Chapter 7 · Section 7.2

7.2 Applications of Linear Systems

Systems of differential equations are not an artificial exercise; they are what you get whenever a device has two or more parts that influence each other. In this section we build three such models from basic physical laws and solve them with the operator method of Section 7.1. The three models are the coupled mass–spring system, the two-loop electric network, and two interconnected tanks. They look different, but the modelling steps are the same each time, and so is the mathematics behind them.

Modelling in a nutshell

Choose the unknowns, one for each part of the device (a displacement for each mass, a current for each loop, an amount of salt for each tank), and fix a sign convention.
Write one balance law per part: Newton's second law for each mass, Kirchhoff's voltage law for each loop, "rate in $-$ rate out" for each tank.
Look for the coupling terms. They always involve the unknowns of two parts: the stretch $x_{2} - x_{1}$ of a spring between two masses, the current $i_{1} - i_{2}$ in a shared branch, the flow from one tank into the other.
Write the system in operator form and compute $Δ$ . Its order tells you how many initial conditions the model needs.
Solve, apply the initial conditions, and check the answer against physical common sense: steady states, conservation laws, signs.

Coupled masses and springs

Two bodies of masses $m_{1}$ and $m_{2}$ slide without friction on a horizontal track. The left body is attached to a wall by a spring of stiffness $k_{1}$ , the two bodies are joined by a spring of stiffness $k_{2}$ , and the right body is attached to a second wall by a spring of stiffness $k_{3}$ (Figure 7.2.1). In the equilibrium position no spring exerts a net force on either body.

Figure 7.2.1. Two masses coupled by three springs. The displacements

x_{1} (t)

x_{2} (t)

are measured from the equilibrium positions, positive to the right. Only the middle spring couples the motions.

Forces on each body

Let $x_{1} (t)$ and $x_{2} (t)$ be the displacements of the bodies from equilibrium, positive to the right. By Hooke's law a spring that is stretched by an amount $s$ pulls on each of its ends with force $k s$ (a negative $s$ means compression and a push).

Left spring. It is stretched by $x_{1}$ and pulls body 1 to the left: force $- k_{1} x_{1}$ .
Middle spring. Its length changes by $x_{2} - x_{1}$ . If $x_{2} > x_{1}$ it is stretched and pulls body 1 to the right and body 2 to the left. So it exerts $+ k_{2} (x_{2} - x_{1})$ on body 1 and $- k_{2} (x_{2} - x_{1})$ on body 2. These formulas stay correct when $x_{2} < x_{1}$ , because then the sign of $x_{2} - x_{1}$ flips the directions automatically.
Right spring. If body 2 moves right by $x_{2}$ , the right spring is compressed by $x_{2}$ and pushes body 2 to the left: force $- k_{3} x_{2}$ .

Newton's second law for each body gives the system

m_{1} x_{1}^{''} m_{2} x_{2}^{''} = - k_{1} x_{1} + k_{2} (x_{2} - x_{1}), = - k_{2} (x_{2} - x_{1}) - k_{3} x_{2} .

In operator form:

(m_{1} D^{2} + k_{1} + k_{2}) x_{1} - k_{2} x_{2} - k_{2} x_{1} + (m_{2} D^{2} + k_{2} + k_{3}) x_{2} = 0, = 0.

The operator determinant $Δ = (m_{1} D^{2} + k_{1} + k_{2}) (m_{2} D^{2} + k_{2} + k_{3}) - k_{2}^{2}$ has order $4$ , so the general solution contains four constants. That matches the physics: we must give both initial positions and both initial velocities. If the middle spring is removed ( $k_{2} = 0$ ), the system splits into two independent oscillators of Chapter 5.

Example 7.2.1 (Energy that moves back and forth)

Take $m_{1} = m_{2} = 2$ kg, $k_{1} = k_{3} = 8$ N/m and a weak coupling spring $k_{2} = 1$ N/m. At $t = 0$ body 1 is pulled $10$ cm to the right, body 2 is held at its equilibrium position, and both are released from rest. Find $x_{1} (t)$ and $x_{2} (t)$ (in cm).

Solution

Equations. With the given numbers, $k_{1} + k_{2} = k_{2} + k_{3} = 9$ : $(2 D^{2} + 9) x_{1} - x_{2} = 0, - x_{1} + (2 D^{2} + 9) x_{2} = 0.$ Because the equations are linear and homogeneous we may measure displacements in centimetres.
Determinant. It is a difference of squares: $Δ = (2 D^{2} + 9)^{2} - 1 = (2 D^{2} + 9 - 1) (2 D^{2} + 9 + 1) = (2 D^{2} + 8) (2 D^{2} + 10) = 4 (D^{2} + 4) (D^{2} + 5) .$ The characteristic roots are $\pm 2 i$ and $\pm 5 i$ .
Solve for $x_{1}$ . $Δ x_{1} = 0$ , so $x_{1} = c_{1} cos 2 t + c_{2} sin 2 t + c_{3} cos 5 t + c_{4} sin 5 t .$
Shortcut for $x_{2}$ . The first equation contains no derivative of $x_{2}$ , so it gives $x_{2}$ directly: $x_{2} = 2 x_{1}^{''} + 9 x_{1}$ . On $cos 2 t$ and $sin 2 t$ the operator $2 D^{2} + 9$ multiplies by $2 (- 4) + 9 = 1$ ; on $cos 5 t$ and $sin 5 t$ by $2 (- 5) + 9 = - 1$ . Hence $x_{2} = c_{1} cos 2 t + c_{2} sin 2 t - c_{3} cos 5 t - c_{4} sin 5 t .$
Initial positions. $x_{1} (0) = c_{1} + c_{3} = 10$ and $x_{2} (0) = c_{1} - c_{3} = 0$ , so $c_{1} = c_{3} = 5$ .
Initial velocities. $x_{1}^{'} (0) = 2 c_{2} + 5 c_{4} = 0$ and $x_{2}^{'} (0) = 2 c_{2} - 5 c_{4} = 0$ , so $c_{2} = c_{4} = 0$ .
Answer. $x_{1} = 5 (cos 2 t + cos 5 t), x_{2} = 5 (cos 2 t - cos 5 t) .$
Interpretation. The sum-to-product identities rewrite the answer as $x_{1} = 10 cos (\frac{5 - 2}{2} t) cos (\frac{5 + 2}{2} t), x_{2} = 10 sin (\frac{5 - 2}{2} t) sin (\frac{5 + 2}{2} t) .$ Each body oscillates fast (angular frequency $\frac{5 + 2}{2} \approx 2.12$ ) inside a slowly varying envelope. The envelope of $x_{1}$ first vanishes when $\frac{5 - 2}{2} t = \frac{π}{2}$ , that is at $t = π / (5 - 2) \approx 13.3$ s. At that moment body 1 is almost at rest and body 2 swings with the full amplitude $10$ cm; then the motion moves back. See Figure 7.2.2.

Figure 7.2.2. Example 7.2.1. The weak middle spring slowly passes the motion from body 1 (top) to body 2 (bottom) and back. Dashed curves: the envelopes

\pm 10 cos (\frac{5 - 2}{2} t)

and

\pm 10 sin (\frac{5 - 2}{2} t)

Normal modes

The two frequencies in Example 7.2.1 have a simple meaning. If the bodies start with equal displacements, $x_{1} = x_{2}$ , the middle spring is never stretched, and each body simply oscillates on its outer spring with $ω^{2} = k_{1} / m_{1} = 4$ , i.e. $ω = 2$ . If they start with opposite displacements, $x_{2} = - x_{1}$ , the middle spring is stretched by $2∣ x_{1} ∣$ and adds the restoring force $- 2 k_{2} x_{1}$ on body 1; then $ω^{2} = (k_{1} + 2 k_{2}) / m_{1} = 5$ . These two special motions, in which both bodies move with one single frequency, are the normal modes of the system. Every other motion is a superposition of the two. You will see the same idea again in Section 7.4, where special solutions $x = A e^{λ t}$ , $y = B e^{λ t}$ play the role of normal modes.

Why engineers care

A multi-storey building, a car body on its suspension, or a turbine shaft with several discs behaves like a chain of masses and springs. Each normal mode has its own natural frequency, and a periodic force near any of them can cause resonance (Section 5.5). Finding normal frequencies is therefore a routine step in structural design.

A two-loop electric network

We now use Kirchhoff's laws from Section 5.6. The voltage drop across a resistor $R$ carrying current $i$ is $R i$ ; across an inductor $L$ it is $L d i / d t$ . Kirchhoff's voltage law says that around any closed loop the voltage drops add up to the applied voltage.

The network in Figure 7.2.3 has three branches between the junctions $P$ and $Q$ : the left branch with a source $E (t)$ and an inductor $L_{1}$ , the middle branch with a resistor $R_{1}$ , and the right branch with an inductor $L_{2}$ and a resistor $R_{2}$ . We use loop currents: $i_{1}$ circulates clockwise in the left loop and $i_{2}$ clockwise in the right loop. The middle branch belongs to both loops, so (by Kirchhoff's current law at $P$ ) it carries $i_{1} - i_{2}$ downward.

Left loop. Drops: $L_{1} i_{1}^{'}$ across the inductor and $R_{1} (i_{1} - i_{2})$ across the shared resistor. They balance the source: $L_{1} i_{1}^{'} + R_{1} (i_{1} - i_{2}) = E (t) .$
Right loop. Going clockwise we cross the middle branch upward, against the current $i_{1} - i_{2}$ , so its drop counts as $R_{1} (i_{2} - i_{1})$ . There is no source: $L_{2} i_{2}^{'} + R_{2} i_{2} + R_{1} (i_{2} - i_{1}) = 0.$

The outer loop gives the sum of these two equations, so it adds no new information. In operator form,

(L_{1} D + R_{1}) i_{1} - R_{1} i_{2} - R_{1} i_{1} + (L_{2} D + R_{1} + R_{2}) i_{2} = E (t), = 0.

$Δ$ has order $2$ : we need the two initial currents, one per inductor. (Current through an inductor cannot jump, which is why these are the natural initial data.)

Example 7.2.2

In the network of Figure 7.2.3 let $L_{1} = L_{2} = 1$ H, $R_{1} = 8 Ω$ , $R_{2} = 12 Ω$ and $E = 48$ V (a battery). The switch is closed at $t = 0$ , when no current flows. Find $i_{1} (t)$ , $i_{2} (t)$ and the current in the middle branch.

Solution

Equations. $(D + 8) i_{1} - 8 i_{2} = 48, - 8 i_{1} + (D + 20) i_{2} = 0.$
Determinant. $Δ = (D + 8) (D + 20) - 64 = D^{2} + 28 D + 96 = (D + 4) (D + 24) .$
Which unknown first? The second equation contains no derivative of $i_{1}$ , so once $i_{2}$ is known it gives $i_{1} = \frac{1}{8} (i_{2}^{'} + 20 i_{2})$ algebraically. So we solve for $i_{2}$ first.
Equation for $i_{2}$ . By (7.1.3), $Δ i_{2} = L_{1} f_{2} - L_{3} f_{1} = (D + 8) [0] - (- 8) [48] = 384$ , i.e. $i_{2}^{''} + 28 i_{2}^{'} + 96 i_{2} = 384.$ A constant particular solution is $384/96 = 4$ , so $i_{2} = k_{1} e^{- 4 t} + k_{2} e^{- 24 t} + 4.$
Find $i_{1}$ . The operator $\frac{1}{8} (D + 20)$ multiplies $e^{- 4 t}$ by $\frac{16}{8} = 2$ , $e^{- 24 t}$ by $\frac{- 4}{8} = - \frac{1}{2}$ , and the constant $4$ by $\frac{20}{8}$ : $i_{1} = 2 k_{1} e^{- 4 t} - \frac{1}{2} k_{2} e^{- 24 t} + 10.$
Initial conditions. $i_{2} (0) = k_{1} + k_{2} + 4 = 0$ and $i_{1} (0) = 2 k_{1} - \frac{1}{2} k_{2} + 10 = 0$ . From the first, $k_{2} = - 4 - k_{1}$ . Substituting: $2 k_{1} + 2 + \frac{1}{2} k_{1} + 10 = 0$ , so $\frac{5}{2} k_{1} = - 12$ , $k_{1} = - \frac{24}{5}$ and $k_{2} = \frac{4}{5}$ .
Answer. $i_{1} i_{2} i_{1} - i_{2} = 10 - \frac{48}{5} e^{- 4 t} - \frac{2}{5} e^{- 24 t}, = 4 - \frac{24}{5} e^{- 4 t} + \frac{4}{5} e^{- 24 t}, = 6 - \frac{24}{5} e^{- 4 t} - \frac{6}{5} e^{- 24 t} .$
Check with physics. After a long time the currents are constant, inductors have no voltage drop, and the full $48$ V appears across both $R_{1}$ and $R_{2}$ . So the middle branch carries $48/8 = 6$ A, the right branch $48/12 = 4$ A, and the battery $6 + 4 = 10$ A. The limits of our formulas agree.

Figure 7.2.3. Left: the two-loop network; the middle branch carries

i_{1} - i_{2}

. Right: the currents of Example 7.2.2 approach the steady values

10

4

and

6

A. The fast term

e^{- 24 t}

shapes only the first fraction of a second; the slow term

e^{- 4 t}

sets the time scale.

Interconnected tanks

In Section 3.3 one tank of brine led to one first-order equation: the rate of change of the amount of salt equals the rate at which salt enters minus the rate at which it leaves. With two tanks connected by pipes, salt can travel in both directions and we get one such balance for each tank. The rate at which salt is carried by a pipe is

(flow rate, L/min) \times (concentration in the tank the pipe comes from, kg/L),

and if the tank is well stirred its concentration is (amount of salt)/(volume).

Figure 7.2.4. The tank system of Example 7.2.3. The volumes stay constant because each tank receives exactly as much liquid as it loses: A gains

5 + 3

and loses

8

L/min; B gains

8

and loses

3 + 5

L/min.

Example 7.2.3

Tank A holds $40$ L of brine containing $8$ kg of dissolved salt; tank B holds $20$ L of pure water. From $t = 0$ : pure water flows into A at $5$ L/min, brine is pumped from A to B at $8$ L/min and from B back to A at $3$ L/min, and brine drains from B at $5$ L/min (Figure 7.2.4). Both tanks are kept well stirred. Find the amounts of salt $x (t)$ in A and $y (t)$ in B. When does B contain the most salt?

Solution

Concentrations. The volumes stay at $40$ L and $20$ L, so the concentrations are $x /40$ and $y /20$ kg/L.
Balance for A. Salt enters only through the pipe from B: $3 \cdot \frac{y}{20}$ kg/min. It leaves through the pipe to B: $8 \cdot \frac{x}{40}$ kg/min. The fresh water brings no salt. So $x^{'} = \frac{3 y}{20} - \frac{8 x}{40} = - \frac{x}{5} + \frac{3 y}{20} .$
Balance for B. Salt enters from A at $8 \cdot \frac{x}{40}$ and leaves through two pipes at a total of $(3 + 5) \cdot \frac{y}{20}$ kg/min: $y^{'} = \frac{x}{5} - \frac{2 y}{5} .$
Operator form without fractions. Multiply the first equation by $20$ and the second by $5$ : $(20 D + 4) x - 3 y = 0, - x + (5 D + 2) y = 0.$
Determinant. $Δ = (20 D + 4) (5 D + 2) - 3 = 100 D^{2} + 60 D + 5 = 5 (20 D^{2} + 12 D + 1) = 5 (10 D + 1) (2 D + 1) .$ The roots are $- \frac{1}{10}$ and $- \frac{1}{2}$ .
Solve. $x = c_{1} e^{- t /10} + c_{2} e^{- t /2}$ . The first equation gives $y = \frac{1}{3} (20 x^{'} + 4 x)$ algebraically. On $e^{- t /10}$ the factor is $\frac{1}{3} (- 2 + 4) = \frac{2}{3}$ ; on $e^{- t /2}$ it is $\frac{1}{3} (- 10 + 4) = - 2$ : $y = \frac{2}{3} c_{1} e^{- t /10} - 2 c_{2} e^{- t /2} .$
Initial conditions. $x (0) = c_{1} + c_{2} = 8$ and $y (0) = \frac{2}{3} c_{1} - 2 c_{2} = 0$ . The second gives $c_{1} = 3 c_{2}$ ; then $4 c_{2} = 8$ , so $c_{2} = 2$ and $c_{1} = 6$ .
Answer. $x = 6 e^{- t /10} + 2 e^{- t /2}, y = 4 e^{- t /10} - 4 e^{- t /2} (kg, t in min) .$
Maximum in B. $y^{'} = - 0.4 e^{- t /10} + 2 e^{- t /2} = 0$ gives $e^{- 0.4 t} = 0.2$ , so $t = \frac{5}{2} ln 5 \approx 4.02$ min. Then $y \approx 4 (0.669 - 0.134) \approx 2.14$ kg.
Check. Both amounts tend to $0$ , as they must when only fresh water enters. The slow exponential $e^{- t /10}$ dominates after a few minutes; in that regime $y / x \to \frac{4}{6}$ , so the concentrations approach the fixed ratio $\frac{y /20}{x /40} = \frac{4}{3}$ .

Figure 7.2.5. Salt in tanks A and B (Example 7.2.3). Tank B first fills with salt from A, peaks at

t = \frac{5}{2} ln 5 \approx 4.02

min, and is then slowly flushed together with A.

One mathematics, three devices

All three models led to constant-coefficient linear systems, and in all three the characteristic roots carried the physics: purely imaginary roots for the undamped springs (oscillation), negative real roots for the circuit and the tanks (decay to a steady state). Sections 7.3 and 7.4 explain this connection in general for systems in normal form. Initial-value problems of this kind can also be solved with the Laplace transform (Section 9.5).

Exercises

Two bodies of mass $1$ kg sit in a line on a frictionless horizontal track: a wall, a spring with $k_{1} = 3$ N/m, body 1, a spring with $k_{2} = 2$ N/m, body 2, and nothing to the right of body 2. (a) Show that $x_{1}^{''} = - 5 x_{1} + 2 x_{2}$ , $x_{2}^{''} = 2 x_{1} - 2 x_{2}$ . (b) Find $Δ$ and the two normal frequencies. (c) Solve with $x_{1} (0) = 1$ , $x_{2} (0) = 2$ , both at rest. (d) Solve with $x_{1} (0) = 0$ , $x_{2} (0) = 5$ , both at rest.
Answer
(b) $Δ = (D^{2} + 5) (D^{2} + 2) - 4 = (D^{2} + 1) (D^{2} + 6)$ ; $ω = 1$ and $ω = 6$ . (c) $x_{1} = cos t$ , $x_{2} = 2 cos t$ (a pure normal mode). (d) $x_{1} = 2 cos t - 2 cos 6 t$ , $x_{2} = 4 cos t + cos 6 t$ .
For the system of Example 7.2.1, find the motion when both bodies start at equilibrium, body 1 with velocity $1$ cm/s and body 2 at rest.
Answer
$x_{1} = \frac{1}{4} sin 2 t + \frac{1}{2 5} sin 5 t$ , $x_{2} = \frac{1}{4} sin 2 t - \frac{1}{2 5} sin 5 t$ .
In Example 7.2.1 show that if $x_{2} (0) = - x_{1} (0) = - a$ and $x_{2}^{'} (0) = - x_{1}^{'} (0) = - b$ , then $x_{2} (t) = - x_{1} (t)$ for all $t$ , and find $x_{1}$ .
Answer
$c_{1} = c_{2} = 0$ , $c_{3} = a$ , $c_{4} = b / 5$ : $x_{1} = a cos 5 t + \frac{b}{5} sin 5 t$ , $x_{2} = - x_{1}$ .
Repeat Example 7.2.2 with $L_{1} = \frac{1}{2}$ H, $L_{2} = 1$ H, $R_{1} = 4 Ω$ , $R_{2} = 8 Ω$ , $E = 16$ V and zero initial currents.
Answer
$Δ \propto (D + 4) (D + 16)$ . $i_{1} = 6 - \frac{16}{3} e^{- 4 t} - \frac{2}{3} e^{- 16 t}$ , $i_{2} = 2 - \frac{8}{3} e^{- 4 t} + \frac{2}{3} e^{- 16 t}$ .
In the network of Example 7.2.2 the currents have reached their steady values $i_{1} = 10$ A, $i_{2} = 4$ A. At $t = 0$ the battery is replaced by a short circuit ( $E = 0$ ). Find $i_{1} (t)$ and $i_{2} (t)$ .
Answer
$i_{1} = \frac{48}{5} e^{- 4 t} + \frac{2}{5} e^{- 24 t}$ , $i_{2} = \frac{24}{5} e^{- 4 t} - \frac{4}{5} e^{- 24 t}$ .
Repeat Example 7.2.3, but now the water entering tank A carries $0.2$ kg of salt per litre and both tanks initially contain pure water. (a) Write the system. (b) Solve it. (c) Find the limiting amounts and explain them.
Answer
(a) $x^{'} = 1 - \frac{x}{5} + \frac{3 y}{20}$ , $y^{'} = \frac{x}{5} - \frac{2 y}{5}$ . (b) $x = 8 - \frac{15}{2} e^{- t /10} - \frac{1}{2} e^{- t /2}$ , $y = 4 - 5 e^{- t /10} + e^{- t /2}$ . (c) $8$ kg and $4$ kg: both tanks end up at the inflow concentration $0.2$ kg/L.
Tank A holds $30$ L and tank B $60$ L. A pump moves $6$ L/min from A to B and another $6$ L/min from B to A; nothing enters or leaves the pair. Initially A contains $9$ kg of salt and B is pure water. Find $x (t)$ and $y (t)$ and show that $x + y$ is constant.
Answer
$x^{'} = - \frac{x}{5} + \frac{y}{10}$ , $y^{'} = \frac{x}{5} - \frac{y}{10}$ ; $x = 3 + 6 e^{- 3 t /10}$ , $y = 6 - 6 e^{- 3 t /10}$ ; $x + y = 9$ .
Tank A ( $100$ L) and tank B ( $50$ L) start with pure water. Brine with $0.1$ kg/L enters A at $2$ L/min; $6$ L/min flow from A to B, $4$ L/min from B back to A, and $2$ L/min drain from B. Write the system and find its constant (equilibrium) solution.
Answer
$x^{'} = \frac{1}{5} - \frac{3 x}{50} + \frac{2 y}{25}$ , $y^{'} = \frac{3 x}{50} - \frac{3 y}{25}$ ; equilibrium $x = 10$ kg, $y = 5$ kg.
In Example 7.2.3 show that the total amount of salt satisfies $(x + y)^{'} = - y /4$ , and check this with the solution found there.
Answer
Adding the equations, the exchange terms cancel and only the drain remains: $(x + y)^{'} = - 5 \cdot \frac{y}{20}$ . With the solution, $(x + y)^{'} = - e^{- t /10} + e^{- t /2} = - \frac{1}{4} (4 e^{- t /10} - 4 e^{- t /2})$ .
Rewrite the system of Example 7.2.1 as four first-order equations in $x_{1}$ , $v_{1} = x_{1}^{'}$ , $x_{2}$ , $v_{2} = x_{2}^{'}$ .
Answer
$x_{1}^{'} = v_{1}$ , $v_{1}^{'} = - \frac{9}{2} x_{1} + \frac{1}{2} x_{2}$ , $x_{2}^{'} = v_{2}$ , $v_{2}^{'} = \frac{1}{2} x_{1} - \frac{9}{2} x_{2}$ .