@C_R Thank you for pointing that out. I mistakenly assumed VariationalDifferentiation was a Maple command, which explains the issue with the output. However, Maple still returns zero for the commutator, leading to zero as the output instead of the expected equation of motion (16). VD1.mw

@acer 

What I'm asking is how to find the form of Z (like in the attached image) using Maple? The form in the image is simple enough to solve by hand, but for more complex expressions, it's difficult to handle manually, so Maple would be useful.

For example, I need to solve an equation like this:

3Z^2 + 2I\ln(\text{RootOf}(3Z^2 + 5Z + 12)) - 4Zx =0

Usually, we don't write something like "RootOf(3Z^2 + 2I\ln(\text{RootOf}(3Z^2 + 5Z + 12)) - 4Zx)" directly in text. Could you please show how to get a solution for Z using Maple? 

@acer How to find the form of Z (like in Eq. (5)) in Maple? ss_root.mw

@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);