First order ODEs and environmental balance

A differential equation connects a state to its rate of change. A balance model starts from a simple accounting principle: rate of stored amount equals input minus output plus internal production. Writing that principle before selecting a solution formula prevents unit errors and missing volume factors.

1 States initial conditions and existence

A first-order ODE has form $y'=F(t,y)$. A solution is a differentiable function satisfying it on an interval. An initial condition $y(t_0)=y_0$ selects a trajectory when uniqueness holds.

A useful local theorem is: if $F$ is continuous near $(t_0,y_0)$ and locally Lipschitz in $y$ there, then a unique local solution exists. Continuous $\partial F/\partial y$ is a sufficient Lipschitz condition on a smaller rectangle by the mean value theorem. This is a local result; solutions can still blow up in finite time, as $y'=y^2$, $y(0)=1$, with solution $1/(1-t)$ shows. A full proof uses contraction of the integral map $y\mapsto y_0+\int_{t_0}^tF(s,y(s))ds$ on a short interval: boundedness keeps it in a closed function ball, and the Lipschitz constant times interval length makes it a contraction. Uniform completeness and geometric error bounds give the unique fixed point, which the fundamental theorem turns into the solution.

2 Linear equations and the integrating factor

For $y'+p(t)y=q(t)$ with continuous coefficients, define $\mu(t)=\exp(\int_{t_0}^tp(s)ds)$. It is positive and $\mu'=p\mu$. The product rule gives $(\mu y)'=\mu q$, hence

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

This derives the formula, proves existence and proves uniqueness directly for linear equations. Homogeneous solutions are multiplied by the decaying or growing factor $\mu^{-1}$.

3 A well-mixed reservoir

A reservoir has constant positive volume $V>0$, input and output volume rate $Q$, incoming concentration $c_{\rm in}$, and first-order removal rate $k\ge0$. Let $M(t)$ be stored contaminant mass and $c=M/V$. Assuming perfect mixing, output mass rate is $Qc$. The balance is

$$M'=Qc_{\rm in}-Q\frac MV-kM.$$

Equivalently,

$$c'=\frac QV c_{\rm in}-\left(\frac QV+k\right)c.$$

If $Q>0$ and the inputs are constant, the equilibrium is $c_*=Qc_{\rm in}/(Q+kV)$ and

$$c(t)=c_*+(c_0-c_*)e^{-(Q/V+k)t}.$$

The exponential rate has units inverse time. The solution approaches $c_*$, not necessarily $c_{\rm in}$ because internal removal may persist. If the volume varies, the identity $M=Vc$ differentiates to $M'=Vc'+V'c$; using the constant-volume concentration equation would then omit a term.

For $V=100$, $Q=10$, $k=0.05$ in consistent units and $c_{\rm in}=6$, the equilibrium is 4 and $c(t)=4+(c_0-4)e^{-0.15t}$. With initially clean water, half the equilibrium level is reached at $t=(\log2)/0.15$.

4 Separable equations and equilibrium caution

For $y'=a y(1-y/K)$ with $a,K>0$, nonzero nonequilibrium solutions separate:

$$\frac{dy}{y(1-y/K)}=a\,dt.$$

Integrating partial fractions and applying a positive initial value gives

$$y(t)=\frac K{1+(K/y_0-1)e^{-at}}.$$

For positive initial value, this formula holds on the maximal interval containing $t=0$ on which its denominator is nonzero. It is valid for every $t\ge0$; if $y_0>K$, its extension into the past has a finite-time pole and is not a solution through that pole. The equilibria $y=0$ and $y=K$ must be recorded separately before dividing by $y(1-y/K)$. For $0<y_0<K$, the solution increases toward $K$. Such a logistic model is a specified phenomenological assumption, not a universal ecological law.

The uniform Lipschitz theorem, including finite real systems

Here is the precise meaning and proof of the local Lipschitz assumption used above. Let $d\ge1$ be finite, let $U\subset\mathbb R\times\mathbb R^d$ be open, and let $F:U\to\mathbb R^d$ be continuous. Around $(t_0,y_0)$ assume there are $a,b>0$ and $L\ge0$ such that the closed cylinder $Q=[t_0-a,t_0+a]\times\{y:\|y-y_0\|\le b\}$ is contained in $U$, and

$$\|F(t,y)-F(t,z)\|\le L\|y-z\|\quad((t,y),(t,z)\in Q).$$

The same $L$ works for every time in the rectangle. Continuity of the derivative with respect to $y$ is one sufficient way to obtain this bound by integrating along the segment in the closed ball, but existence of that derivative is not required in the theorem.

Continuity and compactness give $\|F\|\le M$ on $Q$. Set $h=\min(a,b/(M+1),1/[2(L+1)])>0$, $I=[t_0-h,t_0+h]$, and $q=hL<1/2$. Consider the continuous functions $u:I\to\mathbb R^d$ with $\sup_I\|u-y_0\|\le b$, in the supremum distance. This space is complete. Indeed, a uniformly Cauchy sequence is pointwise Cauchy in each of its finitely many real coordinates; real completeness provides pointwise limits. Passing to the limit in a common Cauchy bound gives uniform convergence in the vector norm. The limit is continuous by the three-term estimate using a fixed continuous member of the sequence, and the closed value bound passes to the limit. This proves completeness of this particular space rather than assuming it for arbitrary targets.

Define $(Tu)(t)=y_0+\int_{t_0}^tF(s,u(s))\,ds$, integrating components. Then

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

For times before $t_0$, reversing the integration limits gives the same estimates. Start at the constant function $u_0=y_0$ and iterate $u_{n+1}=Tu_n$. If $q=0$, $T$ is constant, so $u_1$ is its fixed point. If $0<q<1$, let $D=\|u_1-u_0\|_\infty$. Induction gives $\|u_{j+1}-u_j\|_\infty\le q^jD$. Hence for $p\ge1$,

$$\|u_{n+p}-u_n\|_\infty\le\frac{q^nD}{1-q}\longrightarrow0.$$

Completeness gives $u_n\to u$. The estimate $\|Tu-u\|_\infty\le q\|u-u_n\|_\infty+\|u_{n+1}-u\|_\infty$ makes $Tu=u$, and letting $p\to\infty$ gives the same geometric error bound to the limit. The fundamental theorem in each coordinate gives $u'=F(t,u)$ on the interior of $I$ and $u(t_0)=y_0$.

Two fixed points satisfy $\|u-v\|_\infty\le q\|u-v\|_\infty$, hence coincide. Any other local solution has its graph in $Q$ on a sufficiently small common closed interval about $t_0$, by continuity at its initial value. Integrating its equation there gives the same fixed-point equation, now with contraction factor at most $q$, so it agrees with $u$. This proves local existence and local uniqueness for the stated uniform Lipschitz condition, including its scalar case. No global existence or nonlinear stability conclusion has been added.

For a physical reservoir interpretation use forward time $t\ge0$, nonnegative flow rates and incoming/initial concentrations, and positive volume. The integrating-factor integral then proves nonnegativity while those hypotheses hold. If volume changes, retain its positive-time interval and the $V'c$ term in the mass derivative.

5 Graded practice

E1 Foundation. Solve $y'+2y=6$ with $y(0)=1$ and verify directly.

E2 Balance. A constant-volume lake has $V=200$, $Q=20$, no internal removal and clean inflow. If $c(0)=5$, find $c(t)$ and its half-life.

E3 Forcing. Solve $y'+y=t$ with $y(0)=0$.

E4 Modelling. A tank has inflow $Q_{\rm in}$ and outflow $Q_{\rm out}$, with incoming concentration $c_{\rm in}$ and no reaction. Derive the concentration equation while volume remains positive.

6 Full solutions

S1. $y=3-2e^{-2t}$. Then $y'=4e^{-2t}$ and $y'+2y=6$; its initial value is 1.

S2. $c'=-0.1c$, so $c(t)=5e^{-0.1t}$ and half-life is $10\log2$. The removal parameter is $Q/V$, not the bare flow rate 20.

S3. The integrating factor is $e^t$. Since $\int_0^tse^sds=(t-1)e^t+1$, $y=t-1+e^{-t}$. Substitution and $y(0)=0$ verify it.

S4. $V'=Q_{\rm in}-Q_{\rm out}$ and $M'=Q_{\rm in}c_{\rm in}-Q_{\rm out}c$. Combining $M'=Vc'+V'c$ yields $Vc'=Q_{\rm in}(c_{\rm in}-c)$. Outflow cancels explicitly from this concentration equation but still affects $V(t)$. Perfect mixing and the stated rates remain assumptions.

Summary

Write a mass balance before solving an ODE. Integrating factors solve linear equations with controlled hypotheses; separation must not discard equilibria. Initial conditions, time units, mixing assumptions and volume evolution are part of the model, not optional interpretation after the calculation.