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FZ

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1 years, 214 days
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These are replies submitted by FZ

@janhardo Most probably. How did yo...

FZ 30
March 08 2025
0 0

@janhardo Most probably. How did you get it? I tried by using 'simplify(Bans)', 'simplify(Bans,size)' , 'simplify(Bans,complex)'.

@mmcdara I just found that 'sim...

FZ 30
February 10 2025
0 0

@mmcdara I just found that 'simply(expr, power)' return simplified form, same as in (13).             

@dharr You can find the details in&...

FZ 30
February 06 2025
0 0

@dharr You can find the details in http://dx.doi.org/10.11948/20190408

@dharr My first equation is somethi...

FZ 30
February 06 2025
0 0

@dharr My first equation is something like (1.7), and the corresponding Hamiltonian is given in (1.12) in the attached image.

@dharr It's not the same form. ...

FZ 30
February 05 2025
0 0

@dharr It's not the same form. The first equation is indeed more complex, and phi(xi) also appears in the denominator, but can we write it in a form similar to the second equation?

@dharr For example,  How can ...

FZ 30
February 05 2025
0 0

@dharr For example, 

How can I systematically compute such a Hamiltonian in Maple for the previous equation?

@nm How to assume phi is an even fu...

FZ 30
February 03 2025
0 0

@nm How to assume phi is an even function and theta is an odd function as in the image? Also how to convert 'D(phi)(-c*t + x) = phi_\xi' as in Eq. (1.5)?

@mmcdara, do you have a file with m...

FZ 30
February 01 2025
0 0

@mmcdara, do you have a file with more than one dependent function? My problem has three dependent functions 'q', 'r', and 's' (see attached file). Since there is no 'G_x' and 'F_x' in the first equation bilinear form, I tried to remove it by integrating eq. (9) w.r.t. 'x' but Maple does not evaluate this integral. How to fix such issues and find the bilinear form as in the attached image after (10) in BE1.mw?   

@mmcdara Yes this question is about...

FZ 30
January 31 2025
0 0

@mmcdara Yes this question is about Hirota bilinear derivative. 

@dharr Why integral w.r.t 'a...

FZ 30
January 29 2025
0 0

@dharr Why integral w.r.t 'a' appear? Since 'a', 'B1', 'B2', 'delta1', and 'delta2' are constants.

@dharr I just modified the explicit...

FZ 30
January 28 2025
0 0

@dharr I just modified the explicit expression of z but the system is  stuck on 'Evaluating' and does not show any output.

> 

restart;
> 

with(PDEtools):
> 

interface(showassumed=0):
> 

assume(x::real);assume(t::real);assume(lambda1::complex);assume(lambda2::complex);assume(A1::real);assume(A2::real);assume(B1::real);assume(B2::real);assume(delta1::real); assume(delta2::real);assume(y::real);assume(a::real);
> 

z := -2*delta2*exp((-2*I*a*delta1^3*y + (x + y + t)^2*(B2 + B1)*delta1^2 + (-2*I*a*y*delta2^2 + t*I)*delta1 + delta2^2*(x + y + t)^2*(B2 + B1))/(delta1^2 + delta2^2))/((exp((2*(delta1^2 + delta2^2)*(x + y + t)^2*B2 + 2*(a*y*delta1^2 + a*y*delta2^2 + 1/2*t)*delta2)/(delta1^2 + delta2^2)) + exp((2*(delta1^2 + delta2^2)*(x + y + t)^2*B1 - 2*(a*y*delta1^2 + a*y*delta2^2 + 1/2*t)*delta2)/(delta1^2 + delta2^2)))*(delta1^2 + delta2^2)):
> 

pde:=diff(u(x,y,t),t)-diff(u(x,y,t),x)=z:
> 

pdsolve(pde,u(x,y,t));
> 

#bc := u(x, y, t) -> 0 as x -> infinity;
> 

sol := pdsolve(pde, u(x, y, t), HINT = real);

Warning: System is inconsistent

  

(1)
> 

ic := u(x, y, 0) = F1(x, y):
sol := pdsolve([pde, ic], u(x, y, t));

NULL

Download pdsolve1.mw

@dharr I used natural boundary/init...

FZ 30
January 25 2025
0 0

@dharr I used natural boundary/initial conditions but the system does not show output and is stuck on 'Evaluating'. 

> 

restart;
> 

with(PDEtools):
> 

z := 32*delta2^3*(exp((2*a*y*delta2^3 + 2*(x + y + t)^3*(B1 + 2*B2)*delta2^2 + (2*a*delta1^2*y + t)*delta2 + 2*delta1^2*(x + y + t)^3*(B1 + 2*B2))/(delta1^2 + delta2^2)) - exp((-2*a*y*delta2^3 + 4*(B1 + B2/2)*(x + y + t)^3*delta2^2 + (-2*a*delta1^2*y - t)*delta2 + 4*delta1^2*(B1 + B2/2)*(x + y + t)^3)/(delta1^2 + delta2^2)))*a/((delta1^2 + delta2^2)^2*(exp((2*(delta1^2 + delta2^2)*(x + y + t)^3*B2 + 2*delta2*(a*(delta1^2 + delta2^2)*y + t/2))/(delta1^2 + delta2^2)) + exp((2*(delta1^2 + delta2^2)*(x + y + t)^3*B1 - 2*delta2*(a*(delta1^2 + delta2^2)*y + t/2))/(delta1^2 + delta2^2)))^3):
> 

pde:=diff(u(x,y,t),t)-diff(u(x,y,t),x)=z:
> 

pdsolve(pde,u(x,y,t));

u(x, y, t) = -32*a*((1/2)*exp(2*(y^3+3*y^2*(x+t)+3*y*(x+t)^2+(x+t)^3)*B1-4*(x+y+t)^3*B1)/(exp(2*(B2*y^3*delta1^2+B2*y^3*delta2^2+3*B2*y^2*delta1^2*(x+t)+3*B2*y^2*delta2^2*(x+t)+3*B2*y*(x+t)^2*delta1^2+3*B2*y*(x+t)^2*delta2^2+B2*(x+t)^3*delta1^2+B2*(x+t)^3*delta2^2+2*y*a*delta1^2*delta2+2*a*y*delta2^3+delta2*(x+t))/(delta1^2+delta2^2))+exp(2*x*delta2/(delta1^2+delta2^2)+2*(x+y+t)^3*B1))-(1/4)*(exp(2*(x+y+t)^3*B1)+exp(2*(y^3+3*y^2*(x+t)+3*y*(x+t)^2+(x+t)^3)*B1))*exp(2*(B2*y^3*delta1^2+B2*y^3*delta2^2+3*B2*y^2*delta1^2*(x+t)+3*B2*y^2*delta2^2*(x+t)+3*B2*y*(x+t)^2*delta1^2+3*B2*y*(x+t)^2*delta2^2+B2*(x+t)^3*delta1^2+B2*(x+t)^3*delta2^2+2*y*a*delta1^2*delta2+2*a*y*delta2^3+delta2*(x+t))/(delta1^2+delta2^2)-4*(x+y+t)^3*B1)/(exp(2*(B2*y^3*delta1^2+B2*y^3*delta2^2+3*B2*y^2*delta1^2*(x+t)+3*B2*y^2*delta2^2*(x+t)+3*B2*y*(x+t)^2*delta1^2+3*B2*y*(x+t)^2*delta2^2+B2*(x+t)^3*delta1^2+B2*(x+t)^3*delta2^2+2*y*a*delta1^2*delta2+2*a*y*delta2^3+delta2*(x+t))/(delta1^2+delta2^2))+exp(2*x*delta2/(delta1^2+delta2^2)+2*(x+y+t)^3*B1))^2)*delta2^2*exp(2*delta2*(x+t)/(delta1^2+delta2^2)+4*a*y*delta2^3/(delta1^2+delta2^2)+4*a*delta1^2*delta2*y/(delta1^2+delta2^2)+2*delta1^2*B1*y^3/(delta1^2+delta2^2)+2*delta2^2*B1*y^3/(delta1^2+delta2^2)+6*delta1^2*B1*y^2*(x+t)/(delta1^2+delta2^2)+6*delta2^2*B1*y^2*(x+t)/(delta1^2+delta2^2)+6*delta1^2*B1*y*(x+t)^2/(delta1^2+delta2^2)+6*delta2^2*B1*y*(x+t)^2/(delta1^2+delta2^2)+2*delta1^2*B1*(x+t)^3/(delta1^2+delta2^2)+2*delta2^2*B1*(x+t)^3/(delta1^2+delta2^2)+2*delta1^2*B2*y^3/(delta1^2+delta2^2)+2*delta2^2*B2*y^3/(delta1^2+delta2^2)+6*delta1^2*B2*y^2*(x+t)/(delta1^2+delta2^2)+6*delta2^2*B2*y^2*(x+t)/(delta1^2+delta2^2)+6*delta1^2*B2*y*(x+t)^2/(delta1^2+delta2^2)+6*delta2^2*B2*y*(x+t)^2/(delta1^2+delta2^2)+2*delta1^2*B2*(x+t)^3/(delta1^2+delta2^2)+2*delta2^2*B2*(x+t)^3/(delta1^2+delta2^2))/(delta1^2+delta2^2)+_F1(y, x+t)

 (1)
> 

#bc := u(x, y, t) -> 0 as x -> infinity;
> 

sol := pdsolve(pde, u(x, y, t), HINT = real);

Warning: System is inconsistent

  

(2)
> 

ic := u(x, y, 0) = F1(x, y):
sol := pdsolve([pde, ic], u(x, y, t), HINT = real);

NULL

Download pdsolve.mw

@Kitonum Since  and I wan...

FZ 30
January 25 2025
0 0

@Kitonum Since z = u_t - u_x and I want to calculate u from the expression of z. I am not sure either can we write it as you suggested. 

simplify(1/(diff(z, t)-diff(z, x)));
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