Higher regularity for solutions of elliptic equations
$begingroup$
Let $Omega$ be a bounded domain in $mathbb{R}^d$. Let $fin L^infty(Omega)$. For the problem
$$-Delta u=fmbox{ in }Omega\
~~~~~~~~~u=0mbox{ on }partialOmega,$$
one could seek solutions in two ways. One is through Lax-Milgram Lemma, since $L^2(Omega)subset L^infty(Omega)$, $fin L^2(Omega)$ and hence a solution $u$ exists to the problem in $H^1_0(Omega)$. However, $fin L^p(Omega)$ for all $pin(1,infty)$, and hence one could mimic the monotone methods which are used to prove existence of solutions to monotone nonlinear problems to prove that the solution also exists in $W^{1,p}_0(Omega)$ for all $pin(1,infty)$. My question is:
By uniqueness, all these solutions are the same functions measure theoretically, hence does this not imply a higher regularity result for the solution of this equation? In fact, since the solution is in all the $W^{1,p}_0$, would this not imply that the solution is in fact Holder continuous?
It feels to me that there is something wrong with my argument, but I am not able to figure it out.
regularity-theory-of-pdes elliptic-equations
asked Nov 10 '18 at 23:30
Tanuj DipshikhaTanuj Dipshikha
197210
$endgroup$
add a comment |
$begingroup$
Let $Omega$ be a bounded domain in $mathbb{R}^d$. Let $fin L^infty(Omega)$. For the problem
$$-Delta u=fmbox{ in }Omega\
~~~~~~~~~u=0mbox{ on }partialOmega,$$
one could seek solutions in two ways. One is through Lax-Milgram Lemma, since $L^2(Omega)subset L^infty(Omega)$, $fin L^2(Omega)$ and hence a solution $u$ exists to the problem in $H^1_0(Omega)$. However, $fin L^p(Omega)$ for all $pin(1,infty)$, and hence one could mimic the monotone methods which are used to prove existence of solutions to monotone nonlinear problems to prove that the solution also exists in $W^{1,p}_0(Omega)$ for all $pin(1,infty)$. My question is:
By uniqueness, all these solutions are the same functions measure theoretically, hence does this not imply a higher regularity result for the solution of this equation? In fact, since the solution is in all the $W^{1,p}_0$, would this not imply that the solution is in fact Holder continuous?
It feels to me that there is something wrong with my argument, but I am not able to figure it out.
regularity-theory-of-pdes elliptic-equations
asked Nov 10 '18 at 23:30
Tanuj DipshikhaTanuj Dipshikha
197210
$endgroup$
add a comment |
$begingroup$
Let $Omega$ be a bounded domain in $mathbb{R}^d$. Let $fin L^infty(Omega)$. For the problem
$$-Delta u=fmbox{ in }Omega\
~~~~~~~~~u=0mbox{ on }partialOmega,$$
one could seek solutions in two ways. One is through Lax-Milgram Lemma, since $L^2(Omega)subset L^infty(Omega)$, $fin L^2(Omega)$ and hence a solution $u$ exists to the problem in $H^1_0(Omega)$. However, $fin L^p(Omega)$ for all $pin(1,infty)$, and hence one could mimic the monotone methods which are used to prove existence of solutions to monotone nonlinear problems to prove that the solution also exists in $W^{1,p}_0(Omega)$ for all $pin(1,infty)$. My question is:
By uniqueness, all these solutions are the same functions measure theoretically, hence does this not imply a higher regularity result for the solution of this equation? In fact, since the solution is in all the $W^{1,p}_0$, would this not imply that the solution is in fact Holder continuous?
It feels to me that there is something wrong with my argument, but I am not able to figure it out.
regularity-theory-of-pdes elliptic-equations
asked Nov 10 '18 at 23:30
Tanuj DipshikhaTanuj Dipshikha
197210
$endgroup$
Let $Omega$ be a bounded domain in $mathbb{R}^d$. Let $fin L^infty(Omega)$. For the problem
$$-Delta u=fmbox{ in }Omega\
~~~~~~~~~u=0mbox{ on }partialOmega,$$
one could seek solutions in two ways. One is through Lax-Milgram Lemma, since $L^2(Omega)subset L^infty(Omega)$, $fin L^2(Omega)$ and hence a solution $u$ exists to the problem in $H^1_0(Omega)$. However, $fin L^p(Omega)$ for all $pin(1,infty)$, and hence one could mimic the monotone methods which are used to prove existence of solutions to monotone nonlinear problems to prove that the solution also exists in $W^{1,p}_0(Omega)$ for all $pin(1,infty)$. My question is:
By uniqueness, all these solutions are the same functions measure theoretically, hence does this not imply a higher regularity result for the solution of this equation? In fact, since the solution is in all the $W^{1,p}_0$, would this not imply that the solution is in fact Holder continuous?
It feels to me that there is something wrong with my argument, but I am not able to figure it out.
regularity-theory-of-pdes elliptic-equations
regularity-theory-of-pdes elliptic-equations
asked Nov 10 '18 at 23:30
Tanuj DipshikhaTanuj Dipshikha
197210
asked Nov 10 '18 at 23:30
Tanuj DipshikhaTanuj Dipshikha
197210
edited Nov 10 '18 at 23:37
Tanuj Dipshikha
asked Nov 10 '18 at 23:30
Tanuj DipshikhaTanuj Dipshikha
197210
asked Nov 10 '18 at 23:30
Tanuj DipshikhaTanuj Dipshikha
197210
asked Nov 10 '18 at 23:30
Tanuj DipshikhaTanuj Dipshikha
197210
197210
add a comment |
add a comment |
1 Answer
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$begingroup$
Your argument is correct, assuming you can show existence in $W^{1,p}_0(Omega).$ They key point is that you have uniqueness of solutions in $H^1_0(Omega),$ so higher regularity solutions automatically coincide with the $H^1_0(Omega)$ solution you obtain from Lax-Milgram. The idea is the same as when you prove elliptic regularity (the case $f in C^{infty}$); you start with a weak solution but end up proving it's actually smooth.
The fact that $u$ is Hölder continuous doesn't contradict anything; the point is that it's second derivative is not Hölder continuous in general, so this is consistent with the fact that $Delta u in L^{infty}.$
In fact, using more sophisticated techniques one can show that $u in W^{2,p}(Omega)$ for all $p < infty.$ In particular, its first derivatives are $alpha$-Hölder continous for all $alpha in (0,1)$ by Sobolev embedding.
answered Dec 4 '18 at 19:32
ktoiktoi
2,3861616
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your argument is correct, assuming you can show existence in $W^{1,p}_0(Omega).$ They key point is that you have uniqueness of solutions in $H^1_0(Omega),$ so higher regularity solutions automatically coincide with the $H^1_0(Omega)$ solution you obtain from Lax-Milgram. The idea is the same as when you prove elliptic regularity (the case $f in C^{infty}$); you start with a weak solution but end up proving it's actually smooth.
The fact that $u$ is Hölder continuous doesn't contradict anything; the point is that it's second derivative is not Hölder continuous in general, so this is consistent with the fact that $Delta u in L^{infty}.$
In fact, using more sophisticated techniques one can show that $u in W^{2,p}(Omega)$ for all $p < infty.$ In particular, its first derivatives are $alpha$-Hölder continous for all $alpha in (0,1)$ by Sobolev embedding.
answered Dec 4 '18 at 19:32
ktoiktoi
2,3861616
$endgroup$
add a comment |
$begingroup$
Your argument is correct, assuming you can show existence in $W^{1,p}_0(Omega).$ They key point is that you have uniqueness of solutions in $H^1_0(Omega),$ so higher regularity solutions automatically coincide with the $H^1_0(Omega)$ solution you obtain from Lax-Milgram. The idea is the same as when you prove elliptic regularity (the case $f in C^{infty}$); you start with a weak solution but end up proving it's actually smooth.
The fact that $u$ is Hölder continuous doesn't contradict anything; the point is that it's second derivative is not Hölder continuous in general, so this is consistent with the fact that $Delta u in L^{infty}.$
In fact, using more sophisticated techniques one can show that $u in W^{2,p}(Omega)$ for all $p < infty.$ In particular, its first derivatives are $alpha$-Hölder continous for all $alpha in (0,1)$ by Sobolev embedding.
answered Dec 4 '18 at 19:32
ktoiktoi
2,3861616
$endgroup$
add a comment |
$begingroup$
Your argument is correct, assuming you can show existence in $W^{1,p}_0(Omega).$ They key point is that you have uniqueness of solutions in $H^1_0(Omega),$ so higher regularity solutions automatically coincide with the $H^1_0(Omega)$ solution you obtain from Lax-Milgram. The idea is the same as when you prove elliptic regularity (the case $f in C^{infty}$); you start with a weak solution but end up proving it's actually smooth.
The fact that $u$ is Hölder continuous doesn't contradict anything; the point is that it's second derivative is not Hölder continuous in general, so this is consistent with the fact that $Delta u in L^{infty}.$
In fact, using more sophisticated techniques one can show that $u in W^{2,p}(Omega)$ for all $p < infty.$ In particular, its first derivatives are $alpha$-Hölder continous for all $alpha in (0,1)$ by Sobolev embedding.
answered Dec 4 '18 at 19:32
ktoiktoi
2,3861616
$endgroup$
Your argument is correct, assuming you can show existence in $W^{1,p}_0(Omega).$ They key point is that you have uniqueness of solutions in $H^1_0(Omega),$ so higher regularity solutions automatically coincide with the $H^1_0(Omega)$ solution you obtain from Lax-Milgram. The idea is the same as when you prove elliptic regularity (the case $f in C^{infty}$); you start with a weak solution but end up proving it's actually smooth.
The fact that $u$ is Hölder continuous doesn't contradict anything; the point is that it's second derivative is not Hölder continuous in general, so this is consistent with the fact that $Delta u in L^{infty}.$
In fact, using more sophisticated techniques one can show that $u in W^{2,p}(Omega)$ for all $p < infty.$ In particular, its first derivatives are $alpha$-Hölder continous for all $alpha in (0,1)$ by Sobolev embedding.
answered Dec 4 '18 at 19:32
ktoiktoi
2,3861616
answered Dec 4 '18 at 19:32
ktoiktoi
2,3861616
answered Dec 4 '18 at 19:32
ktoiktoi
2,3861616
answered Dec 4 '18 at 19:32
ktoiktoi
2,3861616
2,3861616
add a comment |
add a comment |
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