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gradient of divergence

Posted 20 avr. 2014, 08:55 UTC−4 Parameters, Variables, & Functions Version 4.4 2 Replies

Tianyu Dong
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Hi all,

Is it possible to represent the gradient of the divergence of a vector field in the Coefficient Form PDE other than using the Weak Form?

Thanks,
Tony
2 Replies Last Post 21 avr. 2014, 06:21 UTC−4
Pu Zhang
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Posted: 1 decade ago 20 avr. 2014, 10:16 UTC−4
Definitely you can. But I think you need to somehow manipulate the form of the expression in order to do that.
Definitely you can. But I think you need to somehow manipulate the form of the expression in order to do that.
Tianyu Dong
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Posted: 1 decade ago 21 avr. 2014, 06:21 UTC−4
Thanks Pu.
We have:
grad div F = [F1xx+F2yx, F1xy+F2yy]; (1)
div(c*grad F) = [c11*(F1xx+F1yy)+c12*(F2xx+F2yy), c21*(F1xx+F1yy)+c22*(F2xx+F2yy)] (2)
where c=[c11, c12; c21, c22], hence, it seems to me that mixed partial derivative terms in (1) never appear no matter how to choose the coefficient c if we use the Coefficient Form PDE to represent the gradient of divergence. Am I correct?

Thanks,
Tony
Thanks Pu. We have: grad div F = [F1xx+F2yx, F1xy+F2yy]; (1) div(c*grad F) = [c11*(F1xx+F1yy)+c12*(F2xx+F2yy), c21*(F1xx+F1yy)+c22*(F2xx+F2yy)] (2) where c=[c11, c12; c21, c22], hence, it seems to me that mixed partial derivative terms in (1) never appear no matter how to choose the coefficient c if we use the Coefficient Form PDE to represent the gradient of divergence. Am I correct? Thanks, Tony

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