Question

i) The initial value problem dy/dx = x^2+y^2, y(0)= has a unique solution in some interval of the form - h<x<h.

Answer 1

Answer on Question #69886 – Math – Differential Equations

Question

i) The initial value problem

$$\frac{dy}{dx} = x^2 + y^2, \quad y(0) = 0$$

has a unique solution in some interval of the form $-h < x < h$.

Solution

Theorem (Existence and uniqueness theorem for First-Order ODE’s):

Let $f(x, y)$ be a real valued function which is continuous on the rectangle

$$R = \{(x, y); |x - x_0| \leq a, |y - y_0| \leq b\}, (a, b > 0).$$

Assume $f$ has a partial derivative with respect to $y$ and that $\frac{\partial f}{\partial y}$ is also continuous on the rectangle $R$.

Then there exists an interval $I = [x_0 - h, x_0 + h]$ with $h \leq a$ such that the initial value problem

$$\left\{ \begin{array}{l} y' = f(x, y), \\ y(x_0) = y_0 \end{array} \right.$$

has a unique solution $y(x)$ defined on the interval $I$.

We have that

$$f(x, y) = x^2 + y^2, \quad x_0 = 0, \quad y_0 = 0$$

The function $f$ is continuously differentiable in $\mathbb{R}^2$ because its partial derivatives

$$\frac{\partial f}{\partial x}(x, y) = 2x, \quad \frac{\partial f}{\partial y}(x, y) = 2y$$

are continuous functions in $\mathbb{R}^2$. Therefore, the initial problem

$$\frac{dy}{dx} = x^2 + y^2, \quad y(0) = 0$$

has a unique solution in an interval

$$- h < x < h, \quad h > 0.$$

Consider the rectangle

$$R = \{(x, y); |x| \leq a, |y| \leq b\}, (a, b > 0).$$

Since $f$ is continuous in a closed and bounded domain, it is necessarily bounded in $R$, i.e., there exists $K > 0$ such that

$$|f(x, y)| \leq K, \quad \forall (x, y) \in R.$$

Set

$$K = \max_{(x, y) \in R} |f(x, y)|.$$

Then the initial value problem has at most one solution $y = y(x)$ in the interval

$$- h < x < h, \quad h > 0,$$

where

$$h = \min \left\{ a, \frac{b}{K} \right\}.$$

We have

$$K = \max_{(x, y) \in R} |f(x, y)| = \max_{(x, y) \in R} (x^2 + y^2) = a^2 + b^2 \Rightarrow h = \min \left\{a, \frac{b}{a^2 + b^2} \right\}$$

Set

$$a = 1$$

and

$$b = 1.$$

Then

$$h = \min \left\{1, \frac{1}{(1)^2 + (1)^2}\right\} = \frac{1}{2}.$$

Hence, we can state that the solution $y = y(x)$ of the initial value problem exists in the interval $-0.5 < x < 0.5$.

**Answer:** The statement is true. There exists a unique solution $y = y(x)$ of the initial value problem in the interval $-0.5 < x < 0.5$.

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