Finite-Time Blow-Up in Separable Equations

When a Solution Becomes Unbounded in Finite Time

A differential equation can have a solution that exists only up to a finite time because its values grow without bound as that time is approached. This is called finite-time blow-up. It is different from a formula merely being undefined at a point: the key issue is whether the solution itself becomes unbounded as it approaches the endpoint of its interval.

Separable equations make this behavior accessible. Their formulas often show both the time at which blow-up occurs and whether it happens forward or backward from the initial time. The equation \(y'=y^2\) is a basic example: positive initial values blow up in forward time, while negative initial values blow up in backward time.

A solution's maximal interval is the largest open interval containing its initial time on which that solution is defined and satisfies the differential equation. If a solution becomes unbounded at a finite endpoint, it cannot be extended through that endpoint as a finite-valued continuous function, and therefore cannot be extended as a differentiable solution.

The Basic Example: \(y'=y^2\)

Consider the initial-value problem \(y'=y^2\), \(y(t_0)=y_0\). If \(y_0\ne 0\), separation gives \(dy/y^2=dt\). Integrating and applying the initial condition produces a formula whose denominator can vanish. That zero determines a candidate finite endpoint, but we must check the formula and its behavior there.

The formula is defined wherever its denominator is nonzero. The denominator vanishes at \(T=t_0+1/y_0\). Whether this is a future or past endpoint depends on the sign of \(y_0\). The solution with \(y_0=0\) must be treated separately: it is the equilibrium solution \(y(t)=0\), since dividing by \(y^2\) would have excluded it.

Blow-Up Can Occur in Backward Time

The same denominator gives a different direction of blow-up when the initial value is negative. For \(y_0<0\), the time \(T=t_0+1/y_0\) lies before \(t_0\). On the interval to the right of \(T\), the denominator is positive, while the numerator is negative. As \(t\to T^+\), the solution therefore tends to \(-\infty\). The finite endpoint is the left endpoint of the maximal interval containing the initial time.

Why an Unbounded Endpoint Cannot Be Crossed

A solution of a differential equation must be differentiable on its interval, and every differentiable function is continuous there. If a solution could be extended through a finite endpoint \(T\), its extension would be continuous at \(T\) and would have a finite value there. Continuity would then require the solution values to approach that finite value as \(t\to T\). That contradicts \(|y(t)|\to\infty\).

This argument establishes why a pole associated with genuine blow-up marks an endpoint of the solution's interval, not just a gap in a convenient algebraic expression. A formula may also have a different branch on the other side of a pole. Such a branch does not provide an extension through the pole: it has no finite value there and cannot join continuously to the original solution.

An Integral Test Along an Increasing Branch

The mechanism behind the \(y'=y^2\) example applies more broadly. Consider an autonomous separable equation \(y'=g(y)\), and suppose \(g\) is continuous and positive for \(y\ge y_0\). Along a solution starting at \(y(t_0)=y_0\), the solution increases. Separating variables relates the time needed to reach a value \(Y>y_0\) to the integral of \(1/g\) from \(y_0\) to \(Y\).

Here is the justification. Define \(H(Y)=\int_{y_0}^{Y}du/g(u)\). Because \(g\) is continuous and positive, \(H\) is differentiable and strictly increasing, with \(H'(Y)=1/g(Y)>0\). Thus it has an inverse on its range. The equation \(H(y)=t-t_0\) defines the increasing branch, and differentiating this relation gives \(H'(y)y'=1\), hence \(y'=g(y)\). The branch reaches each finite \(Y\) at time \(t=t_0+H(Y)\).

If \(L\) is finite, then \(H(Y)\to L\) as \(Y\to\infty\), so the endpoint time is \(T=t_0+L\). For any fixed \(M>y_0\), once \(t>t_0+H(M)\), the increasing branch has \(y(t)>M\). Since this holds for every such \(M\), \(y(t)\to+\infty\) as \(t\to T^-\). The unbounded limit prevents extension through \(T\). If instead the integral diverges, this increasing branch takes arbitrarily long to reach arbitrarily large values; this test does not by itself describe branches that do not increase toward \(+\infty\).

A Different Separable Equation

Finite-time blow-up is not limited to \(y'=y^2\). Consider \(y'=y(y-1)\) with \(y(0)=2\). Along this solution, \(y>1\), so separation and partial fractions are valid. The resulting formula has a finite pole, and differentiation verifies both the equation and the initial condition.

A Common Sign Error

When solving \(y'=y^2\), it is easy to lose a minus sign while integrating \(y^{-2}\). For \(y(0)=2\), an incorrect candidate is \(y(t)=2/(1+2t)\). It has the correct initial value, but that alone does not make it a solution. Differentiation gives

These expressions are not equal, so the candidate fails the differential equation. The correct integration is \(\int y^{-2}\,dy=-1/y\), which gives \(y(t)=2/(1-2t)\). Its derivative is \(4/(1-2t)^2\), equal to its square. Checking both the initial condition and the differential equation is essential before using a formula to locate blow-up.

How to Analyze a Candidate Blow-Up

For a separable initial-value problem, use the following checks to distinguish genuine finite-time blow-up from a misleading algebraic feature.

A denominator becoming zero is a useful signal, but not the complete argument. The endpoint must be approached from within the solution interval, and the one-sided limit must be checked. The integral test gives another way to identify a finite endpoint when the solution increases along a branch toward arbitrarily large values.