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Question: How to obtain a solution for bvp

Question:How to obtain a solution for bvp

666 jvbasha 130 Maple 18
ode dsolve numeric differential-equations bvp duplicate-question
July 18 2020
0

Dear maple users,

Greetings.


JVB.mw
 

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restart; with(plots)
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fcns := {f(x), g(x), t(x)}
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m := .2; pa := 3.14*(1/3); aa := .1; bb := .3; ta := .2; kt := .4; h2 := 1+m*ax+bb*sin((2*3.14)*(ax-ta)); h1 := -1-m*ax-aa*sin((2*3.14)*(ax-ta)+pa); a2 := kt+aa*sin((2*3.14)*(ax-ta))+bb*sin((2*3.14)*(ax-ta)+pa)

1+.2*ax+.3*sin(6.28*ax-1.256)

  

-1-.2*ax-.1*sin(6.28*ax-.209333333)

  

.4+.1*sin(6.28*ax-1.256)+.3*sin(6.28*ax-.209333333)

 (1)
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eq1 := 1.5*(diff(f(x), `$`(x, 4)))-.40*(diff((diff(f(x), `$`(x, 2)))^3, `$`(x, 2)))-.20*(diff(f(x), `$`(x, 2)))+.20*(diff(t(x), `$`(x, 1)))+.20*(diff(g(x), `$`(x, 1))) = 0

1.5*(diff(diff(diff(diff(f(x), x), x), x), x))-2.40*(diff(diff(f(x), x), x))*(diff(diff(diff(f(x), x), x), x))^2-1.20*(diff(diff(f(x), x), x))^2*(diff(diff(diff(diff(f(x), x), x), x), x))-.20*(diff(diff(f(x), x), x))+.20*(diff(t(x), x))+.20*(diff(g(x), x)) = 0

 (2)
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eq2 := 1.75*(diff(t(x), `$`(x, 2)))+(diff(t(x), `$`(x, 1)))*(diff(g(x), `$`(x, 1)))+2*(diff(t(x), `$`(x, 1)))^2+.7*((diff(f(x), `$`(x, 2)))^2-(diff(f(x), `$`(x, 2)))^4)+.1*(diff(f(x), `$`(x, 1)))^2+.1*t(x) = 0

1.75*(diff(diff(t(x), x), x))+(diff(t(x), x))*(diff(g(x), x))+2*(diff(t(x), x))^2+.7*(diff(diff(f(x), x), x))^2-.7*(diff(diff(f(x), x), x))^4+.1*(diff(f(x), x))^2+.1*t(x) = 0

 (3)
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eq3 := diff(g(x), `$`(x, 2))+2*(diff(t(x), `$`(x, 2)))-.5*g(x) = 0

diff(diff(g(x), x), x)+2*(diff(diff(t(x), x), x))-.5*g(x) = 0

 (4)
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bc := f(h2) = (1/2)*a2, (D(f))(h2) = 0, f(h1) = -(1/2)*a2, (D(f))(h1) = 0, t(h2) = 1, t(h1) = 0, g(h2) = 1, g(h1) = 0

f(1+.2*ax+.3*sin(6.28*ax-1.256)) = .2000000000+0.5000000000e-1*sin(6.28*ax-1.256)+.1500000000*sin(6.28*ax-.209333333), (D(f))(1+.2*ax+.3*sin(6.28*ax-1.256)) = 0, f(-1-.2*ax-.1*sin(6.28*ax-.209333333)) = -.2000000000-0.5000000000e-1*sin(6.28*ax-1.256)-.1500000000*sin(6.28*ax-.209333333), (D(f))(-1-.2*ax-.1*sin(6.28*ax-.209333333)) = 0, t(1+.2*ax+.3*sin(6.28*ax-1.256)) = 1, t(-1-.2*ax-.1*sin(6.28*ax-.209333333)) = 0, g(1+.2*ax+.3*sin(6.28*ax-1.256)) = 1, g(-1-.2*ax-.1*sin(6.28*ax-.209333333)) = 0

 (5)
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L := [.1]; AP := NULL; NN := nops(L); for k to NN do R := dsolve(eval({bc, eq1, eq2, eq3}, ax = L[k]), fcns, type = numeric, method = bvp[midrich], maxmesh = 1000, abserr = 10^(-4), AP); AP := approxsoln = R; X1 || k := rhs(R(0)[4]) end do; ba3 := [(X1 || (1 .. NN))(0)]

[.1]

  

1

  

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

  

R(0)

 (6)
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``


 

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How to obtain a solution for various values of "ax"

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