Applications, Examples and Libraries

Share your work here

The complete Maple Conference 2024 program is now available online. You can view it HERE.
Thank you to all those who voted for the "Audience Choice" sessions. The winning topics are listed in the program.

Registration is still open and free of charge.  

We hope to see you there!

Just finished an exciting lecture for undergraduate students on Euler methods for solving ordinary differential equations (ODEs)! 📝 We explored the Euler method and the Improved Euler method (a.k.a. Heun's method) and discussed how these fundamental techniques work for approximating solutions to ODEs.

To make things practical and hands-on, I demonstrated how to implement both methods using Maple—a fantastic tool for experimenting with numerical methods. Seeing these concepts come to life with visualizations helps understand the pros and cons of each method.

The Euler method is the simplest numerical approach, where we approximate the solution by stepping along the slope given by the ODE. But there's a catch: using a large step size can lead to large errors that accumulate quickly, causing significant deviation from the real solution.

Plot Results:

  • With a larger step size (h=0.5h = 0.5h=0.5), the solution tends to drift away significantly from the actual curve.
  • Reducing the step size improves accuracy, but it also means more steps and more computation.

Next, we discussed the Improved Euler method, which is like Euler's smarter sibling. Instead of blindly following the initial slope, it:

  1. Estimates the slope at the start.
  2. Uses that slope to predict an intermediate value.
  3. Then, it takes another slope at the intermediate point.
  4. Averages these two slopes for a better approximation.

This technique makes a big difference in accuracy and stability, especially with larger step sizes.

Plot Results:

  • Using the Improved Euler method, we found that even with a larger step size, the solution is smoother and closer to the true path.
  • The average slope helps mitigate the inaccuracies that arise from only using the beginning point's derivative, effectively reducing the local error.
  • The standard Euler method can produce solutions that oscillate or diverge, especially for larger step sizes.
  • The Improved Euler method follows the actual trend of the solution more faithfully, even with fewer steps. This makes it a more efficient choice for balancing computational effort and accuracy.
  • The Euler method is great for its simplicity but often requires very small step sizes to be accurate, leading to more computational effort.
  • The Improved Euler method—which we implemented and visualized on Maple—proved to be more reliable and accurate, especially for larger step sizes.

It was amazing to see the students engage with these foundational methods, and implementing them on Maple brought a deeper understanding of numerical analysis and the challenges of solving differential equations.


 


restart

with(plots)

with(DEtools)

ODE1 := diff(y(x), x) = -2*x*y(x)/(x^2+1)

diff(y(x), x) = -2*x*y(x)/(x^2+1)

(1)

NULL

We calculate the general solution to the ODE

NULL

dsolve(ODE1, y(x))

y(x) = c__1/(x^2+1)

(2)

NULL

Now let's solve the problem for the following two inital conditions

y(0) = 2 and y(0) = 1/2

dsolve({ODE1, y(0) = 2}, y(x))

y(x) = 2/(x^2+1)

(3)

dsolve({ODE1, y(0) = 1/2}, y(x))

y(x) = 1/(2*x^2+2)

(4)

Then, we are going to plot the solutions to the IVP's together with the slope field corresponding to the ODE.

 

NULLdfieldplot(ODE1, [y(x)], x = -5 .. 5, y = -5 .. 5, color = blue, scaling = constrained, axes = boxed)

 

 

DEplot(ODE1, y(x), x = -5 .. 5, y = -5 .. 5, [[y(0) = 2], [y(0) = 4]], linecolor = red, color = blue, scaling = constrained, axes = boxed)

 

Problem 2 (EULER METHOD)

NULL

NULL

ODE2 := diff(y(x), x) = sin(x*y(x))

diff(y(x), x) = sin(x*y(x))

(5)

 

 

dsolve(ODE2, y(x))

NULL

Maple returned no output! That means Maple is unable to solve the equation.

NULL

If you are curious what steps Maple went through to  find a solution before failing
to do so, you can ask to see the steps using the command "infolevel". The levels
of information that you can request range from 0 to 5.

 

````

dsolve(ODE2, y(x))

soln := dsolve({ODE2, y(0) = 3}, y(x), numeric)

proc (x_rkf45) local _res, _dat, _vars, _solnproc, _xout, _ndsol, _pars, _n, _i; option `Copyright (c) 2000 by Waterloo Maple Inc. All rights reserved.`; if 1 < nargs then error "invalid input: too many arguments" end if; _EnvDSNumericSaveDigits := Digits; Digits := 15; if _EnvInFsolve = true then _xout := evalf[_EnvDSNumericSaveDigits](x_rkf45) else _xout := evalf(x_rkf45) end if; _dat := Array(1..4, {(1) = proc (_xin) local _xout, _dtbl, _dat, _vmap, _x0, _y0, _val, _dig, _n, _ne, _nd, _nv, _pars, _ini, _par, _i, _j, _k, _src; option `Copyright (c) 2002 by Waterloo Maple Inc. All rights reserved.`; table( [( "complex" ) = false ] ) _xout := _xin; _pars := []; _dtbl := array( 1 .. 4, [( 1 ) = (array( 1 .. 28, [( 1 ) = (datatype = float[8], order = C_order, storage = rectangular), ( 2 ) = (datatype = float[8], order = C_order, storage = rectangular), ( 3 ) = ([0, 0, 0, Array(1..0, 1..2, {}, datatype = float[8], order = C_order)]), ( 4 ) = (Array(1..65, {(1) = 1, (2) = 1, (3) = 0, (4) = 0, (5) = 0, (6) = 0, (7) = 1, (8) = 0, (9) = 0, (10) = 0, (11) = 0, (12) = 0, (13) = 0, (14) = 0, (15) = 0, (16) = 0, (17) = 0, (18) = 21, (19) = 30000, (20) = 0, (21) = 0, (22) = 1, (23) = 4, (24) = 0, (25) = 1, (26) = 15, (27) = 1, (28) = 0, (29) = 1, (30) = 3, (31) = 3, (32) = 0, (33) = 1, (34) = 0, (35) = 0, (36) = 0, (37) = 0, (38) = 0, (39) = 0, (40) = 0, (41) = 0, (42) = 0, (43) = 1, (44) = 0, (45) = 0, (46) = 0, (47) = 0, (48) = 0, (49) = 0, (50) = 50, (51) = 1, (52) = 0, (53) = 0, (54) = 0, (55) = 0, (56) = 0, (57) = 0, (58) = 0, (59) = 10000, (60) = 0, (61) = 1000, (62) = 0, (63) = 0, (64) = -1, (65) = 0}, datatype = integer[8])), ( 5 ) = (Array(1..28, {(1) = .0, (2) = 0.10e-5, (3) = .0, (4) = 0.500001e-14, (5) = .0, (6) = .16290418802201975, (7) = .0, (8) = 0.10e-5, (9) = .0, (10) = .0, (11) = .0, (12) = .0, (13) = 1.0, (14) = .0, (15) = .49999999999999, (16) = .0, (17) = 1.0, (18) = 1.0, (19) = .0, (20) = .0, (21) = 1.0, (22) = 1.0, (23) = .0, (24) = .0, (25) = 0.10e-14, (26) = .0, (27) = .0, (28) = .0}, datatype = float[8], order = C_order)), ( 6 ) = (Array(1..1, {(1) = 3.0}, datatype = float[8], order = C_order)), ( 7 ) = ([Array(1..4, 1..7, {(1, 1) = .0, (1, 2) = .203125, (1, 3) = .3046875, (1, 4) = .75, (1, 5) = .8125, (1, 6) = .40625, (1, 7) = .8125, (2, 1) = 0.6378173828125e-1, (2, 2) = .0, (2, 3) = .279296875, (2, 4) = .27237892150878906, (2, 5) = -0.9686851501464844e-1, (2, 6) = 0.1956939697265625e-1, (2, 7) = .5381584167480469, (3, 1) = 0.31890869140625e-1, (3, 2) = .0, (3, 3) = -.34375, (3, 4) = -.335235595703125, (3, 5) = .2296142578125, (3, 6) = .41748046875, (3, 7) = 11.480712890625, (4, 1) = 0.9710520505905151e-1, (4, 2) = .0, (4, 3) = .40350341796875, (4, 4) = 0.20297467708587646e-1, (4, 5) = -0.6054282188415527e-2, (4, 6) = -0.4770040512084961e-1, (4, 7) = .77858567237854}, datatype = float[8], order = C_order), Array(1..6, 1..6, {(1, 1) = .0, (1, 2) = .0, (1, 3) = .0, (1, 4) = .0, (1, 5) = .0, (1, 6) = 1.0, (2, 1) = .25, (2, 2) = .0, (2, 3) = .0, (2, 4) = .0, (2, 5) = .0, (2, 6) = 1.0, (3, 1) = .1875, (3, 2) = .5625, (3, 3) = .0, (3, 4) = .0, (3, 5) = .0, (3, 6) = 2.0, (4, 1) = .23583984375, (4, 2) = -.87890625, (4, 3) = .890625, (4, 4) = .0, (4, 5) = .0, (4, 6) = .2681884765625, (5, 1) = .1272735595703125, (5, 2) = -.5009765625, (5, 3) = .44921875, (5, 4) = -0.128936767578125e-1, (5, 5) = .0, (5, 6) = 0.626220703125e-1, (6, 1) = -0.927734375e-1, (6, 2) = .626220703125, (6, 3) = -.4326171875, (6, 4) = .1418304443359375, (6, 5) = -0.861053466796875e-1, (6, 6) = .3131103515625}, datatype = float[8], order = C_order), Array(1..6, {(1) = .0, (2) = .386, (3) = .21, (4) = .63, (5) = 1.0, (6) = 1.0}, datatype = float[8], order = C_order), Array(1..6, {(1) = .25, (2) = -.1043, (3) = .1035, (4) = -0.362e-1, (5) = .0, (6) = .0}, datatype = float[8], order = C_order), Array(1..6, 1..5, {(1, 1) = .0, (1, 2) = .0, (1, 3) = .0, (1, 4) = .0, (1, 5) = .0, (2, 1) = 1.544, (2, 2) = .0, (2, 3) = .0, (2, 4) = .0, (2, 5) = .0, (3, 1) = .9466785280815533, (3, 2) = .25570116989825814, (3, 3) = .0, (3, 4) = .0, (3, 5) = .0, (4, 1) = 3.3148251870684886, (4, 2) = 2.896124015972123, (4, 3) = .9986419139977808, (4, 4) = .0, (4, 5) = .0, (5, 1) = 1.2212245092262748, (5, 2) = 6.019134481287752, (5, 3) = 12.537083329320874, (5, 4) = -.687886036105895, (5, 5) = .0, (6, 1) = 1.2212245092262748, (6, 2) = 6.019134481287752, (6, 3) = 12.537083329320874, (6, 4) = -.687886036105895, (6, 5) = 1.0}, datatype = float[8], order = C_order), Array(1..6, 1..5, {(1, 1) = .0, (1, 2) = .0, (1, 3) = .0, (1, 4) = .0, (1, 5) = .0, (2, 1) = -5.6688, (2, 2) = .0, (2, 3) = .0, (2, 4) = .0, (2, 5) = .0, (3, 1) = -2.4300933568337584, (3, 2) = -.20635991570891224, (3, 3) = .0, (3, 4) = .0, (3, 5) = .0, (4, 1) = -.10735290581452621, (4, 2) = -9.594562251021896, (4, 3) = -20.470286148096154, (4, 4) = .0, (4, 5) = .0, (5, 1) = 7.496443313968615, (5, 2) = -10.246804314641219, (5, 3) = -33.99990352819906, (5, 4) = 11.708908932061595, (5, 5) = .0, (6, 1) = 8.083246795922411, (6, 2) = -7.981132988062785, (6, 3) = -31.52159432874373, (6, 4) = 16.319305431231363, (6, 5) = -6.0588182388340535}, datatype = float[8], order = C_order), Array(1..3, 1..5, {(1, 1) = .0, (1, 2) = .0, (1, 3) = .0, (1, 4) = .0, (1, 5) = .0, (2, 1) = 10.126235083446911, (2, 2) = -7.487995877607633, (2, 3) = -34.800918615557414, (2, 4) = -7.9927717075687275, (2, 5) = 1.0251377232956207, (3, 1) = -.6762803392806898, (3, 2) = 6.087714651678606, (3, 3) = 16.43084320892463, (3, 4) = 24.767225114183653, (3, 5) = -6.5943891257167815}, datatype = float[8], order = C_order)]), ( 9 ) = ([Array(1..1, {(1) = .1}, datatype = float[8], order = C_order), Array(1..1, {(1) = .0}, datatype = float[8], order = C_order), Array(1..1, {(1) = .0}, datatype = float[8], order = C_order), Array(1..1, {(1) = .0}, datatype = float[8], order = C_order), Array(1..1, {(1) = .0}, datatype = float[8], order = C_order), Array(1..1, 1..1, {(1, 1) = .0}, datatype = float[8], order = C_order), Array(1..1, 1..1, {(1, 1) = .0}, datatype = float[8], order = C_order), Array(1..1, {(1) = .0}, datatype = float[8], order = C_order), Array(1..1, 1..1, {(1, 1) = .0}, datatype = float[8], order = C_order), Array(1..1, 1..6, {(1, 1) = .0, (1, 2) = .0, (1, 3) = .0, (1, 4) = .0, (1, 5) = .0, (1, 6) = .0}, datatype = float[8], order = C_order), Array(1..1, {(1) = 0}, datatype = integer[8]), Array(1..1, {(1) = .0}, datatype = float[8], order = C_order), Array(1..1, {(1) = .0}, datatype = float[8], order = C_order), Array(1..1, {(1) = .0}, datatype = float[8], order = C_order), Array(1..1, {(1) = .0}, datatype = float[8], order = C_order), Array(1..1, {(1) = .0}, datatype = float[8], order = C_order), Array(1..2, {(1) = .0, (2) = .0}, datatype = float[8], order = C_order), Array(1..1, {(1) = 0}, datatype = integer[8])]), ( 8 ) = ([Array(1..1, {(1) = 3.0}, datatype = float[8], order = C_order), Array(1..1, {(1) = .0}, datatype = float[8], order = C_order), Array(1..1, {(1) = .0}, datatype = float[8], order = C_order), 0, 0]), ( 11 ) = (Array(1..6, 0..1, {(1, 1) = .0, (2, 0) = .0, (2, 1) = .0, (3, 0) = .0, (3, 1) = .0, (4, 0) = .0, (4, 1) = .0, (5, 0) = .0, (5, 1) = .0, (6, 0) = .0, (6, 1) = .0}, datatype = float[8], order = C_order)), ( 10 ) = ([proc (N, X, Y, YP) option `[Y[1] = y(x)]`; YP[1] := sin(X*Y[1]); 0 end proc, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]), ( 13 ) = (), ( 12 ) = (), ( 15 ) = ("rkf45"), ( 14 ) = ([0, 0]), ( 18 ) = ([]), ( 19 ) = (0), ( 16 ) = ([0, 0, 0, 0, 0, 0, []]), ( 17 ) = ([proc (N, X, Y, YP) option `[Y[1] = y(x)]`; YP[1] := sin(X*Y[1]); 0 end proc, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]), ( 22 ) = (0), ( 23 ) = (0), ( 20 ) = ([]), ( 21 ) = (0), ( 27 ) = (""), ( 26 ) = (Array(1..0, {})), ( 25 ) = (Array(1..0, {})), ( 24 ) = (0), ( 28 ) = (0)  ] ))  ] ); _y0 := Array(0..1, {(1) = 0.}); _vmap := array( 1 .. 1, [( 1 ) = (1)  ] ); _x0 := _dtbl[1][5][5]; _n := _dtbl[1][4][1]; _ne := _dtbl[1][4][3]; _nd := _dtbl[1][4][4]; _nv := _dtbl[1][4][16]; if not type(_xout, 'numeric') then if member(_xout, ["start", "left", "right"]) then if _Env_smart_dsolve_numeric = true or _dtbl[1][4][10] = 1 then if _xout = "left" then if type(_dtbl[2], 'table') then return _dtbl[2][5][1] end if elif _xout = "right" then if type(_dtbl[3], 'table') then return _dtbl[3][5][1] end if end if end if; return _dtbl[1][5][5] elif _xout = "method" then return _dtbl[1][15] elif _xout = "storage" then return evalb(_dtbl[1][4][10] = 1) elif _xout = "leftdata" then if not type(_dtbl[2], 'array') then return NULL else return eval(_dtbl[2]) end if elif _xout = "rightdata" then if not type(_dtbl[3], 'array') then return NULL else return eval(_dtbl[3]) end if elif _xout = "enginedata" then return eval(_dtbl[1]) elif _xout = "enginereset" then _dtbl[2] := evaln(_dtbl[2]); _dtbl[3] := evaln(_dtbl[3]); return NULL elif _xout = "initial" then return procname(_y0[0]) elif _xout = "laxtol" then return _dtbl[`if`(member(_dtbl[4], {2, 3}), _dtbl[4], 1)][5][18] elif _xout = "numfun" then return `if`(member(_dtbl[4], {2, 3}), _dtbl[_dtbl[4]][4][18], 0) elif _xout = "parameters" then return [seq(_y0[_n+_i], _i = 1 .. nops(_pars))] elif _xout = "initial_and_parameters" then return procname(_y0[0]), [seq(_y0[_n+_i], _i = 1 .. nops(_pars))] elif _xout = "last" then if _dtbl[4] <> 2 and _dtbl[4] <> 3 or _x0-_dtbl[_dtbl[4]][5][1] = 0. then error "no information is available on last computed point" else _xout := _dtbl[_dtbl[4]][5][1] end if elif _xout = "function" then if _dtbl[1][4][33]-2. = 0 then return eval(_dtbl[1][10], 1) else return eval(_dtbl[1][10][1], 1) end if elif _xout = "map" then return copy(_vmap) elif type(_xin, `=`) and type(rhs(_xin), 'list') and member(lhs(_xin), {"initial", "parameters", "initial_and_parameters"}) then _ini, _par := [], []; if lhs(_xin) = "initial" then _ini := rhs(_xin) elif lhs(_xin) = "parameters" then _par := rhs(_xin) elif select(type, rhs(_xin), `=`) <> [] then _par, _ini := selectremove(type, rhs(_xin), `=`) elif nops(rhs(_xin)) < nops(_pars)+1 then error "insufficient data for specification of initial and parameters" else _par := rhs(_xin)[-nops(_pars) .. -1]; _ini := rhs(_xin)[1 .. -nops(_pars)-1] end if; _xout := lhs(_xout); _i := false; if _par <> [] then _i := `dsolve/numeric/process_parameters`(_n, _pars, _par, _y0) end if; if _ini <> [] then _i := `dsolve/numeric/process_initial`(_n-_ne, _ini, _y0, _pars, _vmap) or _i end if; if _i then `dsolve/numeric/SC/reinitialize`(_dtbl, _y0, _n, procname, _pars); if _Env_smart_dsolve_numeric = true and type(_y0[0], 'numeric') and _dtbl[1][4][10] <> 1 then procname("right") := _y0[0]; procname("left") := _y0[0] end if end if; if _xout = "initial" then return [_y0[0], seq(_y0[_vmap[_i]], _i = 1 .. _n-_ne)] elif _xout = "parameters" then return [seq(_y0[_n+_i], _i = 1 .. nops(_pars))] else return [_y0[0], seq(_y0[_vmap[_i]], _i = 1 .. _n-_ne)], [seq(_y0[_n+_i], _i = 1 .. nops(_pars))] end if elif _xin = "eventstop" then if _nv = 0 then error "this solution has no events" end if; _i := _dtbl[4]; if _i <> 2 and _i <> 3 then return 0 end if; if _dtbl[_i][4][10] = 1 and assigned(_dtbl[5-_i]) and _dtbl[_i][4][9] < 100 and 100 <= _dtbl[5-_i][4][9] then _i := 5-_i; _dtbl[4] := _i; _j := round(_dtbl[_i][4][17]); return round(_dtbl[_i][3][1][_j, 1]) elif 100 <= _dtbl[_i][4][9] then _j := round(_dtbl[_i][4][17]); return round(_dtbl[_i][3][1][_j, 1]) else return 0 end if elif _xin = "eventstatus" then if _nv = 0 then error "this solution has no events" end if; _i := [selectremove(proc (a) options operator, arrow; _dtbl[1][3][1][a, 7] = 1 end proc, {seq(_j, _j = 1 .. round(_dtbl[1][3][1][_nv+1, 1]))})]; return ':-enabled' = _i[1], ':-disabled' = _i[2] elif _xin = "eventclear" then if _nv = 0 then error "this solution has no events" end if; _i := _dtbl[4]; if _i <> 2 and _i <> 3 then error "no events to clear" end if; if _dtbl[_i][4][10] = 1 and assigned(_dtbl[5-_i]) and _dtbl[_i][4][9] < 100 and 100 < _dtbl[5-_i][4][9] then _dtbl[4] := 5-_i; _i := 5-_i end if; if _dtbl[_i][4][9] < 100 then error "no events to clear" elif _nv < _dtbl[_i][4][9]-100 then error "event error condition cannot be cleared" else _j := _dtbl[_i][4][9]-100; if irem(round(_dtbl[_i][3][1][_j, 4]), 2) = 1 then error "retriggerable events cannot be cleared" end if; _j := round(_dtbl[_i][3][1][_j, 1]); for _k to _nv do if _dtbl[_i][3][1][_k, 1] = _j then if _dtbl[_i][3][1][_k, 2] = 3 then error "range events cannot be cleared" end if; _dtbl[_i][3][1][_k, 8] := _dtbl[_i][3][1][_nv+1, 8] end if end do; _dtbl[_i][4][17] := 0; _dtbl[_i][4][9] := 0; if _dtbl[1][4][10] = 1 then if _i = 2 then try procname(procname("left")) catch:  end try else try procname(procname("right")) catch:  end try end if end if end if; return  elif type(_xin, `=`) and member(lhs(_xin), {"eventdisable", "eventenable"}) then if _nv = 0 then error "this solution has no events" end if; if type(rhs(_xin), {('list')('posint'), ('set')('posint')}) then _i := {op(rhs(_xin))} elif type(rhs(_xin), 'posint') then _i := {rhs(_xin)} else error "event identifiers must be integers in the range 1..%1", round(_dtbl[1][3][1][_nv+1, 1]) end if; if select(proc (a) options operator, arrow; _nv < a end proc, _i) <> {} then error "event identifiers must be integers in the range 1..%1", round(_dtbl[1][3][1][_nv+1, 1]) end if; _k := {}; for _j to _nv do if member(round(_dtbl[1][3][1][_j, 1]), _i) then _k := `union`(_k, {_j}) end if end do; _i := _k; if lhs(_xin) = "eventdisable" then _dtbl[4] := 0; _j := [evalb(assigned(_dtbl[2]) and member(_dtbl[2][4][17], _i)), evalb(assigned(_dtbl[3]) and member(_dtbl[3][4][17], _i))]; for _k in _i do _dtbl[1][3][1][_k, 7] := 0; if assigned(_dtbl[2]) then _dtbl[2][3][1][_k, 7] := 0 end if; if assigned(_dtbl[3]) then _dtbl[3][3][1][_k, 7] := 0 end if end do; if _j[1] then for _k to _nv+1 do if _k <= _nv and not type(_dtbl[2][3][4][_k, 1], 'undefined') then userinfo(3, {'events', 'eventreset'}, `reinit #2, event code `, _k, ` to defined init `, _dtbl[2][3][4][_k, 1]); _dtbl[2][3][1][_k, 8] := _dtbl[2][3][4][_k, 1] elif _dtbl[2][3][1][_k, 2] = 0 and irem(iquo(round(_dtbl[2][3][1][_k, 4]), 32), 2) = 1 then userinfo(3, {'events', 'eventreset'}, `reinit #2, event code `, _k, ` to rate hysteresis init `, _dtbl[2][5][24]); _dtbl[2][3][1][_k, 8] := _dtbl[2][5][24] elif _dtbl[2][3][1][_k, 2] = 0 and irem(iquo(round(_dtbl[2][3][1][_k, 4]), 2), 2) = 0 then userinfo(3, {'events', 'eventreset'}, `reinit #2, event code `, _k, ` to initial init `, _x0); _dtbl[2][3][1][_k, 8] := _x0 else userinfo(3, {'events', 'eventreset'}, `reinit #2, event code `, _k, ` to fireinitial init `, _x0-1); _dtbl[2][3][1][_k, 8] := _x0-1 end if end do; _dtbl[2][4][17] := 0; _dtbl[2][4][9] := 0; if _dtbl[1][4][10] = 1 then procname(procname("left")) end if end if; if _j[2] then for _k to _nv+1 do if _k <= _nv and not type(_dtbl[3][3][4][_k, 2], 'undefined') then userinfo(3, {'events', 'eventreset'}, `reinit #3, event code `, _k, ` to defined init `, _dtbl[3][3][4][_k, 2]); _dtbl[3][3][1][_k, 8] := _dtbl[3][3][4][_k, 2] elif _dtbl[3][3][1][_k, 2] = 0 and irem(iquo(round(_dtbl[3][3][1][_k, 4]), 32), 2) = 1 then userinfo(3, {'events', 'eventreset'}, `reinit #3, event code `, _k, ` to rate hysteresis init `, _dtbl[3][5][24]); _dtbl[3][3][1][_k, 8] := _dtbl[3][5][24] elif _dtbl[3][3][1][_k, 2] = 0 and irem(iquo(round(_dtbl[3][3][1][_k, 4]), 2), 2) = 0 then userinfo(3, {'events', 'eventreset'}, `reinit #3, event code `, _k, ` to initial init `, _x0); _dtbl[3][3][1][_k, 8] := _x0 else userinfo(3, {'events', 'eventreset'}, `reinit #3, event code `, _k, ` to fireinitial init `, _x0+1); _dtbl[3][3][1][_k, 8] := _x0+1 end if end do; _dtbl[3][4][17] := 0; _dtbl[3][4][9] := 0; if _dtbl[1][4][10] = 1 then procname(procname("right")) end if end if else for _k in _i do _dtbl[1][3][1][_k, 7] := 1 end do; _dtbl[2] := evaln(_dtbl[2]); _dtbl[3] := evaln(_dtbl[3]); _dtbl[4] := 0; if _dtbl[1][4][10] = 1 then if _x0 <= procname("right") then try procname(procname("right")) catch:  end try end if; if procname("left") <= _x0 then try procname(procname("left")) catch:  end try end if end if end if; return  elif type(_xin, `=`) and lhs(_xin) = "eventfired" then if not type(rhs(_xin), 'list') then error "'eventfired' must be specified as a list" end if; if _nv = 0 then error "this solution has no events" end if; if _dtbl[4] <> 2 and _dtbl[4] <> 3 then error "'direction' must be set prior to calling/setting 'eventfired'" end if; _i := _dtbl[4]; _val := NULL; if not assigned(_EnvEventRetriggerWarned) then _EnvEventRetriggerWarned := false end if; for _k in rhs(_xin) do if type(_k, 'integer') then _src := _k elif type(_k, 'integer' = 'anything') and type(evalf(rhs(_k)), 'numeric') then _k := lhs(_k) = evalf[max(Digits, 18)](rhs(_k)); _src := lhs(_k) else error "'eventfired' entry is not valid: %1", _k end if; if _src < 1 or round(_dtbl[1][3][1][_nv+1, 1]) < _src then error "event identifiers must be integers in the range 1..%1", round(_dtbl[1][3][1][_nv+1, 1]) end if; _src := {seq(`if`(_dtbl[1][3][1][_j, 1]-_src = 0., _j, NULL), _j = 1 .. _nv)}; if nops(_src) <> 1 then error "'eventfired' can only be set/queried for root-finding events and time/interval events" end if; _src := _src[1]; if _dtbl[1][3][1][_src, 2] <> 0. and _dtbl[1][3][1][_src, 2]-2. <> 0. then error "'eventfired' can only be set/queried for root-finding events and time/interval events" elif irem(round(_dtbl[1][3][1][_src, 4]), 2) = 1 then if _EnvEventRetriggerWarned = false then WARNING(`'eventfired' has no effect on events that retrigger`) end if; _EnvEventRetriggerWarned := true end if; if _dtbl[_i][3][1][_src, 2] = 0 and irem(iquo(round(_dtbl[_i][3][1][_src, 4]), 32), 2) = 1 then _val := _val, undefined elif type(_dtbl[_i][3][4][_src, _i-1], 'undefined') or _i = 2 and _dtbl[2][3][1][_src, 8] < _dtbl[2][3][4][_src, 1] or _i = 3 and _dtbl[3][3][4][_src, 2] < _dtbl[3][3][1][_src, 8] then _val := _val, _dtbl[_i][3][1][_src, 8] else _val := _val, _dtbl[_i][3][4][_src, _i-1] end if; if type(_k, `=`) then if _dtbl[_i][3][1][_src, 2] = 0 and irem(iquo(round(_dtbl[_i][3][1][_src, 4]), 32), 2) = 1 then error "cannot set event code for a rate hysteresis event" end if; userinfo(3, {'events', 'eventreset'}, `manual set event code `, _src, ` to value `, rhs(_k)); _dtbl[_i][3][1][_src, 8] := rhs(_k); _dtbl[_i][3][4][_src, _i-1] := rhs(_k) end if end do; return [_val] elif type(_xin, `=`) and lhs(_xin) = "direction" then if not member(rhs(_xin), {-1, 1, ':-left', ':-right'}) then error "'direction' must be specified as either '1' or 'right' (positive) or '-1' or 'left' (negative)" end if; _src := `if`(_dtbl[4] = 2, -1, `if`(_dtbl[4] = 3, 1, undefined)); _i := `if`(member(rhs(_xin), {1, ':-right'}), 3, 2); _dtbl[4] := _i; _dtbl[_i] := `dsolve/numeric/SC/IVPdcopy`(_dtbl[1], `if`(assigned(_dtbl[_i]), _dtbl[_i], NULL)); if 0 < _nv then for _j to _nv+1 do if _j <= _nv and not type(_dtbl[_i][3][4][_j, _i-1], 'undefined') then userinfo(3, {'events', 'eventreset'}, `reinit #4, event code `, _j, ` to defined init `, _dtbl[_i][3][4][_j, _i-1]); _dtbl[_i][3][1][_j, 8] := _dtbl[_i][3][4][_j, _i-1] elif _dtbl[_i][3][1][_j, 2] = 0 and irem(iquo(round(_dtbl[_i][3][1][_j, 4]), 32), 2) = 1 then userinfo(3, {'events', 'eventreset'}, `reinit #4, event code `, _j, ` to rate hysteresis init `, _dtbl[_i][5][24]); _dtbl[_i][3][1][_j, 8] := _dtbl[_i][5][24] elif _dtbl[_i][3][1][_j, 2] = 0 and irem(iquo(round(_dtbl[_i][3][1][_j, 4]), 2), 2) = 0 then userinfo(3, {'events', 'eventreset'}, `reinit #4, event code `, _j, ` to initial init `, _x0); _dtbl[_i][3][1][_j, 8] := _x0 else userinfo(3, {'events', 'eventreset'}, `reinit #4, event code `, _j, ` to fireinitial init `, _x0-2*_i+5.0); _dtbl[_i][3][1][_j, 8] := _x0-2*_i+5.0 end if end do end if; return _src elif _xin = "eventcount" then if _dtbl[1][3][1] = 0 or _dtbl[4] <> 2 and _dtbl[4] <> 3 then return 0 else return round(_dtbl[_dtbl[4]][3][1][_nv+1, 12]) end if elif type(_xin, `=`) and lhs(_xin) = "setdatacallback" then if not type(rhs(_xin), 'nonegint') then error "data callback must be a nonnegative integer (address)" end if; _dtbl[1][28] := rhs(_xin) else return "procname" end if end if; if _xout = _x0 then return [_x0, seq(evalf(_dtbl[1][6][_vmap[_i]]), _i = 1 .. _n-_ne)] end if; _i := `if`(_x0 <= _xout, 3, 2); if _xin = "last" and 0 < _dtbl[_i][4][9] and _dtbl[_i][4][9] < 100 then _dat := eval(_dtbl[_i], 2); _j := _dat[4][20]; return [_dat[11][_j, 0], seq(_dat[11][_j, _vmap[_i]], _i = 1 .. _n-_ne-_nd), seq(_dat[8][1][_vmap[_i]], _i = _n-_ne-_nd+1 .. _n-_ne)] end if; if not type(_dtbl[_i], 'array') then _dtbl[_i] := `dsolve/numeric/SC/IVPdcopy`(_dtbl[1], `if`(assigned(_dtbl[_i]), _dtbl[_i], NULL)); if 0 < _nv then for _j to _nv+1 do if _j <= _nv and not type(_dtbl[_i][3][4][_j, _i-1], 'undefined') then userinfo(3, {'events', 'eventreset'}, `reinit #5, event code `, _j, ` to defined init `, _dtbl[_i][3][4][_j, _i-1]); _dtbl[_i][3][1][_j, 8] := _dtbl[_i][3][4][_j, _i-1] elif _dtbl[_i][3][1][_j, 2] = 0 and irem(iquo(round(_dtbl[_i][3][1][_j, 4]), 32), 2) = 1 then userinfo(3, {'events', 'eventreset'}, `reinit #5, event code `, _j, ` to rate hysteresis init `, _dtbl[_i][5][24]); _dtbl[_i][3][1][_j, 8] := _dtbl[_i][5][24] elif _dtbl[_i][3][1][_j, 2] = 0 and irem(iquo(round(_dtbl[_i][3][1][_j, 4]), 2), 2) = 0 then userinfo(3, {'events', 'eventreset'}, `reinit #5, event code `, _j, ` to initial init `, _x0); _dtbl[_i][3][1][_j, 8] := _x0 else userinfo(3, {'events', 'eventreset'}, `reinit #5, event code `, _j, ` to fireinitial init `, _x0-2*_i+5.0); _dtbl[_i][3][1][_j, 8] := _x0-2*_i+5.0 end if end do end if end if; if _xin <> "last" then if 0 < 0 then if `dsolve/numeric/checkglobals`(op(_dtbl[1][14]), _pars, _n, _y0) then `dsolve/numeric/SC/reinitialize`(_dtbl, _y0, _n, procname, _pars, _i) end if end if; if _dtbl[1][4][7] = 0 then error "parameters must be initialized before solution can be computed" end if end if; _dat := eval(_dtbl[_i], 2); _dtbl[4] := _i; try _src := `dsolve/numeric/SC/IVPrun`(_dat, _xout) catch: userinfo(2, `dsolve/debug`, print(`Exception in solnproc:`, [lastexception][2 .. -1])); error  end try; if _dat[17] <> _dtbl[1][17] then _dtbl[1][17] := _dat[17]; _dtbl[1][10] := _dat[10] end if; if _src = 0 and 100 < _dat[4][9] then _val := _dat[3][1][_nv+1, 8] else _val := _dat[11][_dat[4][20], 0] end if; if _src <> 0 or _dat[4][9] <= 0 then _dtbl[1][5][1] := _xout else _dtbl[1][5][1] := _val end if; if _i = 3 and _val < _xout then Rounding := -infinity; if _dat[4][9] = 1 then error "cannot evaluate the solution further right of %1, probably a singularity", evalf[8](_val) elif _dat[4][9] = 2 then error "cannot evaluate the solution further right of %1, maxfun limit exceeded (see <a href='http://www.maplesoft.com/support/help/search.aspx?term=dsolve,maxfun' target='_new'>?dsolve,maxfun</a> for details)", evalf[8](_val) elif _dat[4][9] = 3 then if _dat[4][25] = 3 then error "cannot evaluate the solution past the initial point, problem may be initially singular or improperly set up" else error "cannot evaluate the solution past the initial point, problem may be complex, initially singular or improperly set up" end if elif _dat[4][9] = 4 then error "cannot evaluate the solution further right of %1, accuracy goal cannot be achieved with specified 'minstep'", evalf[8](_val) elif _dat[4][9] = 5 then error "cannot evaluate the solution further right of %1, too many step failures, tolerances may be too loose for problem", evalf[8](_val) elif _dat[4][9] = 6 then error "cannot evaluate the solution further right of %1, cannot downgrade delay storage for problems with delay derivative order > 1, try increasing delaypts", evalf[8](_val) elif _dat[4][9] = 10 then error "cannot evaluate the solution further right of %1, interrupt requested", evalf[8](_val) elif 100 < _dat[4][9] then if _dat[4][9]-100 = _nv+1 then error "constraint projection failure on event at t=%1", evalf[8](_val) elif _dat[4][9]-100 = _nv+2 then error "index-1 and derivative evaluation failure on event at t=%1", evalf[8](_val) elif _dat[4][9]-100 = _nv+3 then error "maximum number of event iterations reached (%1) at t=%2", round(_dat[3][1][_nv+1, 3]), evalf[8](_val) else if _Env_dsolve_nowarnstop <> true then `dsolve/numeric/warning`(StringTools:-FormatMessage("cannot evaluate the solution further right of %1, event #%2 triggered a halt", evalf[8](_val), round(_dat[3][1][_dat[4][9]-100, 1]))) end if; Rounding := 'nearest'; _xout := _val end if else error "cannot evaluate the solution further right of %1", evalf[8](_val) end if elif _i = 2 and _xout < _val then Rounding := infinity; if _dat[4][9] = 1 then error "cannot evaluate the solution further left of %1, probably a singularity", evalf[8](_val) elif _dat[4][9] = 2 then error "cannot evaluate the solution further left of %1, maxfun limit exceeded (see <a href='http://www.maplesoft.com/support/help/search.aspx?term=dsolve,maxfun' target='_new'>?dsolve,maxfun</a> for details)", evalf[8](_val) elif _dat[4][9] = 3 then if _dat[4][25] = 3 then error "cannot evaluate the solution past the initial point, problem may be initially singular or improperly set up" else error "cannot evaluate the solution past the initial point, problem may be complex, initially singular or improperly set up" end if elif _dat[4][9] = 4 then error "cannot evaluate the solution further left of %1, accuracy goal cannot be achieved with specified 'minstep'", evalf[8](_val) elif _dat[4][9] = 5 then error "cannot evaluate the solution further left of %1, too many step failures, tolerances may be too loose for problem", evalf[8](_val) elif _dat[4][9] = 6 then error "cannot evaluate the solution further left of %1, cannot downgrade delay storage for problems with delay derivative order > 1, try increasing delaypts", evalf[8](_val) elif _dat[4][9] = 10 then error "cannot evaluate the solution further right of %1, interrupt requested", evalf[8](_val) elif 100 < _dat[4][9] then if _dat[4][9]-100 = _nv+1 then error "constraint projection failure on event at t=%1", evalf[8](_val) elif _dat[4][9]-100 = _nv+2 then error "index-1 and derivative evaluation failure on event at t=%1", evalf[8](_val) elif _dat[4][9]-100 = _nv+3 then error "maximum number of event iterations reached (%1) at t=%2", round(_dat[3][1][_nv+1, 3]), evalf[8](_val) else if _Env_dsolve_nowarnstop <> true then `dsolve/numeric/warning`(StringTools:-FormatMessage("cannot evaluate the solution further left of %1, event #%2 triggered a halt", evalf[8](_val), round(_dat[3][1][_dat[4][9]-100, 1]))) end if; Rounding := 'nearest'; _xout := _val end if else error "cannot evaluate the solution further left of %1", evalf[8](_val) end if end if; if _EnvInFsolve = true then _dig := _dat[4][26]; if type(_EnvDSNumericSaveDigits, 'posint') then _dat[4][26] := _EnvDSNumericSaveDigits else _dat[4][26] := Digits end if; _Env_dsolve_SC_native := true; if _dat[4][25] = 1 then _i := 1; _dat[4][25] := 2 else _i := _dat[4][25] end if; _val := `dsolve/numeric/SC/IVPval`(_dat, _xout, _src); _dat[4][25] := _i; _dat[4][26] := _dig; [_xout, seq(_val[_vmap[_i]], _i = 1 .. _n-_ne)] else Digits := _dat[4][26]; _val := `dsolve/numeric/SC/IVPval`(eval(_dat, 2), _xout, _src); [_xout, seq(_val[_vmap[_i]], _i = 1 .. _n-_ne)] end if end proc, (2) = Array(0..0, {}), (3) = [x, y(x)], (4) = []}); _vars := _dat[3]; _pars := map(lhs, _dat[4]); _n := nops(_vars)-1; _solnproc := _dat[1]; if not type(_xout, 'numeric') then if member(x_rkf45, ["start", 'start', "method", 'method', "left", 'left', "right", 'right', "leftdata", "rightdata", "enginedata", "eventstop", 'eventstop', "eventclear", 'eventclear', "eventstatus", 'eventstatus', "eventcount", 'eventcount', "laxtol", 'laxtol', "numfun", 'numfun', NULL]) then _res := _solnproc(convert(x_rkf45, 'string')); if 1 < nops([_res]) then return _res elif type(_res, 'array') then return eval(_res, 1) elif _res <> "procname" then return _res end if elif member(x_rkf45, ["last", 'last', "initial", 'initial', "parameters", 'parameters', "initial_and_parameters", 'initial_and_parameters', NULL]) then _xout := convert(x_rkf45, 'string'); _res := _solnproc(_xout); if _xout = "parameters" then return [seq(_pars[_i] = _res[_i], _i = 1 .. nops(_pars))] elif _xout = "initial_and_parameters" then return [seq(_vars[_i+1] = [_res][1][_i+1], _i = 0 .. _n), seq(_pars[_i] = [_res][2][_i], _i = 1 .. nops(_pars))] else return [seq(_vars[_i+1] = _res[_i+1], _i = 0 .. _n)] end if elif type(_xout, `=`) and member(lhs(_xout), ["initial", 'initial', "parameters", 'parameters', "initial_and_parameters", 'initial_and_parameters', NULL]) then _xout := convert(lhs(x_rkf45), 'string') = rhs(x_rkf45); if type(rhs(_xout), 'list') then _res := _solnproc(_xout) else error "initial and/or parameter values must be specified in a list" end if; if lhs(_xout) = "initial" then return [seq(_vars[_i+1] = _res[_i+1], _i = 0 .. _n)] elif lhs(_xout) = "parameters" then return [seq(_pars[_i] = _res[_i], _i = 1 .. nops(_pars))] else return [seq(_vars[_i+1] = [_res][1][_i+1], _i = 0 .. _n), seq(_pars[_i] = [_res][2][_i], _i = 1 .. nops(_pars))] end if elif type(_xout, `=`) and member(lhs(_xout), ["eventdisable", 'eventdisable', "eventenable", 'eventenable', "eventfired", 'eventfired', "direction", 'direction', NULL]) then return _solnproc(convert(lhs(x_rkf45), 'string') = rhs(x_rkf45)) elif _xout = "solnprocedure" then return eval(_solnproc) elif _xout = "sysvars" then return _vars end if; if procname <> unknown then return ('procname')(x_rkf45) else _ndsol := 1; _ndsol := _ndsol; _ndsol := pointto(_dat[2][0]); return ('_ndsol')(x_rkf45) end if end if; try _res := _solnproc(_xout); [seq(_vars[_i+1] = _res[_i+1], _i = 0 .. _n)] catch: error  end try end proc

(6)

[soln(1), soln(1.5), soln(2), soln(2.5), soln(3), soln(3.5), soln(4), soln(4.5), soln(5)]

[[x = 1., y(x) = HFloat(3.5280317971877286)], [x = 1.5, y(x) = HFloat(3.1226400064202693)], [x = 2., y(x) = HFloat(2.653970420106217)], [x = 2.5, y(x) = HFloat(2.323175669304077)], [x = 3., y(x) = HFloat(2.3019187052877816)], [x = 3.5, y(x) = HFloat(2.6587033583656714)], [x = 4., y(x) = HFloat(2.497876146260152)], [x = 4.5, y(x) = HFloat(2.220305976552359)], [x = 5., y(x) = HFloat(1.9758297601444128)]]

(7)

p1 := odeplot(soln, [x, y(x)], x = 0 .. 5, color = magenta, thickness = 2, scaling = constrained, view = [0 .. 5, 0 .. 5])

 

p2 := dfieldplot(ODE2, [y(x)], x = 0 .. 5, y = 0 .. 5, color = blue, scaling = constrained)

display(p1, p2)

 

DEplot(ODE2, y(x), x = 0 .. 5, y = 0 .. 5, [[y(0) = 3]], linecolor = red, color = blue, scaling = constrained, axes = boxed)

 

Construct approximate solutions for x from 0 to 5 to the initial problem 2 using Euler's method with the three different step sizes Δx=0.5, 0.25, 0.125

f := proc (x, y) options operator, arrow; sin(x*y) end proc

proc (x, y) options operator, arrow; sin(y*x) end proc

(8)

Eulermethod := proc (f, x_start, y_start, dx, n_total) local x, y, Y, i, k; x[1] := x_start; y[1] := y_start; for i to n_total do y[i+1] := y[i]+f(x[i], y[i])*dx; x[i+1] := x[1]+i*dx end do; Y := [seq([x[k], y[k]], k = 1 .. n_total)]; return Y end proc

proc (f, x_start, y_start, dx, n_total) local x, y, Y, i, k; x[1] := x_start; y[1] := y_start; for i to n_total do y[i+1] := y[i]+f(x[i], y[i])*dx; x[i+1] := x[1]+i*dx end do; Y := [seq([x[k], y[k]], k = 1 .. n_total)]; return Y end proc

(9)

NULL

NULL

NULL

NULL

A1 := Eulermethod(f, 0, 3, .5, 10)

[[0, 3], [.5, 3.], [1.0, 3.498747493], [1.5, 3.323942476], [2.0, 2.842530099], [2.5, 2.560983065], [3.0, 2.620477947], [3.5, 3.120464063], [4.0, 2.621830593], [4.5, 2.185032319]]

(10)

A2 := Eulermethod(f, 0, 3, .25, 20)

[[0, 3], [.25, 3.], [.50, 3.170409690], [.75, 3.420383740], [1.00, 3.556616077], [1.25, 3.455813236], [1.50, 3.224836021], [1.75, 2.976782400], [2.00, 2.757025820], [2.25, 2.583147563], [2.50, 2.469680146], [2.75, 2.442487816], [3.00, 2.547535655], [3.25, 2.791971512], [3.50, 2.877900373], [3.75, 2.727027361], [4.00, 2.547414295], [4.25, 2.374301829], [4.50, 2.219839448], [4.75, 2.086091178]]

(11)

A3 := Eulermethod(f, 0, 3, .125, 40)

[[0, 3], [.125, 3.], [.250, 3.045784066], [.375, 3.132030172], [.500, 3.247342837], [.625, 3.372168142], [.750, 3.479586272], [.875, 3.542983060], [1.000, 3.548166882], [1.125, 3.498733738], [1.250, 3.409546073], [1.375, 3.297015011], [1.500, 3.174012084], [1.625, 3.049159854], [1.750, 2.927817142], [1.875, 2.813241461], [2.000, 2.707496816], [2.125, 2.612101612], [2.250, 2.528513147], [2.375, 2.458549923], [2.500, 2.404840953], [2.625, 2.371369082], [2.750, 2.364080535], [2.875, 2.391119623], [3.000, 2.460798019], [3.125, 2.572154041], [3.250, 2.695044010], [3.375, 2.772263417], [3.500, 2.780805371], [3.625, 2.742906337], [3.750, 2.680985435], [3.875, 2.607451728], [4.000, 2.528940353], [4.125, 2.449278434], [4.250, 2.370825619], [4.375, 2.295054886], [4.500, 2.222824117], [4.625, 2.154537647], [4.750, 2.090275081], [4.875, 2.029905439]]

(12)

NULL

p3 := pointplot(A1, color = green, scaling = constrained, symbol = circle)

p4 := pointplot(A2, color = black, scaling = constrained, symbol = asterisk)

p5 := pointplot(A3, color = red, scaling = constrained, symbol = diamond)

NULL

display([p1, p3, p4, p5])

 

NULL

In this plot, the differences between the lines visually represent how using different step sizes affects the overall solution accuracy. The red points are closest to what a more accurate numerical solution would look like, while the green points are more of a rough approximation.

 

 

Problem 3 (IMPROVED EULER METHOD

``

 

 

 

ImprovedEulermethod := proc (f, x_start, y_start, dx, n_total) local x, y, Y, i, k, k1, k2; x[1] := x_start; y[1] := y_start; for i to n_total do k1 := f(x[i], y[i]); k2 := f(x[i]+dx, y[i]+k1*dx); y[i+1] := y[i]+(1/2)*dx*(k1+k2); x[i+1] := x[1]+i*dx end do; Y := [seq([x[k], y[k]], k = 1 .. n_total+1)]; return Y end proc

proc (f, x_start, y_start, dx, n_total) local x, y, Y, i, k, k1, k2; x[1] := x_start; y[1] := y_start; for i to n_total do k1 := f(x[i], y[i]); k2 := f(x[i]+dx, y[i]+k1*dx); y[i+1] := y[i]+(1/2)*dx*(k1+k2); x[i+1] := x[1]+i*dx end do; Y := [seq([x[k], y[k]], k = 1 .. n_total+1)]; return Y end proc

(13)

A := ImprovedEulermethod(f, 0, 3, .5, 10); B := ImprovedEulermethod(f, 0, 3, .25, 20); C := ImprovedEulermethod(f, 0, 3, .125, 40)

[[0, 3], [.125, 3.022892033], [.250, 3.089335383], [.375, 3.190999573], [.500, 3.311460714], [.625, 3.426126738], [.750, 3.508312296], [.875, 3.539992283], [1.000, 3.518184043], [1.125, 3.451931748], [1.250, 3.354946855], [1.375, 3.240089444], [1.500, 3.117181611], [1.625, 2.992925708], [1.750, 2.871597591], [1.875, 2.755823574], [2.000, 2.647215450], [2.125, 2.546845751], [2.250, 2.455612702], [2.375, 2.374560360], [2.500, 2.305227222], [2.625, 2.250113264], [2.750, 2.213376654], [2.875, 2.201814459], [3.000, 2.225589190], [3.125, 2.295322044], [3.250, 2.406184220], [3.375, 2.516909932], [3.500, 2.583378006], [3.625, 2.599915321], [3.750, 2.579967129], [3.875, 2.537160528], [4.000, 2.481052245], [4.125, 2.417781798], [4.250, 2.351265142], [4.375, 2.284028571], [4.500, 2.217702881], [4.625, 2.153313287], [4.750, 2.091461577], [4.875, 2.032452273], [5.000, 1.976386380]]

(14)

NULL

plot2 := pointplot(A, color = green, scaling = constrained, symbol = circle)

plot3 := pointplot(B, color = black, scaling = constrained, symbol = asterisk)

plot4 := pointplot(C, color = red, scaling = constrained, symbol = diamond)

NULL

display([p1, plot2, plot3, plot4])

 

The Improved Euler method, by averaging slopes, provides a significant improvement over the basic Euler method, particularly when the step size is relatively large.


Download Diff_equations.mw

This is a task from one forum:  “Let's mark an arbitrary point on the circle. Let's draw a segment from this point, perpendicular to the diameter, and draw a circle, the center of which is at this point, and the radius is equal to this segment. Let's mark the intersection point of the segment connecting the intersection points of the circles with the perpendicular segment. Prove that the locus of all such points is an ellipse.”
I wanted to get a picture of a numerically animated "proof" using Maple tools.

МАTH_HЕLP_PLANET.mw
 And in fact, it turned out that AB=2AC, or AC=BC.

Source codes (seen in the pdf file) for the paper "Maximal Gap Among Integers Having a Common Divisor with an Odd Semi-prime".

MaxGapTheorem2.pdf

The flag of Germany on the strip of the German mathematician August Ferdinand Möbius. Basically, it's just one way to represent flags of a certain type. It seemed that the flag looked good on the Mobius strip.
FLAG.mw

Hello!

I present a simple work-up of rolling a plane curve along a fixed plane curve in 2d space. Maple sources are attached.

Kind regards!

Source.zip

Today in class, we presented an exercise based on the paper titled "Analysis of regular and chaotic dynamics in a stochastic eco-epidemiological model" by Bashkirtseva, Ryashko, and Ryazanova (2020). In this exercise, we kept all parameters of the model the same as in the paper, but we varied the parameter β, which represents the rate of infection spread. The goal was to observe how changes in β impact the system's dynamics, particularly focusing on the transition between regular and chaotic behavior.

This exercise involves studying a mathematical model that appears in eco-epidemiology. The model is described by the following set of equations:

dx/dt = rx-bx^2-cxy-`&beta;xz`/(a+x)-a[1]*yz/(e+y)

dy/dt = -`&mu;y`+`&beta;xy`/(a+x)-a[2]*yz/(d+y)

" (dz)/(dt)=-mz+((c[1 ]a[1])[ ]xz)/(e[]+x)+((c[1 ]a[2])[ ]yz)/(d+y)"

 

where r, b, c, β, α,a[1],a[2], e, d, m, c[1], c[2], μ>0 are given parameters. This model generalizes the classic predator-prey system by incorporating disease dynamics within the prey population. The populations are divided into the following groups:

 

• 

Susceptible prey population (x): Individuals in the prey population that are healthy but can become infected by a disease.

• 

Infected prey population (y): Individuals in the prey population that are infected and can transmit the disease to others.

• 

Predator population (z): The predator population that feeds on both susceptible (x) and infected (y) prey.

 

The initial conditions are always x(0)=0.2, y(0)=0.05, z(0)=0.05,  and we will vary the parameter β.;

 

For this exercise, the parameters are fixed as follows:

 

"r=1,` b`=1,` c`=0.01, a=0.36 ,` a`[1]=0.01,` a`[2]=0.05,` e`[]=15,` m`=0.01,` d`=0.5,` c`[1]=2,` `c[2]==1,` mu`=0.4."

NULL

Task (a)

• 

Solve the system numerically for the given parameter values and initial conditions with β=0.6 over the time interval t2[0,20000].

• 

Plot the solutions x(t), y(t), and  z(t) over this time interval.

• 

Comment on the model's predictions, keeping in mind that the time units are usually days.

• 

Also, plot the trajectory in the 3D space (x,y,z).

 

restart

r := 1; b := 1; f := 0.1e-1; alpha := .36; a[1] := 0.1e-1; a[2] := 0.5e-1; e := 15; m := 0.1e-1; d := .5; c[1] := 2; c[2] := 1; mu := .4; beta := .6

sys := {diff(x(t), t) = r*x(t)-b*x(t)^2-f*x(t)*y(t)-beta*x(t)*y(t)/(alpha+x(t))-a[1]*x(t)*z(t)/(e+x(t)), diff(y(t), t) = -mu*y(t)+beta*x(t)*y(t)/(alpha+x(t))-a[2]*y(t)*z(t)/(d+y(t)), diff(z(t), t) = -m*z(t)+c[1]*a[1]*x(t)*z(t)/(e+x(t))+c[2]*a[2]*y(t)*z(t)/(d+y(t))}

{diff(x(t), t) = x(t)-x(t)^2-0.1e-1*x(t)*y(t)-.6*x(t)*y(t)/(.36+x(t))-0.1e-1*x(t)*z(t)/(15+x(t)), diff(y(t), t) = -.4*y(t)+.6*x(t)*y(t)/(.36+x(t))-0.5e-1*y(t)*z(t)/(.5+y(t)), diff(z(t), t) = -0.1e-1*z(t)+0.2e-1*x(t)*z(t)/(15+x(t))+0.5e-1*y(t)*z(t)/(.5+y(t))}

(1)

ics := {x(0) = .2, y(0) = 0.5e-1, z(0) = 0.5e-1}

{x(0) = .2, y(0) = 0.5e-1, z(0) = 0.5e-1}

(2)

NULL

sol := dsolve(`union`(sys, ics), {x(t), y(t), z(t)}, numeric, range = 0 .. 20000, maxfun = 0, output = listprocedure, abserr = 0.1e-7, relerr = 0.1e-7)

`[Length of output exceeds limit of 1000000]`

(3)

X := subs(sol, x(t)); Y := subs(sol, y(t)); Z := subs(sol, z(t))

``

plot('[X(t)]', t = 0 .. 20000, numpoints = 350, title = "Trajectory of x(t)", axes = boxed, gridlines, color = ["#40e0d0"])

 

plot('[Y(t)]', t = 0 .. 20000, numpoints = 350, title = "Trajectory", axes = boxed, gridlines, title = "Trajectory of y(t)", color = ["SteelBlue"])

 

``

plot('[Z(t)]', t = 0 .. 20000, numpoints = 350, title = "Trajectory", axes = boxed, gridlines, title = "Trajectory of Z(t)", color = "Black"); with(DEtools)

 

with(DEtools)

DEplot3d(sys, {x(t), y(t), z(t)}, t = 0 .. 20000, [[x(0) = .2, y(0) = 0.5e-1, z(0) = 0.5e-1]], numpoints = 35000, color = blue, thickness = 1, linestyle = solid)

 

Task (b)

• 

Repeat the study in part (a) with the same initial conditions but set β=0.61.

NULL

restart

r := 1; b := 1; f := 0.1e-1; alpha := .36; a[1] := 0.1e-1; a[2] := 0.5e-1; e := 15; m := 0.1e-1; d := .5; c[1] := 2; c[2] := 1; mu := .4; beta := .61

NULL

sys := {diff(x(t), t) = r*x(t)-b*x(t)^2-f*x(t)*y(t)-beta*x(t)*y(t)/(alpha+x(t))-a[1]*x(t)*z(t)/(e+x(t)), diff(y(t), t) = -mu*y(t)+beta*x(t)*y(t)/(alpha+x(t))-a[2]*y(t)*z(t)/(d+y(t)), diff(z(t), t) = -m*z(t)+c[1]*a[1]*x(t)*z(t)/(e+x(t))+c[2]*a[2]*y(t)*z(t)/(d+y(t))}

{diff(x(t), t) = x(t)-x(t)^2-0.1e-1*x(t)*y(t)-.61*x(t)*y(t)/(.36+x(t))-0.1e-1*x(t)*z(t)/(15+x(t)), diff(y(t), t) = -.4*y(t)+.61*x(t)*y(t)/(.36+x(t))-0.5e-1*y(t)*z(t)/(.5+y(t)), diff(z(t), t) = -0.1e-1*z(t)+0.2e-1*x(t)*z(t)/(15+x(t))+0.5e-1*y(t)*z(t)/(.5+y(t))}

(4)

NULL

ics := {x(0) = .2, y(0) = 0.5e-1, z(0) = 0.5e-1}

{x(0) = .2, y(0) = 0.5e-1, z(0) = 0.5e-1}

(5)

sol := dsolve(`union`(sys, ics), {x(t), y(t), z(t)}, numeric, range = 0 .. 20000, maxfun = 0, output = listprocedure, abserr = 0.1e-7, relerr = 0.1e-7)

`[Length of output exceeds limit of 1000000]`

(6)

X := subs(sol, x(t)); Y := subs(sol, y(t)); Z := subs(sol, z(t))

NULL

plot('[X(t)]', t = 0 .. 20000, numpoints = 350, title = "Trajectory of x(t)", axes = boxed, gridlines, color = ["Blue"])

 

plot('[Y(t)]', t = 0 .. 20000, numpoints = 350, title = "Trajectory of  Y(t)", axes = boxed, gridlines, color = "Red")

 

plot('[Z(t)]', t = 0 .. 20000, numpoints = 350, title = "Trajectory of  Y(t)", axes = boxed, gridlines, color = "Black")

 

NULL

with(DEtools)

DEplot3d(sys, {x(t), y(t), z(t)}, t = 0 .. 20000, [[x(0) = .2, y(0) = 0.5e-1, z(0) = 0.5e-1]], maxfun = 0, numpoints = 35000, color = blue, thickness = 1, linestyle = solid)

 

The rate of the infection spread is affected by the average number of contacts each person has (β=0.6) and increases depending on the degree of transmission within the population, in particular within specific subpopulations (such as those in rural areas). A detailed epidemiological study showed that the spread of infection is most significant in urban areas, where population density is higher, while in rural areas, the rate of infection remains relatively low. This suggests that additional public health measures are needed to reduce transmission in densely populated areas, particularly in regions with high population density such as cities

``

Download math_model_eco-epidemiology.mw

This is another attempt to tell about one way to solve the problem of inverse kinematics of a manipulator.  
We have a flat three-link manipulator. Its movement is determined by changing three angles - these are three control parameters. 1. the first link rotates around the black fixed point, 2. the second link rotates around the extreme movable point of the first link, 3. the third link − around the last point of the second link. These movable points are red. (The order of the links is from thick to thin.) The working point is green. For example, we need it to move along a circle. But the manipulator has one extra mobility (degree of freedom), that is, the problem has an infinite number of solutions. We have the ability to remove this extra degree of freedom mathematically. And this can also be done in an infinite number of ways.
Let us give two examples where the same manipulator performs the same movement of the working point in different ways. In one case the last red point moves in a straight line, and in the other case it moves in an ellipse. The result is the same. In the corresponding program texts, the manipulator model is described by a system of nonlinear equations f1, f2, f3, f4, f5 relative to the coordinates of the ends of the links (very easy to understand). The specific additional connection that takes away one degree of freedom is highlighted in blue. Equation of a circle in red color.

1.mw

2.mw


And as an elective. The same circle was obtained using a spatial 3-link manipulator with 5 degrees of freedom. In the last text, blue and red colors perform the same functions as in the previous texts.
3.mw

 

VerifyTools is a package that has been available in Maple for roughly 24 years, but until now it has never been documented, as it was originally intended for internal use only. Documentation for it will be included in the next release of Maple. Here is a preview:

VerifyTools is similar to the TypeTools package. A type is essentially a predicate that a single expression can either satisfy or not. Analogously, a verification is a predicate that applies to a pair of expressions, comparing them. Just as types can be combined to produce compound types, verifications can also be combined to produce compund verifications. New types can be created, retrieved, queried, or deleted using the commands AddType, GetType (or GetTypes), Exists, and RemoveType, respectively. Similarly in the VerifyTools package we can create, retrieve, query or delete verifications using AddVerification, GetVerification (or GetVerifications), Exists, and RemoveVerification.

The package command VerifyTools:-Verify is also available as the top-level Maple command verify which should already be familiar to expert Maple users. Similarly, the command VerifyTools:-IsVerification is also available as a type, that is,

VerifyTools:-IsVerification(ver);

will return the same as

type(ver, 'verification');

The following examples show what can be done with these commands. Note that in each example where the Verify command is used, it is equivalent to the top-level Maple command verify. (Also note that VerifyTools commands shown below will be slightly different compared to the Maple2024 version):

with(VerifyTools):

Suppose we want to create a verification which will checks that the length of a result has not increased compared to the expected result. We can do this using the AddVerification command:

AddVerification(length_not_increased, (a, b) -> evalb(length(a) <= length(b)));

First, we can check the existence of our new verification and get its value:

Exists(length_not_increased);

true

GetVerification(length_not_increased);

proc (a, b) options operator, arrow; evalb(length(a) <= length(b)) end proc

For named verifications, IsVerification is equivalent to Exists (since names are only recognized as verifications if an entry exists for them in the verification database):

IsVerification(length_not_increased);

true

On the other hand, a nontrivial structured verification can be checked with IsVerification,

IsVerification(boolean = length_not_increased);

true

whereas Exists only accepts names:

Exists(boolean = length_not_increased);

Error, invalid input: VerifyTools:-Exists expects its 1st argument, x, to be of type symbol, but received boolean = length_not_increased

The preceding command using Exists is also equivalent to the following type call:

type(boolean = length_not_increased, verification);

true

Now, let's use the new verification:

Verify(x + 1/x, (x^2 + 1)/x, length_not_increased);

true

Verify((x^2 + 1)/x, x + 1/x, length_not_increased);

false

Finally, let's remove the verification:

RemoveVerification(length_not_increased);

Exists(length_not_increased);

false

GetVerification(length_not_increased);

Error, (in VerifyTools:-GetVerification) length_not_increased is not a recognized verification

GetVerifications returns the list of all verifications known to the system:

GetVerifications();

[Array, FAIL, FrobeniusGroupId, Global, Matrix, MultiSet, PermGroup, RootOf, SmallGroupId, Vector, address, after, approx, array, as_list, as_multiset, as_set, attributes, boolean, box, cbox, curve, curves, dataframe, dataseries, default, default, dummyvariable, equal, evala, evalc, expand, false, float, function, function_bounds, function_curve, function_shells, greater_equal, greater_than, in_convex_polygon, indef_int, interval, less_equal, less_than, list, listlist, matrix, member, multiset, neighborhood, neighbourhood, normal, permute_elements, plot, plot3d, plot_distance, plotthing_compile_result, polynom, procedure, ptbox, range, rational, record, relation, reverse, rifset, rifsimp, rtable, set, sign, simplify, sublist, `subset`, subtype, superlist, superset, supertype, symbol, table, table_indices, testeq, text, true, truefalse, type, undefined, units, vector, verifyfunc, wildcard, xmltree, xvm]

Download VerificationTools_blogpost.mw

Austin Roche
Software Architect
Mathematical Software
Maplesoft

Circles inscribed between curves can be specified by a system of equations relative to the coordinates of the center of the circle and the coordinates of the tangent points. Such a system can have 5 or 6 equations and 6 variables, which are mentioned above.
In the case of 5 equations, we can immediately obtain an infinite set of solutions by selecting the ones we need from it. 
(See the attached text for more details.)
The 1st equation is responsible for the belonging of the point of tangency to one of the curves.
The 2nd equation is responsible for the belonging of the point of tangency to another curve.
In the 3rd equation, the points of tangency on the curves belong to the inscribed circle.
In the 4th and 5th equations, the condition is satisfied that the tangents to the curves are perpendicular to the radii of the circle at the points of contact.
The 6th equation serves either to find a specific inscribed circle or to find an infinite set of solutions. It is selected based on the type of curves and their mutual arrangement.

In this example, we search for a subset of the solution set using the Draghilev method by solving the first five equations of the system: we inscribe circles in two "angles" formed by the intersection of the exponent and the ellipse.
The text of this example, its solution in the form of a picture,"big" option and pictures of similar examples.

INSCRIBED_CIRCLES.mw


 


Addition 09/01/24, 
One curve for the first two equations in coordinates x1,x2 and x3,x4
f1:=
 x1^2 - 2.5*x1*x2 + 3*x2^2 - 1;
f2:=
 x3^2 - 2.5*x3*x4 + 3*x4^2 - 1;


This post is inspired by minhthien2016's question.

The problem, denoted 2/N/1, for reasons that will appear clearly further on, is to pack N disks into the unit square in such a way that the sum of their radii is maximum.

I replied this problem using Optimization-NLPSolve for N from 1 (obvious solution) to 16, which motivated a few questions, in particular:

  • @Carl Love: "Can we confirm that the maxima are global (NLPSolve tends to return local optima)?
    Using NLPSolve indeed does not guarantee that the solution found is the (a?) global maximum. In fact packing problems are generaly tackled by using evolutionnary algorithms, greedy algorithms, or specific heuristic strategies.
    Nevertheless, running NLPSolve a large number of times from different initial points may provide several different optima whose the largest one can be reasonably considered as the global maximum you are looking for.
    Of course this may come to a large price in term of computational time.

     
  • @acer: "How tight are [some constraints], at different N? Are they always equality?"
    The fact some inequality constraints type always end to equality constraints (which means that in an optimal packing each disk touches at least one other annd, possibly the boundary of the box) seems hard to prove mathematically, but I gave here a sketch of informal proof.



I found 2/N/1 funny enough to spend some time digging into it and looking to some generalizations I will refer to as D/N/M:  How to pack N D-hypersheres into the unit D-hypercube such that the sum of the M-th power of their radii is maximum?
For the sake of simplicity I will say ball instead of disk/sphere/hypersphere and box instead of square/cube/hypercube.

The first point is that problems D/N/1 do not have a unique solution as soon as N > 1 , indeed any solution can be transformed into another one using symmetries with respect to medians and diagonals of the box. Hereafter I use this convention:

Two solutions and s' are fundamental solutions if:

  1. the ordered lists of radii and s'  contain are identical but there is no composition of symmetries from to s',
  2. or, the ordered lists of radii and s'  contain are not identical.
     

It is easy to prove that 2/2/1 and 3/2/1, and likely D/2/1, have an infinity of fundamental solutions: see directory FOCUS__2|2|1_and_3|2|1 in the attached zip file..
At the same time 2/N/2, 3/N/3, and likely D/N/D, have only one fundamental solution (see directory FOCUS__2|N|2 for more details and a simple way to characterize these solutions

 (Indeed the strategy ito find the solution of D/N/D  in placing the biggest possible ball in the largest void D/N-1/D contains. Unfortunately this characterization is not algorithmically constructive in the sense that findind this biggest void is a very complex geometrical and combinatorial problem.
 it requires finding the largest void  in a pack of balls)


Let Md, 1(N)  the maximum value of the sum of balls radii for problem d/N/1.
The first question I asked myself is: How does Md, 1(N) grows with N?

 

(Md, 1(N) is obviously a strictly increasing function of N: indeed the solution of problem d/N/1 contains several voids where a ball of strictly positive radius can be placed, then  Md+1, 1(N) > Md, 1(N) )


The answer seems amazing as intensive numerical computations suggest that
                                      

See D|N|M__Growth_law.mw in the attached sip file.
This formula fits very well the set of points  { [n, Sd, 1(n) , n=1..48) } for d=2..6.
I have the feeling that this conjecture might be proven (rejected?) by rigourous mathematical arguments.


Fundamental solutions raise several open problems:

  • Are D/2/1 problems the only one with more than one fundamental solutions?

    The truth is that I have not been capable to find any other example (which does not mean they do not exist).
    A quite strange thing is the behaviour of NLPSolve: as all the solutions of D/2/1 are equally likely, the histogram of the solutions provided by a large number of NLPSolve runs from different initial points is far from being uniform.
    F
    or more detail refer ro directory FOCUS__2|2|1_and_3|2|1
     in the attached zip file
    I do not understand where this bias comes from: is it due to the implementation of SQP in NLPSolve, or to SQP itself?

     
  • For some couples (D, N) the solution of D/N/1 is made of balls of same radius.
    For N from 1 to 48 this is (numerically)
     the case for 2/1/1 and2/2/1, but the three dimensional case is reacher as it contains  3/1/13/2/1,  3/3/1,  3/4/1 and 3/14/1 (this latter being quite surprising).
    Is there only a finite number of values 
    N such that D/N/1 is made of balls with identical radii?
    If it is so, is this number increasing with
     D?
    It is worth noting that those values of
    N mean that the solution of problems D/N/1 are identical to those of a more classic packing problem: "What is the largest radius N identical balls packed in a unit bow may have?".
    For an exhaustive survey of this latter problem see
    Packomania.

     
  • A related question is "How does the number of different radii evolves as N increases dor given values of D and M?
    Displays of 2D and 3D packings show that the set of radii has significantly less elements than
    N... at least for values of N not too large. So might we expect that solution of, let us say, 2/100/1 can contain 100 balls of 10 different radii, or it is more reasonable to expect it contains 100 balls of 100 different radii?

     
  • At the opposite numerical investigations of  2/N/1 and  3/N/1 suggest that the number of different radii a fundamental solution contains increases with N (more a trend than a continuous growth).
    So, is it true that very large values of N correspond to solutions where the number of different radii is also very large?

    Or could it be that the growth of the number of different radii I observed is simply the consequence of partially converged results?
     
  • Numerical investigations show that for a given dimension d and a given number of balls n,  solutions of d/n/1 and d/n/M (1 < M < d) problems are rather often the same. Is this a rule with a few exceptions or a false impression due to the fact that I did not pushed the simulations to values of n large enough to draw a more solid picture)?


It is likely that some of the questions above could be adressed by using a more powerful machine than mine.


All the codes and results are gathered in  a zip file you can download from OneDrive Google  (link at the end of this post, 262 Mb, 570 Mb when unzipped, 1119 files).
Install this zip file in the directory of your choice and unzip it to get a directory named
PACKING
Within it:

  • README.mw contains a description of the different codes and directories
  • Repository.rtf must contain a string repesenting the absolute path of directory PACKING


Follow this link OneDrive Google


 

An attractor is called strange if it has a fractal structure, that is if it has a non-integer Hausdorff dimension. This is often the case when the dynamics on it are chaotic, but strange nonchaotic attractors also exist.  If a strange attractor is chaotic, exhibiting sensitive dependence on initial conditions, then any two arbitrarily close alternative initial points on the  attractor, after any of various numbers of iterations, will lead to  points that are arbitrarily far apart (subject to the confines of the attractor), and after any of various other numbers of iterations will  lead to points that are arbitrarily close together. Thus a dynamic system with a chaotic attractor is locally unstable yet globally stable: once some sequences have entered the attractor, nearby points diverge  from one another but never depart from the attractor.


The term strange attractor was coined by David Ruelle and Floris Takens to describe the attractor resulting from a series of bifurcations of a system describing fluid flow. Strange attractors are often differentiable in a few directions, but some are like a Cantor dust, and therefore not differentiable. Strange attractors may also be found  in the presence of noise, where they may be shown to support invariant  random probability measures of Sinai–Ruelle–Bowen type.


Examples of strange attractors include the  Rössler attractor, and Lorenz attractor.

 

 

THOMAS``with(plots); b := .20; sys := diff(x(t), t) = sin(y(t))-b*x(t), diff(y(t), t) = sin(z(t))-b*y(t), diff(z(t), t) = sin(x(t))-b*z(t); sol := dsolve({sys, x(0) = 1.1, y(0) = 1.1, z(0) = -0.1e-1}, {x(t), y(t), z(t)}, numeric); odeplot(sol, [x(t), y(t), z(t)], t = 0 .. 600, axes = boxed, numpoints = 50000, labels = [x, y, z], title = "Thomas Attractor")

 

 

 

Dabras``

with(plots); a := 3.00; b := 2.7; c := 1.7; d := 2.00; e := 9.00; sys := diff(x(t), t) = y(t)-a*x(t)+b*y(t)*z(t), diff(y(t), t) = c*y(t)-x(t)*z(t)+z(t), diff(z(t), t) = d*x(t)*y(t)-e*z(t); sol := dsolve({sys, x(0) = 1.1, y(0) = 2.1, z(0) = -2.00}, {x(t), y(t), z(t)}, numeric); odeplot(sol, [x(t), y(t), z(t)], t = 0 .. 100, axes = boxed, numpoints = 35000, labels = [x, y, z], title = "Dabras Attractor")

 

Halvorsen

NULLwith(plots); a := 1.89; sys := diff(x(t), t) = -a*x(t)-4*y(t)-4*z(t)-y(t)^2, diff(y(t), t) = -a*y(t)-4*z(t)-4*x(t)-z(t)^2, diff(z(t), t) = -a*z(t)-4*x(t)-4*y(t)-x(t)^2; sol := dsolve({sys, x(0) = -1.48, y(0) = -1.51, z(0) = 2.04}, {x(t), y(t), z(t)}, numeric, maxfun = 300000); odeplot(sol, [x(t), y(t), z(t)], t = 0 .. 600, axes = boxed, numpoints = 35000, labels = [x, y, z], title = "Halvorsen Attractor")

 

Chen

 

 

with(plots); alpha := 5.00; beta := -10.00; delta := -.38; sys := diff(x(t), t) = alpha*x(t)-y(t)*z(t), diff(y(t), t) = beta*y(t)+x(t)*z(t), diff(z(t), t) = delta*z(t)+(1/3)*x(t)*y(t); sol := dsolve({sys, x(0) = -7.00, y(0) = -5.00, z(0) = -10.00}, {x(t), y(t), z(t)}, numeric); odeplot(sol, [x(t), y(t), z(t)], t = 0 .. 100, axes = boxed, numpoints = 35000, labels = [x, y, z], title = "Chen Attractor")

 

References

1. 

https://www.dynamicmath.xyz/strange-attractors/

2. 

https://en.wikipedia.org/wiki/Attractor#Strange_attractor

``


 

Download Attractors.mw


 

This project discusses predator-prey system, particularly the Lotka-Volterra equations,which model the interaction between two sprecies: prey and predators. Let's solve the Lotka-Volterra equations numerically and visualize the results.

NULL

NULL

alpha := 1.0; beta := .1; g := 1.5; delta := 0.75e-1; ode1 := diff(x(t), t) = alpha*x(t)-beta*x(t)*y(t); ode2 := diff(y(t), t) = delta*x(t)*y(t)-g*y(t); eq1 := -beta*x*y+alpha*x = 0; eq2 := delta*x*y-g*y = 0; equilibria := solve({eq1, eq2}, {x, y}); print("Equilibrium Points: ", equilibria); initial_conditions := x(0) = 40, y(0) = 9; sol := dsolve({ode1, ode2, initial_conditions}, {x(t), y(t)}, numeric); eq_points := [seq([rhs(eq[1]), rhs(eq[2])], `in`(eq, equilibria))]

[[0., 0.], [20., 10.]]

plots[odeplot](sol, [[t, x(t)], [t, y(t)]], t = 0 .. 100, legend = ["Rabbits", "Wolves"], title = "Prey-Predator Dynamics", labels = ["Time", "Population"])

NULL

NULL

NULL

sol_plot := plots:-odeplot(sol, [[x(t), y(t)]], 0 .. 100, color = "blue"); equilibrium_plot := plots:-pointplot(eq_points, color = "red", symbol = solidcircle, symbolsize = 15); plots:-display([sol_plot, equilibrium_plot], title = "Phase Portrait with Equilibrium Points", labels = ["Rabbits", "Wolves"])

Now, we need to handle a modified version of the Lotka-Volterra equations. These modified equations incorporate logistic growth fot the prey population.

 

 

restart

alpha := 1.0; beta := .1; g := 1.5; delta := 0.75e-1; k := 100; ode1 := diff(x(t), t) = alpha*x(t)*(1-x(t)/k)-beta*x(t)*y(t); ode2 := diff(y(t), t) = delta*x(t)*y(t)-g*y(t); eq1 := alpha*x*(1-x/k)-beta*x*y = 0; eq2 := delta*x*y-g*y = 0; equilibria := solve({eq1, eq2}, {x, y}); print("Equilibrium Points: ", equilibria); initial_conditions := x(0) = 40, y(0) = 9; sol := dsolve({ode1, ode2, initial_conditions}, {x(t), y(t)}, numeric); eq_points := [seq([rhs(eq[1]), rhs(eq[2])], `in`(eq, equilibria))]

[[0., 0.], [100., 0.], [20., 8.]]

plots[odeplot](sol, [[t, x(t)], [t, y(t)]], t = 0 .. 100, legend = ["Rabbits", "Wolves"], title = "Prey-Predator Dynamics", labels = ["Time", "Population"])

NULL

plots:-odeplot(sol, [[x(t), y(t)]], 0 .. 50, color = "blue"); equilibrium_plot := plots:-pointplot(eq_points, color = "red", symbol = solidcircle, symbolsize = 15); plots:-display([plots:-odeplot(sol, [[x(t), y(t)]], 0 .. 50, color = "blue"), equilibrium_plot], title = "Phase Portrait with Equilibrium Points", labels = ["Rabbits", "Wolves"])

NULL


 

Download predator_prey2.mw

We are pleased to announce that the registration for the Maple Conference 2024 is now open.

Like the last few years, this year’s conference will be a free virtual event. Please visit the conference page for more information on how to register.

This year we are offering a number of new sessions, including more product training options and an Audience Choice session.
You can find an overview of the program on the Sessions page. Those who register before September 10th, 2024 will have a chance to vote for the topics they want to learn more about during the Audience Choice session.

We hope to see you there!

This is a reminder that presentation applications for the Maple Conference are due July 17, 2024.

The conference is a a free virtual event and will be held on October 24 and 25, 2024.

We are inviting submissions of presentation proposals on a range of topics related to Maple, including Maple in education, algorithms and software, and applications. We also encourage submission of proposals related to Maple Learn. You can find more information about the themes of the conference and how to submit a presentation proposal at the Call for Participation page.

I encourage all of you here in the Maple Primes community to consider joining us for this event, whether as a presenter or an attendee!

Kaska Kowalska
Contributed Program Co-Chair

1 2 3 4 5 6 7 Last Page 1 of 73