yangzb's profile - activity

All right

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.

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.

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.

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All right:)

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.

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.

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

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?

Here is what I get.

var('y t s k')
y=function('y',x,t)
de=diff(y,t,2)-diff(y,x,2)==k*sin(pi*x)
de.laplace(t,s)

gives

s^2*laplace(y(x, t), t, s) - D[0](y)(x, t)^2*D[0, 0](laplace)(y(x, t), t, s) - s*y(x, 0) - D[0](laplace)(y(x, t), t, s)*D[0, 0](y)(x, t) - D[1](y)(x, 0) == k*sin(pi*x)/s

Within this expression, the meanings are:

$\frac{\partial y}{\partial t}$ = D[1](y)(x, t)

$\frac{\partial y}{\partial x}$ = D[0](y)(x, t)

$\frac{\partial^2 y}{\partial t^2}$ = D[1, 1](y)(x, t)

$\frac{\partial^2 y}{\partial x^2}$ = D[0, 0](y)(x, t)

So, D[0](y)(0, x) is $\frac{\partial y}{\partial t}|_{t=0}$.

You can check the first one above by doing diff(y,t).

What you're seeing is the displayed notation for symbolic partial derivatives. D[0] means "apply the derivative operator with respect to variable number 0" and D[0,0] represents two successive derivatives with respect to the variable. Observe:

sage: f = function('f', x)
sage: f.diff(x)
D[0](f)(x)
sage: f.diff(x).diff(x)
D[0, 0](f)(x)

Excuse me, but I have a small question about the code that is :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 you help me?

I got it. Thanks!

I think they are different between -D[0](Laplace)(y(t,x),t,s)*D[1,1](y)(t,x)-D[0,0](laplace)(y(t,x),t,s)*D[1](y)(t,x)^2 and -D[1,1]L(y(t,x),t,s).

But,shouldn't it be :s^2*L(y(t,x),t,s)-s*y(0,x)-D[0](y)(0,x)-D[1,1]L(y(t,x),t,s). on the left side?

In sage, I performed the laplace transformation:

de=diff(y,t,2)-diff(y,x,2)==k*sin(pi*x);de.laplace(t,s)

and got such a code:

s^2L(y(x,t),t,s)-D[0](y)(x,t)2D[0,0]L(y(x,t),t,s)?sy(x,0)?D[0]L(y(x,t),t,s)D[0,0](y)(x,t)?D[1](y)(x,0)=sksin(?x)

Can someone tells me what D[0](y)(x,t)2D[0,0]L(y(x,t),t,s) and D[0]L(y(x,t),t,s)D[0,0](y)(x,t) mean?
Thanks!