Advanced Search

WA573

80 Reputation

4 Badges

2 years, 131 days
Contact WA573

MaplePrimes Activity


These are questions asked by WA573

'n' and 'm' are real...

WA573 80 Maple 2022
syntax substitution assumptions
January 05 2024
0 7

As I assumed 'n' and 'm' are real, eta is complex. But still, there is a bar on these discrete independent variables. Secondly, the substitution of (8) applies in some terms of 'r2', and the remaining terms remain as is it.

> 

restart
> 

with(LinearAlgebra); with(PDEtools); with(plots); with(LREtools)
> 

setup(mathematicalnotation = true)

setup(mathematicalnotation = true)

 (1)
> 

assume(n::real); assume(m::real)
> 

A := proc (n, m) options operator, arrow; Matrix([[eta*phi(n, m), conjugate(eta)*conjugate(psi(n, m))], [phi(n, m), conjugate(psi(n, m))]]) end proc; Adet := Determinant(A(n, m))

eta*phi(n, m)*conjugate(psi(n, m))-conjugate(eta)*conjugate(psi(n, m))*phi(n, m)

 (2)
> 

B := proc (n, m) options operator, arrow; Matrix([[phi(n, m), conjugate(psi(n, m))], [-psi(n, m), conjugate(phi(n, m))]]) end proc; Bdet := Determinant(B(n, m))

phi(n, m)*conjugate(phi(n, m))+conjugate(psi(n, m))*psi(n, m)

 (3)
> 

r := Adet/Bdet

(eta*phi(n, m)*conjugate(psi(n, m))-conjugate(eta)*conjugate(psi(n, m))*phi(n, m))/(phi(n, m)*conjugate(phi(n, m))+conjugate(psi(n, m))*psi(n, m))

 (4)
> 

p := {eta = 1+I, phi(n, m) = (1+I*a*eta)^n*(1+I*b*eta^2)^m, psi(n, m) = (1-I*a*eta)^n*(1-I*b*eta^2)^m, conjugate(eta) = 1-I, conjugate(phi(n, m)) = (1-I*a*conjugate(eta))^n*(1-I*b*conjugate(eta)^2)^m, conjugate(phi(n, m)) = (1+I*a*conjugate(eta))^n*(1+I*b*conjugate(eta)^2)^m}

{eta = 1+I, phi(n, m) = (1+I*a*eta)^n*(1+I*b*eta^2)^m, psi(n, m) = (1-I*a*eta)^n*(1-I*b*eta^2)^m, conjugate(eta) = 1-I, conjugate(phi(n, m)) = (1-I*a*conjugate(eta))^n*(1-I*b*conjugate(eta)^2)^m, conjugate(phi(n, m)) = (1+I*a*conjugate(eta))^n*(1+I*b*conjugate(eta)^2)^m}

 (5)
> 

r1 := simplify(subs(p, r))

(2*I)*(1+I*a*eta)^n*(1+I*b*eta^2)^m*conjugate((1-I*a*eta)^n*(1-I*b*eta^2)^m)/((1+I*a*eta)^n*(1+I*b*eta^2)^m*(1-I*a*conjugate(eta))^n*(1-I*b*conjugate(eta)^2)^m+abs(-1+I*a*eta)^(2*n)*abs(I*b*eta^2-1)^(2*m))

 (6)
> 

r2 := 1-I*delta(r1, n)

1-I*((2*I)*(1+I*a*eta)^(n+1)*(1+I*b*eta^2)^m*conjugate((1-I*a*eta)^(n+1)*(1-I*b*eta^2)^m)/((1+I*a*eta)^(n+1)*(1+I*b*eta^2)^m*(1-I*a*conjugate(eta))^(n+1)*(1-I*b*conjugate(eta)^2)^m+abs(-1+I*a*eta)^(2*n+2)*abs(I*b*eta^2-1)^(2*m))-(2*I)*(1+I*a*eta)^n*(1+I*b*eta^2)^m*conjugate((1-I*a*eta)^n*(1-I*b*eta^2)^m)/((1+I*a*eta)^n*(1+I*b*eta^2)^m*(1-I*a*conjugate(eta))^n*(1-I*b*conjugate(eta)^2)^m+abs(-1+I*a*eta)^(2*n)*abs(I*b*eta^2-1)^(2*m)))

 (7)
> 

exp_expr := subs({(1+I*b*eta^2)^m = exp(I*eta^2*t)}, r2)

1-I*((2*I)*(1+I*a*eta)^(n+1)*exp(I*eta^2*t)*conjugate((1-I*a*eta)^(n+1)*(1-I*b*eta^2)^m)/((1+I*a*eta)^(n+1)*exp(I*eta^2*t)*(1-I*a*conjugate(eta))^(n+1)*(1-I*b*conjugate(eta)^2)^m+abs(-1+I*a*eta)^(2*n+2)*abs(I*b*eta^2-1)^(2*m))-(2*I)*(1+I*a*eta)^n*exp(I*eta^2*t)*conjugate((1-I*a*eta)^n*(1-I*b*eta^2)^m)/((1+I*a*eta)^n*exp(I*eta^2*t)*(1-I*a*conjugate(eta))^n*(1-I*b*conjugate(eta)^2)^m+abs(-1+I*a*eta)^(2*n)*abs(I*b*eta^2-1)^(2*m)))

 (8)
> 

``
> 

NULL
> 

NULL
> 

NULL
> 

plot3d(abs(exp_expr), n = -5 .. 5, t = -5 .. 5, eta = 1+I)

Error, (in plot3d) unexpected option: eta = 1+I

  
 

Download soldis.mw

derivative from finite difference...

WA573 80 Maple 2022
derivative differential-equations finite-difference
November 28 2023
1 2

Since (1/h)[f(i+1,t)-f(i,t)]=f(x,t)_{x} as h goes to zero, 'i' is the discrete index along x-axis. How to do it in Maple? How to reduce Eq. (5) into continuous derivatives?

> 

restart
> 

with(LinearAlgebra)
> 

with(PDEtools)
> 

with(Physics)
> 

with(plots)
> 

Setup(mathematicalnotation = true)

[mathematicalnotation = true]

 (1)
> 

``
> 

U := proc (i, t) options operator, arrow; Matrix([[1+I*(q(i+1, t)-q(i, t))/lambda, I*(r(i+1, t)-r(i, t))/lambda], [I*(r(i+1, t)-r(i, t))/lambda, 1-I*(q(i+1, t)-q(i, t))/lambda]]) end proc

proc (i, t) options operator, arrow; Matrix([[1+Physics:-`*`(Physics:-`*`(I, q(i+1, t)-q(i, t)), Physics:-`^`(lambda, -1)), Physics:-`*`(Physics:-`*`(I, r(i+1, t)-r(i, t)), Physics:-`^`(lambda, -1))], [Physics:-`*`(Physics:-`*`(I, r(i+1, t)-r(i, t)), Physics:-`^`(lambda, -1)), 1-Physics:-`*`(Physics:-`*`(I, q(i+1, t)-q(i, t)), Physics:-`^`(lambda, -1))]]) end proc

 (2)
> 

``
> 

V := proc (i, t) options operator, arrow; Matrix([[-((1/2)*I)*lambda, -r(i, t)], [r(i, t), ((1/2)*I)*lambda]]) end proc

proc (i, t) options operator, arrow; Matrix([[Physics:-`*`(Physics:-`*`(Physics:-`*`(Physics:-`*`(1, Physics:-`^`(2, -1)), I), lambda), -1), Physics:-`*`(r(i, t), -1)], [r(i, t), Physics:-`*`(Physics:-`*`(Physics:-`*`(1, Physics:-`^`(2, -1)), I), lambda)]]) end proc

 (3)
> 

NULL
> 

z := diff(U(i, t), t)+U(i, t).V(i, t)-V(i+1, t).U(i, t)

Matrix(%id = 4525182530)

 (4)
> 

z11 := simplify(lambda*z[1, 1]/h, size) = 0

I*(r(i+1, t)^2-r(i, t)^2+(D[2](q))(i+1, t)-(diff(q(i, t), t)))/h = 0

 (5)
> 

NULL

Download limit.mw

Residue for a function...

WA573 80 Maple 2022
residue physics
November 17 2023
0 0

Since C2=D1.D1inv should be equal to I. But return is just an expression (see attached). Further, how to obtain residue for a function C2?

residue.mw

How do I solve a differential equation i...

WA573 80 Maple 2022
eigenvalues differential-equations
January 10 2023
2 2
> 

restart
> 

with(DEtools); with(LinearAlgebra)
> 

diff(u(x, t), t) = [Matrix([[0, (1/2)*mu*k^2], [2*A^2-(1/2)*mu*k^2, 0]])]*u(x, t)

diff(u(x, t), t) = [Matrix(%id = 36893489823894642308)]*u(x, t)

 (1)
> 

"where u(x,t)=[u1 u2]^(T) is a vector. The solution of differential equation (1) is u=v*exp(w*t)."
> 

where*w^2 = -(1/4)*mu^2*k^4+mu*k^2*A^2
> 

"How can we solve differntial equation*(1) on Maple"?""

NULL

Download dsol.mw

Why the expression is too large?...

WA573 80 Maple 2022
solve
November 27 2022
1 0

Why the expression of lambda[1] is too large? One of the possible values of lambda[1] should be "sqrt(1/(4*omega^2 - 4))*a[-1]". 

JCN.mw 

1 2 3 4 5 6 Page 3 of 6