Separable and first order differential equations

Multivariable geometry describes trajectories, but a differential equation can determine the trajectory from a rate law. We begin with separable and linear first-order equations, preserving equilibrium solutions and the intervals on which each formula makes sense.

1 An equation and an initial condition

A solution of $y'=F(t,y)$ is a differentiable function satisfying it throughout an interval. An initial condition $y(t_0)=y_0$ supplies a state, not a guarantee of global existence or uniqueness. Continuous $F$ with a continuous partial derivative in $y$ near the initial point is a standard sufficient local existence-and-uniqueness hypothesis. It gives no promise that the solution continues forever.

For $y'=y^2$, $y(0)=1$, separation gives $y=1/(1-t)$ on $t<1$. The blow-up at 1 cannot be crossed by announcing the same algebraic expression a solution through that point; it is undefined there. Initial-value solutions have intervals as part of their answer.

2 Separation without losing equilibria

For $y'=p(t)q(y)$, first record every constant solution with $q(y_*)=0$. On a region where $q(y)\ne0$, separation gives

$$\int\frac{dy}{q(y)}=\int p(t)dt+C.$$

This follows by applying the chain rule to an antiderivative of $1/q(y)$. The resulting implicit formula is valid on a branch where the divisions and antiderivatives are defined. Solving it algebraically may introduce additional domain restrictions.

Example. For $y'=y(1-y)$, equilibria are 0 and 1. Other positive solutions satisfy $\log|y/(1-y)|=t+C$, giving

$$y(t)=\frac1{1+Ae^{-t}}.$$

For $y(0)=1/4$, $A=3$ and the solution increases toward 1. Dividing by $y(1-y)$ without first checking zeros would lose both equilibrium solutions.

If the rate law is not locally Lipschitz, separate branches can sometimes join an equilibrium and uniqueness can fail. For $y'=2\sqrt y$, $y\ge0$, the function equal to zero up to a time $T\ge0$ and $(t-T)^2$ afterward solves the initial condition $y(0)=0$. It is differentiable at $T$. This is why an implicit integration calculation alone is not a general uniqueness proof.

3 Linear first-order equations and integrating factors

For $y'+p(t)y=q(t)$ with continuous coefficients on an interval, let

$$\mu(t)=\exp\left(\int_{t_0}^tp(s)ds\right).$$

It is positive and satisfies $\mu'=p\mu$. Therefore $(\mu y)'=\mu q$, so the unique solution with $y(t_0)=y_0$ is

$$y(t)=\mu(t)^{-1}\left[y_0+\int_{t_0}^t\mu(s)q(s)ds\right].$$

Proof of uniqueness. The difference of two solutions satisfies $(\mu(y_1-y_2))'=0$ and has zero initial value, so it vanishes. The formula itself verifies existence by the fundamental theorem and product rule. $\square$

Unlike separation, this formula does not require dividing by $y$, so it handles solutions passing through zero naturally. The coefficient interval must still avoid singularities in $p,q$.

4 Motion and changing environments

A particle with one-dimensional velocity $v$ under constant driving acceleration $a$ and linear drag coefficient $k>0$ satisfies $v'+kv=a$. Its velocity is

$$v(t)=\frac ak+\left(v_0-\frac ak\right)e^{-kt}.$$

With initial position $x_0$, integrate again:

$$x(t)=x_0+\frac ak t+\frac{v_0-a/k}{k}(1-e^{-kt}).$$

Here $k$ has inverse-time units and $a/k$ has velocity units. The terminal velocity is a model consequence, not a claim that every physical drag law is linear.

For a changing ambient temperature $T_{\rm env}=20+t$ and rate law $T'=-(T-T_{\rm env})$, $T(0)=30$, the equation is $T'+T=20+t$. A particular solution is $19+t$, so

$$T(t)=19+t+11e^{-t}.$$

The object does not generally approach the moving ambient curve exactly; the late-time difference approaches $-1$. A constant-ambient exponential formula would solve a different equation.

The exact domain behind separation

Take continuous real $p$ on an interval $I$ and continuous real $q$ on an interval $J$. First retain every equilibrium $q(y_*)=0$. On a connected open subinterval $J_0\subset J$ where $q$ has no zeros, choose $Q$ with $Q'(y)=1/q(y)$. Its derivative has constant sign, so $Q$ is strictly monotone. On the interval $Q(J_0)$ its inverse is differentiable with derivative $q(Q^{-1}(u))$: this follows by dividing the difference quotient for $Q$ and using its nonzero derivative. For times with $P(t)+C\in Q(J_0)$, where $P'=p$, the function $y=Q^{-1}(P+C)$ therefore satisfies $y'=p(t)q(y)$. Conversely any solution staying in $J_0$ satisfies $(Q\circ y)'=p$, hence the implicit formula. This is a statement on that branch and time interval; it does not assert uniqueness at a zero of $q$.

A complete local existence and uniqueness proof for the stated sufficient condition

The theorem here assumes continuous $F$ and continuous $F_y$ on an open neighborhood of the initial point. The following full proof uses exactly that sufficient condition, and includes the real uniform-limit facts on which the iteration depends.

1. The local theorem and its assumptions

A solution of $y'=f(t,y)$ is a differentiable real function on an interval whose graph lies in the domain of $f$ and satisfies the equation there. The initial condition is $y(t_0)=y_0$. Local uniqueness means that any two such solutions with this initial value agree on some common neighborhood of $t_0$.

Theorem. Let $U\subset\mathbb R^2$ be open and contain $(t_0,y_0)$. Suppose $f:U\to\mathbb R$ is continuous, its partial derivative $f_y$ exists on $U$, and $f_y$ is continuous there. Then the initial-value problem has a solution on an open interval about $t_0$, unique in the local sense just defined. No derivative with respect to $t$ is required.

2. Two uniform-limit facts

For a nonempty compact interval $I$, write $C(I)$ for the continuous real functions on $I$ and $\|u-v\|_\infty=\sup_{t\in I}|u(t)-v(t)|$. Uniform convergence is convergence in this distance. A Cauchy sequence in this distance is uniformly Cauchy.

Theorem: uniform completeness

A sequence of real-valued functions is uniformly convergent if and only if it is uniformly Cauchy.

Proof. Uniform convergence gives the Cauchy estimate by two errors of size $\varepsilon/2$. Conversely, uniform Cauchy implies that for each fixed $x$, $(f_n(x))$ is a real Cauchy sequence, so define $f(x)=\lim_n f_n(x)$. Fix $\varepsilon>0$ and use uniform Cauchy with tolerance $\varepsilon/2$. For every $x$ and $n\ge N$, pass $m\to\infty$ in the estimate to get $|f_n(x)-f(x)|\le\varepsilon/2<\varepsilon$. The same $N$ works everywhere, proving uniform convergence. Completeness of the target $\mathbb R$ supplied the pointwise limits.

Theorem

Uniform limit preserves continuity

If each $f_n$ is continuous at $a\in A$ relative to $A$ and $f_n\to f$ uniformly on $A$, then $f$ is continuous at $a$.

Proof

The three-term estimate

Given $\varepsilon>0$, choose $N$ so $|f_N(x)-f(x)|<\varepsilon/3$ for all $x\in A$. By continuity of the fixed function $f_N$ at $a$, choose $\delta$ so $|f_N(x)-f_N(a)|<\varepsilon/3$ whenever $|x-a|<\delta$. Then

$$|f(x)-f(a)|\le|f(x)-f_N(x)|+|f_N(x)-f_N(a)|+|f_N(a)-f(a)|<\varepsilon.$$

These facts use completeness of the real numbers. They will also preserve the closed value constraint imposed below.

3. Proof by an integral iteration

Rectangle and quantitative constants. Choose $a,b>0$ so the closed rectangle $Q=[t_0-a,t_0+a]\times[y_0-b,y_0+b]$ lies in $U$. The extreme-value theorem supplies finite constants $M,L\ge0$ with $|f|\le M$ and $|f_y|\le L$ on $Q$. The one-variable mean value theorem in the $y$ coordinate gives

$$|f(t,y)-f(t,z)|\le L|y-z|\qquad(t\in[t_0-a,t_0+a],\ y,z\in[y_0-b,y_0+b]).$$

All intervals in this use lie in $Q$; no differentiability in $t$ has been assumed. Put

$$h=\min\left\{a,\frac b{M+1},\frac1{2(L+1)}\right\}>0,\quad I=[t_0-h,t_0+h],\quad q=hL<\frac12.$$

Then $hM<b$. The added ones avoid division by zero when $M=0$ or $L=0$.

A complete space of permitted functions. Let $X=\{u\in C(I):\|u-y_0\|_\infty\le b\}$, where $y_0$ denotes the constant function. If $(u_n)$ is Cauchy in this distance, uniform completeness gives a uniform limit $u$. The uniform-limit continuity theorem makes $u$ continuous, and passage to the limit in $|u_n(t)-y_0|\le b$ gives $|u(t)-y_0|\le b$ for every $t$. Thus $u\in X$ and $X$ is complete.

The integral map. Define

$$(Tu)(t)=y_0+\int_{t_0}^{t}f(s,u(s))\,ds.$$

The integrand is continuous, since the graph of $u$ stays in $Q$. Thus $Tu$ is continuous, and

$$\|Tu-y_0\|_\infty\le hM<b,\qquad\|Tu-Tv\|_\infty\le hL\|u-v\|_\infty=q\|u-v\|_\infty.$$

For $t<t_0$ these bounds follow by reversing the integral limits before taking absolute values. Therefore $T$ maps $X$ into itself and contracts distances.

Convergence and a fixed point. Start with $u_0(t)=y_0$ and set $u_{n+1}=Tu_n$. If $q=0$, the distance estimate says $Tu=Tv$ for all $u,v\in X$; hence $Tu_1=Tu_0=u_1$, already a fixed point. Suppose $0<q<1$ and put $d=\|u_1-u_0\|_\infty$. Repeated application of the contraction estimate gives $\|u_{j+1}-u_j\|_\infty\le q^jd$ for every integer $j\ge0$. Consequently, for $p\ge1$,

$$\|u_{n+p}-u_n\|_\infty\le\sum_{j=n}^{n+p-1}q^jd\le\frac{q^n d}{1-q}\longrightarrow0\quad(n\to\infty).$$

Completeness gives $u_n\to u\in X$. Since

$$\|Tu-u\|_\infty\le q\|u-u_n\|_\infty+\|u_{n+1}-u\|_\infty\longrightarrow0,$$

we have $Tu=u$. Passing $p\to\infty$ above also proves $\|u-u_n\|_\infty\le q^nd/(1-q)$. For $q=0$, $u_1$ is exact instead.

The differential equation and uniqueness. The fixed-point equation and the fundamental theorem of calculus give $u'(t)=f(t,u(t))$ for interior $t\in I$, and $u(t_0)=y_0$. Restricting to the interior of $I$ supplies the claimed local solution. If $u,v\in X$ are both fixed points, $\|u-v\|_\infty\le q\|u-v\|_\infty$ and $q<1$ force equality of the functions.

Finally, any other local solution is continuous at $t_0$ and has value $y_0$, so its graph stays in $Q$ on a sufficiently small common closed interval $J=[t_0-\eta,t_0+\eta]\subset I$, with $\eta>0$. The differential equation integrated from $t_0$ gives the same integral equation there. On $J$ the contraction estimate has factor $\eta L\le q<1$, so it agrees with $u$ on $J$. This establishes local uniqueness without assuming an alternative solution remains in $Q$ for all future time. $\square$

Local existence does not assert global continuation. A later domain boundary or unbounded growth requires a separate continuation argument. The derivative condition is sufficient, not a necessary condition for every uniquely solvable equation.

5 Graded practice

E1 Foundation. Solve $y'=3t^2y$ with $y(0)=2$, and identify the zero solution separately.

E2 Branch and interval. Solve $y'=y^2$, $y(0)=-1$, and give the maximal interval containing zero on which the formula is defined.

E3 Integrating factor. Solve $y'+2y=2t$ with $y(0)=1$.

E4 Motion. Solve $v'+2v=6$, $v(0)=0$, then find position with $x(0)=0$.

6 Full solutions

S1. Separation for nonzero $y$ gives $\log|y|=t^3+C$, so the specified solution is $y=2e^{t^3}$. The constant zero solution exists but does not satisfy this initial value. The displayed solution is defined for every real $t$.

S2. The solution is $y=-1/(1+t)$. Its maximal interval containing zero is $(-1,\infty)$; the denominator vanishes at $t=-1$. The branch on $(-\infty,-1)$ is not part of the same initial-value interval.

S3. A particular solution is $t-1/2$, and the homogeneous solution is $Ce^{-2t}$. The initial value gives $C=3/2$, hence $y=t-1/2+(3/2)e^{-2t}$. Substituting gives the right-hand side $2t$.

S4. Velocity is $v=3(1-e^{-2t})$. Integrating from zero gives $x=3t-(3/2)(1-e^{-2t})$. Differentiating recovers velocity, and both initial conditions hold.

Summary

Separation solves product-form rate laws on appropriate nonzero branches, while equilibrium solutions must be checked first. Integrating factors solve linear equations without dividing by the unknown. Initial conditions, uniqueness hypotheses, coefficient domains and maximal intervals distinguish a complete solution from an isolated antiderivative calculation.