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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


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

    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

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


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


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


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

    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

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


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

    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?

    Contributors and Attributions

    This page titled 2.8: Theory of Existence and Uniqueness is shared under a not declared license and was authored, remixed, and/or curated by Larry Green.