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These are replies submitted by WA573

@dharr red k/2 is -k/2. In the paper, the system of differential equations has variable coefficients, so by applying a transformation (that makes different) I wrote them into constant coefficients. How to choose the explicit form of A and B?

@dharr @mmcdara I am trying to solve the system of equations (1) and (2) in XintoL.mw by using eigenvalue problems, which can be done if one can write 'X' in terms of 'L' (these matrices are constant). The solution can be found in [https://doi.org/10.1016/j.aml.2020.106859], see (2.1) and (3.5).

@mmcdara By following your comments. I tied for another equation, but h(n+1,t) remains as it is at the output. (In this case, h(n,t)=r1sol, and f(n,t)=q1sol, and shift(q(n, t), n) - q(n, t) = a) verif1_May2024.mw

@acer yes, correct.

@acer Still no substitution is done for D[2](s)(n + 1, t) in (10), D[2] = (∂)/(∂ t). Also, when we define s(n, t) = delta1*r(sigma1*n, sigma2*t), it means n -> sigma1*n, and t -> sigma2*t which implies (∂)/(∂ (sigma2*t))=1/(sigma2)(∂)/(∂ t), sigma2 is constant. How can I do it on Maple?

@acer Now I got this error.

@Axel Vogt In the posted image, eta_{1}=eta_{1}(t;lambda1) and eta_{2}=eta_{2}(t;lambda2), t is variable and lambda complex parameter. When we say eta_{2} -> infinity and eta_{1} -> O(1), then terms with eta_{1} survive (since in the posted image a simple example is considered, so just one term survives) and terms with eta_{2} vanishes or canceled. But when I tried it in limit.mw to find the behavior of y4 (see ref. (3)) by imposing certain assumptions the output remained the same (just constant) for various assumptions. My question is how to get the final expression with some exp(...) terms after applying the limit. I also did it by using asympt command, but no change.

@Rouben Rostamian  Not in a worksheet. 

@Rouben Rostamian  When t->infinity, one of the expressions survives by imposing conditions on lambda1, lambda2 (but this is not happening in limit.mw). As you can see the attached Fig. when eta1->O(1), eta2->infinity, e^{eta1} survive and e^{-eta2} zero.

@Rouben Rostamian  I have also used Im(lambda1) > Im(lambda2), Re(lambda1) > Re(lambda2) and got same answer. 

@sursumCorda It works. But why there is too much difference between Continuous and Discrete plots? Since product(1 - I*sin(n_values[i]*h) is just exp(-I*sin(x)), when x=nh and h -> 0.

@sursumCorda Thanks. I plotted the curves by using plot, but there are no curves for Continuos -Real and Continuous - Imag. 

p1 := plot(x_values, cont_real, 'color' = "blue", 'legend' = "Continuous - Real");
p2 := plot(x_values, disc_real, 'color' = "green", 'legend' = "Discrete - Real");
p3 := plot(x_values, cont_imag, 'color' = "red", 'legend' = "Continuous - Imag");
p4 := plot(x_values, disc_imag, 'color' = "orange", 'legend' = "Discrete - Imag");
display(p1, p2, p3, p4);

@dharr @Carl Love How to use simple notation in soldis.mw? I want 'n~, m~' should be 'n, m' and 'psi(n~,m~), phi(n~,m~)' should be 'psi,phi'. When I use 'declare', then get 'Error, (in PDEtools:-declare) invalid input: conjugate expects 1 argument, but received 2' 

@dharr I am trying to plot the solution for the discrete values of 'n' and continuous values of 't' by using pointplot3d. But encountered an error. How to fix it?


expression := 1-I*((2*I)*I^(n+1)*exp(-2*t)*(2+I)^(n+1)*exp(2*t)/(I^(n+1)*(exp(-2*t))^2*(-I)^(n+1)+(2+I)^(n+1)*(exp(2*t))^2*(2-I)^(n+1))-(2*I)*I^n*exp(-2*t)*(2+I)^n*exp(2*t)/(I^n*(exp(-2*t))^2*(-I)^n+(2+I)^n*(exp(2*t))^2*(2-I)^n)); n_values := [1, 2, 3, 4, 5]; t_values := [-2 .. 2]; with(plots); pointplot3d([seq([n_val, t_val, evalf(subs([n, t], [n_val, t_val], expression))], `in`(n_val, n_values), `in`(t_val, t_values))], style = point, symbol = cross)



[1, 2, 3, 4, 5]


[-2 .. 2]


Error, invalid input: seq expects its 3rd argument, step, to be of type numeric, but received t_val in [-2 .. 2]


Download plot_dis.mw

@Carl Love As in (5), the bar is over eta but in (6)-(8) some terms have bar on 'n' and 'm'.

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