Equilibria, Motion, and Initial Conditions
An autonomous differential equation has a right-hand side that depends on the dependent variable but not explicitly on the independent variable. Its standard form is \(y'=F(y)\). When \(F\) has several zeros, the equation has several constant solutions, and those solutions divide the number line into intervals where the direction of motion may differ.
Solving an initial-value problem in this setting takes more than separating variables. First identify every equilibrium, then determine the sign of \(F\) between equilibria. If the initial value is an equilibrium, the constant solution must be kept. Otherwise, separation is valid only on the interval containing the initial value, where \(F\) does not vanish.
Factor First and Read the Phase Line
For a factored equation, the equilibrium values are the roots of \(F(y)\). Between consecutive roots, \(F\) has a constant sign if it is continuous and has no additional zeros there. The sign tells the direction of motion: when \(F(y)>0\), solutions increase; when \(F(y)<0\), they decrease.
A phase line records this information on the \(y\)-axis. Mark each equilibrium and place an upward arrow on intervals where \(F>0\), and a downward arrow where \(F<0\). This gives a useful qualitative check on a formula obtained by separation. It also helps determine which interval can contain a solution with a specified initial value.
| Sign of \(F(y)\) | What a solution does there |
|---|---|
| \(F(y)>0\) | It increases as \(t\) increases. |
| \(F(y)<0\) | It decreases as \(t\) increases. |
| \(F(y)=0\) | The corresponding constant function is an equilibrium solution. |
The phase line describes direction, not the exact speed or a complete formula. Separation provides the formula when the initial value lies away from an equilibrium.
Why Separation Works on One Branch
Suppose \(F\) is continuous on an interval \(I\) and is nonzero throughout \(I\). A solution whose values lie in \(I\) can be separated because division by \(F(y)\) is then legitimate. Define an antiderivative on \(I\) by \(G(y)=\int_{y_0}^{y} \frac{du}{F(u)}\). The chain rule explains why this integral appears:
Consequently, \(G(y(t))=t-t_0\) for a solution with \(y(t_0)=y_0\). In integral form, this is
This argument justifies separation on an interval where \(F\neq 0\). It does not justify division at an equilibrium, because \(1/F(y)\) is undefined there. The initial value must therefore be checked against the equilibria before separating.
Worked Example: A Logistic-Type Equation with Two Equilibria
Worked Example: A Logistic-Type Equation with Two Equilibria
Solve \(y'=y(2-y)\), \(y(0)=1\), and use the signs of the factors to check the direction of motion.
The factored right-hand side vanishes at \(y=0\) and \(y=2\). Since the initial value \(1\) is between them, \(F(y)\) is nonzero on that branch and separation is allowed there. For \(0<y<2\), both factors \(y\) and \(2-y\) are positive, so \(y'>0\).
Separate and integrate using partial fractions:
Because \(\int \frac{1}{2(2-y)}\,dy=-\frac12\ln|2-y|\), integration gives
On the branch \(0<y<2\), the ratio \(y/(2-y)\) is positive. Multiply by \(2\), combine the logarithms, and exponentiate:
Use \(y(0)=1\): \(1/(2-1)=A e^0\), so \(A=1\). Solving for \(y\) gives
Check the initial condition: \(y(0)=2/(1+1)=1\). To check the differential equation, differentiate the formula. Writing \(q=e^{-2t}\), we have \(q'=-2q\), so
Thus the formula satisfies both the equation and the initial condition. It increases while remaining between the equilibria, as the phase-line sign predicts.
Worked Example: Three Equilibria and Branch Selection
Worked Example: Three Equilibria and Branch Selection
Solve \(y'=y(1-y^2)\), \(y(0)=\frac12\).
Factor \(F(y)=y(1-y)(1+y)\). The equilibria are \(-1,0,1\). For \(0<y<1\), all three factors in \(y(1-y)(1+y)\) are positive, so solutions increase on this interval. The initial value is in this branch, so separation does not cross a zero of \(F\).
Separate variables. With \(u=y^2\), \(du=2y\,dy\), and therefore
Returning to \(y\), the separated equation becomes
where the logarithm has a positive argument on \(0<y<1\). At \(t=0\), \(y=\frac12\), so \(y^2/(1-y^2)=(1/4)/(3/4)=1/3\). Hence \(A=1/3\). Rearranging gives
The branch condition \(0<y<1\) selects the positive square root:
At \(t=0\), this gives \(1/\sqrt{4}=1/2\). To verify the equation, put \(q=3e^{-2t}\), so \(q'=-2q\) and \(y=(1+q)^{-1/2}\). Then
The derivative agrees with the right-hand side. The solution increases on the interval \(0<y<1\), consistent with the phase line.
Worked Example: An Initial Value at an Equilibrium
Worked Example: An Initial Value at an Equilibrium
Solve \(y'=(y+2)y(1-y)\), \(y(0)=-2\).
The factors show equilibria at \(-2\), \(0\), and \(1\). Since the initial value is an equilibrium, substitute the constant function \(y(t)=-2\): its derivative is \(0\), and the right-hand side is \(({-2}+2)(-2)(1-(-2))=0\). Therefore \(y(t)=-2\) is a solution satisfying the initial condition.
There is no need, and no permission, to divide by \(y+2\) at this initial value. Moreover, the right-hand side is a polynomial, so it is smooth. The local existence and uniqueness result for smooth equations implies that the solution through \((0,-2)\) is unique on its interval of existence. Since the constant function is one such solution, it is the solution of this initial-value problem.
The phase-line signs provide a qualitative check. For \(y<-2\), the factors have signs \((-)(-)(+)\), so \(y'>0\). For \(-2<y<0\), they have signs \((+)(-)(+)\), so \(y'<0\). The arrows point toward \(-2\) from both sides, but a solution starting exactly at \(-2\) remains there.
A Reliable Procedure
For an autonomous initial-value problem with a factored right-hand side, use the following order. It keeps equilibrium solutions visible and makes the chosen branch explicit.
Factor \(F(y)\) and solve \(F(y)=0\). Record each constant solution before dividing by any factor.
If \(y_0\) is an equilibrium, test the corresponding constant function in the original equation. If not, identify the interval between equilibria containing \(y_0\).
Determine the sign of \(F\) on each interval. Use it to predict whether the solution increases or decreases there.
When \(F\neq0\) on the branch, integrate \(dy/F(y)=dt\), using partial fractions or another appropriate algebraic rearrangement.
Determine the integration constant and any sign or branch choice from the initial value. Substitute the resulting formula into the original equation and check the initial condition.
Common Pitfall: Dividing Away an Equilibrium
Consider \(y'=y(2-y)\). A tempting first step is to divide both sides by \(y(2-y)\). That produces an equation involving \(1/[y(2-y)]\), but it is defined only when \(y\neq0\) and \(y\neq2\). The division has excluded the equilibrium solutions \(y(t)=0\) and \(y(t)=2\).
The incorrect conclusion would be: “The separated formula gives all solutions, so there are no constant solutions.” The correction is to check the roots before dividing. Both constants satisfy the original equation: at \(y=0\), \(y(2-y)=0\), and at \(y=2\), \(y(2-y)=0\). A separated formula derived on an interval away from these roots describes non-equilibrium solutions on that branch; it does not replace the equilibrium solutions.
Check Your Understanding
For each problem, identify what the equilibria and signs tell you before attempting a calculation.
- For \(y'=y(3-y)\), find the equilibrium values and determine the sign of \(y'\) when \(0<y<3\).
- For \(y'=y(1-y^2)\), \(y(0)=-\frac12\), which interval between equilibria contains the initial value?
- Why is dividing by \(F(y)\) invalid when the initial value is an equilibrium?
- For \(y'=(y-1)(y+2)\), \(y(0)=1\), what candidate solution should be checked first?
- After obtaining an explicit formula by separation, what two checks should be made against the original initial-value problem?
