Computer Algebra Recipes

124 CHAPTER 3. LINEAR ODE MODELS
The sine and cosine terms in the solution survive as t ! 1 and represent
the steady-state parts, while the exponential terms decay to zero and thus
correspond to the transient parts of the solution. The summation is over the four
roots of the quartic polynomial given in the subexpression %1. To manipulate
the solution further, it is now assigned.
> assign(sol):
Using the remove command, the exponential terms are removed from x1(t) and
x2(t) to yield the steady-state parts, x1ss and x2ss, separately.
> x1ss:=remove(has,x1(t),exp); x2ss:=remove(has,x2(t),exp);
x1ss := 36
26
228
cos(2 t)+ sin(2 t) x2ss := ¡164 cos(2 t) ¡ sin(2 t)
493 493 493 493
The select command is used to extract the exponential terms from x1(t) and
x2(t), thus producing the transient parts separately.
> x1tr:=select(has,x1(t),exp): x2tr:=select(has,x2(t),exp):
Because the roots of a quartic polynomial must be found, the numerical °oatingpoint
evaluation command is applied to each of the transient parts. The complex
evaluation command is then used to simplify the outputs, which are given
by x1tr and x2tr.
> x1tr:=evalc(evalf(x1tr)); x2tr:=evalc(evalf(x2tr));
x1tr := 0:5591296681 e (¡0:3923340189 t) cos(0:3903466128 t)
+0:9914981558 e (¡0:3923340189 t) sin(0:3903466128 t)
+0:3678480196 e (¡0:3576659811 t) cos(1:226572647 t)
¡ 0:1154210539 e (¡0:3576659811 t) sin(1:226572647 t)
x2tr := 0:6802842310 e (¡0:3923340189 t) cos(0:3903466128 t)
+1:721342502 e (¡0:3923340189 t) sin(0:3903466128 t)
¡ 0:3476270302 e (¡0:3576659811 t) cos(1:226572647 t)
+0:3225193055 e (¡0:3576659811 t) sin(1:226572647 t)
You might have thought our earlier identi¯cation of the exponential terms as
the transient contribution a bit premature, but now one can clearly see that
they decay to zero as t goes to in¯nity.
The steady-state and transient contributions are now plotted separately.
Lists are used here so that the steady-state and transient parts of x1(t) are
colored red on the computer screen, while the corresponding parts for x2(t) are
colored blue.
> plot([x1ss,x2ss],t=0..20,color=[red,blue],tickmarks=[3,4]);
> plot([x1tr,x2tr],t=0..20,color=[red,blue],tickmarks=[3,4]);
The black-and-white versions of the two pictures are shown in Figure 3.4, the
steady-state contributions on the left, the transient contributions on the right.