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Elliptic regularity for $-nabla cdot(a(x)nabla u)$

1

1

$begingroup$

Define $Du= -nabla cdot (a(x)nabla u)$ where $a$ is a smooth function which is away from $0$ and bounded.

Suppose I have $u in H^1(Omega)$ on a smooth bounded domain and I also know that $Du in L^2(Omega)$.

Does this imply that $Delta u in L^2(Omega)$ and $u in H^2(Omega)$?

For the Laplacian case: we get immediately that the Laplacian is in $L^2$ and then elliptic regularity theory gives $H^2$ for $u$. But this case, I don't know????

functional-analysis pde regularity-theory-of-pdes

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asked Apr 2 at 14:52

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1

1

$begingroup$

Define $Du= -nabla cdot (a(x)nabla u)$ where $a$ is a smooth function which is away from $0$ and bounded.

Suppose I have $u in H^1(Omega)$ on a smooth bounded domain and I also know that $Du in L^2(Omega)$.

Does this imply that $Delta u in L^2(Omega)$ and $u in H^2(Omega)$?

For the Laplacian case: we get immediately that the Laplacian is in $L^2$ and then elliptic regularity theory gives $H^2$ for $u$. But this case, I don't know????

functional-analysis pde regularity-theory-of-pdes

share|cite|improve this question

asked Apr 2 at 14:52

StopUsingFacebookStopUsingFacebook

31718

$endgroup$

add a comment | 

$begingroup$

Define $Du= -nabla cdot (a(x)nabla u)$ where $a$ is a smooth function which is away from $0$ and bounded.

Suppose I have $u in H^1(Omega)$ on a smooth bounded domain and I also know that $Du in L^2(Omega)$.

Does this imply that $Delta u in L^2(Omega)$ and $u in H^2(Omega)$?

For the Laplacian case: we get immediately that the Laplacian is in $L^2$ and then elliptic regularity theory gives $H^2$ for $u$. But this case, I don't know????

functional-analysis pde regularity-theory-of-pdes

share|cite|improve this question

asked Apr 2 at 14:52

StopUsingFacebookStopUsingFacebook

31718

$endgroup$

share|cite|improve this question

asked Apr 2 at 14:52

StopUsingFacebookStopUsingFacebook

31718

share|cite|improve this question

asked Apr 2 at 14:52

StopUsingFacebookStopUsingFacebook

31718

asked Apr 2 at 14:52

StopUsingFacebookStopUsingFacebook

31718

asked Apr 2 at 14:52

StopUsingFacebookStopUsingFacebook

31718

1 Answer
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$begingroup$

If you expand the divergence in $Du$, you get
$$- nabla a cdot nabla u - a Delta u = Du$$
Rearranging yields
$$Delta u(x) = - frac1a(x) nabla a(x) cdot nabla u(x) - frac1a(x) Du(x)$$
Since $a$ is smooth and bounded away from $0$, we conclude using Holder's inequality that both terms on the right are in $L^2$. In particular, their norms in $L^2$ are controlled by
$$leftlVert frac1a(x) nabla a(x) cdot nabla u(x) rightrVert_L^2 leq leftlVert frac1a rightrVert_L^infty lVert nabla a rVert_L^infty lVert u rVert_H^1,$$
$$leftlVert frac1a(x) Du(x) rightrVert_L^2 leq leftlVert frac1a rightrVert_L^infty lVert Du rVert_L^2$$
Thus, $Delta u in L^2(Omega)$, and elliptic regularity puts $u in H^2$.

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answered Apr 3 at 15:57

StrantsStrants

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0

$begingroup$

If you expand the divergence in $Du$, you get
$$- nabla a cdot nabla u - a Delta u = Du$$
Rearranging yields
$$Delta u(x) = - frac1a(x) nabla a(x) cdot nabla u(x) - frac1a(x) Du(x)$$
Since $a$ is smooth and bounded away from $0$, we conclude using Holder's inequality that both terms on the right are in $L^2$. In particular, their norms in $L^2$ are controlled by
$$leftlVert frac1a(x) nabla a(x) cdot nabla u(x) rightrVert_L^2 leq leftlVert frac1a rightrVert_L^infty lVert nabla a rVert_L^infty lVert u rVert_H^1,$$
$$leftlVert frac1a(x) Du(x) rightrVert_L^2 leq leftlVert frac1a rightrVert_L^infty lVert Du rVert_L^2$$
Thus, $Delta u in L^2(Omega)$, and elliptic regularity puts $u in H^2$.

share|cite|improve this answer

answered Apr 3 at 15:57

StrantsStrants

5,89921736

$endgroup$

add a comment | 

0

$begingroup$

If you expand the divergence in $Du$, you get
$$- nabla a cdot nabla u - a Delta u = Du$$
Rearranging yields
$$Delta u(x) = - frac1a(x) nabla a(x) cdot nabla u(x) - frac1a(x) Du(x)$$
Since $a$ is smooth and bounded away from $0$, we conclude using Holder's inequality that both terms on the right are in $L^2$. In particular, their norms in $L^2$ are controlled by
$$leftlVert frac1a(x) nabla a(x) cdot nabla u(x) rightrVert_L^2 leq leftlVert frac1a rightrVert_L^infty lVert nabla a rVert_L^infty lVert u rVert_H^1,$$
$$leftlVert frac1a(x) Du(x) rightrVert_L^2 leq leftlVert frac1a rightrVert_L^infty lVert Du rVert_L^2$$
Thus, $Delta u in L^2(Omega)$, and elliptic regularity puts $u in H^2$.

share|cite|improve this answer

answered Apr 3 at 15:57

StrantsStrants

5,89921736

$endgroup$

add a comment | 

$begingroup$

If you expand the divergence in $Du$, you get
$$- nabla a cdot nabla u - a Delta u = Du$$
Rearranging yields
$$Delta u(x) = - frac1a(x) nabla a(x) cdot nabla u(x) - frac1a(x) Du(x)$$
Since $a$ is smooth and bounded away from $0$, we conclude using Holder's inequality that both terms on the right are in $L^2$. In particular, their norms in $L^2$ are controlled by
$$leftlVert frac1a(x) nabla a(x) cdot nabla u(x) rightrVert_L^2 leq leftlVert frac1a rightrVert_L^infty lVert nabla a rVert_L^infty lVert u rVert_H^1,$$
$$leftlVert frac1a(x) Du(x) rightrVert_L^2 leq leftlVert frac1a rightrVert_L^infty lVert Du rVert_L^2$$
Thus, $Delta u in L^2(Omega)$, and elliptic regularity puts $u in H^2$.

share|cite|improve this answer

answered Apr 3 at 15:57

StrantsStrants

5,89921736

$endgroup$

share|cite|improve this answer

answered Apr 3 at 15:57

StrantsStrants

5,89921736

answered Apr 3 at 15:57

StrantsStrants

5,89921736

answered Apr 3 at 15:57

StrantsStrants

5,89921736