Continuations of Solutions to x' = h(t)g(x)

# Continuations of Solutions to x' = h(t)g(x)

 Theorem 1: If $x' = h(t)g(x)$ with $x(\tau) = \xi$ ($\xi > 0$) be an IVP such that: 1) $h$ is positive and continuous on the interval $(t_0, \infty)$. 2) $g$ is positive and continuous on the interval $(0, \infty)$. 3) For any $A > 0$ we have that $\displaystyle{\lim_{B \to \infty} \int_A^b \frac{1}{g(x)} \: dx = \infty}$. Then any solution $\phi$ to this IVP can be continued to the right over the interval $(\tau, \infty)$.

## Example 1

Consider the IVP $x' = x^2$ with $x(1) = \xi$. Determine whether the above theorem can be applied.

Let $h(t) = 1$ and $g(x) = x^2$. Then $x' = h(t)g(x)$. $h$ is a positive and continuous function on $(-\infty, \infty)$, in particular, $(0, \infty)$, and $g$ is a positive continuous function on $(0, \infty)$. Let $A > 0$, and consider the integral:

(1)
\begin{align} \quad \lim_{B \to \infty} \int_A^B \frac{1}{g(x)} \: dx &= \lim_{B \to \infty} \int_A^B \frac{1}{x^2} \: dx \\ &= \lim_{B \to \infty} \int_A^B x^{-2} \: dx \\ &= \lim_{B \to \infty} - \frac{1}{x} \biggr \lvert_{x=A}^{x=B} \\ &= \lim_{B \to \infty} - \frac{1}{B} + \frac{1}{A} \\ &= \frac{1}{A} \\ & < \infty \end{align}

Since for all $A > 0$ we have that $\displaystyle{\lim_{B \to \infty} \int_A^B \frac{1}{g(x)} \: dx < \infty}$ we cannot apply the theorem above.

## Example 2

Let $\lambda > 0$ and consider the IVP $x' = \lambda x$ with $x(0) = \xi$ ($\xi > 0$). Determine whether the above theorem can be applied.

Let $h(t) = \lambda$ and $g(x) = x$. Then $h$ and $g$ are both positive functions on $(0, \infty)$. For $A > 0$ we have that:

(2)
\begin{align} \quad \lim_{B \to \infty} \int_A^B \frac{1}{g(x)} &= \lim_{B \to \infty} \int_A^B \frac{1}{x} \: dx \\ &= \lim_{B \to \infty} \ln(x) |_{x=A}^{x=B} \\ &= \lim_{B \to \infty} \ln(B) - \ln(A) \\ &= \infty \end{align}

So the theorem above can be applied. That is, any solution $\phi$ to the IVP $x' = h(t)g(x)$ with $x(\tau) = \xi$ can be continued to the right over the interval $(\tau, \infty) = (0, \infty)$.

We can find the solution to the IVP as follows. Since $x' = \lambda x$ is a separable differential equation we have that:

(3)
\begin{align} \quad \int_{\phi(0)}^{\phi(t)} \frac{1}{x} \: dx = \int_0^t \lambda \: dt \quad \Leftrightarrow \quad x = \xi e^{\lambda t} \end{align}

Clearly the solution $x = \xi e^{\lambda t}$ is a solution defined for $t \in (0, \infty)$.