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Finding extreme points: cannot solve the equation

Posted 9 months ago

First, attach the code, and I will raise my questions after the code.

{r = 22, l = 2 10^-1, c = 1 10^-4, vi = 24, initvalueil = 0, 
  initvaluevc = 0, tstart = 0, tend = 0.08, limitdt = 0.0001, 
  error = 0.0019152};
\[CapitalDelta]vi = 0;
A = {{0, -1/l}, {1/c, -1/(r c)}};
G = Inverse[A];
B = {1/l, 0};
S1 = Inverse[DiagonalMatrix[{1, 1}] - dt*A + 1/2*dt*dt*A . A];
S2 = Inverse[DiagonalMatrix[{1, 1}] - dt*A];
STAR1 = S1 . G . (B*vi + G . B*\[CapitalDelta]vi/dt) - 
   G . (B*(vi + \[CapitalDelta]vi) + G . B*\[CapitalDelta]vi/dt);
STAR2 = S2 . G . (B*vi + G . B*\[CapitalDelta]vi/dt) - 
   G . (B*(vi + \[CapitalDelta]vi) + G . B*\[CapitalDelta]vi/dt);
X0 = {0, 0};
X[t_] := 
  Simplify[(MatrixPower[S1, i] - MatrixPower[S2, i]) . 
      X0 + (DiagonalMatrix[{1, 1}] . (DiagonalMatrix[{1, 1}] - 
          MatrixPower[S1, i]) . 
        Inverse[DiagonalMatrix[{1, 1}] - S1]) . 
      STAR1 - (DiagonalMatrix[{1, 1}] . (DiagonalMatrix[{1, 1}] - 
          MatrixPower[S2, i]) . 
        Inverse[DiagonalMatrix[{1, 1}] - S2]) . STAR2] /. {i -> 
     t/dt};
ilerror = X[t][[1]];
ContourPlot[
 Abs[ilerror] == error, {t, tstart, tend}, {dt, 0, limitdt}, 
 ContourShading -> None, PlotPoints -> 50, FrameLabel -> Automatic, 
 Epilog -> {PointSize[Large], Red, 
   Point[{t, dt} /. {t -> 0.00899448, dt -> 0.00006}]}]
condequal = Abs[ilerror] == error;
dilt = D[condequal, t];
Solve[dilt == 0, t]

Since error has been assigned a value of 0.0019152, condequal can be considered as the relationship between dt and t. Therefore, I want to take the derivative of t through condequal to obtain the value of the corresponding extreme point t when dt obtains its extreme value. But Solve instruction cannot solve correctly in the code, how can I modify the code?

POSTED BY: James James
7 Replies
Posted 9 months ago

Not yet. I tried using FindRoot, but it seems that the program was unable to output the results. So I try to simplify the formula in the subsequent problems to make it easier to solve.

POSTED BY: James James

This works fine for me:

{r = 22, l = 2 10^-1, c = 1 10^-4, vi = 24, initvalueil = 0, 
  initvaluevc = 0, tstart = 0, tend = 0.08, limitdt = 0.0001, 
  error = 0.0019152};
\[CapitalDelta]vi = 0;
A = {{0, -1/l}, {1/c, -1/(r c)}};
G = Inverse[A];
B = {1/l, 0};
S1 = Inverse[DiagonalMatrix[{1, 1}] - dt*A + 1/2*dt*dt*A . A];
S2 = Inverse[DiagonalMatrix[{1, 1}] - dt*A];
STAR1 = S1 . G . (B*vi + G . B*\[CapitalDelta]vi/dt) - 
   G . (B*(vi + \[CapitalDelta]vi) + G . B*\[CapitalDelta]vi/dt);
STAR2 = S2 . G . (B*vi + G . B*\[CapitalDelta]vi/dt) - 
   G . (B*(vi + \[CapitalDelta]vi) + G . B*\[CapitalDelta]vi/dt);
X0 = {0, 0};
X[t_] := 
  Simplify[(MatrixPower[S1, i] - MatrixPower[S2, i]) . 
      X0 + (DiagonalMatrix[{1, 1}] . (DiagonalMatrix[{1, 1}] - 
          MatrixPower[S1, i]) . 
        Inverse[DiagonalMatrix[{1, 1}] - S1]) . 
      STAR1 - (DiagonalMatrix[{1, 1}] . (DiagonalMatrix[{1, 1}] - 
          MatrixPower[S2, i]) . 
        Inverse[DiagonalMatrix[{1, 1}] - S2]) . STAR2] /. {i -> 
     t/dt};
ilerror = X[t][[1]];
ContourPlot[
 Abs[ilerror] == error, {t, tstart, tend}, {dt, 0, limitdt}, 
 ContourShading -> None, PlotPoints -> 50, FrameLabel -> Automatic, 
 Epilog -> {PointSize[Large], Red, 
   Point[{t, dt} /. {t -> 0.00899448, dt -> 0.00006}]}]
condequal = ilerror == error;
dilt = D[condequal, t];
FindRoot[{condequal, dilt}, {{t, 0.01}, {dt, 0.00005}}]
POSTED BY: Gianluca Gorni
Posted 9 months ago

First, attach the code, and I will raise my questions after the code.

{r = 22, l = 2 10^-1, c = 1 10^-4, vi = 24, initvalueil = 0, 
  initvaluevc = 0, tstart = 0, tend = 0.08, limitdt = 0.0001, 
  error = 0.0019152};
\[CapitalDelta]vi = 0;
A = {{0, -1/l}, {1/c, -1/(r c)}};
G = Inverse[A];
B = {1/l, 0};
S1 = Inverse[DiagonalMatrix[{1, 1}] - dt*A + 1/2*dt*dt*A . A];
S2 = Inverse[DiagonalMatrix[{1, 1}] - dt*A];
STAR1 = S1 . G . (B*vi + G . B*\[CapitalDelta]vi/dt) - 
   G . (B*(vi + \[CapitalDelta]vi) + G . B*\[CapitalDelta]vi/dt);
STAR2 = S2 . G . (B*vi + G . B*\[CapitalDelta]vi/dt) - 
   G . (B*(vi + \[CapitalDelta]vi) + G . B*\[CapitalDelta]vi/dt);
X0 = {0, 0};
X[t_] := 
  Simplify[(MatrixPower[S1, i] - MatrixPower[S2, i]) . 
      X0 + (DiagonalMatrix[{1, 1}] . (DiagonalMatrix[{1, 1}] - 
          MatrixPower[S1, i]) . 
        Inverse[DiagonalMatrix[{1, 1}] - S1]) . 
      STAR1 - (DiagonalMatrix[{1, 1}] . (DiagonalMatrix[{1, 1}] - 
          MatrixPower[S2, i]) . 
        Inverse[DiagonalMatrix[{1, 1}] - S2]) . STAR2] /. {i -> 
     t/dt};
ilerror = X[t][[1]];
ContourPlot[
 Abs[ilerror] == error, {t, tstart, tend}, {dt, 0, limitdt}, 
 ContourShading -> None, PlotPoints -> 50, FrameLabel -> Automatic, 
 Epilog -> {PointSize[Large], Red, 
   Point[{t, dt} /. {t -> 0.00899448, dt -> 0.00006}]}]
condequal = Abs[ilerror] == error;
dilt = D[condequal, t];
Solve[dilt == 0, t]

Since error has been assigned a value of 0.0019152, condequal can be considered as the relationship between dt and t. Therefore, I want to take the derivative of t through condequal to obtain the value of the corresponding extreme point t when dt obtains its extreme value. But Solve instruction cannot solve correctly in the code, how can I modify the code?

POSTED BY: James James

Your function ilerror is positive in your domain. You can remove the Abs, which gives trouble when you take the derivative. Instead of Solve you can use FindRoot:

condequal = ilerror == error;
dilt = D[condequal, t];
FindRoot[{condequal, dilt},
 {{t, 0.01}, {dt, 0.00006}}]

which uses numerical methods to solve the system.

POSTED BY: Gianluca Gorni

Correction: dilt alone is already a relation. Writing dilt == 0 will make Solve complain.

POSTED BY: Gianluca Gorni

The way you set it up, dilt == 0 is also a relationship between t and dt, and if you Solve it, it will give a function, not a single point. You can try

condequal = Abs[ilerror] == error;
dilt = D[condequal, t];
Solve[{condequal, dilt}, {t, dt}]

but I doubt that Solve will manage such a complicated system.

POSTED BY: Gianluca Gorni
Posted 9 months ago
condequal = Abs[ilerror] == error;
dilt = D[condequal, t];
Solve[{condequal, dilt}, {t, dt}]

Yes, it seems that after modifying to the above code, Solve cannot solve either. enter image description here

What if I want to obtain the minimum value (red dot) of the curve in the graph by taking a derivative equal to 0 to obtain the pole? Because the previous issue used FindMinimum, but I don't want to rely entirely on MMA instructions...

POSTED BY: James James
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