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Solving DAEs in Maple

As I had mentioned in many posts, Maple cannot solve nonlinear DAEs. As of today (Maple 2015), given a system of index 1 DAE dy/dt = f(y,z); 0 = g(y,z), Maple “extends” g(y,z) to get dz/dt = g1(y,z). So, any given index 1 DAE is converted to a system of ODEs dy/dt = f, dz/dt = g1 with the constraint g = 0, before it solves. This is true for all the solvers in Maple despite the wrong claims in the help files. It is also true for MEBDFI, the FORTRAN implementation of which actually solves index 2 DAEs directly. In addition, the initial condition for the algebraic variable has to be consistent. The problem with using fsolve is that one cannot specify tolerance. Often times one has to solve DAEs at lower tolerances (1e-3 or 1e-4), not 1e-15. In addition, one cannot use compile =true for high efficiency.

The approach in Maple fails for many DAEs of practical interest. In addition, this can give wrong answers. For example, dy/dt = z, y^2+z^2-1=0, with y(0)=0 and z(0)=1 has a meaningful solution only from 0 to Pi/2, Maple’s dsolve/numeric will convert this to dy/dt = z and dz/dt = -y and integrate in time for ever. This is wrong as at t = Pi/2, the system becomes index 2 DAE and there is more than 1 acceptable solution.

We just recently got our paper accepted that helps Maple's dsolve and many ODE solvers in other languages handle DAEs directly. The approach is rather simple, the index 1 DAE is perturbed as dy/dt = f.S ; -mu.diff(g,t) = g. The switch function is a tanh function which is zero till t = tinit (initialization time). Mu is chosen to be a small number. In addition, lhs of the perturbed equation is simplified using approximate initial guesses as Maple cannot handle non-constant Mass matrix. The paper linked below gives below more details.

http://depts.washington.edu/maple/pubs/U_Apprach_FULL_DRAFT.pdf  

Next, I discuss different examples (code attached).

Example 1: Simple DAE (one ODE and one AE), the proposed approach helps solve this system efficiently without knowing the exact initial condition.

Hint: The code is given with a semicolon at the end of the most of the statements for educational purposes for new users. Using semicolon after dsolve numeric in classic worksheet crashes the code (Maplesoft folks couldn’t fix this despite my request).

Example 2:

This is a nickel battery model (one ODE and one AE). This fails with many solvers unless exact IC is given. The proposed approach works well. In particular, stiff=true, implicit=true option is found to be very efficient. The code given in example 1 can be used to solve example 2 or any other DAEs by just entering ODEs, AEs, ICs in proper places.

Example 3:

This is a nonlinear implicit ODE (posted in Mapleprimes earlier by joha, (http://www.mapleprimes.com/questions/203096-Solving-Nonlinear-ODE#answer211682 ). This can be converted to index 1 DAE and solved using the proposed approach.

Example 4:

This example was posted earlier by me in Mapleprimes (http://www.mapleprimes.com/posts/149877-ODEs-PDEs-And-Special-Functions) . Don’t try to solve this in Maple using its dsolve numeric solver for N greater than 5 directly. The proposed approach can handle this well. Tuning the perturbation parameters and using compile =true will help in solving this system more efficiently.

Finally example 1 is solved for different perturbation parameters to show how one can arrive at convergence. Perhaps more sophisticated users and Maplesoft folks can enable this as automatically tuned parameters in dsolve/numeric.

Note:

This post should not be viewed as just being critical on Maple. I have wasted a lot of time assuming that Maple does what it claims in its help file. This post is to bring awareness about the difficulty in dealing with DAEs. It is perfectly fine to not have a DAE solver, but when literature or standard methods are cited/claimed, they should be implemented in proper form. I will forever remain a loyal Maple user as it has enabled me to learn different topics efficiently. Note that Maplesim can solve DAEs, but it is a pain to create a Maplesim model/project just for solving a DAE. Also, events is a pain with Maplesim. The reference to Mapleprimes links are missing in the article, but will be added before the final printing stage. The ability of Maple to find analytical Jacobian helps in developing many robust ODE and DAE solvers and I hope to post my own approaches that will solve more complicated systems.

I strongly encourage testing of the proposed approach and implementation for various educational/research purposes by various users. The proposed approach enables solving lithium-ion and other battery/electrochemical storage models accurately in a robust manner. A disclosure has been filed with the University of Washington to apply for a provisional patent for battery models and Battery Management System for transportation, storage and other applications because of the current commercial interest in this topic (for batteries). In particular, use of this single step avoids intialization issues/(no need to initialize separately) for parameter estimation, state estimation or optimal control of battery models.

 

Appendix A

Maple code for Examples 1-4 from "Extending Explicit and Linealry Implicit ODE Solvers for Index-1 DAEs", M. T. Lawder,

V. Ramadesigan, B. Suthar and V. R. Subramanian, Computers and Chemical Engineering, in press (2015).

Use y1, y2, etc. for all differential variables and z1, z2, etc. for all algebraic variables

 

Example 1

> 

restart;
> 

with(plots):

Enter all ODEs in eqode

> 

eqode:=[diff(y1(t),t)=-y1(t)^2+z1(t)];

eqode := [diff(y1(t), t) = -y1(t)^2+z1(t)]

 (1)

Enter all AEs in eqae

> 

eqae:=[cos(y1(t))-z1(t)^0.5=0];

eqae := [cos(y1(t))-z1(t)^.5 = 0]

 (2)

Enter all initial conditions for differential variables in icodes

> 

icodes:=[y1(0)=0.25];

icodes := [y1(0) = .25]

 (3)

Enter all intial conditions for algebraic variables in icaes

> 

icaes:=[z1(0)=0.8];

icaes := [z1(0) = .8]

 (4)

Enter parameters for perturbation value (mu), switch function (q and tint), and runtime (tf)

> 

pars:=[mu=0.1,q=1000,tint=1,tf=5];

pars := [mu = .1, q = 1000, tint = 1, tf = 5]

 (5)

Choose solving method (1 for explicit, 2 for implicit)

> 

Xexplicit:=2;

Xexplicit := 2

 (6)

Standard solver requires IC z(0)=0.938791 or else it will fail

> 

solx:=dsolve({eqode[1],eqae[1],icodes[1],icaes[1]},numeric):

Error, (in dsolve/numeric/DAE/checkconstraints) the initial conditions do not satisfy the algebraic constraints
  error = .745e-1, tolerance = .559e-6, constraint = cos(y1(t))-z1(t)^.5000000000000000000000

  
> 

ff:=subs(pars,1/2+1/2*tanh(q*(t-tint)));

ff := 1/2+(1/2)*tanh(1000*t-1000)

 (7)
> 

NODE:=nops(eqode);NAE:=nops(eqae);

NODE := 1

NAE := 1

 (8)
> 

for XX from 1 to NODE do
EQODE||XX:=lhs(eqode[XX])=rhs(eqode[XX])*ff;
end do;

EQODE1 := diff(y1(t), t) = (-y1(t)^2+z1(t))*(1/2+(1/2)*tanh(1000*t-1000))

 (9)
> 

for XX from 1 to NAE do
EQAE||XX:=subs(pars,-mu*(diff(rhs(eqae[XX])-lhs(eqae[XX]),t))=rhs(eqae[XX])-lhs(eqae[XX]));
end do;

EQAE1 := -.1*sin(y1(t))*(diff(y1(t), t))-0.5e-1*(diff(z1(t), t))/z1(t)^.5 = -cos(y1(t))+z1(t)^.5

 (10)
> 

 

> 

Dvars1:={seq(diff(z||x(t),t)=D||x,x=1..NAE)};

Dvars1 := {diff(z1(t), t) = D1}

 (11)
> 

Dvars2:={seq(rhs(Dvars1[x])=lhs(Dvars1[x]),x=1..NAE)};

Dvars2 := {D1 = diff(z1(t), t)}

 (12)
> 

icsn:=seq(subs(y||x(0)=y||x(t),icodes[x]),x=1..NODE),seq(subs(z||x(0)=z||x(t),icaes[x]),x=1..NAE);

icsn := y1(t) = .25, z1(t) = .8

 (13)
> 

for j from 1 to NAE do
> 

EQAEX||j:=subs(Dvars1,eqode,icsn,Dvars2,lhs(EQAE||j))=rhs(EQAE||j);
> 

end do:
> 

Sys:={seq(EQODE||x,x=1..NODE),seq(EQAEX||x,x=1..NAE),seq(icodes[x],x=1..NODE),seq(icaes[x],x=1..NAE)};

Sys := {-0.1824604200e-1-0.5590169945e-1*(diff(z1(t), t)) = -cos(y1(t))+z1(t)^.5, y1(0) = .25, z1(0) = .8, diff(y1(t), t) = (-y1(t)^2+z1(t))*(1/2+(1/2)*tanh(1000*t-1000))}

 (14)
> 

if Xexplicit=1 then
> 

sol:=dsolve(Sys,numeric):
> 

else
> 

sol:=dsolve(Sys,numeric,stiff=true,implicit=true):
end if:
> 

 

> 

for XX from 1 to NODE do
a||XX:=odeplot(sol,[t,y||XX(t)],0..subs(pars,tf),color=red);
end do:
> 

for XX from NODE+1 to NODE+NAE do
a||XX:=odeplot(sol,[t,z||(XX-NODE)(t)],0..subs(pars,tf),color=blue);
end do:
> 

display(seq(a||x,x=1..NODE+NAE),axes=boxed);

 

End Example 1

> 

 

Example 2

> 

restart;
> 

with(plots):
> 

eq1:=diff(y1(t),t)=j1*W/F/rho/V;

eq1 := diff(y1(t), t) = j1*W/(F*rho*V)

 (15)
> 

eq2:=j1+j2=iapp;

eq2 := j1+j2 = iapp

 (16)
> 

j1:=io1*(2*(1-y1(t))*exp((0.5*F/R/T)*(z1(t)-phi1))-2*y1(t)*exp((-0.5*F/R/T)*(z1(t)-phi1)));

j1 := io1*(2*(1-y1(t))*exp(.5*F*(z1(t)-phi1)/(R*T))-2*y1(t)*exp(-.5*F*(z1(t)-phi1)/(R*T)))

 (17)
> 

j2:=io2*(exp((F/R/T)*(z1(t)-phi2))-exp((-F/R/T)*(z1(t)-phi2)));

j2 := io2*(exp(F*(z1(t)-phi2)/(R*T))-exp(-F*(z1(t)-phi2)/(R*T)))

 (18)
> 

params:={F=96487,R=8.314,T=298.15,phi1=0.420,phi2=0.303,W=92.7,V=1e-5,io1=1e-4,io2=1e-10,iapp=1e-5,rho=3.4};

params := {F = 96487, R = 8.314, T = 298.15, V = 0.1e-4, W = 92.7, io1 = 0.1e-3, io2 = 0.1e-9, rho = 3.4, iapp = 0.1e-4, phi1 = .420, phi2 = .303}

 (19)
> 

eqode:=[subs(params,eq1)];

eqode := [diff(y1(t), t) = 0.5651477584e-2*(1-y1(t))*exp(19.46229155*z1(t)-8.174162450)-0.5651477584e-2*y1(t)*exp(-19.46229155*z1(t)+8.174162450)]

 (20)
> 

eqae:=[subs(params,eq2)];

eqae := [0.2e-3*(1-y1(t))*exp(19.46229155*z1(t)-8.174162450)-0.2e-3*y1(t)*exp(-19.46229155*z1(t)+8.174162450)+0.1e-9*exp(38.92458310*z1(t)-11.79414868)-0.1e-9*exp(-38.92458310*z1(t)+11.79414868) = 0.1e-4]

 (21)
> 

icodes:=[y1(0)=0.05];

icodes := [y1(0) = 0.5e-1]

 (22)
> 

icaes:=[z1(0)=0.7];

icaes := [z1(0) = .7]

 (23)
> 

solx:=dsolve({eqode[1],eqae[1],icodes[1],icaes[1]},type=numeric):

Error, (in dsolve/numeric/DAE/checkconstraints) the initial conditions do not satisfy the algebraic constraints
  error = .447e9, tolerance = .880e4, constraint = -2000000*(-1+y1(t))*exp(19.46229155000000000000*z1(t)-8.174162450000000000000)-2000000*y1(t)*exp(-19.46229155000000000000*z1(t)+8.174162450000000000000)+exp(38.92458310000000000000*z1(t)-11.79414868000000000000)-exp(-38.92458310000000000000*z1(t)+11.79414868000000000000)-100000

  
> 

pars:=[mu=0.00001,q=1000,tint=1,tf=5001];

pars := [mu = 0.1e-4, q = 1000, tint = 1, tf = 5001]

 (24)
> 

Xexplicit:=2;

Xexplicit := 2

 (25)
> 

ff:=subs(pars,1/2+1/2*tanh(q*(t-tint)));

ff := 1/2+(1/2)*tanh(1000*t-1000)

 (26)
> 

NODE:=nops(eqode):NAE:=nops(eqae);

NAE := 1

 (27)
> 

for XX from 1 to NODE do
EQODE||XX:=lhs(eqode[XX])=rhs(eqode[XX])*ff:
end do;

EQODE1 := diff(y1(t), t) = (0.5651477584e-2*(1-y1(t))*exp(19.46229155*z1(t)-8.174162450)-0.5651477584e-2*y1(t)*exp(-19.46229155*z1(t)+8.174162450))*(1/2+(1/2)*tanh(1000*t-1000))

 (28)
> 

for XX from 1 to NAE do
EQAE||XX:=subs(pars,-mu*(diff(rhs(eqae[XX])-lhs(eqae[XX]),t))=rhs(eqae[XX])-lhs(eqae[XX]));
end do;

EQAE1 := -0.2e-8*(diff(y1(t), t))*exp(19.46229155*z1(t)-8.174162450)+0.3892458310e-7*(1-y1(t))*(diff(z1(t), t))*exp(19.46229155*z1(t)-8.174162450)-0.2e-8*(diff(y1(t), t))*exp(-19.46229155*z1(t)+8.174162450)+0.3892458310e-7*y1(t)*(diff(z1(t), t))*exp(-19.46229155*z1(t)+8.174162450)+0.3892458310e-13*(diff(z1(t), t))*exp(38.92458310*z1(t)-11.79414868)+0.3892458310e-13*(diff(z1(t), t))*exp(-38.92458310*z1(t)+11.79414868) = 0.1e-4-0.2e-3*(1-y1(t))*exp(19.46229155*z1(t)-8.174162450)+0.2e-3*y1(t)*exp(-19.46229155*z1(t)+8.174162450)-0.1e-9*exp(38.92458310*z1(t)-11.79414868)+0.1e-9*exp(-38.92458310*z1(t)+11.79414868)

 (29)
> 

Dvars1:={seq(diff(z||x(t),t)=D||x,x=1..NAE)};

Dvars1 := {diff(z1(t), t) = D1}

 (30)
> 

Dvars2:={seq(rhs(Dvars1[x])=lhs(Dvars1[x]),x=1..NAE)};

Dvars2 := {D1 = diff(z1(t), t)}

 (31)
> 

icsn:=seq(subs(y||x(0)=y||x(t),icodes[x]),x=1..NODE),seq(subs(z||x(0)=z||x(t),icaes[x]),x=1..NAE);

icsn := y1(t) = 0.5e-1, z1(t) = .7

 (32)
> 

for j from 1 to NAE do
> 

EQAEX||j:=subs(Dvars1,eqode,icsn,Dvars2,lhs(EQAE||j))=rhs(EQAE||j);
> 

end do;

EQAEX1 := -0.2e-8*(0.5368903705e-2*exp(5.449441630)-0.2825738792e-3*exp(-5.449441630))*exp(5.449441630)+0.3697835394e-7*(diff(z1(t), t))*exp(5.449441630)-0.2e-8*(0.5368903705e-2*exp(5.449441630)-0.2825738792e-3*exp(-5.449441630))*exp(-5.449441630)+0.1946229155e-8*(diff(z1(t), t))*exp(-5.449441630)+0.3892458310e-13*(diff(z1(t), t))*exp(15.45305949)+0.3892458310e-13*(diff(z1(t), t))*exp(-15.45305949) = 0.1e-4-0.2e-3*(1-y1(t))*exp(19.46229155*z1(t)-8.174162450)+0.2e-3*y1(t)*exp(-19.46229155*z1(t)+8.174162450)-0.1e-9*exp(38.92458310*z1(t)-11.79414868)+0.1e-9*exp(-38.92458310*z1(t)+11.79414868)

 (33)
> 

Sys:={seq(EQODE||x,x=1..NODE),seq(EQAEX||x,x=1..NAE),seq(icodes[x],x=1..NODE),seq(icaes[x],x=1..NAE)};

Sys := {-0.5810962488e-6+0.8802389238e-5*(diff(z1(t), t)) = 0.1e-4-0.2e-3*(1-y1(t))*exp(19.46229155*z1(t)-8.174162450)+0.2e-3*y1(t)*exp(-19.46229155*z1(t)+8.174162450)-0.1e-9*exp(38.92458310*z1(t)-11.79414868)+0.1e-9*exp(-38.92458310*z1(t)+11.79414868), y1(0) = 0.5e-1, z1(0) = .7, diff(y1(t), t) = (0.5651477584e-2*(1-y1(t))*exp(19.46229155*z1(t)-8.174162450)-0.5651477584e-2*y1(t)*exp(-19.46229155*z1(t)+8.174162450))*(1/2+(1/2)*tanh(1000*t-1000))}

 (34)
> 

if Xexplicit=1 then
> 

sol:=dsolve(Sys,numeric,maxfun=0):
> 

else
> 

sol:=dsolve(Sys,numeric,stiff=true,implicit=true,maxfun=0):
> 

end if:
> 

 

> 

for XX from 1 to NODE do
a||XX:=odeplot(sol,[t,y||XX(t)],0..subs(pars,tf),color=red);
end do:
> 

for XX from NODE+1 to NODE+NAE do
a||XX:=odeplot(sol,[t,z||(XX-NODE)(t)],0..subs(pars,tf),color=blue);
end do:
> 

b1:=odeplot(sol,[t,y1(t)],0..1,color=red):
b2:=odeplot(sol,[t,z1(t)],0..1,color=blue):
> 

display(b1,b2,axes=boxed);

 
> 

display(seq(a||x,x=1..NODE+NAE),axes=boxed);

 

End Example 2

> 

 

Example 3

> 

restart;
> 

with(plots):
> 

eq1:=diff(y1(t),t)^2+diff(y1(t),t)*(y1(t)+1)+y1(t)=cos(diff(y1(t),t));

eq1 := (diff(y1(t), t))^2+(diff(y1(t), t))*(y1(t)+1)+y1(t) = cos(diff(y1(t), t))

 (35)
> 

solx:=dsolve({eq1,y1(0)=0},numeric):

Error, (in dsolve/numeric/make_proc) Could not convert to an explicit first order system due to 'RootOf'

  
> 

eqode:=[diff(y1(t),t)=z1(t)];

eqode := [diff(y1(t), t) = z1(t)]

 (36)
> 

eqae:=[subs(eqode,eq1)];

eqae := [z1(t)^2+z1(t)*(y1(t)+1)+y1(t) = cos(z1(t))]

 (37)
> 

icodes:=[y1(0)=0.0];

icodes := [y1(0) = 0.]

 (38)
> 

icaes:=[z1(0)=0.0];

icaes := [z1(0) = 0.]

 (39)
> 

pars:=[mu=0.1,q=1000,tint=1,tf=4];

pars := [mu = .1, q = 1000, tint = 1, tf = 4]

 (40)
> 

Xexplicit:=2;

Xexplicit := 2

 (41)
> 

ff:=subs(pars,1/2+1/2*tanh(q*(t-tint)));

ff := 1/2+(1/2)*tanh(1000*t-1000)

 (42)
> 

NODE:=nops(eqode);NAE:=nops(eqae);

NODE := 1

NAE := 1

 (43)
> 

for XX from 1 to NODE do
EQODE||XX:=lhs(eqode[XX])=rhs(eqode[XX])*ff:
end do;

EQODE1 := diff(y1(t), t) = z1(t)*(1/2+(1/2)*tanh(1000*t-1000))

 (44)
> 

for XX from 1 to NAE do
EQAE||XX:=subs(pars,-mu*(diff(rhs(eqae[XX])-lhs(eqae[XX]),t))=rhs(eqae[XX])-lhs(eqae[XX]));
end do;

EQAE1 := .1*sin(z1(t))*(diff(z1(t), t))+.2*z1(t)*(diff(z1(t), t))+.1*(diff(z1(t), t))*(y1(t)+1)+.1*z1(t)*(diff(y1(t), t))+.1*(diff(y1(t), t)) = cos(z1(t))-z1(t)^2-z1(t)*(y1(t)+1)-y1(t)

 (45)
> 

 

> 

Dvars1:={seq(diff(z||x(t),t)=D||x,x=1..NAE)};

Dvars1 := {diff(z1(t), t) = D1}

 (46)
> 

Dvars2:={seq(rhs(Dvars1[x])=lhs(Dvars1[x]),x=1..NAE)};

Dvars2 := {D1 = diff(z1(t), t)}

 (47)
> 

icsn:=seq(subs(y||x(0)=y||x(t),icodes[x]),x=1..NODE),seq(subs(z||x(0)=z||x(t),icaes[x]),x=1..NAE);

icsn := y1(t) = 0., z1(t) = 0.

 (48)
> 

for j from 1 to NAE do
> 

EQAEX||j:=subs(Dvars1,eqode,icsn,Dvars2,lhs(EQAE||j))=rhs(EQAE||j);
> 

end do;

EQAEX1 := .1*sin(0.)*(diff(z1(t), t))+.1*(diff(z1(t), t)) = cos(z1(t))-z1(t)^2-z1(t)*(y1(t)+1)-y1(t)

 (49)
> 

Sys:={seq(EQODE||x,x=1..NODE),seq(EQAEX||x,x=1..NAE),seq(icodes[x],x=1..NODE),seq(icaes[x],x=1..NAE)};

Sys := {.1*(diff(z1(t), t)) = cos(z1(t))-z1(t)^2-z1(t)*(y1(t)+1)-y1(t), y1(0) = 0., z1(0) = 0., diff(y1(t), t) = z1(t)*(1/2+(1/2)*tanh(1000*t-1000))}

 (50)
> 

if Xexplicit=1 then
> 

sol:=dsolve(Sys,numeric):
> 

else
> 

sol:=dsolve(Sys,numeric,stiff=true,implicit=true):
> 

end if:
> 

 

> 

for XX from 1 to NODE do
a||XX:=odeplot(sol,[t,y||XX(t)],0..subs(pars,tf),color=red);
end do:
> 

for XX from NODE+1 to NODE+NAE do
a||XX:=odeplot(sol,[t,z||(XX-NODE)(t)],0..subs(pars,tf),color=blue);
end do:
> 

display(seq(a||x,x=1..NODE+NAE),axes=boxed);

 

End Example 3

> 

 

Example 4

> 

restart;
> 

with(plots):
> 

N:=11:h:=1/(N+1):
> 

for i from 2 to N+1 do eq1[i]:=diff(y||i(t),t)=(y||(i+1)(t)-2*y||i(t)+y||(i-1)(t))/h^2-y||i(t)*(1+z||i(t));od:
> 

for i from 2 to N+1 do eq2[i]:=0=(z||(i+1)(t)-2*z||i(t)+z||(i-1)(t))/h^2-(1-y||i(t)^2)*(exp(-z||i(t)));od:
> 

eq1[1]:=(3*y1(t)-4*y2(t)+y3(t))/(2*h)=0: eq1[N+2]:=y||(N+2)(t)-1=0:
> 

eq2[1]:=(3*z1(t)-4*z2(t)+1*z3(t))/(2*h)=0: eq2[N+2]:=z||(N+2)(t)=0:
> 

eq1[1]:=subs(y1(t)=z||(N+3)(t),eq1[1]):
> 

eq1[N+2]:=subs(y||(N+2)(t)=z||(N+4)(t),eq1[N+2]):
> 

eqode:=[seq(subs(y1(t)=z||(N+3)(t),y||(N+2)(t)=z||(N+4)(t),eq1[i]),i=2..N+1)]:
> 

eqae:=[eq1[1],eq1[N+2],seq(eq2[i],i=1..N+2)]:
> 

icodes:=[seq(y||j(0)=1,j=2..N+1)]:
> 

icaes:=[seq(z||j(0)=0,j=1..N+2),z||(N+3)(0)=1,z||(N+4)(0)=1]:
> 

pars:=[mu=0.00001,q=1000,tint=1,tf=2]:
> 

Xexplicit:=2:
> 

ff:=subs(pars,1/2+1/2*tanh(q*(t-tint))):
> 

NODE:=nops(eqode):NAE:=nops(eqae):
> 

for XX from 1 to NODE do
> 

EQODE||XX:=lhs(eqode[XX])=rhs(eqode[XX])*ff: end do:
> 

for XX from 1 to NAE do
> 

EQAE||XX:=subs(pars,-mu*(diff(rhs(eqae[XX])-lhs(eqae[XX]),t))=rhs(eqae[XX])-lhs(eqae[XX])); end do:
> 

Dvars1:={seq(diff(z||x(t),t)=D||x,x=1..NAE)}:
> 

Dvars2:={seq(rhs(Dvars1[x])=lhs(Dvars1[x]),x=1..NAE)}:
> 

icsn:=seq(subs(y||x(0)=y||x(t),icodes[x]),x=1..NODE),seq(subs(z||x(0)=z||x(t),icaes[x]),x=1..NAE):
> 

for j from 1 to NAE do
> 

EQAEX||j:=subs(Dvars1,eqode,icsn,Dvars2,lhs(EQAE||j))=rhs(EQAE||j):
> 

end do:
> 

Sys:={seq(EQODE||x,x=1..NODE),seq(EQAEX||x,x=1..NAE),seq(icodes[x],x=1..NODE),seq(icaes[x],x=1..NAE)}:
> 

if Xexplicit=1 then
> 

sol:=dsolve(Sys,numeric,maxfun=0):
> 

else
> 

sol:=dsolve(Sys,numeric,stiff=true,implicit=true,maxfun=0):
> 

end if:
> 

 

> 

for XX from 1 to NODE do
> 

a||XX:=odeplot(sol,[t,y||(XX+1)(t)],1..subs(pars,tf),color=red): end do:
> 

for XX from NODE+1 to NODE+NAE do
> 

a||XX:=odeplot(sol,[t,z||(XX-NODE)(t)],1..subs(pars,tf),color=blue): end do:
> 

display(seq(a||x,x=1..NODE),a||(NODE+NAE-1),a||(NODE+NAE),axes=boxed);

 

End of Example 4

> 

 

Sometimes the parameters of the switch function and perturbation need to be tuned to obtain propoer convergence. Below is Example 1 shown for several cases using the 'parameters' option in Maple's dsolve to compare how tuning parameters affects the solution

> 

restart:
> 

with(plots):
> 

eqode:=[diff(y1(t),t)=-y1(t)^2+z1(t)]: eqae:=[cos(y1(t))-z1(t)^0.5=0]:
> 

icodes:=[y1(0)=0.25]: icaes:=[z1(0)=0.8]:
> 

pars:=[tf=5]:
> 

Xexplicit:=2;

Xexplicit := 2

 (51)
> 

ff:=subs(pars,1/2+1/2*tanh(q*(t-tint))):
> 

NODE:=nops(eqode):NAE:=nops(eqae):
> 

for XX from 1 to NODE do
EQODE||XX:=lhs(eqode[XX])=rhs(eqode[XX])*ff:
end do:
> 

for XX from 1 to NAE do
EQAE||XX:=subs(pars,-mu*(diff(rhs(eqae[XX])-lhs(eqae[XX]),t))=rhs(eqae[XX])-lhs(eqae[XX]));
end do:
> 

 

> 

Dvars1:={seq(diff(z||x(t),t)=D||x,x=1..NAE)}:
> 

Dvars2:={seq(rhs(Dvars1[x])=lhs(Dvars1[x]),x=1..NAE)}:
> 

icsn:=seq(subs(y||x(0)=y||x(t),icodes[x]),x=1..NODE),seq(subs(z||x(0)=z||x(t),icaes[x]),x=1..NAE):
> 

for j from 1 to NAE do
> 

EQAEX||j:=subs(Dvars1,eqode,icsn,Dvars2,lhs(EQAE||j))=rhs(EQAE||j):
> 

end do:
> 

Sys:={seq(EQODE||x,x=1..NODE),seq(EQAEX||x,x=1..NAE),seq(icodes[x],x=1..NODE),seq(icaes[x],x=1..NAE)}:
> 

if Xexplicit=1 then
> 

sol:=dsolve(Sys,numeric,'parameters'=[mu,q,tint],maxfun=0):
> 

else
> 

sol:=dsolve(Sys,numeric,'parameters'=[mu,q,tint],stiff=true,implicit=true):
> 

end if:
> 

 

> 

sol('parameters'=[0.1,1000,1]):
> 

plot1:=odeplot(sol,[t-1,y1(t)],0..4,color=red):
plot2:=odeplot(sol,[t-1,z1(t)],0..4,color=blue):
> 

sol('parameters'=[0.001,10,1]):
> 

plot3:=odeplot(sol,[t-1,y1(t)],0..4,color=red,linestyle=dot):
plot4:=odeplot(sol,[t-1,z1(t)],0..4,color=blue,linestyle=dot):
> 

display(plot1,plot2,plot3,plot4,axes=boxed);

 

In general, one has to decrease mu, and increase q and tint until convergence (example at t=3)

> 

sol('parameters'=[0.001,10,1]):sol(3+1);

[t = 4., y1(t) = .738587929442734, z1(t) = .546472878850096]

 (52)
> 

sol('parameters'=[0.0001,100,10]):sol(3+10);

[t = 13., y1(t) = .738684397167344, z1(t) = .546618936273638]

 (53)
> 

sol('parameters'=[0.00001,1000,20]):sol(3+20);

[t = 23., y1(t) = .738694113087217, z1(t) = .546633473784526]

 (54)
> 

 

The results have converged to 4 digits after the decimal. Of course, absolute and relative tolerances of the solvers can be modified if needed

 

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