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Khair Muhammad Saraz

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maple code...

Khair Muhammad Saraz 5
March 06 2024
0 0
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restart
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PDEtools[declare](f(x), prime = x)

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

  

`derivatives with respect to`*x*`of functions of one variable will now be displayed with '`

 (1)
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N := 3
 
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"g(x):=f[0](x)+p f[1](x)+p^2 f[2](x)+p^3 f[3](x)"

proc (x) options operator, arrow, function_assign; f[0](x)+p*f[1](x)+p^2*f[2](x)+p^3*f[3](x) end proc

 (2)
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NULL
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HPMeq := (1-p)*(diff(g(x), x, x, x))+p*(diff(g(x), x, x, x)+(1/2)*(diff(g(x), x, x))*g(x))

(1-p)*(diff(diff(diff(f[0](x), x), x), x)+p*(diff(diff(diff(f[1](x), x), x), x))+p^2*(diff(diff(diff(f[2](x), x), x), x))+p^3*(diff(diff(diff(f[3](x), x), x), x)))+p*(diff(diff(diff(f[0](x), x), x), x)+p*(diff(diff(diff(f[1](x), x), x), x))+p^2*(diff(diff(diff(f[2](x), x), x), x))+p^3*(diff(diff(diff(f[3](x), x), x), x))+(1/2)*(diff(diff(f[0](x), x), x)+p*(diff(diff(f[1](x), x), x))+p^2*(diff(diff(f[2](x), x), x))+p^3*(diff(diff(f[3](x), x), x)))*(f[0](x)+p*f[1](x)+p^2*f[2](x)+p^3*f[3](x)))

 (3)
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for i from 0 to N do equ[1][i] := coeff(HPMeq, p, i) = 0 end do

diff(diff(diff(f[0](x), x), x), x) = 0

  

diff(diff(diff(f[1](x), x), x), x)+(1/2)*(diff(diff(f[0](x), x), x))*f[0](x) = 0

  

diff(diff(diff(f[2](x), x), x), x)+(1/2)*(diff(diff(f[0](x), x), x))*f[1](x)+(1/2)*(diff(diff(f[1](x), x), x))*f[0](x) = 0

  

diff(diff(diff(f[3](x), x), x), x)+(1/2)*(diff(diff(f[0](x), x), x))*f[2](x)+(1/2)*(diff(diff(f[1](x), x), x))*f[1](x)+(1/2)*(diff(diff(f[2](x), x), x))*f[0](x) = 0

 (4)
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cond[1][0] := f[0](0) = 0, (D(f[0]))(0) = 0, (D(f[0]))(5) = 1

f[0](0) = 0, (D(f[0]))(0) = 0, (D(f[0]))(5) = 1

 (5)
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for j to N do cond[1][j] := f[j](0) = 0, (D(f[j]))(0) = 0, (D(f[j]))(5) = 0 end do

f[1](0) = 0, (D(f[1]))(0) = 0, (D(f[1]))(5) = 0

  

f[2](0) = 0, (D(f[2]))(0) = 0, (D(f[2]))(5) = 0

  

f[3](0) = 0, (D(f[3]))(0) = 0, (D(f[3]))(5) = 0

 (6)
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"for i from 0 to N do dsolve({equ[1][i], cond[1][i]}, f[i](x)); f[i](x):=rhs(`%`) end do; "

f[0](x) = (1/10)*x^2

  

(1/10)*x^2

  

{f[1](x) = -(1/6000)*x^5+(1/2)*_C1*x^2+_C2*x}

  

Error, invalid input: rhs received {f[1](x) = -1/6000*x^5+1/2*_C1*x^2+_C2*x}, which is not valid for its 1st argument, expr

  
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