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I'm giving this old problem a shot with the pdesolve built-in function. Yes, this problem has an analytical solution, and most likely, it can be solved using dimensional arrays and finite differences too. I might be missing something obvious as to how I have set it up. Please, take a look. Any feedback is greatly appreciated. Thanks,
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Gus Larsen wrote:
I'm giving this old problem a shot with the pdesolve built-in function. Yes, this problem has an analytical solution, and most likely, it can be solved using dimensional arrays and finite differences too. I might be missing something obvious as to how I have set it up. Please, take a look. Any feedback is greatly appreciated. Thanks,
You're covering the range Y = 0..1, but the second expression in the parentheses contains 1/Y, hence the error. I'm not sure why it doesn't throw an error at the other end of Y range, as there is a 1/(1-Y^2), which should also give the divide by zero error.
Stuart
Gus Larsen wrote:
I'm giving this old problem a shot with the pdesolve built-in function. Yes, this problem has an analytical solution, and most likely, it can be solved using dimensional arrays and finite differences too. I might be missing something obvious as to how I have set it up. Please, take a look. Any feedback is greatly appreciated. Thanks,
You're covering the range Y = 0..1, but the second expression in the parentheses contains 1/Y, hence the error. I'm not sure why it doesn't throw an error at the other end of Y range, as there is a 1/(1-Y^2), which should also give the divide by zero error.
Stuart
Excellent! Thanks, Stuart. I assumed that somehow the software was going to be able to handle the "0/0" at Y=0 by some numerical scheme of the L'Hopital's rule (as Sy also vanishes for the second term). I will keep these limitations in mind. Best, Gus