1   Flow-in, Flow-out Example

Example (B&D). At $t=0$ a tank contains $Q_0$ lb of salt dissolved in 100 gal of water. Salted water with concentration $\tfrac{1}{4}$ lb/gal enters at rate $r$ gal/min; the well-stirred mixture drains at the same rate. Find:

1. The amount of salt $Q(t)$ at any time.
2. The limiting amount $Q_L$ after a very long time.
3. If $r=3$ and $Q_0 = 2Q_L$, the time $T$ for salt to be within 2% of $Q_L$.
4. The flow rate needed so that $T \leq 45$ min.

Since we are dealing with DEs, the key observation is to express every rate of change relevant to the tank flow:

$$\frac{dQ}{dt} = \text{(rate of salt in)} - \text{(rate of salt out)}.$$

We determine:

• Salt entering per minute: $\;\dfrac{1}{4}\;\text{lb/gal} \times r\;\text{gal/min} = \dfrac{r}{4}\;\text{lb/min}$
• Salt draining per minute: $\;r\;\text{gal/min} \times \dfrac{Q}{100}\;\text{lb/gal} = \dfrac{rQ}{100}\;\text{lb/min}$

Hence the IVP is:

$$\frac{dQ}{dt} = \frac{r}{4} - \frac{rQ}{100} = \frac{25r - rQ}{100}, \qquad Q(0) = Q_0.$$

(i) Separation of variables.

$$\int \frac{dQ}{Q - 25} = \int \frac{-r}{100}\,dt.$$

Hence

$$\ln|Q - 25| = -\frac{rt}{100} + c.$$

So $Q - 25 = Ce^{-rt/100}$ where $C = e^{\pm c}$. Therefore

$$Q(t) = 25 + Ce^{-rt/100}.$$

The initial condition $Q(0) = Q_0$ gives $C = Q_0 - 25$:

(ii) As $t \to \infty$, $\;Q(t) \to 25 = Q_L$.

(iii) With $Q_0 = 50 = 2Q_L$ and $r = 3$. We need $Q(T)$ within 2% of $Q_L = 25$, i.e. $Q(T) = 25.5$:

$$25.5 = 25 + (50 - 25)e^{-3T/100} = 25 + 25\,e^{-T/50}.$$

So $e^{-T/50} = 0.5/25 = 0.02$, giving

$$T = \frac{\ln 50}{0.03} \approx 130.4 \text{ min.}$$

(iv) For $T \leq 45$ min:

$$25.5 = 25 + 25\,e^{-45r/100} \implies r = \frac{100}{45}\ln 50 \approx 8.69 \text{ gal/min.}$$

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1.1   Oscillating Concentration (B&D §2.3, Example 3)

If the salt concentration flowing into the tank varies as $\gamma(t) = 2 + \sin 2t$ g/L and drains at the same rate, the DE becomes

$$\frac{dq}{dt} + \frac{1}{2}q = 10 + 5\sin 2t,$$

which has the solution

$$q(t) = 20 - \frac{10}{17}\cos 2t + \frac{40}{17}\sin 2t + ce^{-t/2}.$$

Notice the solution also oscillates as the source that flows in. See Boyce & DiPrima for details.

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2   Newton's Law of Cooling

Newton published his finding in a short article in the Philosophical Transactions of the Royal Society anonymously in 1701 (in Latin). It states that the temperature of an object changes at a rate proportional to the difference between the object's temperature and its surroundings.

The "law" is valid provided the temperature difference is small — Newton was well aware of this limitation.

3   Continuous Compounding with Deposit/Withdrawal

Discrete compounding $m$ times over time $t$ at rate $r$: $\;S_0\!\left(1 + \dfrac{r}{m}\right)^{mt}$. Taking $m \to \infty$:

$$S(t) = \lim_{m\to\infty} S_0\left(1 + \frac{r}{m}\right)^{mt} = S_0\,e^{rt}.$$

Example. Continuous compounding means the IVP $\;dS/dt = rS,\; S(0) = S_0$, with unique solution $S(t) = S_0 e^{rt}$.

4   Autonomous (Logistic) Equations

An autonomous DE has the form $\dfrac{dy}{dt} = f(y)$, where $f$ depends on $y$ only — no $t$ appears explicitly. The simplest case $y' = ry$ describes exponential growth, but this is not realistic. In population growth, although initial behaviour is exponential, growth is restricted by physical conditions (food, space, etc.).

A more realistic model assumes $f(y) = y\,h(y)$ where $h(y) \approx r$ when $y$ is small. The simplest such example is $h(y) = r - ay$, giving:

This equation is nonlinear but separable. Sign analysis:

$$\frac{dy}{dt} \begin{cases} > 0 & \text{if } 0 < y < K, \\ = 0 & \text{if } y = 0 \text{ or } y = K, \\ < 0 & \text{if } y > K. \end{cases}$$

The solutions $y(t) = 0$ and $y(t) = K$ are equilibrium solutions.

4.1   Case Analysis

Example (Logistic equation). By separation of variables and partial fractions:

$$rt + c = \int r\,dt = \int \frac{dy}{(1 - y/K)\,y} = \int \frac{1}{y}\,dy + \int \frac{1/K}{1 - y/K}\,dy = \ln|y| - \ln|1 - y/K|.$$

Case I: $0 < y_0 < K$.

$$rt + c = \ln\frac{y}{1 - y/K}, \qquad\text{so}\qquad \frac{y}{1 - y/K} = Ce^{rt}, \quad C = e^c.$$

The initial condition $y(0) = y_0$ gives $C = \dfrac{y_0}{1 - y_0/K}$. Solving for $y$:

Case II: $y_0 = 0$ or $K$. These are the equilibrium solutions.

Case III: $y_0 > K$. The same formula from Case I remains valid. The only difference is that now $y(t) \to K$ from above as $t \to \infty$ (since $K - y_0 < 0$, the exponential term is negative and decaying).

4.2   Concavity

Using the chain rule:

$$\frac{d^2y}{dt^2} = \frac{d}{dt}f(y) = f'(y)\frac{dy}{dt} = f(y)\,f'(y).$$

Computing explicitly:

$$\frac{d^2y}{dt^2} = r^2 y\!\left(1 - \frac{y}{K}\right)\!\left(1 - \frac{2y}{K}\right).$$

Hence:

$$\frac{d^2y}{dt^2} \begin{cases} > 0 & \text{if } 0 < y < K/2 \text{ (concave up)}, \\ = 0 & \text{if } y = K/2 \text{ (inflection point)}, \\ < 0 & \text{if } K/2 < y < K \text{ (concave down)}. \end{cases}$$

The solution $y = K$ is stable and $y = 0$ is unstable. The inflection point occurs at $y = K/2$.

4.3   Modified Logistic Equation

Consider the equation with the opposite sign:

$$\frac{dy}{dt} = -r\left(1 - \frac{y}{T}\right)y.$$

The sign analysis is reversed:

$$\frac{dy}{dt} \begin{cases} < 0 & \text{if } 0 < y < T, \\ = 0 & \text{if } y = 0 \text{ or } y = T, \\ > 0 & \text{if } y > T. \end{cases}$$

A similar procedure gives for $y_0 < T$:

$$y(t) = \frac{y_0 T}{y_0 + (T - y_0)e^{rt}} \to 0 \quad\text{as } t\to\infty.$$

If $y_0 > T$, then there exists a finite time $t^*$ at which the solution blows up:

$$y_0 + (T - y_0)e^{rt^*} = 0 \implies t^* = \frac{1}{r}\ln\frac{y_0}{y_0 - T} > 0.$$

In fact $y(t^*) = +\infty$. The concavity analysis gives the same structure as the original logistic:

$$\frac{d^2y}{dt^2} = r^2 y\!\left(1 - \frac{y}{T}\right)\!\left(1 - \frac{2y}{T}\right).$$

5   Existence and Uniqueness

We have seen solution curves with different initial values that do not intersect. The following theorem justifies this.

5.1   Example

Consider the IVP

$$\frac{dy}{dx} = \frac{3x^2 + 4x + 2}{2(y-1)}, \qquad y(0) = -1.$$

We identify

$$f(x,y) = \frac{3x^2 + 4x + 2}{2(y-1)}, \qquad \frac{\partial f}{\partial y} = -\frac{3x^2 + 4x + 2}{2(y-1)^2}.$$

The $\partial f/\partial y$ is continuous everywhere except $y = 1$. Since our initial condition is $y_0 = -1 \neq 1$, there is an interval about $y = -1$ (away from $y=1$) in which the IVP has a unique solution:

$$y = 1 - \sqrt{x^3 + 2x^2 + 2x}.$$

5.2   Counter-example

Consider the IVP

$$\frac{dy}{dx} = y^{1/3}, \qquad y(0) = 0.$$

Here $f(x,y) = y^{1/3}$ and $\dfrac{\partial f}{\partial y} = \dfrac{1}{3}y^{-2/3}$. Although $f$ is continuous everywhere, $\partial f/\partial y$ is not continuous at $y = 0$ — which is exactly where the initial condition lies. So no conclusion can be drawn about uniqueness.

The DE is separable:

$$\frac{3}{2}y^{2/3} = x + c.$$

The condition $y(0) = 0$ gives $c = 0$. So

$$\phi_1(x) = \left(\frac{2}{3}x\right)^{3/2}, \qquad x \geq 0$$

is a solution. But so is

$$\phi_2(x) = -\left(\frac{2}{3}x\right)^{3/2}, \qquad x \geq 0$$

and even $\psi(x) \equiv 0$. One can construct infinitely many solutions:

$$\chi(x) = \begin{cases} 0, & 0 \leq x < x_0, \\ \pm\!\left(\frac{2}{3}(x - x_0)\right)^{3/2}, & x \geq x_0. \end{cases}$$

One can verify that $\chi$ is continuous and differentiable even at $x_0$.

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— End of Modelling Notes —