Fahad D.

asked • 08/05/16

nabla^3 u=0

Is it true or not?

Steven W.

tutor
Hi Fahad:

I think more context is needed. I have not found a consistent definition for nabla, or del, cubed. It is related to the Laplacian operator?
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08/05/16

Arturo O.

Fahad,
 
What is nabla?
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08/05/16

Steven W.

tutor
Hi Arturo.  Nabla is the upside-down triangle operator often called "del," which symbolizes the gradient.  The Laplacian operator is nabla- or del-squared (the divergence of the gradient), that appears -- among other places -- in Laplace's equation.  But I am not sure what nabla- or del-cubed is meant to represent.  Perhaps the divergence of the divergence of the gradient?
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08/05/16

Arturo O.

Hello Steve.
 
Then to answer Fahad's question, I think we need to know if the argument u is a scalar or a vector.
 
If u is a scalar, the following triple-del operations are defined:
 
grad (del2 u) = vector
del2 (grad u) = vector
 
I do not know if del2 and grad commute, without doing a detailed derivation.  But I am inclined to think that in general, the triple-del operations do not produce the (0,0,0) vector.
 
If u is a vector, then the following triple-del operations are defined:
 
del2 (div u) = scalar
div (del2 u) = scalar
 
I do not know if div and del2 commute, without doing a detailed derivation.  But I am inclined to think that in general, the triple-del operations do not produce 0.
 
We need to know more about the argument u or u.  Maybe there is a vector subspace over which the triple operation gives zero.  What do you think?
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08/05/16

Steven W.

tutor
This is maybe where I am running up against my own knowledge limit, but what does it mean to take the gradient of a vector?  Gradient produces a vector, but I am not sure how its vector definition would work against the vector components of a vector it was operating on. 
 
So I am not sure (perhaps die to my own lack of knowledge) what del2(grad u) would mean, if u were a scalar, since grad u would produce a vector, and the first operation of the Laplacian is to take the gradient of that vector. 
 
And if div(del2 u) is defined as del3, for u as a vector, that also seems to require taking the gradient of a vector. 
 
I could not find any confirmation of those definitions.  So I would need to be sure I understand what del3 represents before I could go on. I was hoping some further parameters could help clarify that. 
 
Immediately, I agree that there appears to be no general case where this is true.  Even del-squared only equals zero under certain conditions. So, I agree that we need to know more about u and the problem. 
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08/05/16

Arturo O.

Steven,
 
Regarding the meaning of ∇2(∇u), where u is a scalar (and consequently ∇u is a vector) think of the example of the wave equation for the electric field:
 
∇2E - (1/c2) ∂2E/∂t2 = 0
 
It has the Laplacian of the vector E.  So the Laplacian of a vector is defined, as long as it is applied as (∇2)A, but not as ∇•(∇A) for some vector A.
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08/05/16

Steven W.

tutor
In fact, the gradient of a vector makes little sense to me, since it determines the directional rate of change of a function.  But a vector defines its own rate of change in the cardinal directions, so I am not sure what a gradient would do to it.
 
But I can definitely get behind:
 
grad (del2 u) = vector  (for u as a scalar)
del2 (div u) = scalar (for u as a vector)
 
But, then again, I would have thought del2 meant applying the gradient twice, so sometimes the definitions are not what I expect, which is why I was hunting for conventional definition of del3 (without success)
 
But even the definitions I can get behind require knowing whether u is scalar or vector.  Hmmm...
 
 
 
 
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08/05/16

Steven W.

tutor
Shoot, that's what I was worried about.  It's a friggin' dyadic tensor product when you take the gradient of a vector.  Ugh...  this is why I became an experimentalist. :D
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08/05/16

Steven W.

tutor
But, with that in mind, I definitely agree there is no general case where del3 operator produces zero.  Even with the electromagnetic wave case, which you very astutely referred to, there is a requirement that space be free of electric potential and current sources to produce some of the zero results it uses.
 
Thanks for reminding me about the EM wave case, though.  The curl of the curl operation is where that comes from.
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08/05/16

Arturo O.

A lot of food for thought (that dyadic tensor product)!  I suppose its components are things like ∂2/∂x2, ∂2/∂x∂y, ∂2/∂y∂z, ..., etc.  Anyway, I guess we cannot answer Fahad's question without more context, as you mentioned previously.
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08/05/16

1 Expert Answer

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Steven W. answered • 08/05/16

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Physics Ph.D., college instructor (calc- and algebra-based)
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Hi Fahad.  I think the conclusion Arturo and I came to, after our exchange above, is that the statement is probably not generally true, but it may depend on whether u is a scalar (0-rank tensor), vector (1st-rank tensor), or a tensor of some other, higher rank.  Do you have any more information about this problem?
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