7. Análisis de resultados
7.3. Análisis de la entrevista final (EF)
7.3.4. Nivel Institucional:
EXAMPLE 31.24
Newton’s Law of Cooling.Let y = y(t) = the temperature of a large room at time t.
Let x = x(t) = the difference in temperature between the room and a small object placed in it.
Let x be positive if the object is hotter than the room. The differential equations given below could reflect the room-object system.
dx
dt = −kx, k > 0 dy dt = 0 For the sake of simplicity, let k = 1.
(a) i. Solve for y in terms of t and sketch the solution in the ty-plane. ii. Solve for x in terms of t and sketch the solution in the tx-plane.
(b) Graph the solutions in the xy-plane as follows: Pick any initial temperatures y(0) = y0
and x(0) = x0for the room and the small object, respectively. Think about what will
happen to x and y as t increases and sketch the appropriate path, drawing an arrow to indicate the direction the path is traveled. (t will not appear explicitly in the sketch.) 8The three examples that follow were originally suggested to me by Stephen DiPippo when we taught together.
Select various other starting points and sketch each path traced out by x and y as t increases. All paths should have arrows to indicate how x and y change with time. (c) If the system is at equilibrium, then both x and y will remain constant as t varies, that
is,
dy
dt = 0 and dx dt = 0.
Under what conditions is the system at equilibrium? How do these equilibrium points relate to the picture drawn in part (a)?
SOLUTION (a) Each differential equation can be solved independently.
i. dydt = 0 ⇒ y = C
If we let y0denote y(0), then y(t) = y0. In the ty-plane, solutions can be repre-
sented as shown in Figure 31.16.
y
t
Figure 31.16
ii. dxdt = −x ⇒ x(t) = x0e−t
In the tx-plane, solutions can be represented as shown in Figure 31.17.
x
t
Figure 31.17
(b) Pick a point (x0, y0)as a starting point. The value of y will remain constant. If x0is
positive, then as t increases x will remain positive but tend toward zero; if x is negative, then as t increases x will remain negative but tend toward zero. (This information can be read from Figure 31.17.) On the following page is a sketch in the xy-plane. Various starting points (corresponding to t = 0) are chosen, each designated by an “X.” Arrows indicate the direction the path is traveled as t increases.
31.5 Systems of Differential Equations 1027
x y
Figure 31.18
(c) If the system is at equilibrium, then dy dt = 0 and dx dt = 0. dy dt is identically 0. dx dt = −x, therefore dx
dt = 0 if and only if x = 0. In other words, the
system is at equilibrium as long as x, the temperature difference, is zero.
When we represent solutions to the system of differential equations in the xy-plane (with no t-axis), an equilibrium solution is simply a point. As t varies, x and y remain constant. (Contrast this with an equilibrium solution in the ty-plane or the tx-plane in which an equilibrium solution is a horizontal line.) In this example, every point along the y-axis is an equilibrium point.
Below representative solution curves are drawn. These curves, or trajectories, are traced by (x(t), y(t)) as t varies. Arrows indicate the direction the curve is traced as t increases. The xy-plane is often referred to as the phase-plane of the solutions. Initial points are usually not explicitly indicated.
y
x
Figure 31.19
REMARKS
1.Trajectories on the right-hand side of the y-axis and the left-hand side of the y-axis are distinct. If x is initially positive, it will always be positive (see Figure 31.17). Likewise, if x is initially negative, it is always negative. In other words, if one starts at a point to the right of the y-axis, the trajectory through that point will never cross the y-axis. In fact, although the trajectory gets arbitrarily close to the y-axis, x never becomes zero
(again, refer to Figure 31.17). Each point on the y-axis is a distinct equilibrium point. (We can think of an equilibrium point as a degenerative trajectory.) Each horizontal line in Figure 31.19 consists of three distinct solution trajectories. Trajectories do not intersect, although they can get arbitrarily close to one another.
2.We determined the direction of the arrows on the trajectories by referring to Figure 31.17, but that information is directly accessible from the differential equations them- selves. dxdt = −x, therefore
for x > 0, dx
dt <0; xdecreases as t increases. For x < 0, dx
dt >0; xincreases as t increases.
We can determine the “shape” of the trajectories in the xy-plane by using the original differential equations and looking at the relative rates of change of y and x as t varies. The slope of the trajectory isdydx=dy/dtdx/dt, where equality follows from the Chain Rule, provided
dx dt = 0. Therefore dy dx = dy dt dx dt = 0 −x = 0 ⇒ dy dx = 0 ⇒ y = C.
These are horizontal lines in the xy-plane. In Remark 1 we point out that each of these horizontal lines actually consists of three distinct trajectories, one to the right of the y-axis, one to the left of the y-axis, and an equilibrium point on the y-axis.
Note:dxdy=dy/dtdx/dt provided that dxdt is nonzero. If, at a certain point P ,dxdt = 0 and dydt = 0, then neither x nor y is changing with t, so P is an equilibrium point. If, at a certain point Q,dxdt = 0 anddydt = 0, then y is changing with t but x is not, so the trajectory is vertical at Q.
EXAMPLE 31.25
Analyze the system of differential equations below, sketching solution trajectories in the phase-plane.dx
dt = −x
dy
dt = −y
SOLUTION Each of these differential equations can be solved independently. They describe exponential
decay. dx dt = −x ⇒ x(t) = x0e −t dy dt = −y ⇒ y(t) = y0e −t
31.5 Systems of Differential Equations 1029 y t x t Figure 31.20
Now consider the trajectories in the xy-plane. As t changes, (x(t), y(t)) traces out a curve in the xy-plane. Wherever we start out in the plane, as t increases, both the x- and y-coordinates go toward zero. At the equilibrium
dx
dt = −x = 0 and dy
dt = −y = 0
simultaneously. So x = 0 and y = 0. The origin is the only equilibrium point.
x y
Figure 31.21
To determine the shape of the trajectories in the xy-plane, observe that for x0= 0
y(t ) x(t )= y0e−t x0e−t = y0 x0= constant.
Thus, the trajectory starting at (x0, y0)when t = 0 satisfiesyx = k, where k is the constant y0
x0. We obtain y = kx, the equation of a straight line through the origin. The coordinates
of the starting point determine the slope of the line. In Figure 31.22 arrows indicate the direction to travel along these lines as t increases.
We need to treat the case x0= 0 separately. If x0= 0 then x(t) = 0. For y0>0, y(t)
decreases as t increases; for y0<0, y(t) increases as t increases. This gives two trajectories
along the y-axis. If both x0and y0are zero, the system is at equilibrium.
Putting everything together, we obtain the picture sketched on the following page. Apart from the origin, each trajectory is a ray approaching the origin.
x y
Figure 31.22
EXERCISE 31.7 In Example 31.25, we can find the shape of the trajectories by looking atdydx.
dy dx = dy dt dx dt =−y −x, or dy dx = y x.
This is a separable differential equation. Solve for y, showing that y = kx, where k is a constant.
Examples 31.24 and 31.25 are special in that we can solve each of the original dif- ferential equations independently to get x and y explicitly in terms of t and we can see precisely what the trajectories look like in the xy- phase-plane. In general, the situation is more complex.