t} - 1) 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.
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