Serwa

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3 years, 307 days

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These are replies submitted by Serwa

@vv 

Thanks. I was thinking about something similiar when I said that I want to find the final solution by having solutions for each equation.

Intersection of S1 and S2 is a null set here. If we consider a S3 set, with all intersection elements and all combination of variables in S1 and S2, then what you said works. But it is not true for other examples. Like:

S1:= (a+A1, a+A2, B1, B2),  S2:=( A1, A2, a+B1, a+B2). The final answer must be:

(-A1+A2, a+A1+B1,a+A1+B2).

@rlopez 

Thanks. I don't want a union of sets. For example, if there are two PDEs:

diff(f(theta[0], theta[A], theta[B], theta[AB]), theta[AB]) = 0

-(diff(f(theta[0], theta[A], theta[B], theta[AB]), theta[0]))+diff(f(theta[0], theta[A], theta[B], theta[AB]), theta[A])+diff(f(theta[0], theta[A], theta[B], theta[AB]), theta[B]) = 0

Then the answer for the first one is f(theta[0], theta[A], theta[B], theta[AB]) = _F1(theta[0], theta[A], theta[B]), and for the second one is f(theta[0], theta[A], theta[B], theta[AB]) = _F1(theta[0]+theta[A], theta[0]+theta[B], theta[AB]). But I need the following result, which is not a union of those two and we only get it if PDEs are solved at the same time:

{f(theta[0], theta[A], theta[B], theta[AB]) = _F1(theta[0]+theta[A], theta[0]+theta[B])}.

....... and now this makes me think it is not possible to get the final answer by having answers for each individual PDE. I guess the only way is to solve them all simultaneously (which is not working for a large number of equations).

@Kitonum 

Thank you so much. 

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