130 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methods
EXA M P L E I Eigenvalue Problem
Find the eigenvalues and eigenvectors of the matrix
f -4.0 4.0l
(ló) A:| |
L-1.6 1.2l
Solution. The characteristic equation is the quadratic equation
|-4-^ 4 l
det[A-^Il :l l:^2+2.8^+1.6:0.
|-1.6 1.2-^l
It has the solutions \: -2 and,\2 : -0.8. These are the eigenvalues of A.
Eigenvectors are obtained from (14*). For i : ir : -2we have from (14*)
(-4.0+2.0)x1 + 4.0x2 :0
-1.6x1 + 0.2 + 2.0)xr: Q.
A solution of the first equation is ,t1 : 2, ,z : 1. This also satisíies the second equation. (Why?). Hence an
eigenvector of A corresponding to .tr1 : -2.0 is
*" :
[i]
similarly, -"' :
[o],]
is an eigenvector of A corresponding to ,tr2 : -0.8, as obtained from (14*) with ,tr : ),2. Verify this. l
(I7)
4.1 Systems of ODEs as Models
We first illustrate with a few typical examples that systems of ODEs can serve as models
in various applications. We further show that a higher order ODE (with the highest
derivative standing alone on one side) can be reduced to a first-order system. Both facts
account for the practical imporlance of these systems.
E X A M P L E 1 Mixing Problem lnvolving Two Tanks
A mixing problem involving a single tank is modeled by a single ODE, and you may first review the
corresponding Example 3 in Sec. 1.3 because the principle of modeling will be the same for two tanks. The
model will be a system of two first-order ODEs.
Tank Z1 andT2 in Fig. 7'7 contain initially 100 gal of water each. In Zl the water is pure, whereas 150 lb of
ferttlizer are dissolved in T2. By circulating liquid at a rate of 2 gallmin and stir:ring (to keep the mixture uniform)
the amounts of fertilizer yr(r) in I and y2(t) in T2 change with time r. How long should we let the liquid circulate
so that Z1 will contain at least half as much fertilizer as there will be left in T2?
SOlUtiOn. Sfup 1. Setting up the model. As for a single tank, the time rate of change yi(r) ot yl(r) equals
inflow minus outflow. Similarly for tank T2. From Fig. 77 we see that
22yi : Inflow/min - Outflow/min :
l00
y, - ňr,
22yj : Inflow/min - Outflow/min :
100
y, - ňr,
Hence the mathematical model of our mixture problem is the system of first-order ODEs
y', : -0.0Zy1 -l 0.02y2
y;: 0.02y1 - 0.02y2
(Tank Z1)
(Tank I2).
(Tank Z1)
(Tank Z2).
SEC. 4.1 Systems of ODEs as Models
(3)
System of tanks o 2] .5 50
Fig.77. Fertilizer content in Tanks | (lower curve) and I,
As a vector equation with column vector, : ["l and matrix A this becomes
Lyr_]
T -0.02 0.02l
y':Ay. where A:| |
L 0.02 -0.02J
Step 2. General solution. As for a single equation. we try an exponential function of t,
y(t)
150
100
75
50
0
y: ctx(l),^'t + crx(z)"^2t : cl[l] - ,r|_'rf Uo*'
I3I
(1) y : xe^í. Then y' : hxe^' : Lxe^'.
Dividing the last equation lxet' : Axe^'by e^' and interchanging the left and right sides, we obtain
Ax : ),x.
We need nontrivial solutions (solutions that are not identically zero). Hence we have to look for eigenvalues
and eigenvectors of A. The eigenvalues are the solutions of the characteristic equation
| -0.02 -
^
0.02 l
(2) det(A-^I): l l:(-0.02 - D2-0.022:
^(^+0.04):0.l 0.02 -0.02 -
^
l
We see that ,tr1 : 0 (which can very well happen-dgn'1 get mixed up-it ls eigenvectors that must not be zero)
and i2 : -0.04. Eigenvectors are obtained from (l4x) in Sec. 4.0 with i : 0 and
^
: -0.04. For our present
A this gives [we need only the first equation in (l4*)]
-0,02x1 -l 0.02x2: Q and (-0,02 + 0,04)x1 * 0,02x2: 0,
respectively. Hence xt: x2 and x1 : -x2, respectively, and we can take xl : x2: 1 and x1 : -x2: I.
This gives two eigenvectors corresponding to i1 : 0 and i2 : -0.04, respectively, namely,
|-ll l- ll
x'r':l l and x'2':l l.
Lr_] L-r_]
From (1) and the superposition principle (which continues to hold for systems of homogeneous linear ODEs)
we thus obtain a solution
where c1 and c2 are aíbitrary constants, Later we shall call this a general solution.
Step 3. Use of initial conditions. The initial conditions are y1(0) : 0 (no fertllizer in tank Z1) and y2(0) : 150,
From this and (3) with r : 0 we obtain
y(0):.,[i] *.,[_]] :
[:i :,,,,1: [,:.]
System of tanks 2],5 50
y
2(t)
2 sallmin
Tl T22 gallmin
| ------_
-|---l --'--'-
132 CHAP. 4 Systems of ODEs. Phase P[ane. Qualitative Methods
In components this is c1 * c2: 0, c1 - c2:
y: 75x(1) - 75xQ)e
In components,
, 150. The solution is c1 : 75, c2
-o.o4t _ ". ['
-l
_ . l- l
-l
-UU+, - r,
L,_]
-,, L_,_] "
: -15. This gives the answer
-o.o4t
Figure 77 shows the exponential increase of y1 and the exponential decrease of y2 to the common limit 75 lb.
Did you expect this for physical reasons? Can you physically explain why the curves look "sYmmetric"? Would
thelimitchangeif Iinitiallycontained 100 lbof fertilizer anďT2contained50lb?
Step 4. Answer. 71 contains half the feríiltzer amount of T2 tf it contains 1l3 of the total amount, that is,
50 lb. Thus
yt:75-,75e-o,o4t
yz: 75 + 75e-o,o4t
yt : f5 *,75e-0,0at : 50, e-0,04t - !,
Hence the fluid should circulate for at least about half an hour.
(Tank 71, lower curve)
(Tank 72, upper curve).
, : (ln 3)10.04 : 2].5.
l
ExAMPLE 2 Electrical Network
Find the currents 11(r) and I2Q) ínthe network in Fig. 78. Assume all currents and charges to be zero at / : 0,
the instant when the switch is closed.
Z=lhenry C=0,25farad
Switch
t=O
fir = 4 ohms
g = 12 volts..:
ú
ftz = 6 ohms
Fig.78. Electrical network in Example 2
Solution. Step 1. Setting up the mathematical model. The model of this network is obtained from
Kirchhoff's voltage law, as in Sec.2.9 (where we considered single circuits). LetIlG) andI2(t) be the currents
in the left and right loops, respectively. In the left loop the voltage drops are LIi :1i tV] over the inductor
and R1(11 - I) : 4(It - 12) tV] over the resistor, the difference because 11 and 12 flow through the resistor
in opposite directions. By Kirchhoff's voltage law the sum of these drops equals the voltage of the battery; that
is, /| + 4(h * Iz) : I2,hence
(4a) I'r: -4h + 4I2 + 12,
In the right loop the voltage drops are RzIz: 6I2IYI and R1(12 - I) : 4(Iz - 1r) [V] over the resistors and
(llC)I 12 dt : 4 I 12 dt [V] over the capacitor, and their sum is zero,
612+ 4(I2-11) + o
I,rdt:o
or IoI2_ 4Il+ 4
[,ror:o.
Division by 10 and differentiation gives t! - O.+t', + 0.4Ir: g.
To simplify the so|ution process, we first get rid of 0.4I'1, which by (4a) equals 0,4(-4II + 4I2 + 12).
Substitution into the present ODE gives
t! : O.+t', - 0.412: 0.4(-4Il + 4I2 + |2) - 0-412
aG4v
!