Items tagged with ode

TQ.mw

Can any one help for finding the solution of these differntial equations and then plotting the graph for differnt values of M

(FILE ATTACHED)
 

eqn1 := (R/(R-theta(eta))+Omega)*(diff(f(eta), `$`(eta, 3)))+f(eta)*(diff(f(eta), `$`(eta, 2)))+R*(diff(f(eta), `$`(eta, 2)))*(diff(theta(eta), eta))/(R-theta(eta))^2+Omega*(diff(g(eta), eta))+lambda*theta(eta)*Cos(alpha)-M*(diff(f(eta), eta))^2 = 0; eqn2 := (R/(R-theta(eta))+(1/2)*Omega)*(diff(g(eta), `$`(eta, 2)))-2*Omega*(2*g(eta)+diff(f(eta), `$`(eta, 2)))+(diff(f(eta), eta))*g(eta)+(diff(g(eta), eta))*f(eta)+R*(diff(g(eta), eta))*(diff(theta(eta), eta))/(R-theta(eta))^2 = 0; eqn3 := (1+`ε`*theta(eta))*(diff(theta(eta), `$`(eta, 2)))+`ε`*(diff(theta(eta), eta))^2+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta))+Q*theta(eta)+L*exp(-eta) = 0

(R/(R-theta(eta))+Omega)*(diff(diff(diff(f(eta), eta), eta), eta))+f(eta)*(diff(diff(f(eta), eta), eta))+R*(diff(diff(f(eta), eta), eta))*(diff(theta(eta), eta))/(R-theta(eta))^2+Omega*(diff(g(eta), eta))+lambda*theta(eta)*Cos(alpha)-M*(diff(f(eta), eta))^2 = 0

 

(R/(R-theta(eta))+(1/2)*Omega)*(diff(diff(g(eta), eta), eta))-2*Omega*(2*g(eta)+diff(diff(f(eta), eta), eta))+(diff(f(eta), eta))*g(eta)+(diff(g(eta), eta))*f(eta)+R*(diff(g(eta), eta))*(diff(theta(eta), eta))/(R-theta(eta))^2 = 0

 

(1+epsilon*theta(eta))*(diff(diff(theta(eta), eta), eta))+epsilon*(diff(theta(eta), eta))^2+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta))+Q*theta(eta)+L*exp(-eta) = 0

(1)

Omega := 2.; M := .5; R := 5; lambda := 20; `ε` := .2; Pr := 1; Q := .5; L := .5; W := .5; n := .1; alpha := (1/6)*Pi

bc := f(0) = W, (D(f))(0) = 0, (D(f))(infinity) = 0, (D(theta))(0) = -1, theta(infinity) = 0, g(0) = -n*(DD(f))(0), g(infinity) = 0

f(0) = W, (D(f))(0) = 0, (D(f))(N) = 0, (D(theta))(0) = -1, theta(N) = 0, g(0) = -n*(DD(f))(0), g(N) = 0

(2)

``


 

Download TQ.mw

 

hello agian i have the following problem. i need to fit data using the following model:

ode_sub := diff(S(t), t) = -k1*S(t)-S(t)/T1_s;

ode_P1 := diff(P1(t), t) = 2*k1*S(t)-k2*(P1(t)-P2(t)/keq)-P1(t)/T1_p1;

ode_P2 := diff(P2(t), t) = -k2*(-keq*P1(t)+P2(t))/keq-k4*P2(t)-P2(t)/T1_p2;

ode_P2e := diff(P2_e(t), t) = k4*P2(t)-P2_e(t)/T1_p2_e;

ode_system := ode_sub, ode_P1, ode_P2, ode_P2e

known paramters: s0 := 10000; k2 := 1000; T1_s := 14; T1_p2_e := 35; T1_p2 := T1_p1

initial conditions: init := S(0) = s0, P1(0) = 0, P2(0) = 0, P2_e(0) = 0

using the following data the fitting is fine:

T := [0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100]

data_s := [10000.00001377746, 7880.718836217512, 6210.572917625112, 4894.377814704496, 3857.121616806618, 3039.689036293312, 2395.493468824973, 1887.821030424934, 1487.738721378254, 1172.445015238064, 923.970970533911, 728.1555148262845, 573.8389058920359, 452.2263410725434, 356.3868133709972, 280.8584641247961, 221.3366863178145, 174.4291648589732, 137.4627967029932, 108.3305246342751, 85.37230370803576, 67.2794974831867, 53.0210755540487, 41.78436556408739, 32.92915589156605, 25.95049906181449, 20.45092590987601, 16.11678944747619, 12.70115104646253, 10.009439492356, 7.888058666357939, 6.216464754698855, 4.899036441885205, 3.860793948052506, 3.042584031379331, 2.397778731385364, 1.889601342133927, 1.489145259940784, 1.173591417634974, .9248694992255094, .7288443588090404, .5743879613705721, .4526024635132348, .3567687570029392, .2810508404011898, .2215650452813882, .174603181767829, .1376243372528356, .1084232889842753, 0.8542884707952822e-1, 0.6738282660463157e-1]

data_p1 := [0.1194335334401124e-4, 244.8746046039949, 374.8721199398692, 430.5392805383767, 439.6598364813143, 421.0353424914179, 387.1842556288343, 346.2646897222593, 303.4377508746471, 261.8283447091155, 223.1996547160051, 188.4213144493491, 157.7924449350029, 131.2622073983344, 108.5771635112278, 89.37951009190863, 73.26979150957087, 59.84578653950572, 48.72563358658898, 39.56010490461378, 32.03855466968536, 25.88922670933322, 20.87834763772145, 16.80708274458702, 13.50774122768974, 10.84014148654258, 8.687656394505874, 6.954093898245485, 5.560224107929433, 4.441209458524726, 3.544128529596104, 2.825755811619965, 2.251247757181308, 1.792233305651086, 1.425861347838012, 1.133566009019768, .90081361320016, .7153496336919163, .5678861241754847, .4505952916932289, .3572989037753538, .2832489239941939, .2244289868248577, .1778450590752305, .1408633578784151, .1114667192753896, 0.8826814044702111e-1, 0.6979315954603076e-1, 0.5526502783606788e-1, 0.4370354298880999e-1, 0.3456334307662573e-1]

data_p2_p2e := [-0.1821397630630296e-4, 1000.40572909871, 1568.064904416198, 1848.900129881268, 1944.147939710225, 1923.352973299286, 1833.705314342611, 1706.726235937363, 1563.036042115902, 1415.741121363331, 1272.825952816517, 1138.833091575137, 1016.03557293539, 905.2470623752856, 806.3754051707843, 718.7979384094563, 641.6091822001032, 573.7825966275556, 514.2718125966452, 462.0710416682647, 416.2491499591012, 375.9656328260581, 340.4748518743264, 309.124348579787, 281.3484612079911, 256.6599441255977, 234.6416876403397, 214.9378461256197, 197.2453333305823, 181.3059798685686, 166.90059218783, 153.8422174795751, 141.9717783871606, 131.1529759058513, 121.2692959115991, 112.2199678247825, 103.9187616370328, 96.29007054510998, 89.26877137303566, 82.79743967408479, 76.82586273439793, 71.30944943617081, 66.2085909515396, 61.48838150744805, 57.11714763242225, 53.06666224006544, 49.31130738219119, 45.82807853990728, 42.59597194910467, 39.59575450632008, 36.81013335261527]

the fitting is done the following way: 

P1fu, P2fu, P2e_fu, Sfu := op(subs(res, [P1(t), P2(t), P2_e(t), S(t)]))

making residuals:

Q := proc (T1_p1, k1, keq, k4) local P1v, P2v, P2e_v, Sv, resid; option remember;

res(parameters = [T1_p1, k1, keq, k4]);

try P1v := `~`[P1fu](T); P2v := `~`[P2fu](T); P2e_v := `~`[P2e_fu](T); Sv := `~`[Sfu](T); resid := [P1v-data_p1, P2v+P2e_v-data_p2_p2e, Sv-data_s]; return [seq(seq(resid[i][j], i = 1 .. 3), j = 1 .. nops(T))] catch: return [1000000$3*nops(T)] end try end proc;
q := [seq(subs(_nn = n, proc (T1_p1, k1, keq, k4) Q(args)[_nn] end proc), n = 1 .. 3*nops(T))];

finding inital point for the LSsolve:

L := fsolve(q[2 .. 5], [10, 0.2e-1, 4, 4])

fitting the data with the intial point: 

solLS := Optimization:-LSSolve(q, initialpoint = L)

this is all good however, when i used the following data it did not turn out so well (using the same approch as above):

data_s := [96304.74567, 77385.03700, 62621.83067, 51239.94333, 42663.82367, 35084.74100, 28480.28367, 23066.01467, 18774.73700, 15179.13700, 12278.50767, 9937.652000, 8046.848333, 6521.242000, 5287.811667, 4277.779000, 3466.518333, 2835.467000, 2297.796333, 1861.249667, 1529.654000, 1235.353000, 999.6626667, 826.2343333, 667.9480000, 559.9230000, 449.2790000, 376.4860000, 289.1203333, 236.1483333]

data_p1 := [0.86e-1, 3.904, 26.975, 31.719, 41.067, 46.779, 52.115, 43.101, 44.344, 41.094, 36.523, 27.742, 26.543, 28.062, 22.178, 21.303, 14.951, 17.871, 11.422, 12.051, 9.232, 6.817, 6.1, .717, 1.215, 6.146, .772, .375, 2.595, .518]

data_p2_p2e := [-3.024, 22.238, 61.731, 103.816, 132.695, 159.069, 167.302, 160.188, 158.398, 152.943, 146.745, 135.22, 132.145, 120.413, 107.864, 95.339, 90.775, 81.828, 71.065, 70.475, 62.872, 49.955, 40.858, 42.938, 41.311, 35.583, 31.573, 29.841, 29.558, 21.762]

the known parameters is the case are:

s0 := 96304.74567; k2 := 10^5; T1_s := 14; T1_p2_e := 35; T1_p2 := T1_p1

additionally the following fuction affects the solution: 

K:=t->cos((1/180)*beta*Pi)^(t/Tr)

i included this by doing this:

P1fu_K := proc (t) options operator, arrow; P1fu(t)*K(t) end proc;

P2fu_K := proc (t) options operator, arrow; P2fu(t)*K(t) end proc;

P2e_fu_K := proc (t) options operator, arrow;

P2e_fu(t)*K(t) end proc;

Sfu_K := proc (t) options operator, arrow; Sfu(t)*K(t) end proc

resulting in the following residuals:

Q := proc (T1_p1, k1, keq, k4) local P1v, P2v, P2e_v, Sv, resid; option remember;

res(parameters = [T1_p1, k1, keq, k4]);

try P1v := `~`[P1fu_K](T); P2v := `~`[P2fu_K](T); P2e_v := `~`[P2e_fu_K](T); Sv := `~`[Sfu_K](T); resid := [P1v-data_p1, P2v+P2e_v-data_p2_p2e, Sv-data_s]; return [seq(seq(resid[i][j], i = 1 .. 3), j = 1 .. nops(T))] catch: return [1000000$3*nops(T)] end try end proc;
q := [seq(subs(_nn = n, proc (T1_p1, k1, keq, k4) Q(args)[_nn] end proc), n = 1 .. 3*nops(T))];

i think my problem is that the inital point in this case is not known. all i know is that all the fitted parameters should be positive and that k1<1, k4>10, keq>1 and T1_p1>100 (more i do not know) - is there a way to determin the inital point without guessing?

i also know the results, which should be close to these values:

k1=0.000438, k4=0.0385, keq=2.7385 and T1_p1=36.8 the output fit should look something like this

where the red curve is (PP2(t)+PP2_e(t))*K(t)

the blue cure is PP1(t)*K(t) 

anyone able to help - i've tried for 2 days now. it might be that  ode_P1 := diff(P1(t), t) = 2*k1*S(t)-k2*(P1(t)-P2(t)/keq)-P1(t)/T1_p1 should be changed into ode_P1 := diff(P1(t), t) = k1*S(t)-k2*(P1(t)-P2(t)/keq)-P1(t)/T1_p1;

i've tried this but it didnt seem to do much 

anyone able to help?:)

NB stiff=true can be used within the dsolve to speed up the process if needed:)

 

 

 

 


 

 

restart;

Digits := 18;
with(LinearAlgebra);
f := proc (n) 3*sin(x[n]) end proc;

g := proc (n) 3*cos(x[n])

end proc;

#problem call.
for n from 0 to 0 do

e1 := expand(-y[n+3/2]+y[n]-3*y[n+1/2]+3*y[n+1]+1/11612160*(5856*h^4*g(n+1/2)-19968*h^4*g(n+3/2)+2343*h^4*g(n)-76356*h^4*g(n+1)-7058*h^4*g(n+2)+608864*h^3*f(n+1/2)+104864*h^3*f(n+3/2)+28489*h^3*f(n)+702864*h^3*f(n+1)+6439*h^3*f(n+2)));

e2 := expand(-y[n+2]+3*y[n]-8*y[n+1/2]+6*y[n+1]+1/5806080*(18768*h^4*g(n+1/2)-32880*h^4*g(n+3/2)+3867*h^4*g(n)-76356*h^4*g(n+1)-2229*h^4*g(n+2)+965728*h^3*f(n+1/2)+461728*h^3*f(n+3/2)+45953*h^3*f(n)+1405728*h^3*f(n+1)+23903*h^3*f(n+2)));

e3 := expand(-z[n]+(1/383201280*(-4207440*h^4*g(n+1/2)-930192*h^4*g(n+3/2)+371973*h^4*g(n)-3631932*h^4*g(n+1)-41259*h^4*g(n+2)+16136096*h^3*f(n+1/2)+3866720*h^3*f(n+3/2)+5752543*h^3*f(n)+5810400*h^3*f(n+1)+367681*h^3*f(n+2))+4*y[n+1/2]-3*y[n]+y[n+1])/h);

e4 := expand(-z[n+1/2]+(1/191600640*(376320*h^4*g(n+1/2)+118896*h^4*g(n+3/2)-29469*h^4*g(n)+532764*h^4*g(n+1)+5079*h^4*g(n+2)-5812112*h^3*f(n+1/2)-508016*h^3*f(n+3/2)-381553*h^3*f(n)-1236168*h^3*f(n+1)-45511*h^3*f(n+2))-y[n]+y[n+1])/h);

e5 := expand(-z[n+1]+(1/383201280*(-31920*h^4*g(n+1/2)-433776*h^4*g(n+3/2)+71547*h^4*g(n)-2519748*h^4*g(n+1)-17493*h^4*g(n+2)+18565216*h^3*f(n+1/2)+1933216*h^3*f(n+3/2)+885665*h^3*f(n)+10391328*h^3*f(n+1)+158015*h^3*f(n+2))-5*y[n+1/2]+y[n]+3*y[n+1])/h);

e6 := expand(-z[n+3/2]+(1/95800320*(250224*h^4*g(n+1/2)-730680*h^4*g(n+3/2)+61266*h^4*g(n)-1526256*h^4*g(n+1)-22044*h^4*g(n+2)+15680504*h^3*f(n+1/2)+4712456*h^3*f(n+3/2)+735469*h^3*f(n)+22576428*h^3*f(n+1)+203623*h^3*f(n+2))-8*y[n+1/2]+3*y[n]+5*y[n+1])/h);

e7 := expand(-z[n+2]+(1/383201280*(3873264*h^4*g(n+1/2)+332976*h^4*g(n+3/2)+497649*h^4*g(n)-1407564*h^4*g(n+1)-720255*h^4*g(n+2)+114710816*h^3*f(n+1/2)+93716192*h^3*f(n+3/2)+5705827*h^3*f(n)+191366496*h^3*f(n+1)+9635389*h^3*f(n+2))-12*y[n+1/2]+5*y[n]+7*y[n+1])/h);

e8 := expand(-p[n]+(1/191600640*(13423440*h^4*g(n+1/2)+3068304*h^4*g(n+3/2)-1621317*h^4*g(n)+11615292*h^4*g(n+1)+137451*h^4*g(n+2)-32503712*h^3*f(n+1/2)-12664928*h^3*f(n+3/2)-32539039*h^3*f(n)-16869600*h^3*f(n+1)-1223041*h^3*f(n+2))-8*y[n+1/2]+4*y[n]+4*y[n+1])/h^2);

e9 := expand(-p[n+1/2]+(1/191600640*(-3053856*h^4*g(n+1/2)-213216*h^4*g(n+3/2)+98049*h^4*g(n)-509436*h^4*g(n+1)-10191*h^4*g(n+2)-1045120*h^3*f(n+1/2)+831104*h^3*f(n+3/2)+1331083*h^3*f(n)-1207008*h^3*f(n+1)+89941*h^3*f(n+2))-8*y[n+1/2]+4*y[n]+4*y[n+1])/h^2);

e10 := expand(-p[n+1]+(1/63866880*(194160*h^4*g(n+1/2)-373968*h^4*g(n+3/2)+52329*h^4*g(n)-2514924*h^4*g(n+1)-14727*h^4*g(n+2)+14006304*h^3*f(n+1/2)+1695712*h^3*f(n+3/2)+634955*h^3*f(n)+15463008*h^3*f(n+1)+133461*h^3*f(n+2))-8*y[n+1/2]+4*y[n]+4*y[n+1])/h^2);

e11 := expand(-p[n+3/2]+(1/191600640*(1491168*h^4*g(n+1/2)-4758240*h^4*g(n+3/2)+190977*h^4*g(n)-509436*h^4*g(n+1)-103119*h^4*g(n+2)+46274944*h^3*f(n+1/2)+48151168*h^3*f(n+3/2)+2215307*h^3*f(n)+93985056*h^3*f(n+1)+974165*h^3*f(n+2))-8*y[n+1/2]+4*y[n]+4*y[n+1])/h^2);

e12 := expand(-p[n+2]+(1/191600640*(4772688*h^4*g(n+1/2)+11719056*h^4*g(n+3/2)+338619*h^4*g(n)+11615292*h^4*g(n+1)-1822485*h^4*g(n+2)+59770976*h^3*f(n+1/2)+79609760*h^3*f(n+3/2)+3528289*h^3*f(n)+109647648*h^3*f(n+1)+34844287*h^3*f(n+2))-8*y[n+1/2]+4*y[n]+4*y[n+1])/h^2) end do;
M := {e || (1 .. 12)};

y_init := 1;

z_init := 0;

p_init := -2;

x_init := 0; A := 0; B := 1; N := 40;

h := evalf((B-A)/N); count := 1;

X := y[k], y[k+1/2], y[k+1], y[k+3/2], z[k], z[k+1/2], z[k+1], z[k+3/2], p[k], p[k+1/2], p[k+1], p[k+3/2];

step := seq(eval(x, x = n*h), n = 1 .. N);

y_exact := ([seq])(eval(3*cos(x)+(1/2)*x^2-2, x = n*h), n = 1 .. N);

z_exact := ([seq])(eval((1/3*(3*x^2+6*x+3))/(x^3+3*x^2+3*x+1), x = n*h), n = 1 .. N);

p_exact := ([seq])(eval((1/3*(6*x+6))/(x^3+3*x^2+3*x+1)-(1/3)*(3*x^2+6*x+3)^2/(x^3+3*x^2+3*x+1)^2, x = n*h), n = 1 .. N);
vars := seq(X, k = 1);
printf("\n%4s%13s%15s%15s\n", "@", "y_Num", "y_Exact", "y_Error");

for q to N do

for ix to 4 do

x[ix] := h*ix+x_init end do;

result := eval(`<,>`(vars), fsolve(eval(M, [x[0] = x_init, x[1/2] = x_init, x[3/2] = x_init, y[0] = y_init, y[1/2] = y_init, y[3/2] = y_init, z[0] = z_init, z[1/2] = z_init, z[3/2] = z_init, p[0] = p_init, p[1/2] = p_init, p[3/2] = p_init]), {vars}));

for k to 4 do

printf("%5.2f %14.15f", step[count], result[k]);

printf("%20.15f %10.18G \n", y_exact[count], abs(result[k]-y_exact[count]));

count := count+1;

P := [result[k]]

end do;

x_init := x[ix-1];

y_init := result[4];

z_init := result[8];

p_init := result[12]

end do;

 

please that is the code i write to solve the problem after using the matrix form to generate the value but is given me error of the form


   @        y_Num        y_Exact        y_Error
Error, invalid input: eval received fsolve({-6398.00004614630940+6400.00000000000000*y[1], -6397.99992849910140+6400.00000000000000*y[1], -6397.99909739580050+6400.00000000000000*y[1], -199.999989717789185+200.000000000000000*y[1], -40.0000000791700798+40.0000000000000000*y[1], -2.99999993737911015+3*y[1], 39.999999768462113+40.0000000000000000*y[1], -p[1]-6399.99961623646730+6400.00000000000000*y[1], -p[2]-6399.99798489466010+6400.00000000000000*y[1], -y[2]-4.99999972458202552+6*y[1], -z[1]-159.999999048856193+120.000000000000000*y[1], -z[2]-279.999973921987948+280.000000000000000*y[1]}, {p[1], p[2], p[3/2], p[5/2], y[1], y[2], y[3/2], y[5/...
 

Hi,

I need your help to classify the follwing set {0}, {1} and [0,1] are local attractor or not and in the case of local attractor how can we determine the bassin of attraction. 

ode:=diff(x(t),t)=sqrt(x(t));

how can we prove using maple which of {0}, {1} and [0,1] are local attarctor or not.

Many thanks

 

Dear All,

I would like to plot the probability density function of a state variable obtained from solving differential equations. I have found that there are functions called "PDF" and "KernelDensityPlot" in the Statistics package, but they really confuse me. Could you please point me out? My code is as follows.

Ps. Is it possible to plot the PDF directly from the solution of dsolve() without discretizing the results? 


restart:
with(plots): with(DEtools): with(plottools):with(LinearAlgebra): with(Statistics):

v1:=1: f:=-4: v2:=2.515: omega:=1: epsilon:=0.001: k:=0:
sys:=diff(u1(t),t)=v1*u1(t)-(omega+k*u2(t)^2)*u2(t)-(u1(t)^2+u2(t)^2+3*z(t)^2)*u1(t),
     diff(u2(t),t)=(omega+k*u1(t)^2)*u1(t)+v1*u2(t)-(u1(t)^2+u2(t)^2+3*z(t)^2)*u2(t),
     diff(z(t),t)=z(t)*(-v1+3*u1(t)^2+3*u2(t)^2+z(t)^2)+epsilon*z(t)*(v2+f*z(t)^4):

t_start:=50: t_end:=300: dt:=0.05: fs:=1/dt:

solA:=dsolve({sys, u1(0)=0.6, u2(0)=0.6, z(0)=0.1},
             {u1(t),u2(t),z(t)},
              type=numeric, method=rkf45, maxfun=0,
              output=Array([seq(i,i=t_start..t_end, dt)])):

u1:=solA[2,1][..,2]:
u2:=solA[2,1][..,3]:
z:=solA[2,1][..,4]:

u0:=sqrt~(u1^~2+u2^~2+z^~2):

Phi:=z/~u0:

# How could I plot the probability density of Phi (y-axis) against Phi(x-axis)?

Probability_density_function_plot.mw

Thank you!

Very kind wishes,

Wang Zhe

Respected member!
Please help me to find the solution of attached problem.

 


> subject to boundary conditions


``

 

 

NULL

restart

alpha := evalf(2*Pi*(1/180)); EP := .2; lambda := .1; HA := 5; RE := 20

ODEforNum := (1+EP)*(((D@@3)(F))(r)+4*alpha^2*(D(F))(r))+2*alpha*RE*F(r)*(D(F))(r)-HA*alpha^2*(D(F))(r)-3*EP*lambda*((1/2)*(D(F))(r)^2*((D@@3)(F))(r)+(D(F))(r)*((D@@2)(F))(r)^2)/alpha^2-EP*lambda*(72*F(r)^2*(D(F))(r)+2*(D(F))(r)^3+32*F(r)*(D(F))(r)*((D@@2)(F))(r)+2*F(r)^2*((D@@3)(F))(r)) = 0

1.2*((D@@3)(F))(r)-0.243693936e-3*(D(F))(r)+1.396263402*F(r)*(D(F))(r)-24.62104762*(D(F))(r)^2*((D@@3)(F))(r)-49.24209525*(D(F))(r)*((D@@2)(F))(r)^2-1.44*F(r)^2*(D(F))(r)-0.4e-1*(D(F))(r)^3-.64*F(r)*(D(F))(r)*((D@@2)(F))(r)-0.4e-1*F(r)^2*((D@@3)(F))(r) = 0

 
 

 

BCSforNum := F(0) = 1, (D(F))(0) = 0, F(1) = 0

Digits := 15

15

(2)

numsol := dsolve({BCSforNum, ODEforNum}, numeric, output = listprocedure)

Error, (in dsolve/numeric/bvp) initial Newton iteration is not converging

 
 

 

 

 

 

 

 

 

Download MapleN.mw

Respected member!
Please help me to find the solution of attached problem,  I am a new user so please forgive any mistakes.maple.mwmaple.mw

The question explores the family of differential equations dy/dx = sqrt(1+(a*x)+(2*y)) for various values of the parameter a.

This figure shows the tangent field in the case a=1.

sketch a tangent field in the case a=-2.

need the following diagram on maple:

 

Hi

I have the ODEs: x'=x

ode:=diff(x(t),t)-x(t);

 

sketch the phase space and extended phase space of previous ode.

 

Hi

Any help will be appreciated

I have a continuous time dynamical system

x in R ( set of real number)

t  a positive real time

and the function f_t(x)=x-t

How can we plot or sketch their behaviour in the phase space and in the extended phase space

 

Many thanks

 

 

hello i have the following set of ode's:

ode_sub := diff(S(t), t) = -k1*S(t)-S(t)/T1_s;
ode_P1 := diff(P1(t), t) = k1*S(t)-k2*(P1(t)-P2(t)/keq)-P1(t)/T1_p1;
ode_P2 := diff(P2(t), t) = -k2*(-keq*P1(t)+P2(t))/keq-k4*P2(t)-P2(t)/T1_p2;
ode_P2e := diff(P2_e(t), t) = k4*P2(t)-P2_e(t)/T1_p2_e;

ode_system := ode_sub, ode_P1, ode_P2, ode_P2e;

with these parameters:
s0 := 10000;
k2 := 1000; T1_s := 14; T1_p2_e := 35; T1_p2 := T1_p1;
 

i want to find the unkown parameters : T1_p1, k1, keq and k4

my idea was this:

init:=S(0)=s0,P1(0)=0,P2(0)=0,P2_e(0)=0

dsolve({ode_system,init})

sol := combine(expand(%));
PS := subs(sol, [S(t), P1(t), P2(t), P2_e(t)]);
 

P1fu := unapply(PS[2],t);
Sfu := unapply(PS[1],t);
P2fu := unapply(PS[3],t);
P2e_fu := unapply(PS[4],t);
P2_total := unapply(P2fu+P2e_fu, t);
 

the following data is given:

T:=<0,2,4,6,8>

S:=<9999.99913146527,8328.870587730016,6937.009129218748,5777.745632133724,4812.209983843559>

P1:=<0.0,67.86790056712294,114.88787098501874,145.95438088662502,164.85650644237887>

P2_P2e:=<0.0,271.68492651947497,461.9130396605823,589.3710176125417,668.9967533337124> # data from P2(t)+P2_e(t)

 

making the rediduals:

RP1 := convert(P1-P1fu~(T), list);
RS := convert(S-Sfu~(T), list);
RP2_P2e := convert(P2_P2_e-P2_total~(T), list);
 

RPs := [op(RS), op(RP2_P2_e), op(RP1)]

res := Optimization:-LSSolve(RPs, k1 = 0 .. 1, keq = 0 .. 10, k4 = 0 .. 1, T1_p1 = 0 .. 100)

i dont know wheter or not the last step work to get the parameters becuase it takes to long to compute. is there a smarter way to obtain the parameters of the ode's? a numeric approch ?

i tried with dsolve({ode_sysytem,init},numeric,'parameters'=[k1,keq,k4,T1_p1]) however it doesnt seem to get my anywhere since i need to know the parameters to use this (i think)

hope someone can help:)

 

 

Hi

We solve the following ode in the interval (0,Pi):

 diff(u(x),x,x)+u(x)=f(x);

bcs=u(0)=0, u(Pi)=0;

Stating one example of many conditions for such equation to have a
valid solution

Many thanks

 

 

Hi

 

I have an ODE which is based on a seperate function, and I would like to make a plot with the information

dsolve([diff(X(W), W) = (0.536000000000000e-3*(1-X(W)))*(1+X(W)), X(0) = 0], numeric)

and

C_A:= C_A0*(1-X(W))*(1+X(W))

which has been used as part of the ODE.

I would really like to plot C_A as a function of W. I have no problem plotting X as a function to W using odeplot. Ideally I would like to plot C_A and X vs W in the same plot.

Regards

How I can solve it for P?

P=i^(2)r+(&DifferentialD;)/(&DifferentialD; t) (1/(2)l(tetha)i^(2))+1/(2)i^(2)(&DifferentialD;l(tetha))/(&DifferentialD; theta) w

attach i(t) "corrente", l(t) "induttanza", theta "angular", w "rotary speed"graph of funtions

thanks

the question is 

ODE5:= y(x)*diff(y(x),x,x) + (diff(y(x),x)^2=0

dsolve({ODE5,y(0)=4,D(y)(0)=7},y(x))

and my answer appears to be an integration! which is wrong

the correct answer : 2*(4+14*x)^(1/2)

Could someone tell me what did I do wrong? And how could I get to this result?

Thanks a lot!

 

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