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Re: How to plot divergence of gradient as contour
*To*: mathgroup at smc.vnet.net
*Subject*: [mg124067] Re: How to plot divergence of gradient as contour
*From*: Murray Eisenberg <murray at math.umass.edu>
*Date*: Sat, 7 Jan 2012 05:25:42 -0500 (EST)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*References*: <201201051100.GAA14785@smc.vnet.net> <201201060919.EAA26925@smc.vnet.net>
*Reply-to*: murray at math.umass.edu
Correction... a standard modern multivariable calculus text would show,
essentially:
phi[x_, y_] := Sin[x + y]
Div[Grad[phi]][x, y]
On 1/6/12 4:19 AM, Murray Eisenberg wrote:
> You need to use the VectorAnalysis package (carefully!), remembering that:
>
> (i) the default coordinate system is Cartesian[Xx, Yy, Zz]; and
> (ii) the result of using that package's Grad, Div, Curl is _not_
> really a "function" of the variables (more about this later).
>
> In your example:
>
>
> Div[Grad[Sin[Xx + Yy]]]
> -2 Sin[Xx + Yy]
>
>
> ContourPlot[
> Div[Grad[Sin[Xx + Yy]]] /. {Xx -> x, Yy -> y},
> {x, -5, 5}, {y, -5, 5}]
>
> (* or first define a function *)
>
> f[x_, y_] := Div[Grad[Sin[Xx + Yy]]] /. {Xx -> x, Yy -> y}
> ContourPlot[f[x, y], {x, -5, 5}, {y, -5, 5}]
>
> I find the whole paradigm of the VectorAnalysis package to be
> aggravatingly annoying. After all, from a mathematical perspective, if
> you're given a scalar-valued function with vector inputs, then its
> gradient is just a well-defined vector field -- and that has nothing
> whatsoever to do with so-called "coordinate systems". That is, one ought
> to be able to do something like this -- just like any standard modern
> text on multivariable calculus would show:
>
> phi[x_, y_] := Sin[x + y]
> Div[Grad[phi[x, y]]]
>
> In Mathematica, with the VectorAnalysis defaults, you'd get 0 as result
> there, since you didn't override the default variable names Xx, Yy, Zz.
>
> Then there's the whole nonsense of the gradient other coordinate
> systems. That, to my mind, is a very odd way to think about things.
> Mathematically, given a scalar field, it has an associated gradient
> vector field. Period. If you want to "express the gradient in terms of"
> another coordinate system, you're no longer finding that gradient;
> you're finding a composite of functions where you go first from that
> other coordinate system to cartesian coordinates, next take the
> gradient, and finally go back from cartesian coordinates in the result
> to the other coordinate system.
>
> (I trust others will disagree with the preceding argument!)
>
>
> On 1/5/12 6:00 AM, Szymon Roziewski wrote:
>> Hello there,
>>
>> I am concerning to plot something like that.
>>
>> ContourPlot[Divergence[Gradient[Sin[x + y]]], {x, -5, 5}, {y, -5, 5}]
>>
>> Plotting divergence of gradient scalar field.
>> How can I manage to do that?
>
--
Murray Eisenberg murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
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