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Laplace of the differential of another variable in sage

asked 2012-08-20 22:47:19 -0500

anonymous user

Anonymous

updated 2012-08-21 02:32:02 -0500

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 2012-08-21 01:05:04 -0500

calc314 gravatar image

updated 2012-08-21 01:05:41 -0500

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 ( 2012-08-21 02:19:40 -0500 )edit

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 ( 2012-08-21 16:04:03 -0500 )edit

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

calc314 gravatar imagecalc314 ( 2012-08-22 09:32:57 -0500 )edit

All right:)

yangzb gravatar imageyangzb ( 2012-08-22 16:38:49 -0500 )edit

@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 ( 2012-08-24 09:39:18 -0500 )edit

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Asked: 2012-08-20 22:47:19 -0500

Seen: 131 times

Last updated: Sep 10 '12