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Laplace of the differential of another variable in sage

asked 12 years ago

anonymous user

Anonymous

updated 12 years ago

Why did the diff(y.laplace(t,s),x) come out to be D[0](laplace)(y(x, t), t, s)*D[0](y)(x, t)? Could someone help me?

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answered 12 years ago

calc314 gravatar image

updated 12 years ago

As best I can tell, it looks like Sage is doing the chain rule here. So, it is treating laplace(y,t,s) as a symbol and gives the derivative w.r.t. x as D[0](laplace(y,t,s))*D[0](y(x,t)). Taking the second derivative gives something that looks like the product rule and the chain rule were used.

I think Sage is using Maxima to do this. So, this might be an issue with how Maxima handles partial derivatives of Laplace transforms.

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Yeah, it can account for such a result. I think maybe there is something in maxima to be improved. Thank you!

yangzb gravatar imageyangzb ( 12 years ago )

But, as I run in Maxima:laplace(diff(f(x,t),x),t,s),it seems to output the correct answer:'diff('laplace(f(x,t),t,s),x,1). Thus there might be something wrong in sage.

yangzb gravatar imageyangzb ( 12 years ago )

Good catch. I'll work on submitting a ticket on this.

calc314 gravatar imagecalc314 ( 12 years ago )

All right:)

yangzb gravatar imageyangzb ( 12 years ago )

@calc314 - Please cc: me if you do get the ticket open. The formal derivative stuff still needs some better integration with Maxima - no pun intended.

kcrisman gravatar imagekcrisman ( 12 years ago )

Sorry, I did not ask for permission from my professor before posting it. He told me the code still needs some work. He will contribute it to Sage as a contributor himself when it works reasonably well.

yangzb gravatar imageyangzb ( 12 years ago )

That would be awesome. Please keep us informed, and if he does post some code, put the ticket number here as well for other people who find this question.

kcrisman gravatar imagekcrisman ( 12 years ago )

All right

yangzb gravatar imageyangzb ( 12 years ago )

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Asked: 12 years ago

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Last updated: Sep 10 '12