Plotting a solution of a differential equation with Sagemath
$begingroup$
I need to solve a differential equation. The solution will depend on $t$ and $q$, and I need to define that $q$ piecewise depending on $t$.
var('k,Tmax,Tmin,w,T0,q'); T=function('T')(t); Te=function('Te')(t);
assume(k>0); assume(Tmax>Tmin); Te(t)=(Tmax+Tmin)/2+(Tmax-Tmin)/2*sin(w*t);
Now this is the differential equation solution:
sol=desolve(diff(T(t),t)-q+k*(T(t)-Te(t)),[T,t],[0,T0]);
The solution with $q=0$ for example would be
sol.subs(Tmax=21.6,Tmin=15.2,k=0.024,q=0,T0=15.6,w=pi/12);
but I need that q to be a model for a heater that's on from 8 AM to 22 PM, and off from 22 PM to 8 AM. So I need to define a $q$ function that if $t mod 24$ is between $8$ and $22$ its value is $0.0504$, and $0$ otherwise. Something like this
$$q(t)=begin{cases}0.0504 quad if quad t mod 24 in[8,22] \ 0 qquad qquad otherwiseend{cases}$$
I've been trying with the piecewise function but it's not plotting anything, I always get error messages.
Thanks for your time.
calculus ordinary-differential-equations graphing-functions piecewise-continuity sagemath
asked Nov 25 '18 at 2:33
RelureRelure
2,1871035
$endgroup$
add a comment |
$begingroup$
I need to solve a differential equation. The solution will depend on $t$ and $q$, and I need to define that $q$ piecewise depending on $t$.
var('k,Tmax,Tmin,w,T0,q'); T=function('T')(t); Te=function('Te')(t);
assume(k>0); assume(Tmax>Tmin); Te(t)=(Tmax+Tmin)/2+(Tmax-Tmin)/2*sin(w*t);
Now this is the differential equation solution:
sol=desolve(diff(T(t),t)-q+k*(T(t)-Te(t)),[T,t],[0,T0]);
The solution with $q=0$ for example would be
sol.subs(Tmax=21.6,Tmin=15.2,k=0.024,q=0,T0=15.6,w=pi/12);
but I need that q to be a model for a heater that's on from 8 AM to 22 PM, and off from 22 PM to 8 AM. So I need to define a $q$ function that if $t mod 24$ is between $8$ and $22$ its value is $0.0504$, and $0$ otherwise. Something like this
$$q(t)=begin{cases}0.0504 quad if quad t mod 24 in[8,22] \ 0 qquad qquad otherwiseend{cases}$$
I've been trying with the piecewise function but it's not plotting anything, I always get error messages.
Thanks for your time.
calculus ordinary-differential-equations graphing-functions piecewise-continuity sagemath
asked Nov 25 '18 at 2:33
RelureRelure
2,1871035
$endgroup$
add a comment |
$begingroup$
I need to solve a differential equation. The solution will depend on $t$ and $q$, and I need to define that $q$ piecewise depending on $t$.
var('k,Tmax,Tmin,w,T0,q'); T=function('T')(t); Te=function('Te')(t);
assume(k>0); assume(Tmax>Tmin); Te(t)=(Tmax+Tmin)/2+(Tmax-Tmin)/2*sin(w*t);
Now this is the differential equation solution:
sol=desolve(diff(T(t),t)-q+k*(T(t)-Te(t)),[T,t],[0,T0]);
The solution with $q=0$ for example would be
sol.subs(Tmax=21.6,Tmin=15.2,k=0.024,q=0,T0=15.6,w=pi/12);
but I need that q to be a model for a heater that's on from 8 AM to 22 PM, and off from 22 PM to 8 AM. So I need to define a $q$ function that if $t mod 24$ is between $8$ and $22$ its value is $0.0504$, and $0$ otherwise. Something like this
$$q(t)=begin{cases}0.0504 quad if quad t mod 24 in[8,22] \ 0 qquad qquad otherwiseend{cases}$$
I've been trying with the piecewise function but it's not plotting anything, I always get error messages.
Thanks for your time.
calculus ordinary-differential-equations graphing-functions piecewise-continuity sagemath
asked Nov 25 '18 at 2:33
RelureRelure
2,1871035
$endgroup$
I need to solve a differential equation. The solution will depend on $t$ and $q$, and I need to define that $q$ piecewise depending on $t$.
var('k,Tmax,Tmin,w,T0,q'); T=function('T')(t); Te=function('Te')(t);
assume(k>0); assume(Tmax>Tmin); Te(t)=(Tmax+Tmin)/2+(Tmax-Tmin)/2*sin(w*t);
Now this is the differential equation solution:
sol=desolve(diff(T(t),t)-q+k*(T(t)-Te(t)),[T,t],[0,T0]);
The solution with $q=0$ for example would be
sol.subs(Tmax=21.6,Tmin=15.2,k=0.024,q=0,T0=15.6,w=pi/12);
but I need that q to be a model for a heater that's on from 8 AM to 22 PM, and off from 22 PM to 8 AM. So I need to define a $q$ function that if $t mod 24$ is between $8$ and $22$ its value is $0.0504$, and $0$ otherwise. Something like this
$$q(t)=begin{cases}0.0504 quad if quad t mod 24 in[8,22] \ 0 qquad qquad otherwiseend{cases}$$
I've been trying with the piecewise function but it's not plotting anything, I always get error messages.
Thanks for your time.
calculus ordinary-differential-equations graphing-functions piecewise-continuity sagemath
calculus ordinary-differential-equations graphing-functions piecewise-continuity sagemath
asked Nov 25 '18 at 2:33
RelureRelure
2,1871035
asked Nov 25 '18 at 2:33
RelureRelure
2,1871035
edited Nov 25 '18 at 6:04
Relure
asked Nov 25 '18 at 2:33
RelureRelure
2,1871035
asked Nov 25 '18 at 2:33
RelureRelure
2,1871035
asked Nov 25 '18 at 2:33
RelureRelure
2,1871035
2,1871035
add a comment |
add a comment |
1 Answer
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$begingroup$
You can express $q(t)$ as a sum of differences of step functions. Also, it's more efficient to solve the differential equation numerically. I assume you want to plot the solution for some number of days (which can be specified in the code).
days=3
Tmax=21.6; Tmin=15.2; k=0.024; T0=15.6; w=pi/12
var('t');
Te=(Tmax+Tmin)/2+(Tmax-Tmin)/2*sin(w*t)
We define $q(t)$:
q = 0.0504*sum(unit_step(t-8-d*24) - unit_step(t-22-d*24) for d in range(days))
plot(q,t,0,days*24)
We define the ODE (also the one with $q(t)=0$, to compare):
T=function('T')(t);
ode0 = diff(T,t) == -k*(T-Te)
ode1 = diff(T,t) == q-k*(T-Te)
Finally we solve and plot:
sol0=desolve_rk4(ode0, T, ivar=t, ics=[0,T0], step=0.1, end_points=[0,days*24], output='plot', xmin=0,xmax=days*24)
sol1=desolve_rk4(ode1, T, ivar=t, ics=[0,T0], step=0.1, end_points=[0,days*24], output='plot', xmin=0,xmax=days*24, color='red')
sol0 + sol1
answered Jan 15 at 17:25
Ricardo BuringRicardo Buring
1,51111334
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can express $q(t)$ as a sum of differences of step functions. Also, it's more efficient to solve the differential equation numerically. I assume you want to plot the solution for some number of days (which can be specified in the code).
days=3
Tmax=21.6; Tmin=15.2; k=0.024; T0=15.6; w=pi/12
var('t');
Te=(Tmax+Tmin)/2+(Tmax-Tmin)/2*sin(w*t)
We define $q(t)$:
q = 0.0504*sum(unit_step(t-8-d*24) - unit_step(t-22-d*24) for d in range(days))
plot(q,t,0,days*24)
We define the ODE (also the one with $q(t)=0$, to compare):
T=function('T')(t);
ode0 = diff(T,t) == -k*(T-Te)
ode1 = diff(T,t) == q-k*(T-Te)
Finally we solve and plot:
sol0=desolve_rk4(ode0, T, ivar=t, ics=[0,T0], step=0.1, end_points=[0,days*24], output='plot', xmin=0,xmax=days*24)
sol1=desolve_rk4(ode1, T, ivar=t, ics=[0,T0], step=0.1, end_points=[0,days*24], output='plot', xmin=0,xmax=days*24, color='red')
sol0 + sol1
answered Jan 15 at 17:25
Ricardo BuringRicardo Buring
1,51111334
$endgroup$
add a comment |
$begingroup$
You can express $q(t)$ as a sum of differences of step functions. Also, it's more efficient to solve the differential equation numerically. I assume you want to plot the solution for some number of days (which can be specified in the code).
days=3
Tmax=21.6; Tmin=15.2; k=0.024; T0=15.6; w=pi/12
var('t');
Te=(Tmax+Tmin)/2+(Tmax-Tmin)/2*sin(w*t)
We define $q(t)$:
q = 0.0504*sum(unit_step(t-8-d*24) - unit_step(t-22-d*24) for d in range(days))
plot(q,t,0,days*24)
We define the ODE (also the one with $q(t)=0$, to compare):
T=function('T')(t);
ode0 = diff(T,t) == -k*(T-Te)
ode1 = diff(T,t) == q-k*(T-Te)
Finally we solve and plot:
sol0=desolve_rk4(ode0, T, ivar=t, ics=[0,T0], step=0.1, end_points=[0,days*24], output='plot', xmin=0,xmax=days*24)
sol1=desolve_rk4(ode1, T, ivar=t, ics=[0,T0], step=0.1, end_points=[0,days*24], output='plot', xmin=0,xmax=days*24, color='red')
sol0 + sol1
answered Jan 15 at 17:25
Ricardo BuringRicardo Buring
1,51111334
$endgroup$
add a comment |
$begingroup$
You can express $q(t)$ as a sum of differences of step functions. Also, it's more efficient to solve the differential equation numerically. I assume you want to plot the solution for some number of days (which can be specified in the code).
days=3
Tmax=21.6; Tmin=15.2; k=0.024; T0=15.6; w=pi/12
var('t');
Te=(Tmax+Tmin)/2+(Tmax-Tmin)/2*sin(w*t)
We define $q(t)$:
q = 0.0504*sum(unit_step(t-8-d*24) - unit_step(t-22-d*24) for d in range(days))
plot(q,t,0,days*24)
We define the ODE (also the one with $q(t)=0$, to compare):
T=function('T')(t);
ode0 = diff(T,t) == -k*(T-Te)
ode1 = diff(T,t) == q-k*(T-Te)
Finally we solve and plot:
sol0=desolve_rk4(ode0, T, ivar=t, ics=[0,T0], step=0.1, end_points=[0,days*24], output='plot', xmin=0,xmax=days*24)
sol1=desolve_rk4(ode1, T, ivar=t, ics=[0,T0], step=0.1, end_points=[0,days*24], output='plot', xmin=0,xmax=days*24, color='red')
sol0 + sol1
answered Jan 15 at 17:25
Ricardo BuringRicardo Buring
1,51111334
$endgroup$
You can express $q(t)$ as a sum of differences of step functions. Also, it's more efficient to solve the differential equation numerically. I assume you want to plot the solution for some number of days (which can be specified in the code).
days=3
Tmax=21.6; Tmin=15.2; k=0.024; T0=15.6; w=pi/12
var('t');
Te=(Tmax+Tmin)/2+(Tmax-Tmin)/2*sin(w*t)
We define $q(t)$:
q = 0.0504*sum(unit_step(t-8-d*24) - unit_step(t-22-d*24) for d in range(days))
plot(q,t,0,days*24)
We define the ODE (also the one with $q(t)=0$, to compare):
T=function('T')(t);
ode0 = diff(T,t) == -k*(T-Te)
ode1 = diff(T,t) == q-k*(T-Te)
Finally we solve and plot:
sol0=desolve_rk4(ode0, T, ivar=t, ics=[0,T0], step=0.1, end_points=[0,days*24], output='plot', xmin=0,xmax=days*24)
sol1=desolve_rk4(ode1, T, ivar=t, ics=[0,T0], step=0.1, end_points=[0,days*24], output='plot', xmin=0,xmax=days*24, color='red')
sol0 + sol1
answered Jan 15 at 17:25
Ricardo BuringRicardo Buring
1,51111334
answered Jan 15 at 17:25
Ricardo BuringRicardo Buring
1,51111334
answered Jan 15 at 17:25
Ricardo BuringRicardo Buring
1,51111334
answered Jan 15 at 17:25
Ricardo BuringRicardo Buring
1,51111334
1,51111334
add a comment |
add a comment |
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