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Question: How to solve this equation by using HPM technique. Kindly help me

Question:How to solve this equation by using HPM technique. Kindly help me

Shameera 15 Maple
ode homework differential-equations duplicate-question
March 12 2022
1

 

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NULL
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"#(1+c1)f^(' ' ' ')-c1*g^(' ' )-Re(f^*f^(' ' ')-f'*f^(' '))-n0*(1+c1)f^(' ' )-m0*f^(' ' )=0"
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#c2"*g^(' ')+c1(f^(' ')-2 g)-c3*Re(f*g'-f'*g)=0"
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NULL
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NULL

#boundary condition

f=-1, f^'=0, g=0, a=1, b=1   at  η=-1    

f=1, f^'=0, g=0, a=0, b=0   at  η=-1

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restart
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PDEtools[declare](f(t), prime = t)

` f`(t)*`will now be displayed as`*f

  

`derivatives with respect to`*t*`of functions of one variable will now be displayed with '`

 (1)
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N := 15

15

 (2)
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"f(t):=sum((p^((i)))*f[i]t,i=0..N)"

proc (t) options operator, arrow, function_assign; sum((diff(p(x), [`$`(x, i)]))*f[i]*t, i = 0 .. N) end proc

 (3)
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"de1:=(1-p)*((1+c1)*diff(f(t), t $ 4) -n0*(1+c1)*diff(f(t), t $ 2) -m0*diff(f(t), t $ 2)))+p(((1+c1)*diff(f(t), t $ 4)-c1*diff(g(t), t $ 2)-Re(f(t)*diff(f(t), t $ 3)-diff(f(t), t $ 1)*diff(f(t), t 2)) -n0 *(1+c1)*diff(f(t), t $ 2) -m0*diff(f(t), t $ 2) )"
> 

"de2=(1-p)*(c2 *diff(g(t), t $ 2)+c1*(diff(f(t), t $ 2)-2*g(t))))+p*(c2 *diff(g(t), t $ 2)+c1*(diff(f(t), t $ 2)-2*g(t))-c3*Re(f(t)*diff(g(t), t $ 1)-diff(f(t), t $ 1)*g(t))));"
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de3 = (1-p)*(diff(a(t), `$`(t, 2))+D*(diff(b(t), `$`(t, 2))))+p*(diff(a(t), `$`(t, 2))+D*(diff(b(t), `$`(t, 2)))+Ph*(diff(f(t), `$`(t, 1)))*a(t)-ph*f(t)*(diff(a(t), `$`(t, 1))))
> 

de4 = (1-p)*(diff(a(t), `$`(t, 2))+D*(diff(b(t), `$`(t, 2))))+p*(diff(b(t), `$`(t, 2))+S*(diff(a(t), `$`(t, 2)))+Pm*(diff(f(t), `$`(t, 1)))*b(t)-ph*f(t)*(diff(b(t), `$`(t, 1))))
> 

sys1 := eval([de1, de2, de3, de4], p = 1)

[de1, de2, de3, de4]

 (4)
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dsolve(sys1)

Error, (in dsolve) required an indication of the solving variables for the given system

  
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n := 4

4

 (5)
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f := unapply(add(g[k](t)*p^k, k = 0 .. n), t)

proc (t) options operator, arrow; g[0](t)+g[1](t)*p+g[2](t)*p^2+g[3](t)*p^3+g[4](t)*p^4 end proc

 (6)
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g := unapply(add(h[k](t)*p^k, k = 0 .. n), t)

proc (t) options operator, arrow; h[0](t)+h[1](t)*p+h[2](t)*p^2+h[3](t)*p^3+h[4](t)*p^4 end proc

 (7)
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a := unapply(add(i[k](t)*p^k, k = 0 .. n), t)

proc (t) options operator, arrow; i[0](t)+i[1](t)*p+i[2](t)*p^2+i[3](t)*p^3+i[4](t)*p^4 end proc

 (8)
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b := unapply(add(j[k](t)*p^k, k = 0 .. n), t)

proc (t) options operator, arrow; j[0](t)+j[1](t)*p+j[2](t)*p^2+j[3](t)*p^3+j[4](t)*p^4 end proc

 (9)
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NULL
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cond[1][0] := f[0](-1) = -1, (D(f[0]))(-1) = 0, f[0](1) = 1, (D(f[0]))(1) = 0; g[0](-1) = 0, g[0](1) = 0, a[0](-1) = 1, a[0](1) = 1, b[0](-1) = 1, b[0](1) = 0

Error, (in anonymous procedure) invalid input: diff received -1, which is not valid for its 2nd argument

  

h[0](-1)+h[1](-1)*p+h[2](-1)*p^2+h[3](-1)*p^3+h[4](-1)*p^4 = 0, h[0](1)+h[1](1)*p+h[2](1)*p^2+h[3](1)*p^3+h[4](1)*p^4 = 0, i[0](-1)+i[1](-1)*p+i[2](-1)*p^2+i[3](-1)*p^3+i[4](-1)*p^4 = 1, i[0](1)+i[1](1)*p+i[2](1)*p^2+i[3](1)*p^3+i[4](1)*p^4 = 1, j[0](-1)+j[1](-1)*p+j[2](-1)*p^2+j[3](-1)*p^3+j[4](-1)*p^4 = 1, j[0](1)+j[1](1)*p+j[2](1)*p^2+j[3](1)*p^3+j[4](1)*p^4 = 0

 (10)
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m0 := 12; n0 := 10; ph := .1; pm := .1; c1 := 1.2; c2 := .8; c3 := .6; D := 0.3e-1; S := 2.0; Re := 1


 

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