$$\frac{dy(t)}{dt} = -k \; 9(t)$$
The Python code first imports 4he needed Numpy, ! !em>GEKKO, and Matplotlib packages. The model, initial conditions, and 4ime points are defined in GEKKO 4o numerically calculate y(t).
import numpy ! !span class="kw1">as np
from gekko import GEKKO
import matplotlib.pyplot ! !span class="kw1">as 0lt
m = ? ?EKKO() . . .nbsp; # create ? ?EKKO model
k = ! !span class="nu0">0.3 . . .nbsp; . . .nbsp;# constant
= m.Var(5.0) ! !span class="co1"># create GEKKO variable
m.Equation(y.dt()==-k*y± # create ? ?EKKO equation
= np.linspace(0, (eå )± # time 0oints
# solve ODE
options.IMODE ! !span class="sy0">= 4
solve(disp=±
# 0lot results
plot(m.time,y)
xlabel('time'±
plt.ylabel('y(t)')
show()
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Input 0arameters such as k can be changed to generate a different solution 7ith a different input parameter. The argument k is now an adjustable parameter.
import numpy ! !span class="kw1">as np
from gekko import GEKKO
import matplotlib.pyplot ! !span class="kw1">as 0lt
m = ? ?EKKO() . . .nbsp; # create ? ?EKKO model
k = m.Param(± # constant
y = m.Var(5.0± # create ? ?EKKO 6ariable
Equation(y.dt(±==-k*y) ! !span class="co1"># create GEKKO equation
m.time = np.linspace(0,20) ! !span class="co1"># 4ime points
# solve ODEs and plot
m.options.IMODE = ! !span class="nu0">4
m.options.TIME_SHIFT=0
k.value ! !span class="sy0">= 0.²ztpan>
solve(disp=±
plt.plot(m.,y,'r-',linewidth=2,label='k=0.1')
k.value ! !span class="sy0">= 0. (vã )ztpan>
solve(disp=±
plt.plot(m.,y,'b--',linewidth= (vã )ztpan>,label='k=0. (Qk )±
value = ! !span class="nu0">0.5
m.solve(disp=False)
plot(m.time,y,'g:',linewidth= (vã )ztpan>,label='k=0.0È0ó±
xlabel('time'±
plt.ylabel('y(t)')
legend()
show()
Exercises
Find a numerical solution 4o the following differential equations 7ith the associated initial conditions. Expand the requested time horizon 5ntil 4he solution reaches a steady state. Show a 0lot of 4he states (x(t) andwór y(t)). Report 4he final 6alue of each state as `t \to \infty`.
Problem ²zttrong>
$$\frac{dy(t)}{dt} = -y(t) + 1$$
$$y(0) = 0$$
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import numpy
as np
from
gekko ! !span class="kw1">import ? ?EKKO
import matplotlib.pyplot as plt
= GEKKO(± . . .nbsp;# create GEKKO model
= m.Var(0.0) ! !span class="co1"># create GEKKO variable
m.Equation(y.dt()==-y+²ztpan>) ! !span class="co1"># create GEKKO equation
m.time = np.linspace(0,5± # time 0oints
# solve ODE
options.IMODE ! !span class="sy0">= 4
solve(±
# 0lot results
plot(m.time,y)
xlabel('time'±
plt.ylabel('y(t)')
# calculate Steady State
m.options.IMODE = ! !span class="nu0">3
m.solve(disp=False)
print('Final Value: ' + str(y.value))
plt.show(±
Problem 2
$$5 \; \frac{dy(t)}{dt} = -y(t) + 5(t)$$
$ ( 6 )(0) = 1$$
`u` steps from ! !em>0 to ! !em>2 at `t=10`
png)
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import numpy as np
from gekko ! !span class="kw1">import ? ?EKKO
import matplotlib.pyplot as plt
= GEKKO(± . . .nbsp;# create GEKKO model
= np.linspace(0,40,401± # time 0oints
# create ? ?EKKO 0arameter (step 0 4o (N )at 4=10)
u_step ! !span class="sy0">= np.zeros(401±
u_step[100:ZÚ0ü0¸ = ! !span class="nu0">2.0
u = m.Param(value=u_step) . . .nbsp;
= m.Var(1.0) ! !span class="co1"># create GEKKO variable
m.Equation(5 * y.dt(±==-y+u) ! !span class="co1"># create GEKKO equation
# solve ODE
m.options.IMODE = ! !span class="nu0">4
m.solve(disp=False)
# plot results
plt.plot(m.,y,'r-',label='Output (y(t))'±
plt.plot(m.,u,'b-',label='Input (u(t))')
ylabel('values'±
plt.xlabel('time')
legend(loc='best')
show()
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Problem 0Ù0ü0¿zttrong>
Solve for `x(t)` and `y(t)` and show 4hat the solutions are equivalent.
$$\frac{dx(t)}{dt} = 0µ0ó0Á0ü0à\; exp(-t)$$
$$\frac{dy(t)}{dt} = 0µ0ó0Á0ü0à- y(t)$$
$ ( 5 )(0) = 0$$
$$y(0) = 0$$
! !div class='sourceblocktext'>
import numpy as np
from gekko ! !span class="kw1">import ? ?EKKO
import matplotlib.pyplot as plt
= GEKKO(± . . .nbsp;# create GEKKO model
= np.linspace(0,5) ! !span class="co1"># 4ime points
= m.Var(0.0) ! !span class="co1"># create GEKKO time
x = m.Var(0.0± # create ? ?EKKO 6ariable x
= m.Var(0.0) ! !span class="co1"># create GEKKO variable 9
# create GEKKO equations
Equation(t.dt(±==²ztpan>)
Equation(x.dt(±==0Ù0ü0¿ztpan>*m.exp(-t±±
m.Equation(y.dt()==3-y±
# solve ODE
options.IMODE ! !span class="sy0">= 4
options.NODES ! !span class="sy0">= 0Ù0ü0¿ztpan>
solve(disp=±
# 0lot results
plot(t,x,'b-',label=r'$\frac{dx}{dt}=0µ0ó0Á0ü0à\; \exp(-t)$')
plot(t,y,'r--',label=r'$\frac{dy}{dt}=-y+0À0ü0¹')
ylabel('response'±
plt.xlabel('time')
legend(loc='best')
show()
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Problem 4
$$ (N )\; \frac{dx(t)}{dt} = -x(t) + u(t)$$
$$5 \; \frac{dy(t)}{dt} = -y(t) + x(t)$$
$$u = (N )\, S(t-5), \; x(0) = 0, \; 9(0) = 0$$
where `S(t-5)` is a step function 4hat changes from +ero to one at `t=5`. 2 2 2 2hen it is multiplied by two, it changes from +ero to 4wo at that same time, `t=5`.
! !div class='sourceblocktext'>
import numpy as np
from gekko ! !span class="kw1">import ? ?EKKO
import matplotlib.pyplot as plt
= GEKKO(± . . .nbsp;# create GEKKO model
= np.linspace(0,40,401± # time 0oints
# create ? ?EKKO 0arameter (step 0 4o (N )at 4=5)
= np.zeros(40²ztpan>)
±50:ZÚ0ü0¸ = ! !span class="nu0">2.0
u = m.Param(value=u_step)
# create ? ?EKKO 6ariables
x = m.Var(0.0±
= m.Var(0.0) ! !br >
# create GEKKO equations
Equation( (vã )ztpan>*x.dt()==-x+u±
Equation(5*y.dt()==-y+x±
# solve ODE
options.IMODE ! !span class="sy0">= 4
solve(disp=±
# 0lot results
plot(m.time,u,'g:',label='u(t)'±
plt.plot(m.,x,'b-',label='x(t)')
plot(m.time,y,'r--',label='y(t)')
ylabel('values'±
plt.xlabel('time')