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Mariusz Iwaniuk

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N=3...

Mariusz Iwaniuk 1476
March 24 2018
0 0

@Math Pi Euler 


 

We obtain a series expansion in terms of a small parameter, of the solution of a second

order nonlinear differential equation.

> 

restart;
> 

de := diff(z(x),x,x) + z(x) - cos(2*x)/(1+delta*z(x)) = 0;

diff(diff(z(x), x), x)+z(x)-cos(2*x)/(1+delta*z(x)) = 0

 (1)
> 

 bc := z(-Pi/4)=0, z(Pi/4)=0

z(-(1/4)*Pi) = 0, z((1/4)*Pi) = 0

 (2)

We are going to expand the solution of this boundary value problem in powers of delta,

and pick the leading N terms like this:

> 

N := 3;  # change as needed
S := z(x) = add(u[n](x)*delta^n, n=0..N-1);

3

  

z(x) = u[0](x)+u[1](x)*delta+u[2](x)*delta^2

 (3)

Substitute the series expression S in the differential equation and simplify:

> 

eval(de, S):
series(lhs(%), delta, N):
collect(%, delata, simplify):
tmp := convert(%, polynom);

diff(diff(u[0](x), x), x)+u[0](x)-cos(2*x)+(diff(diff(u[1](x), x), x)+u[1](x)+cos(2*x)*u[0](x))*delta+(diff(diff(u[2](x), x), x)+(-u[0](x)^2+u[1](x))*cos(2*x)+u[2](x))*delta^2

 (4)

This is supposed to hold for all delta, therefore the coefficient of each power of delta is zero.
This gives us a system of N coupled differential equations in N unknowns:

> 

DEs := coeffs(tmp, delta);

diff(diff(u[0](x), x), x)+u[0](x)-cos(2*x), diff(diff(u[2](x), x), x)+(-u[0](x)^2+u[1](x))*cos(2*x)+u[2](x), diff(diff(u[1](x), x), x)+u[1](x)+cos(2*x)*u[0](x)

 (5)

Apply the given boundary conditions to generate boundary conditions for the new unknowns:

> 

BCs := seq(subs(z=u[n], [bc])[], n=0..N-1);

u[0](-(1/4)*Pi) = 0, u[0]((1/4)*Pi) = 0, u[1](-(1/4)*Pi) = 0, u[1]((1/4)*Pi) = 0, u[2](-(1/4)*Pi) = 0, u[2]((1/4)*Pi) = 0

 (6)
> 

sol := simplify(value(dsolve({DEs},`build`)));

{u[0](x) = sin(x)*_C6+cos(x)*_C5-(2/3)*cos(x)^2+1/3, u[1](x) = -(4/45)*cos(x)^4+(1/4)*_C5*cos(x)^3+(1/720)*(180*sin(x)*_C6+64)*cos(x)^2+(1/720)*(-180*_C6*x+720*_C3-225*_C5)*cos(x)+7/45+(1/720)*(-180*_C5*x+720*_C4+45*_C6)*sin(x), u[2](x) = -(16/525)*cos(x)^6+(19/144)*_C5*cos(x)^5+(1/302400)*(-40320*_C5^2+40320*_C6^2+39900*sin(x)*_C6+13824)*cos(x)^4+(1/302400)*(-80640*_C5*_C6*sin(x)-18900*_C6*x+75600*_C3-48300*_C5)*cos(x)^3+(1/302400)*((-18900*_C5*x+75600*_C4-31500*_C6)*sin(x)-60480*_C5^2-141120*_C6^2+10496)*cos(x)^2+(1/302400)*(40320*_C5*_C6*sin(x)+(9450*x^2-16275)*_C5-75600*_C4*x+69300*_C6*x+302400*_C1-94500*_C3)*cos(x)+(1/302400)*(-50400*_C5*x+(9450*x^2-24675)*_C6-75600*_C3*x+302400*_C2+18900*_C4)*sin(x)+(2/5)*_C5^2-(1/15)*_C6^2-118/4725}

 (7)

Apply the given boundary conditions.Now build the final solution:

> 

sol2 := solve([eval(rhs(sol[1]), x = -(1/4)*Pi) = 0, eval(rhs(sol[1]), x = (1/4)*Pi) = 0, eval(rhs(sol[2]), x = -(1/4)*Pi) = 0, eval(rhs(sol[2]), x = (1/4)*Pi) = 0, eval(rhs(sol[3]), x = -(1/4)*Pi) = 0, eval(rhs(sol[3]), x = (1/4)*Pi) = 0], {_C1, _C2, _C3, _C4, _C5, _C6})

{_C1 = -(1/90)*2^(1/2)*Pi-(1/30)*2^(1/2), _C2 = 0, _C3 = -(8/45)*2^(1/2), _C4 = 0, _C5 = 0, _C6 = 0}

 (8)
> 

sol3:=eval(S, eval(sol, sol2)):
sol4:=sort(sol3, delta, ascending);

z(x) = 1/3-(2/3)*cos(x)^2+(-(4/45)*cos(x)^4+7/45+(4/45)*cos(x)^2-(8/45)*2^(1/2)*cos(x))*delta+(-(16/525)*cos(x)^6-118/4725+(8/175)*cos(x)^4-(2/45)*2^(1/2)*cos(x)^3+(164/4725)*cos(x)^2+(1/302400)*(-3360*2^(1/2)*Pi+6720*2^(1/2))*cos(x)+(2/45)*2^(1/2)*x*sin(x))*delta^2

 (9)
> 

sol5 := combine(rhs(sol4))

-(1/3)*cos(2*x)-(8/45)*2^(1/2)*cos(x)*delta-(1/90)*cos(4*x)*delta+(1/6)*delta-(1/90)*cos(x)*2^(1/2)*Pi*delta^2+(2/45)*2^(1/2)*x*sin(x)*delta^2-(1/90)*2^(1/2)*cos(x)*delta^2-(1/90)*2^(1/2)*cos(3*x)*delta^2-(1/1050)*cos(6*x)*delta^2+(7/270)*cos(2*x)*delta^2

 (10)
> 

z(x) = sort(collect(sol5, delta), delta, ascending)

z(x) = -(1/3)*cos(2*x)+(-(1/90)*cos(4*x)+1/6-(8/45)*2^(1/2)*cos(x))*delta+(-(1/1050)*cos(6*x)+(7/270)*cos(2*x)-(1/90)*2^(1/2)*cos(x)-(1/90)*2^(1/2)*cos(3*x)-(1/90)*cos(x)*2^(1/2)*Pi+(2/45)*2^(1/2)*x*sin(x))*delta^2

 (11)
> 

``

Remark: There is no impediment, in principle, to calculating higher order terms by increasing N.

Trying N = 3, however, reveals a Maple bug.  The solution fails at the dsolve() step.  We may

solve the system manually without much difficulty, because the coupling of the system's equations

is in triangular form and the solution may be achieved by solving the equations in sequence, a single

equation at a time.


 

Download dsolve-perturbation_solution_ver2.mw

 

....

Mariusz Iwaniuk 1476
March 21 2018
0 0

@Rouben Rostamian  

pde := diff(u(x, t), t)+u(x, t)*(diff(u(x, t), x)) = 0;
ic := u(x, 0) = piecewise(x <= 0, 0, 1);
sol := `assuming`([pdsolve([pde, ic], HINT = f(x)/f(t))], [t > 0, x > 0])

Maple give:

# sol := u(x, t) = 1

 

....

Mariusz Iwaniuk 1476
March 21 2018
0 0

Execute code:

?solve

Or:

?fsolve

 

....

Mariusz Iwaniuk 1476
March 21 2018
0 0

If you want a strange result execute code:

 

restart:
pde := diff(u(x, t), t)+u(x, t)*(diff(u(x, t), x)) = mu*(diff(u(x, t), x$2));
ic := u(x, 0) = piecewise(x >= 1, 0, x < 0, 1);
sol := `assuming`([pdsolve([pde, ic])], [mu > 0, t > 0, x in real])

 

....

Mariusz Iwaniuk 1476
March 19 2018
0 0

@Kitonum 

Ohh Yes.I thought Maple did not have any bugs.That's why i did not checked.

 

Thanks for info.

MaplePrimes AI....

Mariusz Iwaniuk 1476
March 09 2018
0 0

The input should be provided in cut-and-pastable form. Please post actual code rather than an image of code.

....

Mariusz Iwaniuk 1476
March 07 2018
0 0

@mmcdara 

If is a standard trick then Maple does no know  about it.

....

Mariusz Iwaniuk 1476
March 07 2018
0 0

Can you provide the matrix in a copy/pasted form, or something? Nobody is going to type a  matrix in by hand!

....

Mariusz Iwaniuk 1476
March 01 2018
0 0

At first you must solve y from eqn and then put to diff(y, x).

See:https://en.wikipedia.org/wiki/Arc_length

....

Mariusz Iwaniuk 1476
February 27 2018
0 0

The input should be provided in cut-and-pastable form. Correct yours syntax errors.

Simpler in MMA...

Mariusz Iwaniuk 1476
February 24 2018
0 0

@Joe Riel 

@Rouben Rostamian  

Thanks.

In Mathematica is simpler:

 

Maple always complicates many things.

....

Mariusz Iwaniuk 1476
February 22 2018
0 0

@vv 

Yes I know identify mayby fails in certain conditions.

If OP want more information about identify function  just need to enter:

?identify

 

....

Mariusz Iwaniuk 1476
February 21 2018
0 0

Please post actual  Maple code rather than web page address.

Unevaluated....

Mariusz Iwaniuk 1476
February 18 2018
0 0

@rlopez 

with(Student[MultivariateCalculus]);
MultiInt(r*exp(-a^2/r^2), [t, r] = Region(0 .. (1/2)*Pi, 0 .. 2*a*cos(t)), output = steps);

it returns it unevaluated.

Second approach:

MultiInt(r*exp(-a^2/r^2), r = 0 .. 2*a*cos(t), t = 0 .. (1/2)*Pi, coordinates = polar[r, t], output = steps);

no either.

Probably this integral has no-close form to known special function.

 

Mathematica gives only a half answer:

A better look....

Mariusz Iwaniuk 1476
February 14 2018
0 0 
n := 0.5e-1;#decrease value for more  contours.
contourplot(A2, z = 0 .. 1, r = -h(z) .. h(z), axes = boxed, contours = [seq(i, i = -1 .. 1, n)], grid = [80, 80], coloring = ["Niagara Azure", "Orange"], transparency = .4);

 

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