4. Differential Equations and Scientific Balances

Learning goals and preparation

We assume derivatives, the chain/product rules, and Chapter 3's substitution and integration by parts. A differential equation describes how an unknown function changes; an initial condition specifies where its trajectory starts. You will solve elementary separable and linear equations, construct conservation equations, and verify intervals and equilibrium cases.

The scientific examples are hypothetical models. A solution of a model is not evidence that its assumptions hold indefinitely; for example, a perfectly mixed tank model must stop when a tank overflows or its volume ceases to be positive.

4.1 Solutions, initial conditions, and intervals

A first-order ordinary differential equation is $y'=f(t,y)$. A differentiable function solves it on an interval when substituting the function and its derivative makes the equation true at every input there. An initial-value problem adds $y(t_0)=y_0$.

For $y'=ky$, differentiating $Ce^{kt}$ verifies a family of solutions; the initial value fixes $C$. Under the usual local existence-and-uniqueness condition that $f$ and $f_y$ are continuous near an initial point, one obtains a unique local solution. “Local” does not promise a finite solution for every future time.

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.

Worked example 1: a finite interval matters. The problem $y'=y^2$, $y(0)=1$ has solution $y=1/(1-t)$, since its derivative is $1/(1-t)^2=y^2$. It has a pole at one; the maximal connected interval containing zero is $(-\infty,1)$. A formula's algebraic existence on the separate interval $(1,\infty)$ does not create one solution through the singular time.

4.2 Separation and equilibrium checks

For $y'=g(t)h(y)$, where $h(y)\ne0$ along the solution interval,

$$\int\frac{dy}{h(y)}=\int g(t)dt.$$

This notation abbreviates an application of the chain rule to an antiderivative of $1/h$. Before dividing, check all zeros of $h$ for constant equilibrium solutions. They may be lost by separation.

Worked example 2: an implicit solution is still a solution. Solve $y'=t/(1+y^2)$, $y(0)=1$. Multiplying and integrating gives

$$y+\frac{y^3}{3}=\frac{t^2}{2}+\frac43.$$

The function $H(y)=y+y^3/3$ is strictly increasing on $\mathbb R$ because $H'=1+y^2>0$, and its range is all real numbers. Thus the relation defines one real $y$ for each $t$. Differentiating the relation recovers $(1+y^2)y'=t$, and the initial condition checks. A radical or cubic formula is unnecessary for a rigorous implicit answer.

Worked example 3: bounded population growth. The model

$$N'=rN(1-N/K),\qquad r,K>0$$

has equilibria $N=0,K$. For a positive initial value $N_0$, separation gives the verified solution

$$N(t)=\frac{K}{1+(K/N_0-1)e^{-rt}}.$$

For $t\ge0$ its denominator is positive. If $0<N_0<K$, growth is positive and the population tends to $K$. If $N_0>K$, growth is negative and the population decreases toward $K$. The zero equilibrium is treated separately. The parameter $K$ is a model's limiting scale, not an empirically guaranteed future population.

4.3 Linear equations and integrating factors

For a linear equation $y'+p(t)y=q(t)$ with continuous coefficients, choose

$$\mu(t)=\exp\left(\int p(t)dt\right).$$

Then $\mu'=p\mu$ and the product rule turns the equation into $(\mu y)'=\mu q$. With initial value at $t_0$,

$$y(t)=\frac{\mu(t_0)y_0+\int_{t_0}^t\mu(s)q(s)ds}{\mu(t)}.$$

The integrating factor is nonzero; multiplying it by a constant does not change the solution formula. The formula's interval must retain continuity of coefficients.

Worked example 4: cooling toward an ambient level. Suppose $T'=-k(T-T_a)$, $k>0$, with constant ambient $T_a$. Setting $u=T-T_a$ gives $u'=-ku$, hence

$$T(t)=T_a+(T_0-T_a)e^{-kt}.$$

If $T_a=20$, $T_0=80$, and $T(10)=50$, then $e^{-10k}=1/2$ and $k=\ln2/10$. Therefore $T(20)=20+60/4=35$. The model halves the difference from ambient every ten time units, not the absolute temperature.

4.4 Constant-volume mixing

A mass balance is “rate in minus rate out.” Assume instantaneous perfect mixing, no chemical reaction, additive volumes, and a specified inflow concentration. If volume $V$ is constant, inflow and outflow are both $q>0$, inflow concentration is $c_{in}$, and salt mass is $S$, then tank concentration is $S/V$ and

$$S'=q c_{in}-q S/V.$$

Every term has mass-per-time units. The outflow concentration equals the tank concentration only because perfect mixing was assumed.

Worked example 5: a constant tank. Let $V=100$ L, $q=2$ L/min, $c_{in}=3$ g/L, and $S(0)=100$ g. Then

$$S'+0.02S=6,\qquad S(t)=300-200e^{-0.02t}.$$

The concentration is $3-2e^{-0.02t}$ g/L. It reaches $2$ g/L when $e^{-0.02t}=1/2$, namely at $50\ln2$ minutes. The concentration remains between its initial and incoming levels, a useful independent check. With no incoming salt, the same model instead gives exponential washout.

4.5 Changing-volume mixing

If inflow and outflow differ, the volume changes and must not be replaced by a constant. Let $V'=q_{in}-q_{out}$ and

$$S'=q_{in}c_{in}-q_{out}S/V(t).$$

Use the model only while volume is positive and within capacity.

Worked example 6: filling as well as mixing. A tank starts with $100$ L of pure water, receives $3$ L/min at concentration $2$ g/L, and drains $2$ L/min. Before overflow, $V(t)=100+t$ and

$$S'+\frac2{100+t}S=6,\qquad S(0)=0.$$

The integrating factor is $(100+t)^2$, so

$$[(100+t)^2S]'=6(100+t)^2.$$

Integrating gives $S=2(100+t)+C/(100+t)^2$. The initial value gives $C=-2\cdot100^3$. Thus concentration is

$$\frac{S}{V}=2\left(1-\left(\frac{100}{100+t}\right)^3\right).$$

It starts at zero, is nonnegative for $t\ge0$, and remains below the incoming concentration. Although the formula has an infinite-time limit, a finite-capacity tank would overflow first; its physical use must stop at that capacity time.

4.6 Checking a model solution

A dependable solution report has four checks:

1. Substitute into the differential equation, including units.
2. Verify the initial condition.
3. State an interval avoiding denominator zeros or coefficient singularities.
4. Check model bounds, such as nonnegative mass, feasible volume, or a temperature range.

An exact equation solution may fail a modeling check. Conversely, an implicit formula can be fully valid without an elementary explicit expression. A numerical graph alone does not replace any of the four checks.

4.7 Graded exercises

Foundation

1. Solve $y'=-0.5y$, $y(0)=8$, and find the time at which $y=2$.
2. List equilibria of $y'=y(3-y)$ before any division.
3. Verify $y=t-1+2e^{-t}$ solves $y'+y=t$ and find its initial value.

Core

4. Solve $y'=ty$, $y(0)=2$, including the zero equilibrium discussion.
5. Solve $y'+3y=12$, $y(0)=1$.
6. Under the cooling model with ambient $10$ and initial temperature $50$, the temperature difference halves every five time units. Find $T(t)$ and $T(10)$.
7. A $50$ L perfectly mixed tank receives and drains $1$ L/min; incoming concentration is $4$ g/L and initial salt mass is zero. Find salt mass and concentration.
8. For logistic parameters $K=1000$, $r=0.1$, $N_0=100$, find the time at which $N=500$.

Synthesis

9. In Worked example 6, if tank capacity is $150$ L, state the valid future time range before overflow and find the concentration at the capacity boundary.
10. Explain why replacing the outflow concentration in a perfectly mixed tank by the fixed incoming concentration is generally wrong. Use a pure-water initial condition with positive incoming concentration to show an immediate contradiction.

4.8 Full solutions

1. $y=8e^{-0.5t}$. Setting it to two gives $e^{-0.5t}=1/4$, so $t=2\ln4$.
2. The constant solutions satisfy $y(3-y)=0$, giving $y=0,3$.
3. $y'=1-2e^{-t}$, so $y'+y=t$. At zero, $y(0)=-1+2=1$.
4. Away from zero, separation yields $\ln|y|=t^2/2+C$, giving $y=2e^{t^2/2}$ for the initial value. The zero function also solves the equation but not this initial condition. The displayed positive solution is defined for all real $t$.
5. The equilibrium is four and the initial deviation is negative three, so $y=4-3e^{-3t}$. Differentiating and adding $3y$ gives twelve, and the initial value is one.
6. The difference is $40e^{-kt}$ with $k=\ln2/5$. Thus $T(t)=10+40\,2^{-t/5}$ and $T(10)=20$.
7. $S'=4-S/50$, $S(0)=0$, so $S=200(1-e^{-t/50})$ g. Concentration is $4(1-e^{-t/50})$ g/L.
8. $N=1000/(1+9e^{-0.1t})$. Setting $N=500$ gives $1+9e^{-0.1t}=2$, so $t=10\ln9$ in the model's time unit.
9. Volume is $100+t$, so before overflow use $0\le t<50$ minutes. At the capacity boundary $t=50$, concentration is $2[1-(2/3)^3]=38/27$ g/L. The limiting value at that boundary is valid under the pre-overflow mixing assumptions; the same differential model cannot simply continue past overflow without an updated outflow rule.
10. With pure water initially, tank concentration at time zero is zero, so salt outflow must initially be zero. Assigning the positive incoming concentration to the outgoing stream predicts salt leaving before any is present. Perfect mixing requires outflow concentration $S(t)/V(t)$, which evolves from zero; it does not set it equal to the incoming value.

Summary and onward map

Differential equations specify change, and initial values select trajectories. Separation needs equilibrium checks; linear equations admit an integrating-factor derivation. Conservation supplies the correct inputs and outputs in mixing models, with changing volume handled explicitly. A solution must satisfy its equation, initial condition, mathematical interval and physical constraints.