       Re: How to evaluate the Laplacian of a function as a

• To: mathgroup at smc.vnet.net
• Subject: [mg114373] Re: How to evaluate the Laplacian of a function as a
• From: Patrick Scheibe <pscheibe at trm.uni-leipzig.de>
• Date: Thu, 2 Dec 2010 05:41:12 -0500 (EST)

```Hi,

to understand whats happening and why this works please check carefully
the difference between Set (=) and SetDelayed (:=)

<< VectorAnalysis`
func[x_, y_] := 3/Pi*(1 - x^2 + y^2)^2
funcLapl[Xx_, Yy_] = Laplacian[func[Xx, Yy]]

funcLapl[a, b]

Cheers
Patrick

On Wed, 2010-12-01 at 02:15 -0500, Iliyan Georgiev wrote:
> Hi,
>
> I have a stupid problem and cannot find a solution anywhere. I have
> the function
>
> func[x_,y_] := 3/Pi*(1 - x^2 + y^2))^2
>
> which I need the Laplacian of:
>
> funcLapl := Laplacian[func[Xx,Yy]]
>
> The Laplacian is computed correctly, but how can I treat the result as
> a function? I want to be able to evaluate/plot the Laplacian. I tried
> many things and I'm frustrated. My current solution is to manually
> copy the derived Laplacian and create a function from that. It's
> obviously not a good solution.
>
> Any suggestions?
>

```

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