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The '''Sobolev conjugate''' of ''p'' for <math>1\leq p <n</math>, where ''n'' is space dimensionality, is
:<math> p^*=\frac{pn}{n-p}>p</math>
This is an important parameter in the [[Sobolev inequality|Sobolev inequalities]].
== Motivation ==
A question arises whether ''u'' from the [[Sobolev space]] <math>W^{1,p}(R^n)</math> belongs to <math>L^q(R^n)</math> for some ''q''>''p''. More specifically, when does <math>\|Du\|_{L^p(R^n)}</math> control <math>\|u\|_{L^q(R^n)}</math>? It is easy to check that
the following inequality
:<math>\|u\|_{L^q(R^n)}\leq C(p,q)\|Du\|_{L^p(R^n)}</math>    (*)
can not be true for arbitrary ''q''. Consider <math>u(x)\in C^\infty_c(R^n)</math>, infinitely differentiable function with compact support. Introduce <math>u_\lambda(x):=u(\lambda x)</math>. We have that
:<math>\|u_\lambda\|_{L^q(R^n)}^q=\int_{R^n}|u(\lambda x)|^qdx=\frac{1}{\lambda^n}\int_{R^n}|u(y)|^qdy=\lambda^{-n}\|u\|_{L^q(R^n)}^q</math>
:<math>\|Du_\lambda\|_{L^p(R^n)}^p=\int_{R^n}|\lambda Du(\lambda x)|^pdx=\frac{\lambda^p}{\lambda^n}\int_{R^n}|Du(y)|^pdy=\lambda^{p-n}\|Du\|_{L^p(R^n)}^p</math>
The inequality (*) for <math>u_\lambda</math> results in the following inequality for <math>u</math>
:<math>\|u\|_{L^q(R^n)}\leq \lambda^{1-n/p+n/q}C(p,q)\|Du\|_{L^p(R^n)}</math>
If <math>1-n/p+n/q\not = 0</math>, then by letting <math>\lambda</math> going to zero or infinity we obtain a contradiction. Thus the inequality (*) could only be true for
:<math>q=\frac{pn}{n-p}</math>,
which is the Sobolev conjugate.
 
==See also==
*[[Sergei Lvovich Sobolev]]
 
==References==
* Lawrence C. Evans. Partial differential equations. Graduate studies in Mathematics, Vol 19. American Mathematical Society. 1998.  ISBN 0-8218-0772-2
 
[[Category:Sobolev spaces]]

Revision as of 20:53, 28 February 2013

The Sobolev conjugate of p for 1p<n, where n is space dimensionality, is

p*=pnnp>p

This is an important parameter in the Sobolev inequalities.

Motivation

A question arises whether u from the Sobolev space W1,p(Rn) belongs to Lq(Rn) for some q>p. More specifically, when does DuLp(Rn) control uLq(Rn)? It is easy to check that the following inequality

uLq(Rn)C(p,q)DuLp(Rn) (*)

can not be true for arbitrary q. Consider u(x)Cc(Rn), infinitely differentiable function with compact support. Introduce uλ(x):=u(λx). We have that

uλLq(Rn)q=Rn|u(λx)|qdx=1λnRn|u(y)|qdy=λnuLq(Rn)q
DuλLp(Rn)p=Rn|λDu(λx)|pdx=λpλnRn|Du(y)|pdy=λpnDuLp(Rn)p

The inequality (*) for uλ results in the following inequality for u

uLq(Rn)λ1n/p+n/qC(p,q)DuLp(Rn)

If 1n/p+n/q=0, then by letting λ going to zero or infinity we obtain a contradiction. Thus the inequality (*) could only be true for

q=pnnp,

which is the Sobolev conjugate.

See also

References

  • Lawrence C. Evans. Partial differential equations. Graduate studies in Mathematics, Vol 19. American Mathematical Society. 1998. ISBN 0-8218-0772-2