dharr

Dr. David Harrington

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21 years, 3 days
University of Victoria
Professor or university staff
Victoria, British Columbia, Canada

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I am a retired professor of chemistry at the University of Victoria, BC, Canada. My research areas are electrochemistry and surface science. I have been a user of Maple since about 1990.

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These are replies submitted by dharr

You define the function theta(y) in terms of y. Then you use dsolve to find u(y), but you give boundary conditions for the known function theta(y), not the unkown function u(y).

@moh111 What do you know about the signs or ranges of k, re and the variables?

@tomleslie The OP specified the 8 unknowns they were interested in. My understanding is that the OP wants these as functions of k and re. Setting infolevel[solve]:=5 shows that solve uses methods for polynomial systems, but they are very slow. (If k and re are specified then fsolve easily gives a solution.)

You should not use both N(x,t) and N[0] together - use N__0 instead of N[0] (looks the same), and the same for omega, E etc.

You set a Matrix equal to scalar zero; I assume that was meant to be the determinant of that Matrix.

If I guess at what substitution you want, then I get the attached, which still has cosines and x and t in. You will need to specify exactly the steps that lead to the result you expect. My guess is that it is not a matter of substitution, but of solving a PDE and getting some eigenvalues for E, but the equations aren't familiar to me, so many more details are required.

d_relation.mw

@bstuan I was just too lazy to write out the 2023 case :-)

Edit: That is, (A.B)^n is not equal to A^n.B^n in general, though there are some cases where it is true e.g., commuting matrices, n=1, n=0.

@bstuan Note (A.B)^2 = A.B.A.B <> A.A.B.B

@Earl It's not clear to me why the loxodromic spiral would be the (2,3) case. There's only a single spiral, so it's more like (2,1), but that doesn't work out quantitatively. Here is separate code for the loxodromic case:

restart;

Loxodromic spiral. Ref: H.S.M. Coxeter, Loxodromic sequences of tangent spheres, Aequationes Mathematicae, 1 (1968) 104–121, doi:10.1007/BF01817563

tau:=1/2*(1+sqrt(5));
x:=tau+sqrt(tau):'x'=evalf(%);
a:=r*(1+x^2):b:=r*(1+x):c:=r*(x+x^2):   #sides of A0,A1,A2 triangle
cosA1:=(b^2+c^2-a^2)/2/b/c: #internal angle fom cosine rule
beta:=Pi-arccos(cosA1):'beta'=evalf(beta)*180/Pi*degrees;
z:=evalf(x*exp(beta*I));
evalf((sqrt(tau)+1)*(-1/sqrt(tau)+I));
r:=fsolve(r*(1+x)=abs(1-z),r);

1/2+(1/2)*5^(1/2)

x = 2.890053638

beta = 128.1727077*degrees

-1.786151379+2.272019648*I

-1.786151378+2.272019650*I

.9241763720

plots:-display(seq(plottools:-circle([Re(z^v),Im(z^v)],r*x^v),v=0..3),scaling=constrained);

 

Download LoxodromicCircles.mw

You can use dsolve for this. See the help page ?dsolve,ICs (BCs are done in the same way). Do you want an analytical or numerical solution? Suggest you try this, and if there is a problem, upload your worksheet with the green up-arrow, and someone will help you.

@PaulNewton In Maple, use 10^(-14); the () are required. To see more of the Matrix use interface(rtablesize=30). 

@Earl The white rectangle needs to be behind the circles, which usually happens when it is first in the display sequence. Somehow it ends up on top if circle is changed to disk. (Using a list doesn't help). The workaround is just to remove the rectangle. As for (2,3), I'll need to think about that some more...

@jetboo Just use [1,1]. You can have [1,1,1,2,2] etc. See the help page ?D

@samiyare I think changing BCs or ICs won't help; My understading is that each PDE has to have only one variable such as T(x,t). Obviously this is quite limiting. To solve this you may have to implement your own scheme, such as method of lines or differential quadrature.

See here for one example. 

https://www.mapleprimes.com/questions/234558-Numerical-Schemes-For-Second-Order-Mixed#answer287994

@panke I thought you wanted to remove the assumption, but maybe you wanted to remove the tilde, which @Rouben Rostamian has told you how to do. I do not understand your point above.

@mmcdara V=~0 is not a solution because of 0/0, but perhaps it is a solution to the original problem before it was manipulated into this form. I agree there is something strange about the system; perhaps for other values of the OP's free parameters it has clearer solutions.

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