Lesson 2 - Existence and Uniqueness Theorem

Consider the IVP:

\frac{d y}{d x} = f (x, y); y (x_{0}) = y_{0}

The Existence and Uniqueness Theorem states:
a) If $f$ is continuous on an open rectangle $R = (x, y) : x \in (a, b), y \in (c, d)$ that contains the point $(x_{0}, y_{0})$ , then the above IVP has at least one solution on some open subinterval of (a,b) that contains $x_{0}$ .
b) If both $f$ and $f_{y}$ are continuous on R then the above IVP has a unique solution in some open subinterval of $(a, b)$ that contains $x_{0}$ .

Steps to analyzing existence and uniqueness.

Analyze the domain of $f (x, y)$ . Determine whether the domain is continuous for all $x, y \in R$ .
Take the derivative of $f (x, y)$ .
Analyze the domain of $f_{y} (x, y)$ . Determine whether the domain is continuous for all $x, y \in R$ .

Consider the following IVP:

\frac{d y}{d x} = \frac{x^{2} + 2 - y^{2}}{3 + 2 x^{2} + 2 y^{2}}; y (x_{0}) = y_{0}

Because the denominator contains a polynomial in the form $a x^{2} + b y^{2} + C$ where $C \neq 0$ , there are no real combinations of $x$ and $y$ that would yield for the denominator to be undefined. This means that $f (x, y)$ is continuous for all $x, y \in R$ . From the theorem stated above, since this is true for part a, we can guarantee that we have at least one solution for any $(x_{0}, y_{0})$ .
Taking the derivative of $f (x, y)$ , we can check if part b is also true.

f_{y} = \frac{(3 + 2 x^{2} + 2 y^{2}) (1) - (x + 2 - y^{2}) (4 x)}{(3 + 2 x^{2} + 2 y^{2})^{2}}

= \frac{(3 + 2 x^{2} + 2 y^{2}) - (4 x^{2} + 8 x - 4 x y^{2})}{(3 + 2 x^{2} + 2 y^{2})^{2}}

Analyzing the domain of $f_{y} (x, y)$ . In the denominator, we have another polynomial in the form $a x^{2} + b y^{2} + C$ where $C \neq 0$ . Therefore, there are no real combinations of $x$ and $y$ that would yield the denominator undefined. This satisfied part b of the theorem above, guaranteeing that there is one unique solution on an interval $(a, b)$ .

Try For Yourself:

Analyze existence and uniqueness of solutions to the IVP for different values of $(x_{0}, y_{0})$ : $\frac{d y}{d x} = y^{\frac{2}{3}}; y (2) = 0$
Analyze existence and uniqueness of solutions to the IVP for different values of $(x_{0}, y_{0})$ : $\frac{d y}{d x} = \frac{x^{2} - y^{2}}{x^{2} + y^{2}}; y (x_{0} = y_{0})$
Analyze the existence and uniqueness of solutions to the IVP for different values of $(x_{0}, y_{0})$ : $y^{'} = \frac{x + y}{x - y}; y (x_{0}) = y_{0}$

Next Lesson Lesson 3 - Direction Fields

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