Mixing
Harnack Inequality for nonnegative subsolutions to uniformly elliptic PDE
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I am trying to prove a Harnack inequality for a nonnegative subsolution $u in H^1(B_2)$ to the PDE $text{div}(A Du) ge 0$,where $A = A(x)$ is uniformly elliptic. The proof outline I am following is from a set of notes by a professor at my university, and the key step is the following inductive scheme:
Set $x_0$ to be a point such that $$u(x_0) = sup_{B_{(0,1/2)}} u,$$ and choose $x_k$ inductively such that $x_{k+1}$ is such that $$u(x_{k+1}) = sup_{B(x_k, r_k)} u$$
for $r_k$ sufficiently small to be chosen in a moment.
I have all of the steps except the following: suppose $$frac{text{sup}_{B_{0,1/4}} u}{ u(0)}$$ is sufficiently large, then we can choose a sequence $r_k$ such that $sum r_k <1/2$ and a $beta>1$ such that $u(x_{k+1}) ge beta u(x_k)$. That this would imply the result is immediate because it would contradict the boundedness of $u$. The preceding step, which I am led to believe is what implies the claim, is the following:
$$u(x_{k+1}) ge frac{u(x_k) - cr_k^{-q} u(0)}{1-theta}$$
where $c$, $q$ are absolute constants, and $1-theta ge text{osc}_{B_1}u>0$ and $0<theta le inf_{B_1} u$. Here $c,q>0$ are absolute constants.
I basically don't know what to do with this. Even if I assume the ratio in question gets very large, the estimate (from the prior step) becomes useless as $r_k to 0$. So it's unclear to me how to use it infinitely many times. I have the Nash-Digiorgi Holder regularity theorem at my disposal. Any hints or references would be much appreciated! I cannot find a similar proof anywhere, and given that I have provided the details for all of the other (numerous) steps, I would like to complete it.
real-analysis functional-analysis analysis pde elliptic-equations
asked Jan 24 at 6:59
David BowmanDavid Bowman
4,3161924
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add a comment |
$begingroup$
I am trying to prove a Harnack inequality for a nonnegative subsolution $u in H^1(B_2)$ to the PDE $text{div}(A Du) ge 0$,where $A = A(x)$ is uniformly elliptic. The proof outline I am following is from a set of notes by a professor at my university, and the key step is the following inductive scheme:
Set $x_0$ to be a point such that $$u(x_0) = sup_{B_{(0,1/2)}} u,$$ and choose $x_k$ inductively such that $x_{k+1}$ is such that $$u(x_{k+1}) = sup_{B(x_k, r_k)} u$$
for $r_k$ sufficiently small to be chosen in a moment.
I have all of the steps except the following: suppose $$frac{text{sup}_{B_{0,1/4}} u}{ u(0)}$$ is sufficiently large, then we can choose a sequence $r_k$ such that $sum r_k <1/2$ and a $beta>1$ such that $u(x_{k+1}) ge beta u(x_k)$. That this would imply the result is immediate because it would contradict the boundedness of $u$. The preceding step, which I am led to believe is what implies the claim, is the following:
$$u(x_{k+1}) ge frac{u(x_k) - cr_k^{-q} u(0)}{1-theta}$$
where $c$, $q$ are absolute constants, and $1-theta ge text{osc}_{B_1}u>0$ and $0<theta le inf_{B_1} u$. Here $c,q>0$ are absolute constants.
I basically don't know what to do with this. Even if I assume the ratio in question gets very large, the estimate (from the prior step) becomes useless as $r_k to 0$. So it's unclear to me how to use it infinitely many times. I have the Nash-Digiorgi Holder regularity theorem at my disposal. Any hints or references would be much appreciated! I cannot find a similar proof anywhere, and given that I have provided the details for all of the other (numerous) steps, I would like to complete it.
real-analysis functional-analysis analysis pde elliptic-equations
asked Jan 24 at 6:59
David BowmanDavid Bowman
4,3161924
$endgroup$
add a comment |
$begingroup$
I am trying to prove a Harnack inequality for a nonnegative subsolution $u in H^1(B_2)$ to the PDE $text{div}(A Du) ge 0$,where $A = A(x)$ is uniformly elliptic. The proof outline I am following is from a set of notes by a professor at my university, and the key step is the following inductive scheme:
Set $x_0$ to be a point such that $$u(x_0) = sup_{B_{(0,1/2)}} u,$$ and choose $x_k$ inductively such that $x_{k+1}$ is such that $$u(x_{k+1}) = sup_{B(x_k, r_k)} u$$
for $r_k$ sufficiently small to be chosen in a moment.
I have all of the steps except the following: suppose $$frac{text{sup}_{B_{0,1/4}} u}{ u(0)}$$ is sufficiently large, then we can choose a sequence $r_k$ such that $sum r_k <1/2$ and a $beta>1$ such that $u(x_{k+1}) ge beta u(x_k)$. That this would imply the result is immediate because it would contradict the boundedness of $u$. The preceding step, which I am led to believe is what implies the claim, is the following:
$$u(x_{k+1}) ge frac{u(x_k) - cr_k^{-q} u(0)}{1-theta}$$
where $c$, $q$ are absolute constants, and $1-theta ge text{osc}_{B_1}u>0$ and $0<theta le inf_{B_1} u$. Here $c,q>0$ are absolute constants.
I basically don't know what to do with this. Even if I assume the ratio in question gets very large, the estimate (from the prior step) becomes useless as $r_k to 0$. So it's unclear to me how to use it infinitely many times. I have the Nash-Digiorgi Holder regularity theorem at my disposal. Any hints or references would be much appreciated! I cannot find a similar proof anywhere, and given that I have provided the details for all of the other (numerous) steps, I would like to complete it.
real-analysis functional-analysis analysis pde elliptic-equations
asked Jan 24 at 6:59
David BowmanDavid Bowman
4,3161924
$endgroup$
I am trying to prove a Harnack inequality for a nonnegative subsolution $u in H^1(B_2)$ to the PDE $text{div}(A Du) ge 0$,where $A = A(x)$ is uniformly elliptic. The proof outline I am following is from a set of notes by a professor at my university, and the key step is the following inductive scheme:
Set $x_0$ to be a point such that $$u(x_0) = sup_{B_{(0,1/2)}} u,$$ and choose $x_k$ inductively such that $x_{k+1}$ is such that $$u(x_{k+1}) = sup_{B(x_k, r_k)} u$$
for $r_k$ sufficiently small to be chosen in a moment.
I have all of the steps except the following: suppose $$frac{text{sup}_{B_{0,1/4}} u}{ u(0)}$$ is sufficiently large, then we can choose a sequence $r_k$ such that $sum r_k <1/2$ and a $beta>1$ such that $u(x_{k+1}) ge beta u(x_k)$. That this would imply the result is immediate because it would contradict the boundedness of $u$. The preceding step, which I am led to believe is what implies the claim, is the following:
$$u(x_{k+1}) ge frac{u(x_k) - cr_k^{-q} u(0)}{1-theta}$$
where $c$, $q$ are absolute constants, and $1-theta ge text{osc}_{B_1}u>0$ and $0<theta le inf_{B_1} u$. Here $c,q>0$ are absolute constants.
I basically don't know what to do with this. Even if I assume the ratio in question gets very large, the estimate (from the prior step) becomes useless as $r_k to 0$. So it's unclear to me how to use it infinitely many times. I have the Nash-Digiorgi Holder regularity theorem at my disposal. Any hints or references would be much appreciated! I cannot find a similar proof anywhere, and given that I have provided the details for all of the other (numerous) steps, I would like to complete it.
real-analysis functional-analysis analysis pde elliptic-equations
real-analysis functional-analysis analysis pde elliptic-equations
asked Jan 24 at 6:59
David BowmanDavid Bowman
4,3161924
asked Jan 24 at 6:59
David BowmanDavid Bowman
4,3161924
edited Jan 24 at 15:20
David Bowman
asked Jan 24 at 6:59
David BowmanDavid Bowman
4,3161924
asked Jan 24 at 6:59
David BowmanDavid Bowman
4,3161924
asked Jan 24 at 6:59
David BowmanDavid Bowman
4,3161924
4,3161924
add a comment |
add a comment |
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