Chapter 7. Applications of Systems of Ordinary Differential Equations
Tải bản đầy đủ - 0trang
568
Chapter 7 Applications of Systems of Ordinary Differential Equations
C
capacitance,
V
voltage, and
L
inductance.
The relationships corresponding to the drops in voltage in the various components
of the circuit that were stated in Chapter 5 are also given in the following table.
Circuit Element
Inductor
Resistor
Capacitor
Voltage Source
Voltage Drop
dI
L
dt
RI
1
Q
C
V t
7.1.2 L–R–C Circuit with One Loop
In determining the drops in voltage around the circuit, we consistently add the
voltages in the clockwise direction. The positive direction is directed from the negative symbol towards the positive symbol associated with the voltage source. In
summing the voltage drops encountered in the circuit, a drop across a component
is added to the sum if the positive direction through the component agrees with
the clockwise direction. Otherwise, this drop is subtracted. In the case of the following L–R–C circuit with one loop involving each type of component, the current
is equal around the circuit by Kirchhoff’s Current Law as illustrated in Figure 7-1.
Also, by Kirchhoff’s Voltage Law, we have the sum
RI
L
dI
dt
1
Q
C
V t
0.
Figure 7-1 A simple L–R–C circuit
7.1 Mechanical and Electrical Problems with First-Order Linear Systems
569
Solving this equation for dI/ dt and using the relationship between I and Q, dQ/ dt
I, we have the following system of differential equations with initial conditions on
charge and current, respectively:
dQ/ dt
dI/ dt
Q0
I
R
I
L
I0 .
1
Q
LC
Q0 , I 0
V t
L
(7.1)
EXAMPLE 7.1.1: Determine the charge and current in an L–R–C circuit
Q0 and I 0
I0 .
with L 1, R 2, C 4/ 3, and V t
e t if Q 0
SOLUTION: We begin by modeling the circuit with the system of
differential equations
dQ/ dt
I
3
4Q
dI/ dt
2I
e
t
which can be written in matrix form as
dQ/ dt
dI/ dt
0 1
3/ 4 2
Q
I
0
.
e t
We solve the initial-value problem with DSolve, naming the result
sol.
In[1438]:= Clear q, i
sol
Out[1438]=
DSolve D q t , t
i t ,
D i t ,t
3/4q t
2i t
Exp t ,
q 0
q0, i 0
i0 , q t , i t , t
q t
1
2
2
3 t/ 2
t
1
4
i t
2
i0
4
q0
3 t/ 2
t
i0
t/ 2
8
3
12
3 q0
t
4
2 i0
q0 ,
16
3
t
t
t/ 2
q0
4
t
6 i0
570
Chapter 7 Applications of Systems of Ordinary Differential Equations
We now select, copy, and paste the formulas obtained in sol for Q
and I, respectively, and then use Expand to distribute the e 3t/ 2 term
through the parentheses.
In[1439]:= Expand
t
2e
3 t/ 2
Out[1439]= 2
t/ 2
t
2e
3
1
2
t
4
3 t/ 2
1
e
4
4 i0
3 t/ 2
t/ 2
2
1
2
i0
In[1440]:= Expand
Out[1440]=
1 3 t/ 2
4 8 et/2 2
e
2
4 i0
q0 3 et 4 q0
4
i0
t/ 2
3
4
q0
16 et/2
3 4
t
i0
t/ 2
3
2
q0
3 t/ 2
3 t/ 2
q0
3
2
t/ 2
3 t/ 2
q0
6
3e
t
3 t/ 2
3
4
4
i0
4
i0
4
q0
i0
t/ 2
q0
The result indicates that limt Q t
limt I t
0 regardless of the
values of Q0 and I0 . This is conﬁrmed by graphing Q t (in black) and
I t (in gray) together (choosing Q 0
I0
1) in Figure 7-2 as well as
parametrically in Figure 7-3.
In[1441]:= Plot Evaluate q t , i t /.sol/.
q0 1, i0 1 , t, 0, 10 ,
PlotStyle
GrayLevel 0 , GrayLevel 0.5
1.25
1
0.75
0.5
0.25
2
4
6
10
8
-0.25
Figure 7-2 Q t (in black) and I t (in gray) for 0
t
10
7.1 Mechanical and Electrical Problems with First-Order Linear Systems
0.4
0.2
0.2 0.4 0.6 0.8
1
1.2 1.4
-0.2
Figure 7-3 Parametric plot of Q versus I for 0
t
10
In[1442]:= ParametricPlot
Evaluate q t , i t /.sol/.
q0 1, i0 1 , t, 0, 10
7.1.3 L–R–C Circuit with Two Loops
The differential equations that model the circuit become more difﬁcult to derive
as the number of loops in the circuit increases. For example, consider the circuit in
Figure 7-4 that contains two loops.
Figure 7-4
A two-loop circuit
571
572
Chapter 7 Applications of Systems of Ordinary Differential Equations
In this case, the current through the capacitor is equivalent to I1
the voltage drops around each loop, we have:
I2 . Summing
1
Q V t
0
R1 I1
C
dI2
1
L
R2 I2
Q 0.
dt
C
(7.2)
1
1
Q and using the
V t
R1
R1C
I2 we have the following system:
Solving the ﬁrst equation for I1 we ﬁnd that I1
relationship dQ/ dt
I
I1
1
1
Q I2
V t
R1C
R1
1
R2
Q
I2 .
LC
L
dQ
dt
dI2
dt
(7.3)
EXAMPLE 7.1.2: Find Q t , I t , I1 t , and I2 t in the L–R–C circuit with
R2
C
1 and V t
e t if Q 0
3 and
two loops given that R1
1.
I2 0
SOLUTION: The nonhomogeneous system that models this circuit is
dQ/ dt
Q
dI2 / dt
Q
I2
e
t
I2
with initial conditions Q 0
3 and I2 0
1. We solve the initial-value
problem with DSolve naming the result sol. We deﬁne Q t and I2 t
to be the results.
In[1443]:= Clear q, i
sol
Out[1443]=
DSolve D q t , t
q t
i2 t
D i2 t , t
q t
i2 t , q 0
i2 0
1 , q t , i2 t , t
q t
i2 t
In[1444]:= q t
In[1445]:= i2 t
t
3
t
sol
sol
Cos t ,
Cos t 2 3 Sin t
1, 1, 2
1, 2, 2
Sin t
Exp
3,
2
t ,
7.1 Mechanical and Electrical Problems with First-Order Linear Systems
573
We verify that these functions satisfy the system by substituting back
into each equation and simplifying the result with Simplify.
In[1446]:= D q t , t
Simplify
Out[1446]= 0
q t
i2 t
In[1447]:= D i2 t , t
q t
i2 t
Exp
t
//
//Simplify
Out[1447]= 0
We use the relationship dQ/ dt
In[1448]:= i t
Out[1448]=
and then I1 t
D q t ,t
t
3
I to ﬁnd I t
Cos t
t
3
Sin t
I2 t to ﬁnd I1 t .
It
In[1449]:= i1 t
t
Out[1449]=
i t
3
t
i2 t
Cos t
We graph Q t , I t , I1 t , and I2 t with Plot and display the result using
Show and GraphicsArray in Figure 7-5.
3
2.5
2
1.5
1
0.5
1
-0.5
-1
-1.5
-2
2
3
4
5
1 2 3 4 5
1 2 3 4 5
-0.5
-1
-1.5
-2
-2.5
-3
1.4
1.2
1
0.8
0.6
0.4
0.2
1
Figure 7-5 Q t , I t , I1 t , and I2 t for 0
2
t
5
3
4
5
574
Chapter 7 Applications of Systems of Ordinary Differential Equations
In[1450]:= p1
Plot q t , t, 0, 5 , PlotRange All,
DisplayFunction Identity
p2 Plot i t , t, 0, 5 , PlotRange All,
DisplayFunction Identity
p3 Plot i1 t , t, 0, 5 , PlotRange All,
DisplayFunction Identity
p4 Plot i2 t , t, 0, 5 , PlotRange All,
DisplayFunction Identity
Show GraphicsArray
p1, p2 , p3, p4
7.1.4 Spring–Mass Systems
The displacement of a mass attached to the end of a spring was modeled with a
second-order linear differential equation with constant coefﬁcients in Chapter 5.
This situation can then be expressed as a system of ﬁrst-order ordinary differential equations as well. Recall that if there is no external forcing function, then the
cx kx 0,
second-order differential equation that models this situation is mx
where m is the mass attached to the end of the spring, c is the damping coefﬁcient,
and k is the spring constant found with Hooke’s law. This equation is transformed
c
k
x
x and then
y so that y
x
into a system of equations by letting x
m
m
solving the differential equation for x . After substitution, we have the system
dx
dt
dy
dt
y
k
x
m
(7.4)
c
y.
m
In previous chapters, the displacement of the spring was illustrated as a function
of time. However, problems of this type may also be investigated using the phase
plane.
EXAMPLE 7.1.3: Solve the system of differential equations to ﬁnd the
displacement of the mass if m 1, c 0, and k 1.
SOLUTION: In this case, the system is
dx/ dt
dy/ dt
y
x
which in matrix
0 1
X. A general solution is found with DSolve and
10
named gensol for later use.
form is X
7.1 Mechanical and Electrical Problems with First-Order Linear Systems
In[1451]:= Clear x, y
gensol
Out[1451]=
x t
y t
DSolve D x t , t
D y t ,t
x t
C 1 Cos t
C 2 Cos t
y t ,
, x t ,y t
,t
C 2 Sin t ,
C 1 Sin t
Note that this system is equivalent to the second-order differential equax 0, which we solved in Chapters 4 and 5. At that time, we
tion x
found a general solution to be x t
c1 cos t c2 sin t which is equivalent
xt
to the ﬁrst component of X
, the result obtained with DSolve.
yt
Also notice that 0, 0 is the equilibrium point of the system. The eigen0 1
values of A
are Λ
i,
10
0 1
1 0
In[1452]:= Eigenvalues
Out[1452]=
,
so we classify the origin as a center.
We graph several members of the phase plane for this system with
ParametricPlot in Figure 7-6.
4
2
-4
-2
2
-2
-4
Figure 7-6 The origin is a center
4
575
576
Chapter 7 Applications of Systems of Ordinary Differential Equations
In[1453]:= toplot
Flatten
Table x t , y t /.gensol/.
C 1
i, C 2
j , i, 3, 3 ,
j, 0, 3 , 2
In[1454]:= ParametricPlot Evaluate toplot ,
t, 0, 2 Π , PlotRange
5 , 5 , 5, 5
AspectRatio 1
,
7.2 Diffusion and Population Problems
with First-Order Linear Systems
7.2.1 Diffusion through a Membrane
Solving problems to determine the diffusion of a substance (such as glucose or salt)
in a medium (like a blood cell) also leads to systems of ﬁrst-order linear ordinary
differential equations. For example, suppose that two solutions of a substance are
separated by a membrane where the amount of the substance that passes through
the membrane is proportional to the difference in the concentrations of the solutions. The constant of proportionality is called the permeability, P, of the membrane. Therefore, if we let x and y represent the concentration of each solution, and
V1 and V2 represent the volume of each solution, respectively, then the system of
differential equations is given by
dx
dt
dy
dt
P
y
V1
P
x
V2
x
(7.5)
y
where the initial concentrations of x and y are given.
EXAMPLE 7.2.1: Suppose that two salt concentrations of equal volume V are separated by a membrane of permeability P. Given that
P V , determine each concentration at time t if x 0
2 and y 0
10.
7.2 Diffusion and Population Problems with First-Order Linear Systems
SOLUTION: In this case, the initial-value problem that models the
situation is
dx/ dt y x
dy/ dt
x
x0
y
2, y 0
10 .
A general solution of the system is found with DSolve and named
gensol.
In[1455]:= Clear x, y
DSolve D x t , t
D y t ,t
x t
x t ,y t ,t
1 2t
2t
1
C 1
2
gensol
Out[1455]=
x t
1
2
1
2
2t
2t
1
2t
x t ,
,
C 2 ,y t
2t
1
y t
y t
1
2
C 1
2t
1
2t
C 2
We then apply the initial conditions and use Solve to determine the
values of the arbitrary constants.
In[1456]:= cvals
Out[1456]=
C 1
Solve
e
2t
C 2 /.t > 0
C 1
e 2t C 1
C 2 /.t > 0
4, C 2
6
2,
10
The solution is obtained by substituting these values back into the
general solution.
In[1457]:= sol
Out[1457]=
gensol/.cvals
x t
y t
2t
3
2
2t
1
2t
1
1
2t
2t
2
3
2t
2t
1
1
,
2t
Of course, DSolve can be used to solve the initial-value problem
directly as well.
In[1458]:= sol
Out[1458]=
DSolve D x t , t
y t
x t ,
D y t ,t
x t
y t ,x 0
2,
y 0
10 , x t , y t , t
x t
2 2t 2 3 2t ,y t
2 2t 2 3 2t
We graph this solution parametrically with ParametricPlot in
Figure 7-7(a). We then graph x t and y t together in Figure 7-7(b). Notice
577
578
Chapter 7 Applications of Systems of Ordinary Differential Equations
10
10
8
6
4
2
8
6
4
2
1
2
4
6
3
4
5
8 10
(a)
Figure 7-7
2
(b)
(a) Parametric plot of x versus y. (b) x t (in black) and y t (in gray)
that each concentration approaches 6 which is the average value of the
two initial concentrations.
In[1459]:= p1
ParametricPlot x t , y t /.sol,
t, 0, 5 , Compiled False,
PlotRange
0, 10 , 0, 10 ,
AspectRatio 1, AxesOrigin
0, 0 ,
DisplayFunction Identity
In[1460]:= p2
Plot Evaluate x t , y t /.sol ,
t, 0, 5 , PlotRange
0, 10 ,
PlotStyle
GrayLevel 0 ,
GrayLevel 0.5 ,
DisplayFunction Identity
In[1461]:= Show GraphicsArray
p1, p2
7.2.2 Diffusion through a Double-Walled Membrane
Next, consider the situation in which two solutions are separated by a doublewalled membrane, where the inner wall has permeability P1 and the outer wall
has permeability P2 with 0 < P1 < P2 . Suppose that the volume of solution within
the inner wall is V1 and that between the two walls is V2 . Let x represent the concentration of the solution within the inner wall and y the concentration between the