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Mathematics Calculus Equation Solving Graphics and Visualization Wolfram Language Symbolic Computations

DSolve has reached the maximum number of calculations

James James

Posted 2 months ago

I reached the maximum number of calculations when using DSolve to calculate differential equations, which makes it impossible to fully solve. How can I expand the number of calculations of DSolve? Here is my code： system = {vi q[t] == l iL'[t] + vC[t], vC'[t] == iL[t]/c - vC[t]/(r c), vC[0] == 0, iL[0] == 0}; control = {q[0] == 1, WhenEvent[Mod[t, \[Tau]] == (2/3) \[Tau], q[t] -> 0], WhenEvent[Mod[t, \[Tau]] == 0, q[t] -> 1]}; pars = {vi -> 24, r -> 22, l -> 2 10^-2, c -> 1 10^-4, \[Tau] -> 2.5 10^-5}; sol = DSolve[{system, control} /. pars, {vC, iL, q}, {t, 0, .2}, DiscreteVariables -> q]; a = Evaluate[iL[t] /. sol]; b = Evaluate[vC[t] /. sol]; Plot[a, {t, 0, 0.2}, AxesLabel -> {"s", "il[t]/A"}, PlotLegends -> {"LinearlyImplicitEuler"}, PlotStyle -> {Red}, PlotRange -> All] Plot[b, {t, 0, 0.2}, AxesLabel -> {"s", "vc[t]/V"}, PlotLegends -> {"LinearlyImplicitEuler"}, PlotStyle -> {Blue}, PlotRange -> All]

I reached the maximum number of calculations when using DSolve to calculate differential equations, which makes it impossible to fully solve. How can I expand the number of calculations of DSolve?

Here is my code：

system = {vi q[t] == l iL'[t] + vC[t], 
   vC'[t] == iL[t]/c - vC[t]/(r c), vC[0] == 0, iL[0] == 0};
control = {q[0] == 1, 
   WhenEvent[Mod[t, \[Tau]] == (2/3) \[Tau], q[t] -> 0], 
   WhenEvent[Mod[t, \[Tau]] == 0, q[t] -> 1]};
pars = {vi -> 24, r -> 22, l -> 2 10^-2, 
   c -> 1 10^-4, \[Tau] -> 2.5 10^-5};
sol = DSolve[{system, control} /. pars, {vC, iL, q}, {t, 0, .2}, 
   DiscreteVariables -> q];
a = Evaluate[iL[t] /. sol];
b = Evaluate[vC[t] /. sol];
Plot[a, {t, 0, 0.2}, AxesLabel -> {"s", "il[t]/A"}, 
 PlotLegends -> {"LinearlyImplicitEuler"}, PlotStyle -> {Red}, 
 PlotRange -> All]
Plot[b, {t, 0, 0.2}, AxesLabel -> {"s", "vc[t]/V"}, 
 PlotLegends -> {"LinearlyImplicitEuler"}, PlotStyle -> {Blue}, 
 PlotRange -> All]

POSTED BY: James James

7 Replies

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Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 2 months ago

Have you tried `NDSolve` instead of `Solve`?

POSTED BY: Gianluca Gorni

James James

Posted 2 months ago

Because I want to obtain an exact solution, I am using DSolve instead of NDSolve.

POSTED BY: James James

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 2 months ago

With this kind of thinking, we have many thousands of pages of analytical solutions. Do you really need an analytical solution? After calculations by DSolve, conclusions can be drawn ,solution looks off than numeric one ,maybe a bug in DSolve. Even for parameter: \[Tau] -> 2.5 10^-5 DSolve probably give you not correct solution.See attached file. Regards M.I Attachments: Diff equation.nb

POSTED BY: Mariusz Iwaniuk

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 2 months ago

In this case I would try `Piecewise` instead of `WhenEvent`: \[Tau] = 1/9; pwQ[t_] = PiecewiseExpand[Boole[0 <= Mod[t, \[Tau]] < (2/3)*\[Tau]], 0 <= t <= 3 \[Tau]]; spSystem = {vi pwQ[t] == l iL'[t] + vC[t], vC'[t] == iL[t]/c - vC[t]/(r c), vC[0] == 0, iL[0] == 0}; pwSol = DSolve[system2, {vC, iL}, {t, 0, 2/10}] // Simplify; Simplify[spSystem /. pwSol[[1]] /. {{t -> 1/100}, {t -> 1/9 - 1/27 + 1/30}, {t -> 1/9 + 1/27}}]

In this case I would try Piecewise instead of WhenEvent:

\[Tau] = 1/9;
pwQ[t_] = PiecewiseExpand[Boole[0 <= Mod[t, \[Tau]] < (2/3)*\[Tau]],
   0 <= t <= 3 \[Tau]];
spSystem = {vi  pwQ[t] == l  iL'[t] + vC[t],
   vC'[t] == iL[t]/c - vC[t]/(r  c),
   vC[0] == 0, iL[0] == 0};
pwSol = DSolve[system2, {vC, iL},
    {t, 0, 2/10}] // Simplify;
Simplify[spSystem /. pwSol[[1]] /.
  {{t -> 1/100}, {t -> 1/9 - 1/27 + 1/30}, {t -> 1/9 + 1/27}}]

POSTED BY: Gianluca Gorni

James James

Posted 2 months ago

When I expanded the t-value range of the function pwQ[t_] and ran the program, the system prompted "PienewiseExpand cannot convert Floor [4000. \ t] to Pienewise form because the required number of segmented cases to search exceeds the internal limit of $MaxPienewiseCases=100". Can I try changing the size of the internal limit $MaxPierceCases? program code： \[Tau] = 2.5 10^-4; pwQ[t_] = PiecewiseExpand[Boole[0 <= Mod[t, \[Tau]] < (2/3)*\[Tau]], 0 <= t <= 3/100]; pars = {vi -> 24, r -> 22, l -> 2 10^-2, c -> 1 10^-4}; spSystem = {vi pwQ[t] == l iL'[t] + vC[t], vC'[t] == iL[t]/c - vC[t]/(r c), vC[0] == 0, iL[0] == 0}; pwSol = DSolve[spSystem /. pars, {vC, iL}, {t, 0, 2/100}] // Simplify; a = Evaluate[iL[t] /. pwSol]; b = Evaluate[vC[t] /. pwSol]; Plot[a, {t, 0, 0.02}, AxesLabel -> {"s", "il[t]/A"}, PlotLegends -> {"LinearlyImplicitEuler"}, PlotStyle -> {Red}, PlotRange -> All] Plot[b, {t, 0, 0.02}, AxesLabel -> {"s", "vc[t]/V"}, PlotLegends -> {"LinearlyImplicitEuler"}, PlotStyle -> {Blue}, PlotRange -> All]

When I expanded the t-value range of the function pwQ[t_] and ran the program, the system prompted "PienewiseExpand cannot convert Floor [4000. \ t] to Pienewise form because the required number of segmented cases to search exceeds the internal limit of $MaxPienewiseCases=100".

Can I try changing the size of the internal limit $MaxPierceCases?

program code：

\[Tau] = 2.5 10^-4;
pwQ[t_] = 
  PiecewiseExpand[Boole[0 <= Mod[t, \[Tau]] < (2/3)*\[Tau]], 
   0 <= t <= 3/100];
pars = {vi -> 24, r -> 22, l -> 2 10^-2, c -> 1 10^-4};
spSystem = {vi pwQ[t] == l iL'[t] + vC[t], 
   vC'[t] == iL[t]/c - vC[t]/(r c), vC[0] == 0, iL[0] == 0};
pwSol = DSolve[spSystem /. pars, {vC, iL}, {t, 0, 2/100}] // 
   Simplify;
a = Evaluate[iL[t] /. pwSol];
b = Evaluate[vC[t] /. pwSol];
Plot[a, {t, 0, 0.02}, AxesLabel -> {"s", "il[t]/A"}, 
 PlotLegends -> {"LinearlyImplicitEuler"}, PlotStyle -> {Red}, 
 PlotRange -> All]
Plot[b, {t, 0, 0.02}, AxesLabel -> {"s", "vc[t]/V"}, 
 PlotLegends -> {"LinearlyImplicitEuler"}, PlotStyle -> {Blue}, 
 PlotRange -> All]

POSTED BY: James James

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 2 months ago

I didn't know about `$MaxPienewiseCases`. You can try increasing it Do you know in advance how many intervals you need? For a full analytical solution I would use exact numbers: \[Tau] = 25/10 10^-4;

POSTED BY: Gianluca Gorni

James James

Posted 2 months ago

Yes, I found the help document for $MaxPickewiseCases and directly set $MaxPickewiseCases=(the range you want). The link to the help document isenter link description here But my code doesn't produce any results... \[Tau] = 25/10 10^-4; pwQ[t_] = PiecewiseExpand[Boole[0 <= Mod[t, \[Tau]] < (2/3)*\[Tau]], 0 <= t <= 3/100]; $MaxPiecewiseCases = 200; pars = {vi -> 24, r -> 22, l -> 2 10^-2, c -> 1 10^-4}; spSystem = {vi pwQ[t] == l iL'[t] + vC[t], vC'[t] == iL[t]/c - vC[t]/(r c), vC[0] == 0, iL[0] == 0}; pwSol = DSolve[spSystem /. pars, {vC, iL}, {t, 0, 2/100}] // Simplify; a = Evaluate[iL[t] /. pwSol]; b = Evaluate[vC[t] /. pwSol]; Plot[a, {t, 0, 0.02}, AxesLabel -> {"s", "il[t]/A"}, PlotLegends -> {"LinearlyImplicitEuler"}, PlotStyle -> {Red}, PlotRange -> All] Plot[b, {t, 0, 0.02}, AxesLabel -> {"s", "vc[t]/V"}, PlotLegends -> {"LinearlyImplicitEuler"}, PlotStyle -> {Blue}, PlotRange -> All]

Yes, I found the help document for

$MaxPickewiseCases

and directly set $MaxPickewiseCases=(the range you want). The link to the help document isenter link description here

But my code doesn't produce any results...

\[Tau] = 25/10 10^-4;
pwQ[t_] = 
  PiecewiseExpand[Boole[0 <= Mod[t, \[Tau]] < (2/3)*\[Tau]], 
   0 <= t <= 3/100];
$MaxPiecewiseCases = 200;
pars = {vi -> 24, r -> 22, l -> 2 10^-2, c -> 1 10^-4};
spSystem = {vi pwQ[t] == l iL'[t] + vC[t], 
   vC'[t] == iL[t]/c - vC[t]/(r c), vC[0] == 0, iL[0] == 0};
pwSol = DSolve[spSystem /. pars, {vC, iL}, {t, 0, 2/100}] // 
   Simplify;
a = Evaluate[iL[t] /. pwSol];
b = Evaluate[vC[t] /. pwSol];
Plot[a, {t, 0, 0.02}, AxesLabel -> {"s", "il[t]/A"}, 
 PlotLegends -> {"LinearlyImplicitEuler"}, PlotStyle -> {Red}, 
 PlotRange -> All]
Plot[b, {t, 0, 0.02}, AxesLabel -> {"s", "vc[t]/V"}, 
 PlotLegends -> {"LinearlyImplicitEuler"}, PlotStyle -> {Blue}, 
 PlotRange -> All]

POSTED BY: James James

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