Existence-Uniqueness Domain vs. Solution Domain — First-Order ODE

For an initial value problem \( y' = f(x,y) \), two different domains play a role: the region \( D \) where the hypotheses of the existence and uniqueness theorem hold, and the region \( D' \) where the solutions themselves are defined. Neither one has to contain the other — the three examples below show \( D \subsetneq D' \), then \( D' \subsetneq D \), and finally a solution whose domain is bounded even though \( D \) is the whole plane.

Example of \( D \subsetneq D' \)

Consider the equation

\( y' - \frac{1}{x}\, y = x. \)

We find an integrating factor:

\( \mu(x) = e^{\int -\frac{1}{x}\,dx} = e^{-\ln x} = \frac{1}{x}. \)

Multiplying by \( \mu(x) \):

\( \frac{1}{x}\, y' - \frac{1}{x^2}\, y = 1. \)

The left-hand side is a derivative:

\( \frac{d}{dx}\!\left( \frac{y}{x} \right) = 1. \)

Integrating gives \( \dfrac{y}{x} = x + C \), so the general solution is

\( y(x) = x^2 + Cx. \)

The solution family \( y = x^2 + Cx \): every solution is defined on all of \( \mathbb{R} \), although the equation excludes the line \( x = 0 \)

Compare the two domains

The domain where the conditions of the existence and uniqueness theorem hold is \( D = \{ (x,y) \in \mathbb{R}^2 \mid x \neq 0 \} \), while the domain of definition of the solutions is \( D' = \mathbb{R}^2 \). Here \( D \) is strictly contained in \( D' \).

Example of \( D' \subsetneq D \)

Solutions of \( y' = -x/y \) are half-circles: each solution lives entirely in \( y > 0 \) or entirely in \( y < 0 \)

Consider the equation

\( y' = -\frac{x}{y}. \)

Multiplying by \( y \) gives \( y\, y' = -x \). The left-hand side is a derivative:

\( \frac{1}{2} \frac{d}{dx}\!\left( y^2 \right) = -x. \)

Integrating: \( \frac{1}{2} y^2 = -\frac{1}{2} x^2 + C \). Multiplying by \( 2 \) and rearranging:

\( x^2 + y^2 = C. \)

The solutions are half-circles, since a solution must be a differentiable function of \( x \).

Compare the two domains

The existence and uniqueness conditions hold on \( D = \{ (x,y) \in \mathbb{R}^2 \mid y \neq 0 \} \), but each individual solution is defined only inside \( \{ y > 0 \} \) or \( \{ y < 0 \} \). Here the solutions occupy strictly less than \( D \).

Asymptotic solutions: \( D = \mathbb{R}^2 \), yet the solution domain is bounded

Consider the equation

\( y' = y^2. \)

For \( y \neq 0 \) we separate variables:

\( \frac{1}{y^2}\, \frac{dy}{dx} = 1 \;\Rightarrow\; \int \frac{dy}{y^2} = \int dx, \)

so \( -\dfrac{1}{y} = x + C \), and hence

\( y(x) = -\frac{1}{x + C}. \)

For the initial condition \( y(3) = 1 \) we get \( C = -4 \), so

\( y(x) = -\frac{1}{x - 4}, \qquad x \in (-\infty,\, 4). \)

The solution of \( y' = y^2 \), \( y(3)=1 \) blows up along the asymptote \( x = 4 \): its domain is \( (-\infty, 4) \)

The theorem is local

The existence and uniqueness conditions hold on all of \( D = \mathbb{R}^2 \), yet the domain of the solution is bounded. For example, for the initial condition \( y(2) = 1 \) the solution is trapped inside the infinite rectangle \( (-\infty, 3) \times (0, \infty) \): bounded on one side along the \( x \)-axis and on one side along the \( y \)-axis.