Differential equations & phase portraits

What does a differential equation determine once an initial state is specified? This bridge develops three complementary answers: a formula that can be checked, a vector field that gives the direction of motion, and a local theorem that guarantees a unique solution. You will use each answer on elementary real equations, then identify exactly why uniqueness and existence for all time are different questions. These are the tools needed to read the oscillator, Toda and travelling-wave calculations.

Required background. Differentiate exponentials and simple polynomials; integrate $1/x$. The entry check repairs substitution into an equation. No Hamiltonian mechanics is required.

Helpful background. The oscillator lesson gives a physical application and a phase portrait to return to after this bridge.

An equation specifies a derivative

For a state $x(t)\in\mathbb R^n$, a first-order ordinary differential equation has the form

$$\dot x(t)=f(t,x(t)),\qquad x(t_0)=x_0.$$

The dot denotes differentiation with respect to real time. The function $f$ assigns a velocity vector to each allowed time and state. An initial-value problem specifies this function, its domain, and the initial data $(t_0,x_0)$.

A solution on an interval $I$ containing $t_0$ is a differentiable function whose graph stays in that domain and satisfies both displayed equalities. A candidate must therefore pass two separate tests: substitution into the differential equation throughout $I$, and substitution into the initial condition. See Teschl 2012, § 1.2, p. 6, equation (1.14), and § 2.2, p. 36, equation (2.10), author PDF.

For example, let $x$ be scalar and

$$\dot x=-\gamma x,\qquad \gamma\gt0.$$

If $t$ has time units, $\gamma$ has inverse-time units. The candidate

$$x(t)=x_0e^{-\gamma(t-t_0)}$$

has derivative $-\gamma x_0e^{-\gamma(t-t_0)}=-\gamma x(t)$ and gives $x(t_0)=x_0$. It is defined for every real $t$. Positive initial values decrease toward zero; negative ones increase toward zero. The equilibrium $x_0=0$ stays at zero.

Entry check and repair

For $\dot x=-x$ with $x(0)=2$, consider

$$a(t)=2e^{-t},\qquad b(t)=2e^{-2t},\qquad c(t)=3e^{-t}.$$

Which satisfy the equation, which satisfy the initial value, and which solve the complete problem?

Repair: check the derivative and the initial value separately

Both $a$ and $c$ obey $\dot x=-x$, but $c(0)=3$ fails the initial condition. Both $a$ and $b$ equal $2$ at zero, but $b'=-2b$ fails the equation. Only $a$ passes both tests. For a retry, $5e^{-3t}$ solves $\dot x=-3x$, $x(0)=5$; its exponent fixes the rate and its prefactor fixes the initial value.

A phase line gives the direction of motion

An equation is autonomous when $f$ depends only on the state. In one dimension, the sign of $f(x)$ tells whether $x$ increases or decreases. A zero of $f$ is an equilibrium state: the constant function at that value is a solution. A trajectory is a time-dependent solution; an equilibrium trajectory is constant.

From here onward all examples use dimensionless variables and time. Consider

$$\dot x=x(1-x),\qquad x(0)=x_0.$$

Its phase line can be read without solving it:

State or interval	Sign of $x(1-x)$	Forward motion
$x\lt0$	Negative	Toward smaller values
$x=0$	Zero	Equilibrium
$0\lt x\lt1$	Positive	Toward larger values
$x=1$	Zero	Equilibrium
$x\gt1$	Negative	Toward smaller values

Thus nearby states move away from $0$ and toward $1$. The arrows describe motion while a solution exists; they do not establish its full time interval. The role of uniqueness in this phase-line reasoning is discussed in Teschl 2012, § 1.5, pp. 20–21, equation (1.61) and Lemma 1.1, PDF.

To find nonconstant solutions, work on an interval where $x\ne0,1$. Separation gives

$$\frac{dx}{x(1-x)}=dt,\qquad \frac1{x(1-x)}=\frac1x+\frac1{1-x}.$$

Integrating yields

$$\log\left|\frac{x}{1-x}\right|=t+C, \qquad \frac{x}{1-x}=C_1e^t,$$

where the nonzero signed constant $C_1$ absorbs the sign of the ratio. Imposing $x(0)=x_0$ and solving for $x$ gives

$$x(t)=\frac{x_0e^t}{1+x_0(e^t-1)}.$$

Use the connected interval containing zero on which the denominator is nonzero. Restore the constant solutions $x=0$ and $x=1$, which division excluded; the final formula also reproduces them.

As an independent substitution check, write $D=1-x_0+x_0e^t$. Then

$$x'=\frac{x_0e^t(1-x_0)}{D^2} =x(1-x).$$

When $0\lt x_0\lt1$, both terms in $D=(1-x_0)+x_0e^t$ are positive for every real $t$. Hence the solution exists for all real time, remains strictly between $0$ and $1$, approaches $0$ as $t\to-\infty$, and approaches $1$ as $t\to+\infty$. The formula establishes these claims; the arrows alone do not give all of them.

A second-order equation becomes a planar flow

For the oscillator equation $\ddot q=-q$, introduce the velocity $v=\dot q$. The state is now the pair $(q,v)$, and

$$\dot q=v,\qquad \dot v=-q.$$

Initial data must include both $q(0)$ and $v(0)$. At the state $(1,0)$, the derivative is $(0,-1)$: with $q$ horizontal and $v$ vertical, the motion initially points downward. The flow goes clockwise around the origin. The origin itself is an equilibrium.

Along a solution,

$$\frac d{dt}(q^2+v^2)=2qv+2v(-q)=0.$$

Thus a nonzero initial state stays on its circle $q^2+v^2=R^2$. The orbit is the set of states visited; the time parametrization says when each state is visited. A circle alone does not specify speed or direction. For example, doubling this vector field preserves its circular orbits while doubling the traversal speed.

The oscillator calculation and phase portrait develop the exact motion further. More generally, a smooth autonomous field supplies a local flow, with its time interval depending on the initial state; see Teschl 2012, § 6.2, pp. 188–189, Theorem 6.1, PDF.

What local uniqueness guarantees

Here is a sufficient theorem for $\dot x=f(t,x)$. Let $U\subset\mathbb R\times\mathbb R^n$ be open, let $(t_0,x_0)\in U$, and let $f:U\to\mathbb R^n$ be continuous. Assume that near every point of $U$ there is a time–state rectangle and a finite constant $L$ such that

$$\|f(t,x)-f(t,y)\|\le L\|x-y\|$$

for all allowed $t,x,y$ in that rectangle. This is local Lipschitz continuity in the state, locally uniform in time. Then the initial-value problem has a unique continuously differentiable solution on some interval around $t_0$. Continuous first partial derivatives of $f$ with respect to the state are a convenient sufficient condition. This is the local result of Teschl 2012, § 2.2, pp. 37–38, equation (2.18), Theorem 2.2 and its following remark, PDF.

Why does a bound on differences help? Every solution satisfies

$$x(t)=x_0+\int_{t_0}^t f(s,x(s))\,ds.$$

For two solutions staying in the rectangle on a sufficiently short common interval, their maximum separation obeys

$$\sup_t\|x(t)-y(t)\| \le L\delta\sup_t\|x(t)-y(t)\|,$$

where $\delta$ is the largest distance from $t_0$ in that interval. If $L\delta\lt1$, the separation must be zero. The same integral map constructs a solution by successive approximation on a suitably small interval. This explains both the local nature and the importance of the uniform constant.

Uniqueness prevents two autonomous solution curves from meeting and then separating: starting at the meeting state, they would solve the same initial-value problem. In particular, a nonconstant logistic solution cannot reach either equilibrium in finite time and then depart or stop there.

Two cautions matter. First, local existence does not mean existence for all real time. Second, failure of this sufficient Lipschitz hypothesis alone does not prove nonuniqueness. The transfer exercise supplies explicit examples of both distinctions.

Exercises

Guided: approach an equilibrium

Solve $\dot x=-2x+3$, $x(0)=1$. Find the equilibrium, check your solution by differentiation, and describe its forward motion.

Hint

Set the right-hand side to zero, then subtract that equilibrium: $y=x-3/2$. What equation and initial value does $y$ satisfy?

Solution

The equilibrium is $3/2$. Since $y'=-2y$ and $y(0)=-1/2$,

$$x(t)=\frac32-\frac12e^{-2t}.$$

Its derivative is $e^{-2t}=-2x+3$, and $x(0)=1$. The solution exists for all real $t$. For forward time it increases toward $3/2$ from below, never reaching that value at finite time.

Independent: a finite arrival time and an infinite one

For $\dot x=x(1-x)$ with $x(0)=1/4$, find the solution and the first time it reaches $3/4$. Can it cross $1$ at a later finite time? Explain using both the formula and uniqueness.

Hint

Insert $x_0=1/4$ into the derived formula and divide numerator and denominator by $e^t/4$. At a hypothetical meeting with $1$, compare with the constant solution.

Solution

The solution is

$$x(t)=\frac1{1+3e^{-t}}.$$

Setting $x=3/4$ gives $e^{-t}=1/9$, hence $t=\log9$. Strict increase makes this the first arrival. For every finite $t$, the denominator is greater than one, so $x(t)\lt1$. Also, $f(x)=x(1-x)$ is smooth. Meeting $1$ would force local agreement with the equilibrium trajectory, contradicting the nonconstant solution. It approaches $1$ only as $t\to+\infty$.

Transfer: separate blow-up from nonuniqueness

Compare two initial-value problems on the full real time–state plane:

$$\dot x=x^2,\quad x(0)=1; \qquad \dot y=2\sqrt{|y|},\quad y(0)=0.$$

Find the first solution and its maximal interval containing zero. For the second, verify the family

$$y_a(t)= \begin{cases} 0,&t\le a,\\ (t-a)^2,&t\ge a, \end{cases} \qquad a\ge0.$$

Check the join at $t=a$ and the Lipschitz condition near $y=0$. State what each example proves.

Hint

Differentiate $1/x$ in the first problem. For the second, compare the difference quotient $|f(y)-f(0)|/|y|$ for $y\gt0$, and compute the derivative at the join from both sides.

Solution

The first equation gives $-1/x=t-1$, hence

$$x(t)=\frac1{1-t},\qquad -\infty\lt t\lt1.$$

Differentiation gives $x'=1/(1-t)^2=x^2$, and $x(0)=1$. As $t\to1^-$, $x\to+\infty$, so no finite, continuous extension through $1$ exists. This is the maximal solution interval containing zero; the same formula on $t\gt1$ does not continue this initial-value solution. The smooth field gives uniqueness, but no global-in-time solution for these data. Compare Teschl 2012, § 1.3, p. 10, equations (1.28)–(1.29), PDF.

For $t\lt a$, both sides of the second equation vanish. For $t\gt a$, $y_a'=2(t-a)=2\sqrt{y_a}$. At the join, the left derivative is zero and the right difference quotient is $(t-a)^2/(t-a)=t-a\to0$. Thus $y_a$ is continuously differentiable and satisfies the equation there too. Since $a\ge0$, every member has $y_a(0)=0$; the identically zero solution also works.

For $y\gt0$,

$$\frac{|2\sqrt y-0|}{|y-0|}=\frac2{\sqrt y}\longrightarrow\infty.$$

No finite local Lipschitz constant exists at zero. Nonuniqueness is established by the distinct checked solutions, not merely by that failed hypothesis. They can wait at an equilibrium and then leave because the uniqueness assumption used in the phase-line argument is absent.

Return to the physical calculation

Use the oscillator lesson to connect a planar flow with energy and period. In the Toda equations, positions and momenta together specify the initial state. In the KdV travelling-wave calculation, the independent variable is the travelling coordinate rather than time; substitution, integration constants and local uniqueness still require the same care.

References

• Teschl, Gerald. Ordinary Differential Equations and Dynamical Systems. Graduate Studies in Mathematics 140. American Mathematical Society, Providence, RI, 2012. DOI. Author’s preliminary version, PDF, with preface dated April 2012; printed page numbers and theorem numbers above refer to this author copy.