tomleslie

In which case...

Answered: tomleslie 13876

October 22 2020

0 0

I'd probably be using the StringTools:-SearchAll() command, which will return a sequence of all the starting indexes of the supplied pattern, from whihc it is trivial to select the starting index for any particular occurrence. For example to obtain the start index of the third occurrence of the pattern, one wold have

[StringTools:-SearchAll("aba", "abababababababababab" )][3];

In order to get...

Answered: tomleslie 13876

October 16 2020

0 0

@AHSAN

accurate answers for this problem you have to use a high setting of DIgits - Set to 30 in the attached

DIsplaying 30 Digits can make the output look a litttle cluttered so you can separately control the amount of digits which are displayed. NB this affects display only: all calculations are done to the precision set by Digits.

See the atached

>

(1)

>

>	newP:=eval(P,[k=0.1]):

>	plot(newP,lambda=-1..1);

>

(2)

>

evalf(eval(newP, lambda = sols[1])); evalf(eval(newP, lambda = sols[2]))

(3)

>

Download resid2.mw

So...

Answered: tomleslie 13876

October 15 2020

0 0

@Maple_lover1

without going into the boring details of how you might want plot shading to be done ( see the options colorscheme, colorfunc and shading as mentioned beore), you basically want something as shown in the attached.

>

  restart;
  with(plots):
  with(plottools):
  display( [ plot3d
             ( sin(x*y),
               x=-3..3,
               y=-3..3,
               gridstyle=triangular,
               shading=zhue
             ),
             transform
             ( (x,y)->[x,y,-2] )
                      ( contourplot
                        ( sin(x*y),
                          x=-3..3,
                          y=-3..3,
                          contours=[-1/2,1/4,1/2,3/4],
                          filledregions=true,
                          coloring=[green,red]
                        )
             )
           ]
         );

>

Download mapPlot3.mw

You really only try to accept two simple statements: when it comes to plots

If you can do it in Maple you can do it in Matlab
If you can do it in Matlab you can do it in Maple

Now out of the thousands of plot possibilities available in both of these packages - will you ever be able to come with something in one package which cannot be replicated in the other - yeah, maybe? But good luck finding it!

CAn only suggest...

Answered: tomleslie 13876

October 15 2020

0 0

@Maple_lover1

that you play with the various plot3d options available, in paticular,

style
gridstyle
shading

( as well as color=colorscheme or color=colorfunc which will allow explicit shading based on coordinate values)

See the the additional execution group in the attached (which only scratches the surface of all the possibilities)

>

restart:
u:=1/(1. + exp(x))^2 + 1/(1. + exp(-5.*t))^2 - 0.2500000000 + x*(1/(1. + exp(1 - 5*t))^2 - 1./((1. + exp(-5*t))^2) + 0.1776705118 + 0.0415431679756514*piecewise(0. <= t and t <= 0.5000000000, 1.732050808, 0.) + 0.00922094377856479*piecewise(0. <= t and t <= 0.5000000000, 30.98386677*t - 7.745966692, 0.) + 0.0603742508215732*piecewise(0.5000000000 <= t and t <= 1., 1.732050808, 0.) - 0.00399645630498528*piecewise(0.5000000000 <= t and t <= 1., 30.98386677*t - 23.23790008, 0.)) + (-0.00243051684581302*piecewise(0. <= x and x <= 0.5000000000, 1.732050808, 0.) - 0.000809061198761621*piecewise(0. <= x and x <= 0.5000000000, 30.98386677*x - 7.745966692, 0.) - 0.0152377552205917*piecewise(0.5000000000 <= x and x <= 1., 1.732050808, 0.) - 0.00195593427342862*piecewise(0.5000000000 <= x and x <= 1., 30.98386677*x - 23.23790008, 0.))*piecewise(0. <= t and t <= 0.5000000000, 1.732050808, 0.) + (-0.000433590063316381*piecewise(0. <= x and x <= 0.5000000000, 1.732050808, 0.) - 0.000146112803263678*piecewise(0. <= x and x <= 0.5000000000, 30.98386677*x - 7.745966692, 0.) - 0.00319022339097685*piecewise(0.5000000000 <= x and x <= 1., 1.732050808, 0.) - 0.000477063086307787*piecewise(0.5000000000 <= x and x <= 1., 30.98386677*x - 23.23790008, 0.))*piecewise(0. <= t and t <= 0.5000000000, 30.98386677*t - 7.745966692, 0.) + (-0.00276114805649180*piecewise(0. <= x and x <= 0.5000000000, 1.732050808, 0.) - 0.000933166016624500*piecewise(0. <= x and x <= 0.5000000000, 30.98386677*x - 7.745966692, 0.) - 0.0207984584912892*piecewise(0.5000000000 <= x and x <= 1., 1.732050808, 0.) - 0.00314360556336114*piecewise(0.5000000000 <= x and x <= 1., 30.98386677*x - 23.23790008, 0.))*piecewise(0.5000000000 <= t and t <= 1., 1.732050808, 0.) + (0.000172746997599710*piecewise(0. <= x and x <= 0.5000000000, 1.732050808, 0.) + 0.0000586775450031145*piecewise(0. <= x and x <= 0.5000000000, 30.98386677*x - 7.745966692, 0.) + 0.00136190009033518*piecewise(0.5000000000 <= x and x <= 1., 1.732050808, 0.) + 0.000211410172315387*piecewise(0.5000000000 <= x and x <= 1., 30.98386677*x - 23.23790008, 0.))*piecewise(0.5000000000 <= t and t <= 1., 30.98386677*t - 23.23790008, 0.):

>	plot3d( u, x=0..1, t=0..1, style=surface, axes=boxed, colorscheme=[yellow, red] );

>	plot3d( u, x=0..1, t=0..1, axes=normal, gridstyle=triangular, style=patch, shading=zhue );

>

Download mapPlot2.mw

You are missing...

Answered: tomleslie 13876

October 14 2020

0 0

@Maple_lover1

the importatnt part of my original response, which is (original typos corrected)

It is not true that Matlab offers better plotting capability - so I wouldn.t bother anyway.

Now if you can explain exactly what plotting capabilyt exists within Matlab but not Maple I'd get interested. By the way I'm currently running Matlab R2020b and Maple 2020.1.1, although I do have several older versions of both packages if you think that "versions" are an issue.

From a plotting viewpoint - tell me what is worng with the following and I will fix it. NB the plot "renders" rather better in a Maple worksheet than it does on this site

>

restart:
u:=1/(1. + exp(x))^2 + 1/(1. + exp(-5.*t))^2 - 0.2500000000 + x*(1/(1. + exp(1 - 5*t))^2 - 1./((1. + exp(-5*t))^2) + 0.1776705118 + 0.0415431679756514*piecewise(0. <= t and t <= 0.5000000000, 1.732050808, 0.) + 0.00922094377856479*piecewise(0. <= t and t <= 0.5000000000, 30.98386677*t - 7.745966692, 0.) + 0.0603742508215732*piecewise(0.5000000000 <= t and t <= 1., 1.732050808, 0.) - 0.00399645630498528*piecewise(0.5000000000 <= t and t <= 1., 30.98386677*t - 23.23790008, 0.)) + (-0.00243051684581302*piecewise(0. <= x and x <= 0.5000000000, 1.732050808, 0.) - 0.000809061198761621*piecewise(0. <= x and x <= 0.5000000000, 30.98386677*x - 7.745966692, 0.) - 0.0152377552205917*piecewise(0.5000000000 <= x and x <= 1., 1.732050808, 0.) - 0.00195593427342862*piecewise(0.5000000000 <= x and x <= 1., 30.98386677*x - 23.23790008, 0.))*piecewise(0. <= t and t <= 0.5000000000, 1.732050808, 0.) + (-0.000433590063316381*piecewise(0. <= x and x <= 0.5000000000, 1.732050808, 0.) - 0.000146112803263678*piecewise(0. <= x and x <= 0.5000000000, 30.98386677*x - 7.745966692, 0.) - 0.00319022339097685*piecewise(0.5000000000 <= x and x <= 1., 1.732050808, 0.) - 0.000477063086307787*piecewise(0.5000000000 <= x and x <= 1., 30.98386677*x - 23.23790008, 0.))*piecewise(0. <= t and t <= 0.5000000000, 30.98386677*t - 7.745966692, 0.) + (-0.00276114805649180*piecewise(0. <= x and x <= 0.5000000000, 1.732050808, 0.) - 0.000933166016624500*piecewise(0. <= x and x <= 0.5000000000, 30.98386677*x - 7.745966692, 0.) - 0.0207984584912892*piecewise(0.5000000000 <= x and x <= 1., 1.732050808, 0.) - 0.00314360556336114*piecewise(0.5000000000 <= x and x <= 1., 30.98386677*x - 23.23790008, 0.))*piecewise(0.5000000000 <= t and t <= 1., 1.732050808, 0.) + (0.000172746997599710*piecewise(0. <= x and x <= 0.5000000000, 1.732050808, 0.) + 0.0000586775450031145*piecewise(0. <= x and x <= 0.5000000000, 30.98386677*x - 7.745966692, 0.) + 0.00136190009033518*piecewise(0.5000000000 <= x and x <= 1., 1.732050808, 0.) + 0.000211410172315387*piecewise(0.5000000000 <= x and x <= 1., 30.98386677*x - 23.23790008, 0.))*piecewise(0.5000000000 <= t and t <= 1., 30.98386677*t - 23.23790008, 0.):

>	plot3d( u, x=0..1, t=0..1, style=surface, axes=boxed, colorscheme=[yellow, red] );

>

Download mapPlot.mw

Hmm -see the attached...

Answered: tomleslie 13876

October 12 2020

0 0

@raj2018

I should probably have use fsolve() in my previous sheet, since you are only really interested in a numerical solution
however even using fsolve() with guesses for the expected ranges of the unknowns - still no solution
tried the DirectSearch() optimiser package ( a free add-on from the Maple Application centre). This *sometimes* works when the built-in fsolve() fails. This does provide "solutions" for a range of values for delta[d] - see the attched
But are these solutions *plausible*?. DirectSearch() will just return the "best solution" it can find, which may or may not be the "actual" solution!!)

Some observations about the solutions found by by DirectSearch() - first of all the *form* of the ouput for each value of delta[d] is

[ mean square residual error, residual error in each equation, values, number of iterations taken to obtain solution]

For delta[d] from 1e-04 to 1e-03, the values of u[i10] and u[i20] don't vary much at all, being around 3 and 1.5 respectively
The value of phi[d0] varies by several orders of magnitude from O(10^-32) to O(10^-24)
When the value of phi[d0] is O(10^-32), then the mean square residual is "pretty good", ie < O(10^-6), When the value of phi[d0] is O(10^-24), then the mean square residual is bad - up to O(10^7).
The biggest contribution to the mean square residual always occurs from the error in solving the third equation in the system - from which one can conclude that this equation is numerically pretty unstable, with respect to the variable phi[d0]

What would I do next???

Are the values obtained for the three unknowns vaguely plausible ie is phi[d0]~10^-30, u[i10]~2, and u[i20]~1.5 a "believable" solution
If the answer to (1) above is "No" - then it means that you have some idea what the solution should be - so appropriately adjust the constraints on the variables in the DirectSearch() command (or possibly even fsolve()?) in order to assist the solution process
If this system of equations represents a "physical situation" rather than a mathematical abstraction, then consider whether an appropriate choice of 'units' for various parameters/variables might reduce numerical instabilities. Any equation system which simultaneously contains coefficients/parameters of O(10^-48) and O(10^36) is just asking for trouble
If (3) above is impossible then consider raisin the setting of Digits, wihc will force numerical calculations to be performed more accurately - althugh if the setting is high enough that hardware floats cannot be used (Digits=15, roughly) then computation times will increase substantially

See the attached - Note that you will not be able to re-execute this worksheet unless you first install the DirectSearch() add-on

>

restart;

Eq1:=-2.356739746*10^(-48)*(4.682096432*10^11*exp(phi[d0])*u[i10]*u[i20]-7.21774261*10^8*u[i10]^2*u[i20]-1.141225310*10^9*u[i10]*u[i20]^2+2.282450620*10^9*phi[d0]*u[i10]+1.443548522*10^9*phi[d0]*u[i20])/(exp(phi[d0])*phi[d0]*u[i10]*u[i20]=delta[d]);

Eq2:=1.839080460*10^(-38)*(8.117630990*10^9*u[i10]^3*u[i20]-1.217644649*10^9*u[i10]*u[i20]^3-3.988316460*10^9*u[i10]+2.490971436*10^10*u[i20])/(phi[d0]*(6.6471941*10^7*u[i10]-4.15161906*10^8*u[i20]))=delta[d]:

Eq3:=3.321295248*10^(-13)*(1.359375000*10^36*delta[d]*phi[d0]+1.5)/u[i10]+1.329438820*10^(-14)*(1.359375000*10^36*delta[d]*phi[d0]+1.5)/u[i20]-(-1.359375000*10^36*delta[d]*phi[d0]-1.5+1.510416667*10^37*delta[d]*phi[d0]^2)*(8.117630990*10^(-15)*u[i10]^2+1.660647624*10^(-14)*(1.359375000*10^36*delta[d]*phi[d0]+1.5)/u[i10])-1.359375000*10^36*delta[d]*phi[d0]*(-6.189239034*10^(-29)+3.634239206*10^(-25)*u[i10]*sqrt(u[i10]^2+2.547770701)*(1-2*phi[d0]/(u[i10]^2+0.4246284503e-1))^1.0+1.395213672*10^(-54)*u[i10]*sqrt(u[i10]^2+2.547770701)*phi[d0]^2*ln((phi[d0]^2/(1000000000000*(u[i10]^2+0.4246284503e-1)^2)+2.005078125*10^14/(1.275000000+2.265625000*10^34*delta[d]*phi[d0]))/(phi[d0]^2/(1000000000000*(u[i10]^2+0.4246284503e-1)^2)+(1/1000000000000)*(1-2*phi[d0]/(u[i10]^2+0.4246284503e-1))^1.0))/(u[i10]^2+0.4246284503e-1)^2+1.817119603*10^(-25)*u[i20]*sqrt(u[i20]^2+2.547770701)*(1-2*phi[d0]/(u[i20]^2+0.8492569001e-1))^1.0+6.976068361*10^(-56)*u[i20]*sqrt(u[i20]^2+2.547770701)*phi[d0]^2*ln((phi[d0]^2/(1000000000000*(u[i20]^2+0.8492569001e-1)^2)+2.005078125*10^14/(1.275000000+2.265625000*10^34*delta[d]*phi[d0]))/(phi[d0]^2/(1000000000000*(u[i20]^2+0.8492569001e-1)^2)+(1/1000000000000)*(1-2*phi[d0]/(u[i20]^2+0.8492569001e-1))^1.0))/(u[i20]^2+0.8492569001e-1)^2)+6.235395017*10^(-20)*delta[d]*(7.217742610*10^14*u[i10]*(1-2*phi[d0]/u[i10]^2)+1.141225310*10^15*u[i20]*(1-2*phi[d0]/u[i20]^2)-2.207161425*10^17*sqrt(2)*(1.359375000*10^36*delta[d]*phi[d0]+1.5)*exp(phi[d0]))= 0;

sys:=[ Eq1,Eq2,Eq3]:

-0.2356739746e-47*(0.4682096432e12*exp(phi[d0])*u[i10]*u[i20]-721774261.0*u[i10]^2*u[i20]-1141225310.*u[i10]*u[i20]^2+2282450620.*phi[d0]*u[i10]+1443548522.*phi[d0]*u[i20])/(exp(phi[d0])*phi[d0]*u[i10]*u[i20]) = -0.2356739746e-47*(0.4682096432e12*exp(phi[d0])*u[i10]*u[i20]-721774261.0*u[i10]^2*u[i20]-1141225310.*u[i10]*u[i20]^2+2282450620.*phi[d0]*u[i10]+1443548522.*phi[d0]*u[i20])/delta[d]

(1)

>

#
# Use fsolve() to try for a numerical solution, using a value
# for delta[d] somewhere in the middle of the required range
#
# Apply ranges to the remaining unknown variables to assist the
# calculation. NB these ranges are complete GUESSWORK on my part.
# Any better knowledge of expected ranges of answers would be
# very useful and *may* allow an answer to be obtained
#
# Hmmmm - still no solution
#
    fsolve( eval(sys, delta[d]=1e-03),
            { phi[d0]=0..10, u[i10]=0..10, u[i20]=0..10}
          )

$fsolve([-0.2356739746e-47*(0.4682096432e12*exp(phi[d0])*u[i10]*u[i20]-721774261.0*u[i10]^2*u[i20]-1141225310.*u[i10]*u[i20]^2+2282450620.*phi[d0]*u[i10]+1443548522.*phi[d0]*u[i20])/(exp(phi[d0])*phi[d0]*u[i10]*u[i20]) = -0.1103448276e-32*exp(phi[d0])*u[i10]*u[i20]+0.1701034089e-35*u[i10]^2*u[i20]+0.2689571047e-35*u[i10]*u[i20]^2-0.5379142094e-35*phi[d0]*u[i10]-0.3402068177e-35*phi[d0]*u[i20], 0.1839080460e-37*(8117630990.*u[i10]^3*u[i20]-1217644649.*u[i10]*u[i20]^3-3988316460.*u[i10]+0.2490971436e11*u[i20])/(phi[d0]*(66471941.00*u[i10]-415161906.0*u[i20])) = 0.1e-2, 0.3321295248e-12*(0.1359375000e34*phi[d0]+1.5)/u[i10]+0.1329438820e-13*(0.1359375000e34*phi[d0]+1.5)/u[i20]-(-0.1359375000e34*phi[d0]-1.5+0.1510416667e35*phi[d0]^2)*(0.8117630990e-14*u[i10]^2+0.1660647624e-13*(0.1359375000e34*phi[d0]+1.5)/u[i10])-0.1359375000e34*phi[d0]*(-0.6189239034e-28+0.3634239206e-24*u[i10]*(u[i10]^2+2.547770701)^(1/2)*(1-2*phi[d0]/(u[i10]^2+0.4246284503e-1))^1.0+0.1395213672e-53*u[i10]*(u[i10]^2+2.547770701)^(1/2)*phi[d0]^2*ln(((1/1000000000000)*phi[d0]^2/(u[i10]^2+0.4246284503e-1)^2+0.2005078125e15/(1.275000000+0.2265625000e32*phi[d0]))/((1/1000000000000)*phi[d0]^2/(u[i10]^2+0.4246284503e-1)^2+(1/1000000000000)*(1-2*phi[d0]/(u[i10]^2+0.4246284503e-1))^1.0))/(u[i10]^2+0.4246284503e-1)^2+0.1817119603e-24*u[i20]*(u[i20]^2+2.547770701)^(1/2)*(1-2*phi[d0]/(u[i20]^2+0.8492569001e-1))^1.0+0.6976068361e-55*u[i20]*(u[i20]^2+2.547770701)^(1/2)*phi[d0]^2*ln(((1/1000000000000)*phi[d0]^2/(u[i20]^2+0.8492569001e-1)^2+0.2005078125e15/(1.275000000+0.2265625000e32*phi[d0]))/((1/1000000000000)*phi[d0]^2/(u[i20]^2+0.8492569001e-1)^2+(1/1000000000000)*(1-2*phi[d0]/(u[i20]^2+0.8492569001e-1))^1.0))/(u[i20]^2+0.8492569001e-1)^2)+0.4500547630e-7*u[i10]*(1-2*phi[d0]/u[i10]^2)+0.7115990611e-7*u[i20]*(1-2*phi[d0]/u[i20]^2)-0.1376252335e-4*2^(1/2)*(0.1359375000e34*phi[d0]+1.5)*exp(phi[d0]) = 0], {phi[d0], u[i10], u[i20]}, {phi[d0] = 0 .. 10, u[i10] = 0 .. 10, u[i20] = 0 .. 10})$

(2)

>

#
# Use the DirectSearch add-on optimisation package with the
# same constraints to determine whether solutions can be obtained
# for various values of delta[d] across the required range
#
  seq( [ DirectSearch:-SolveEquations
                       ( eval(sys, delta[d]=j),
                         [ phi[d0]=0..10, u[i10]=0..10, u[i20]=0..10 ]
                       )
       ],
       j= 1e-04..1e-03, 1e-04
     );

$[[3.19460803484165*10^(-8), Vector(3, {(1) = -0.2487088061e-5, (2) = -0.133061047429682e-3, (3) = -0.119309062515421e-3}), [phi[d0] = 4.39999368528266*10^(-31), u[i10] = 2.98360668673353, u[i20] = 1.50800349885015], 771]], [[379008.267349024, Vector(3, {(1) = -0.6500013806e-12, (2) = -0.200000008640510e-3, (3) = -615.636473374494}), [phi[d0] = 1.68356130893318*10^(-24), u[i10] = 2.98360623089539, u[i20] = 1.50800427090504], 494]], [[2.59642027567093*10^(-7), Vector(3, {(1) = -0.101865773046122e-4, (2) = -0.435410816534470e-3, (3) = -0.264491364801759e-3}), [phi[d0] = 1.07427366636149*10^(-31), u[i10] = 2.98360616776824, u[i20] = 1.50800438081475], 773]], [[4.97583555741889*10^(-7), Vector(3, {(1) = -0.104446673180020e-4, (2) = -0.538841500887854e-3, (3) = -0.455109109541250e-3}), [phi[d0] = 1.04772812768309*10^(-31), u[i10] = 2.98360563725389, u[i20] = 1.50800528660184], 828]], [[1.22611706447686*10^(-6), Vector(3, {(1) = -0.445466164536864e-4, (2) = -0.109216128236589e-2, (3) = -0.176964393990126e-3}), [phi[d0] = 2.45656631825036*10^(-32), u[i10] = 2.98360646440409, u[i20] = 1.50800387665753], 820]], [[1.27212378604264*10^(-6), Vector(3, {(1) = -0.129541832316612e-4, (2) = -0.772200695800547e-3, (3) = -0.821986654748474e-3}), [phi[d0] = 8.44759684568413*10^(-32), u[i10] = 2.98360588229276, u[i20] = 1.50800486182296], 771]], [[1.50550033690128*10^(-6), Vector(3, {(1) = -0.332753272328057e-4, (2) = -0.114233248967109e-2, (3) = -0.446620165846419e-3}), [phi[d0] = 3.28867443728915*10^(-32), u[i10] = 2.98360814940436, u[i20] = 1.50800098608150], 811]], [[7.95992034093232*10^6, Vector(3, {(1) = -0.4572396327e-12, (2) = -0.800000006078140e-3, (3) = -2821.33307869377}), [phi[d0] = 2.39331217921800*10^(-24), u[i10] = 2.98360855438924, u[i20] = 1.50800028790066], 601]], [[2.34387556435520*10^(-6), Vector(3, {(1) = -0.295772909747718e-4, (2) = -0.129317346188895e-2, (3) = -0.818964679140646e-3}), [phi[d0] = 3.69985601029048*10^(-32), u[i10] = 2.98360717829695, u[i20] = 1.50800264816345], 899]], [[3.27345103272978*10^(-6), Vector(3, {(1) = -0.540071925813580e-4, (2) = -0.171792077937664e-2, (3) = -0.565050839894256e-3}), [phi[d0] = 2.02624339828116*10^(-32), u[i10] = 2.98360595919939, u[i20] = 1.50800473331530], 912]]$

(3)

>

Download solveSys2.mw

Well...

Answered: tomleslie 13876

October 10 2020

0 0

@raj2018

If I had three equations in three unknowns (x,y,z) and a "parameter " such as "tau", and I wanted to obtain values for (x,y,z) as tau varied over a substantial range (say 10^-2.. 10^3), I'd probably be using a semilogplot().

Something like the attached "toy" example

Download solveSys.mw

Give a coherent explanation...

Answered: tomleslie 13876

October 09 2020

0 0

of why you would choose to set the options

minstep=500,maxstep=1000

when the independent variable range in which you are interesed appears to be 0..10

No idea...

Answered: tomleslie 13876

October 09 2020

0 0

@raj2018

since I can't read the paper (which you don't provide)

I can only state the mathematically obvious - that no-one can obtain numerical solutions from three equations in four unknowns

Ignoting Eq1,...

Answered: tomleslie 13876

October 09 2020

0 0

@raj2018

from the last comment in my original worksheet - you now have three equations in four unknowns

#
# Just one problem: having inserted all of these parameters,
# just what is Eq1? It evaluates to 0=0, which is seriously,
# non-useful!! Means one is still left with three *meaningful*
# equations (either [Eq2a, Eq3, Eq4], or [Eq2b, Eq3, Eq4]
# in four unknowns - NOT SOLVABLE
#

Hmmm...

Answered: tomleslie 13876

October 05 2020

0 0

@Reshu Gupta

Sometimes fsolve() fails because it cannot come up with a solution whihc statisfies the problem with sufficient accuracy. In such a circumstance, I use the (free) Maple add-on package called DirecctSearxh(), available here

https://www.maplesoft.com/applications/view.aspx?SID=87637

This uses optimization techniques (a bit like fsolve), but one difference is that it will generally produce the "best" solution it can come u with, even if the solution is not "very good" and leave it to the user to decide whether the obtained solution is "acceptable". Applying DirectectSearch-:SolveEquations() to your problem results in a solution, with residual error of order 10^-7. Whether or not this level of residual error is acceptable is something only you can decide!!

The attached shows the obtained solution, but note that you will not be able to execute this worksheet unless you have the DirectSeaarch package downloaded/installed

Download fsol2.mw

This comment...

Answered: tomleslie 13876

October 03 2020

0 0

@radaar

makes no sense

Don't understand this comment...

Answered: tomleslie 13876

September 24 2020

0 0

@pengcheng Xue

In the attached I have added a couple of commands (highlighted in red) which check the value of y1_15 in solnumeric1 and solnumeric2. These are -0.1 and -0.2 respectively - so I don't see the justification for your comment

it seems not right. In solnumeric2, lambda equal to 0.2, but in the solution y1_15 still equal to -0.1? it should equal to -0.2, doesn't it?

Download toughODE2.mw

I've just checked...

Answered: tomleslie 13876

September 24 2020

0 0

@acer

my original answer (developed in Maple2020.1 in Maple 18 - It seems to work OK

Well...

Answered: tomleslie 13876

September 23 2020

0 0

@sand15

If you posted your worksheet, I could try it on my machine.

If I can reproduce your problem then it is a Maple/worksheet issue, which may be possible to correct or workaround.
If I can't reproduce then it is most likely to be an installation issue on your machine

The above determination would represent some progress in assisting with your problem

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