Mixing
On the proof of the maximum principle for elliptic equations
$begingroup$
From Renardy - "An Introductionto Partial Differential Equations".
Let
$$Lu=a_{ij}(x)frac{partial^2 u}{partial x_ipartial x_j}+b_i(x)frac{partial u}{partial x_i}+c(x)u, xinOmegasubsetmathbb{R}^n$$
The maximum principle states that for $Luge0$ in a bounded domain $Omega$ with $c(x)=0$ the maximum of $u$ is achieved on $partial Omega$.
The proof is done by contradiction. Suppose $Lu>0$ then $u$ cannot achieve its maximum anywhere in $Omega$. Suppose it did at a point $x_0$. Then all first derivatives must vanish and one can show that $Lu(x_0)le 0$, contradiction.
Now (for the case $Lu=0$) an approximation argument is used. Let $u_epsilon=u+epsilonexp(gamma x_1)$. We obtain
$$Lu_epsilon = Lu + epsilon(gamma^2a_11 + gamma b_1)exp(gamma x_1)$$We can then choose $gamma$ large enough s.t. $gamma^2a_11 + gamma b_1ge0$
Then $Lu_epsilon >0$ and we have that $$max_{overline Omega}u_epsilon = max_{partial Omega}u_epsilon$$ for every $epsilon>0$. The theorem follows from $epsilon rightarrow 0$.
Now my question is: Why can I assume that this maximum property holds for the limes? It might be possible that when taking the limit, that the limit function does not satisfy this relation anymore.
pde proof-explanation
edited Feb 1 at 9:46
YuiTo Cheng
2,3694937
asked Feb 1 at 9:37
EpsilonDeltaEpsilonDelta
7301615
$endgroup$
add a comment |
$begingroup$
From Renardy - "An Introductionto Partial Differential Equations".
Let
$$Lu=a_{ij}(x)frac{partial^2 u}{partial x_ipartial x_j}+b_i(x)frac{partial u}{partial x_i}+c(x)u, xinOmegasubsetmathbb{R}^n$$
The maximum principle states that for $Luge0$ in a bounded domain $Omega$ with $c(x)=0$ the maximum of $u$ is achieved on $partial Omega$.
The proof is done by contradiction. Suppose $Lu>0$ then $u$ cannot achieve its maximum anywhere in $Omega$. Suppose it did at a point $x_0$. Then all first derivatives must vanish and one can show that $Lu(x_0)le 0$, contradiction.
Now (for the case $Lu=0$) an approximation argument is used. Let $u_epsilon=u+epsilonexp(gamma x_1)$. We obtain
$$Lu_epsilon = Lu + epsilon(gamma^2a_11 + gamma b_1)exp(gamma x_1)$$We can then choose $gamma$ large enough s.t. $gamma^2a_11 + gamma b_1ge0$
Then $Lu_epsilon >0$ and we have that $$max_{overline Omega}u_epsilon = max_{partial Omega}u_epsilon$$ for every $epsilon>0$. The theorem follows from $epsilon rightarrow 0$.
Now my question is: Why can I assume that this maximum property holds for the limes? It might be possible that when taking the limit, that the limit function does not satisfy this relation anymore.
pde proof-explanation
edited Feb 1 at 9:46
YuiTo Cheng
2,3694937
asked Feb 1 at 9:37
EpsilonDeltaEpsilonDelta
7301615
$endgroup$
add a comment |
$begingroup$
From Renardy - "An Introductionto Partial Differential Equations".
Let
$$Lu=a_{ij}(x)frac{partial^2 u}{partial x_ipartial x_j}+b_i(x)frac{partial u}{partial x_i}+c(x)u, xinOmegasubsetmathbb{R}^n$$
The maximum principle states that for $Luge0$ in a bounded domain $Omega$ with $c(x)=0$ the maximum of $u$ is achieved on $partial Omega$.
The proof is done by contradiction. Suppose $Lu>0$ then $u$ cannot achieve its maximum anywhere in $Omega$. Suppose it did at a point $x_0$. Then all first derivatives must vanish and one can show that $Lu(x_0)le 0$, contradiction.
Now (for the case $Lu=0$) an approximation argument is used. Let $u_epsilon=u+epsilonexp(gamma x_1)$. We obtain
$$Lu_epsilon = Lu + epsilon(gamma^2a_11 + gamma b_1)exp(gamma x_1)$$We can then choose $gamma$ large enough s.t. $gamma^2a_11 + gamma b_1ge0$
Then $Lu_epsilon >0$ and we have that $$max_{overline Omega}u_epsilon = max_{partial Omega}u_epsilon$$ for every $epsilon>0$. The theorem follows from $epsilon rightarrow 0$.
Now my question is: Why can I assume that this maximum property holds for the limes? It might be possible that when taking the limit, that the limit function does not satisfy this relation anymore.
pde proof-explanation
edited Feb 1 at 9:46
YuiTo Cheng
2,3694937
asked Feb 1 at 9:37
EpsilonDeltaEpsilonDelta
7301615
$endgroup$
From Renardy - "An Introductionto Partial Differential Equations".
Let
$$Lu=a_{ij}(x)frac{partial^2 u}{partial x_ipartial x_j}+b_i(x)frac{partial u}{partial x_i}+c(x)u, xinOmegasubsetmathbb{R}^n$$
The maximum principle states that for $Luge0$ in a bounded domain $Omega$ with $c(x)=0$ the maximum of $u$ is achieved on $partial Omega$.
The proof is done by contradiction. Suppose $Lu>0$ then $u$ cannot achieve its maximum anywhere in $Omega$. Suppose it did at a point $x_0$. Then all first derivatives must vanish and one can show that $Lu(x_0)le 0$, contradiction.
Now (for the case $Lu=0$) an approximation argument is used. Let $u_epsilon=u+epsilonexp(gamma x_1)$. We obtain
$$Lu_epsilon = Lu + epsilon(gamma^2a_11 + gamma b_1)exp(gamma x_1)$$We can then choose $gamma$ large enough s.t. $gamma^2a_11 + gamma b_1ge0$
Then $Lu_epsilon >0$ and we have that $$max_{overline Omega}u_epsilon = max_{partial Omega}u_epsilon$$ for every $epsilon>0$. The theorem follows from $epsilon rightarrow 0$.
Now my question is: Why can I assume that this maximum property holds for the limes? It might be possible that when taking the limit, that the limit function does not satisfy this relation anymore.
pde proof-explanation
pde proof-explanation
edited Feb 1 at 9:46
YuiTo Cheng
2,3694937
asked Feb 1 at 9:37
EpsilonDeltaEpsilonDelta
7301615
edited Feb 1 at 9:46
YuiTo Cheng
2,3694937
asked Feb 1 at 9:37
EpsilonDeltaEpsilonDelta
7301615
edited Feb 1 at 9:46
YuiTo Cheng
2,3694937
edited Feb 1 at 9:46
YuiTo Cheng
2,3694937
edited Feb 1 at 9:46
YuiTo Cheng
2,3694937
2,3694937
asked Feb 1 at 9:37
EpsilonDeltaEpsilonDelta
7301615
asked Feb 1 at 9:37
EpsilonDeltaEpsilonDelta
7301615
asked Feb 1 at 9:37
EpsilonDeltaEpsilonDelta
7301615
7301615
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Due to $Omega$ is a bounded domain, hence the added "error part" is uniformly (to $varepsilon$) bounded (denoted as M). Hence we have the following dominate (note that $u_{varepsilon}>u$):$$
max_{Omega}uleq max_{Omega}u_{varepsilon}
=max_{partial_{Omega}}u_{varepsilon}leq max_{partialOmega}+varepsiloncdot M.
$$
The last step comes from $max(a+b)leqmax(a)+max(b)$. Now we take limit and the result follows immediately.
This is the reason why we demaned $Omega$ is a bounded domain. This proof doesn't holds for unbounded $Omega$. I hope I stated the reason clear.;)
answered Feb 1 at 13:51
Zixiao_LiuZixiao_Liu
1038
$endgroup$
$begingroup$
But since $Omega$ is open, we need know if the error part does attain its supremum. But since the error part is $exp(gamma x_1)$ we can continue this function on a closed set, is this correct?
$endgroup$
– EpsilonDelta
Feb 1 at 14:41
$begingroup$
@EpsilonDelta Well, due to $Omega$ is bounded, hence the supremum of the "error part" can be dominated by some constant (depending on $overline{Omega}$). And it is true that we can do the same process on $overline{Omega}$, there are no big differences here. :)
$endgroup$
– Zixiao_Liu
Feb 2 at 0:10
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Due to $Omega$ is a bounded domain, hence the added "error part" is uniformly (to $varepsilon$) bounded (denoted as M). Hence we have the following dominate (note that $u_{varepsilon}>u$):$$
max_{Omega}uleq max_{Omega}u_{varepsilon}
=max_{partial_{Omega}}u_{varepsilon}leq max_{partialOmega}+varepsiloncdot M.
$$
The last step comes from $max(a+b)leqmax(a)+max(b)$. Now we take limit and the result follows immediately.
This is the reason why we demaned $Omega$ is a bounded domain. This proof doesn't holds for unbounded $Omega$. I hope I stated the reason clear.;)
answered Feb 1 at 13:51
Zixiao_LiuZixiao_Liu
1038
$endgroup$
$begingroup$
But since $Omega$ is open, we need know if the error part does attain its supremum. But since the error part is $exp(gamma x_1)$ we can continue this function on a closed set, is this correct?
$endgroup$
– EpsilonDelta
Feb 1 at 14:41
$begingroup$
@EpsilonDelta Well, due to $Omega$ is bounded, hence the supremum of the "error part" can be dominated by some constant (depending on $overline{Omega}$). And it is true that we can do the same process on $overline{Omega}$, there are no big differences here. :)
$endgroup$
– Zixiao_Liu
Feb 2 at 0:10
add a comment |
$begingroup$
Due to $Omega$ is a bounded domain, hence the added "error part" is uniformly (to $varepsilon$) bounded (denoted as M). Hence we have the following dominate (note that $u_{varepsilon}>u$):$$
max_{Omega}uleq max_{Omega}u_{varepsilon}
=max_{partial_{Omega}}u_{varepsilon}leq max_{partialOmega}+varepsiloncdot M.
$$
The last step comes from $max(a+b)leqmax(a)+max(b)$. Now we take limit and the result follows immediately.
This is the reason why we demaned $Omega$ is a bounded domain. This proof doesn't holds for unbounded $Omega$. I hope I stated the reason clear.;)
answered Feb 1 at 13:51
Zixiao_LiuZixiao_Liu
1038
$endgroup$
$begingroup$
But since $Omega$ is open, we need know if the error part does attain its supremum. But since the error part is $exp(gamma x_1)$ we can continue this function on a closed set, is this correct?
$endgroup$
– EpsilonDelta
Feb 1 at 14:41
$begingroup$
@EpsilonDelta Well, due to $Omega$ is bounded, hence the supremum of the "error part" can be dominated by some constant (depending on $overline{Omega}$). And it is true that we can do the same process on $overline{Omega}$, there are no big differences here. :)
$endgroup$
– Zixiao_Liu
Feb 2 at 0:10
add a comment |
$begingroup$
Due to $Omega$ is a bounded domain, hence the added "error part" is uniformly (to $varepsilon$) bounded (denoted as M). Hence we have the following dominate (note that $u_{varepsilon}>u$):$$
max_{Omega}uleq max_{Omega}u_{varepsilon}
=max_{partial_{Omega}}u_{varepsilon}leq max_{partialOmega}+varepsiloncdot M.
$$
The last step comes from $max(a+b)leqmax(a)+max(b)$. Now we take limit and the result follows immediately.
This is the reason why we demaned $Omega$ is a bounded domain. This proof doesn't holds for unbounded $Omega$. I hope I stated the reason clear.;)
answered Feb 1 at 13:51
Zixiao_LiuZixiao_Liu
1038
$endgroup$
Due to $Omega$ is a bounded domain, hence the added "error part" is uniformly (to $varepsilon$) bounded (denoted as M). Hence we have the following dominate (note that $u_{varepsilon}>u$):$$
max_{Omega}uleq max_{Omega}u_{varepsilon}
=max_{partial_{Omega}}u_{varepsilon}leq max_{partialOmega}+varepsiloncdot M.
$$
The last step comes from $max(a+b)leqmax(a)+max(b)$. Now we take limit and the result follows immediately.
This is the reason why we demaned $Omega$ is a bounded domain. This proof doesn't holds for unbounded $Omega$. I hope I stated the reason clear.;)
answered Feb 1 at 13:51
Zixiao_LiuZixiao_Liu
1038
answered Feb 1 at 13:51
Zixiao_LiuZixiao_Liu
1038
answered Feb 1 at 13:51
Zixiao_LiuZixiao_Liu
1038
answered Feb 1 at 13:51
Zixiao_LiuZixiao_Liu
1038
1038
$begingroup$
But since $Omega$ is open, we need know if the error part does attain its supremum. But since the error part is $exp(gamma x_1)$ we can continue this function on a closed set, is this correct?
$endgroup$
– EpsilonDelta
Feb 1 at 14:41
$begingroup$
@EpsilonDelta Well, due to $Omega$ is bounded, hence the supremum of the "error part" can be dominated by some constant (depending on $overline{Omega}$). And it is true that we can do the same process on $overline{Omega}$, there are no big differences here. :)
$endgroup$
– Zixiao_Liu
Feb 2 at 0:10
add a comment |
$begingroup$
But since $Omega$ is open, we need know if the error part does attain its supremum. But since the error part is $exp(gamma x_1)$ we can continue this function on a closed set, is this correct?
$endgroup$
– EpsilonDelta
Feb 1 at 14:41
$begingroup$
@EpsilonDelta Well, due to $Omega$ is bounded, hence the supremum of the "error part" can be dominated by some constant (depending on $overline{Omega}$). And it is true that we can do the same process on $overline{Omega}$, there are no big differences here. :)
$endgroup$
– Zixiao_Liu
Feb 2 at 0:10
$begingroup$
But since $Omega$ is open, we need know if the error part does attain its supremum. But since the error part is $exp(gamma x_1)$ we can continue this function on a closed set, is this correct?
$endgroup$
– EpsilonDelta
Feb 1 at 14:41
$begingroup$
But since $Omega$ is open, we need know if the error part does attain its supremum. But since the error part is $exp(gamma x_1)$ we can continue this function on a closed set, is this correct?
$endgroup$
– EpsilonDelta
Feb 1 at 14:41
$begingroup$
@EpsilonDelta Well, due to $Omega$ is bounded, hence the supremum of the "error part" can be dominated by some constant (depending on $overline{Omega}$). And it is true that we can do the same process on $overline{Omega}$, there are no big differences here. :)
$endgroup$
– Zixiao_Liu
Feb 2 at 0:10
$begingroup$
@EpsilonDelta Well, due to $Omega$ is bounded, hence the supremum of the "error part" can be dominated by some constant (depending on $overline{Omega}$). And it is true that we can do the same process on $overline{Omega}$, there are no big differences here. :)
$endgroup$
– Zixiao_Liu
Feb 2 at 0:10
add a comment |
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