dharr

Dr. David Harrington

6765 Reputation

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20 years, 101 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

@mmead Glad you liked it. I modified it to also handle indexed variables, where the unit is the same for any index. See the edit in my original answer.

@tomleslie I should have noticed that! Vote up.

Please specify exactly what you want to do, and what the question is.

Please upload your worksheet using the green up-arrow (insert contents isn't working but insert link is).

Please upload the worksheet by using the green up-arrow, and let us know what the question is.

@jrive You used eval(subs(s = I*omega, xfer_in_s)). Simpler is to evaluate at s = I*omega: eval(xfer_in_s, s = I*omega). 
eval does automatic simplification where subs does not, and is usually what you want.

@tomleslie I just used that because the OP said "abs (x1(j)-x1(j-1)) < 10^-4 ". I also didn't understand the iteration requirement in relationship to the code given.

@awass Yes looks like you have to use the original PDE and work out the derivatives numerically from the function. See here for a post about that.

 

The error message is because you now have a system of two pdes, so need two initial conditions. Since u is your time variable, and your only initial condition is f(0,x)=g; you also need h(0,x)=something. The other conditions are boundary conditions. I don't use the numeric pde solver enough to know how to solve your problem.

@ I think the constant one is varepsilon, so not the same.

@okokoabraham No I didn't solve it. Hope someone else can help you. [Edit: see below]

@matmxhu For transendental equations it will always be hard, which I guess is why allsolutions only says "more solutions". The tryhard option doesn't help here either. 

Edit: I guess for these types of second order odes, Sturm-Liouville theory tells you the number of nodes go up by one for each successive eigenfunction, so if you only find half the solutions, you can tell some are missing.

If you assume(n::integer) at the beginning (if that is what you want), then the B..H quantities are much simpler. Still your system is not solving quickly; it's perhaps too hard but I didn't wait that long.

@tomleslie Unfortunately, if that is a bug, then there are many others. In general, one needs to be an experienced user to use "type", "op", "has" and others that require knowing the underlying structure, since in many cases things are not what they seem. For example, from looking at the output of series(exp(x),x) and x/3 one would expect that type(series(exp(x),x),`+`) and has(x,3) would both return true, but they both return false.

 

@abdulganiy This is more a math question that a Maple question. I don't understand what you want to do.

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