Equilibrium of a Differential Equation¶

Definition¶

Consider the autonomous first-order differential equation

where $x'(t)=\dfrac{\mathrm{d}x}{\mathrm{d}t}(t)$ and $f$ is a given real function. A number $x^*$ is an equilibrium (steady state) if $f(x^*)=0$$0$. Equivalently, the constant function $x(t)\equiv x^*$ is a solution.

For a vector equation $\boldsymbol{x}'(t)=\boldsymbol{f}\big(\boldsymbol{x}(t)\big)$ with $\boldsymbol{x}(t)\in\mathbb{R}^n$${\mathbb{R}}^n$, an equilibrium is a vector $\boldsymbol{x}^*$ with $\boldsymbol{f}(\boldsymbol{x}^*)=\boldsymbol{0}$$0$.

Interpretation¶

At an equilibrium the rate of change is zero, so a system that starts there stays there for ever. Finding equilibria is an algebraic task: they are the zeros of $f$; no differential equation needs to be solved.

The figure shows the logistic example $x'=x(1-x)$$x)$. On the left, the equilibria are the zeros $0$ and $1$ of $f(x)=x(1-x)$$x)$, and the arrows on the axis form the phase line: to the right where $f>0$$0$, to the left where $f<0$$0$. On the right, the equilibria appear as constant solutions (dashed). The solution from $x(0)=0.2$ rises towards $1$, and a solution from $x(0)=1.5$ falls towards $1$; neither crosses an equilibrium line.

The sign of $f$ between consecutive equilibria shows the direction of motion: where $f(x)>0$ solutions increase, where $f(x)<0$ they decrease. This one-dimensional picture is called the phase line. Whether nearby solutions approach the equilibrium is a separate question, treated in stability of an equilibrium.

Two misunderstandings occur often. An equilibrium is a value of the state $x$, not a time $t$. And an equilibrium need not be reached: in $x'=-x$ the solution $x_0\mathrm{e}^{-t}$ approaches $0$ but never equals it when $x_0\neq0$$0$. For a non-autonomous equation $x'=f(t,x)$$f(t,x)$, a constant $x^*$ is a solution only if $f(t,x^*)=0$ for every $t$.

Key Results¶

• Equilibria of $x'=f(x)$ are exactly the constant solutions.
• For $x'=ax+b$ with $a\neq0$ there is a unique equilibrium $x^*=-b/a$$-b/a$, and every solution satisfies $x(t)-x^*=\mathrm{e}^{at}\big(x(0)-x^*\big)$$x^{\ast })$.
• For $x'=ax+b$ with $a=0$$0$: every point is an equilibrium if $b=0$$0$; there is none if $b\neq0$$0$.
• If $f$ is continuously differentiable, solutions cannot cross or touch an equilibrium (uniqueness of solutions), so a solution starting between two equilibria stays between them and is monotone.
• For $\boldsymbol{x}'=A\boldsymbol{x}+\boldsymbol{b}$ with $A$ invertible, the unique equilibrium is $\boldsymbol{x}^*=-A^{-1}\boldsymbol{b}$$-A^{-1}\mathbf{b}$, using the inverse matrix.

Examples¶

Linear. $x'=4-2x$$2x$: $f(x^*)=0$ gives $x^*=2$$2$. The solution with $x(0)=5$ is $x(t)=2+3\mathrm{e}^{-2t}$$3{\mathrm{e}}^{-2t}$.

Logistic equation. $x'=x(1-x)$ has equilibria $0$ and $1$. Since $f(x)>0$ on $(0,1)$, a solution with $x(0)=0.2$ increases and stays in $(0,1)$; it tends to $1$. For $x>1$$1$, $f(x)<0$ and solutions decrease towards $1$.

No equilibrium. $x'=1+x^2$ has $f(x)\geq1>0$ for every $x$, so there is no equilibrium and every solution is strictly increasing.

Vector case. For $A=\begin{pmatrix}-1&1\\0&-2\end{pmatrix}$ and $\boldsymbol{b}=\begin{pmatrix}1\\4\end{pmatrix}$$\left(\begin{array}{c}1\\ 4\end{array}\right)$, solve $A\boldsymbol{x}^*=-\boldsymbol{b}$$-\mathbf{b}$: the second row gives $-2x_2^*=-4$$-4$, so $x_2^*=2$$2$; the first row gives $-x_1^*+2=-1$$-1$, so $x_1^*=3$$3$. Thus $\boldsymbol{x}^*=(3,2)^{\top}$$(3,2{)}^{\top }$.

Prerequisites¶

• First-Order Differential Equation
• Zero of a Function
• Constant Function

Related Concepts¶

• Stability of an Equilibrium
• Linear Vector ODE
• Particular Solution of an Inhomogeneous Linear ODE
• Stability of a Scalar Difference Equation

Sources¶

• Math V: Advanced Linear Algebra · Week 3

Aliases: equilibrium point, steady state, rest point, equilibrium solution, equilibrium state Keywords: constant solution, f(x)=0, x=-b/a, autonomous equation, phase line, logistic equation Symbols: element of, real numbers

Last updated: 17 October 2026 21:53