Plotting a solution of a differential equation with Sagemath












1












$begingroup$


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







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 25 '18 at 6:04







      Relure

















      asked Nov 25 '18 at 2:33









      RelureRelure

      2,1871035




      2,1871035






















          1 Answer
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          2












          $begingroup$

          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)


          q(t)



          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


          T(t)






          share|cite|improve this answer









          $endgroup$














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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            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)


            q(t)



            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


            T(t)






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              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)


              q(t)



              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


              T(t)






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                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)


                q(t)



                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


                T(t)






                share|cite|improve this answer









                $endgroup$



                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)


                q(t)



                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


                T(t)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 15 at 17:25









                Ricardo BuringRicardo Buring

                1,51111334




                1,51111334






























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