Parameter Effects on Equilibria and Solutions

Why the Parameter Matters

A differential equation can keep the same algebraic form while its solutions change substantially as a parameter varies. For the family

\(a\) is a fixed real number, and \(y\) is the quantity that changes with time \(t\). The parameter affects both where solutions can remain constant and whether nearby solutions move toward or away from those constant values.

The central comparison is between \(a>0\), \(a=0\), and \(a<0\). When \(a\ne 0\), the equation has two equilibrium values, \(0\) and \(a\). At \(a=0\), those values coincide, and the resulting equation has a different pattern of solution behavior.

Equilibria and Phase-Line Directions

To find the equilibria, set the right-hand side equal to zero:

Thus \(y=0\) or \(y=a\). If \(a\ne 0\), these are distinct equilibrium values. If \(a=0\), they are the same value, so there is only one distinct equilibrium: \(y=0\).

Between equilibrium values, the sign of \(y(a-y)\) determines the direction of change. A positive value means \(y\) increases as \(t\) increases; a negative value means \(y\) decreases. The sign must be checked separately on each interval, especially when \(a<0\).

For example, when \(a<0\) and \(y<a\), we have \(y<0\) and \(a-y>0\), so \(y(a-y)<0\). In the interval \(a<y<0\), both \(y\) and \(a-y\) are negative, so their product is positive. Keeping the order of \(a\) and \(0\) in mind prevents sign errors.

What the Phase Line Says About Stability

An equilibrium is called attracting if solutions starting sufficiently close to it move toward it as time increases. It is called repelling if nearby solutions move away from it. The phase-line directions show these tendencies without requiring an explicit solution formula.

If \(a>0\), solutions just below \(0\) decrease, while solutions just above \(0\) increase. Both directions point away from \(0\), so \(0\) is repelling. On either side of \(a\), solutions move toward \(a\): between \(0\) and \(a\) they increase, and above \(a\) they decrease. Thus \(a\) is attracting.

If \(a<0\), the order of the equilibria reverses. For \(y<a\), solutions decrease, moving away from \(a\). For \(a<y<0\), solutions increase, also moving away from \(a\). Therefore \(a\) is repelling. On both sides of \(0\), solutions move toward \(0\), so \(0\) is attracting.

Explicit Solutions When \(a\ne 0\)

The phase line describes directions. A separated formula gives more detail about the trajectory. Suppose \(a\ne 0\), and consider a non-equilibrium initial value \(y(t_0)=y_0\), with \(y_0\ne 0\) and \(y_0\ne a\). Separation gives

The partial-fraction identity

holds because the right-hand side simplifies to \(1/[y(a-y)]\). Integrating both sides, using \(\int 1/(a-y)\,dy=-\ln|a-y|\), yields

Exponentiating and incorporating the sign into a constant gives \(y/(a-y)=K e^{at}\). Applying \(y(t_0)=y_0\) and solving for \(y\) produces

This formula applies on intervals where its denominator is nonzero. The separation divided by \(y(a-y)\), so the equilibrium solutions must also be included separately: \(y(t)=0\) and \(y(t)=a\). In particular, dividing by the right-hand side without checking for equilibria would omit constant solutions.

Worked Examples

What Changes as the Parameter Passes Through Zero?

For \(a>0\), the equilibrium \(a\) is positive and attracts solutions between \(0\) and \(a\), as well as solutions above \(a\). For \(a<0\), the equilibrium \(a\) is negative and repelling, while \(0\) attracts solutions on either side. At \(a=0\), the two equilibrium values coincide.

The equation at \(a=0\), \(y'=-y^2\), has an asymmetric pattern. If \(y>0\), then \(y'<0\), and the formula for a positive initial value \(y_0\) is \(y(t)=y_0/(1+y_0t)\); it tends to \(0\) as \(t\to\infty\). If \(y<0\), then \(y'<0\) as well, and the same formula has denominator zero at \(t=-1/y_0>0\). Thus negative initial values move away from zero and reach an unbounded value in finite positive time. The equilibrium at zero attracts from above but not from both sides.

The parameter also changes the initial slope at a fixed starting value. For example, at \(y=1\), the slope is \(a-1\): it is \(1\) when \(a=2\), \(-1\) when \(a=0\), and \(-3\) when \(a=-2\). This comparison shows that parameter changes affect not only the equilibrium locations but also the immediate direction and rate of change.

A Common Sign Mistake

A sign analysis must respect the order of the values \(a\), \(0\), and \(y\). For instance, take \(a=-2\) and \(y=-3\). It is incorrect to say that both \(y\) and \(a-y\) are negative merely because \(y<a\). In fact,

The corrected method is to determine the sign of each factor separately, then multiply those signs. Here the solution decreases. A reliable phase-line analysis never infers the sign of a product from the position of \(y\) alone.

Why These Parameter Comparisons Matter

Equilibria describe the constant states of a model, while phase-line directions describe how nearby states change. Studying parameter effects connects these two features: a parameter may shift an equilibrium, reverse the direction of motion on an interval, or cause distinct equilibria to merge. The explicit formulas then show whether the predicted direction corresponds to approach to an equilibrium or, for some initial values, to a finite-time singularity.

For this equation, the decisive steps are to find the roots of \(y(a-y)\), order them correctly for the sign of \(a\), and test the factors on each interval. The special case \(a=0\) must be checked separately because the two roots are no longer distinct. These habits apply broadly when comparing parameter-dependent differential equations.