Weak derivative question
0
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I am confused of this example in Evans. Why is $|Du|$ calculated in this example?
pde weak-derivatives
edited Dec 30 '18 at 3:25
dmtri
1,7652521
asked Dec 30 '18 at 2:14
math_novicemath_novice
367
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add a comment |
0
$begingroup$
I am confused of this example in Evans. Why is $|Du|$ calculated in this example?
pde weak-derivatives
edited Dec 30 '18 at 3:25
dmtri
1,7652521
asked Dec 30 '18 at 2:14
math_novicemath_novice
367
$endgroup$
add a comment |
0
0
0
1
$begingroup$
I am confused of this example in Evans. Why is $|Du|$ calculated in this example?
pde weak-derivatives
edited Dec 30 '18 at 3:25
dmtri
1,7652521
asked Dec 30 '18 at 2:14
math_novicemath_novice
367
$endgroup$
I am confused of this example in Evans. Why is $|Du|$ calculated in this example?
pde weak-derivatives
pde weak-derivatives
edited Dec 30 '18 at 3:25
dmtri
1,7652521
asked Dec 30 '18 at 2:14
math_novicemath_novice
367
edited Dec 30 '18 at 3:25
dmtri
1,7652521
asked Dec 30 '18 at 2:14
math_novicemath_novice
367
edited Dec 30 '18 at 3:25
dmtri
1,7652521
edited Dec 30 '18 at 3:25
dmtri
1,7652521
edited Dec 30 '18 at 3:25
dmtri
1,7652521
1,7652521
asked Dec 30 '18 at 2:14
math_novicemath_novice
367
asked Dec 30 '18 at 2:14
math_novicemath_novice
367
asked Dec 30 '18 at 2:14
math_novicemath_novice
367
367
add a comment |
add a comment |
1 Answer
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Because you need to show that $|Du|$ is in $L^p$.
answered Dec 30 '18 at 6:18
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
Not to piggyback on this question, but why is the boundary in the second integral after performing integration by parts $partial B(0,epsilon)$ and not $partial(U - B(0, epsilon))$, is this because of the compact support that the test function has?
$endgroup$
– DaveNine
Dec 30 '18 at 7:15
$begingroup$
Because $phi$ has compact support in $U$ and vanishes on the boundary of $U$.
$endgroup$
– Julián Aguirre
Dec 30 '18 at 17:09
$begingroup$
Thank you for the answer! One more question: why do we want $lvert Durvert$ in $L^1$ when we calculate the integral of the boundary therm?
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– math_novice
Dec 30 '18 at 23:02
$begingroup$
In $L^p$, not in $L^1$. Is the definition of $W^{1,p}$.
$endgroup$
– Julián Aguirre
Dec 31 '18 at 4:03
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
Because you need to show that $|Du|$ is in $L^p$.
answered Dec 30 '18 at 6:18
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
Not to piggyback on this question, but why is the boundary in the second integral after performing integration by parts $partial B(0,epsilon)$ and not $partial(U - B(0, epsilon))$, is this because of the compact support that the test function has?
$endgroup$
– DaveNine
Dec 30 '18 at 7:15
$begingroup$
Because $phi$ has compact support in $U$ and vanishes on the boundary of $U$.
$endgroup$
– Julián Aguirre
Dec 30 '18 at 17:09
$begingroup$
Thank you for the answer! One more question: why do we want $lvert Durvert$ in $L^1$ when we calculate the integral of the boundary therm?
$endgroup$
– math_novice
Dec 30 '18 at 23:02
$begingroup$
In $L^p$, not in $L^1$. Is the definition of $W^{1,p}$.
$endgroup$
– Julián Aguirre
Dec 31 '18 at 4:03
add a comment |
3
$begingroup$
Because you need to show that $|Du|$ is in $L^p$.
answered Dec 30 '18 at 6:18
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
Not to piggyback on this question, but why is the boundary in the second integral after performing integration by parts $partial B(0,epsilon)$ and not $partial(U - B(0, epsilon))$, is this because of the compact support that the test function has?
$endgroup$
– DaveNine
Dec 30 '18 at 7:15
$begingroup$
Because $phi$ has compact support in $U$ and vanishes on the boundary of $U$.
$endgroup$
– Julián Aguirre
Dec 30 '18 at 17:09
$begingroup$
Thank you for the answer! One more question: why do we want $lvert Durvert$ in $L^1$ when we calculate the integral of the boundary therm?
$endgroup$
– math_novice
Dec 30 '18 at 23:02
$begingroup$
In $L^p$, not in $L^1$. Is the definition of $W^{1,p}$.
$endgroup$
– Julián Aguirre
Dec 31 '18 at 4:03
add a comment |
3
3
3
$begingroup$
Because you need to show that $|Du|$ is in $L^p$.
answered Dec 30 '18 at 6:18
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
Because you need to show that $|Du|$ is in $L^p$.
answered Dec 30 '18 at 6:18
Julián AguirreJulián Aguirre
69.5k24297
answered Dec 30 '18 at 6:18
Julián AguirreJulián Aguirre
69.5k24297
answered Dec 30 '18 at 6:18
Julián AguirreJulián Aguirre
69.5k24297
answered Dec 30 '18 at 6:18
Julián AguirreJulián Aguirre
69.5k24297
69.5k24297
$begingroup$
Not to piggyback on this question, but why is the boundary in the second integral after performing integration by parts $partial B(0,epsilon)$ and not $partial(U - B(0, epsilon))$, is this because of the compact support that the test function has?
$endgroup$
– DaveNine
Dec 30 '18 at 7:15
$begingroup$
Because $phi$ has compact support in $U$ and vanishes on the boundary of $U$.
$endgroup$
– Julián Aguirre
Dec 30 '18 at 17:09
$begingroup$
Thank you for the answer! One more question: why do we want $lvert Durvert$ in $L^1$ when we calculate the integral of the boundary therm?
$endgroup$
– math_novice
Dec 30 '18 at 23:02
$begingroup$
In $L^p$, not in $L^1$. Is the definition of $W^{1,p}$.
$endgroup$
– Julián Aguirre
Dec 31 '18 at 4:03
add a comment |
$begingroup$
Not to piggyback on this question, but why is the boundary in the second integral after performing integration by parts $partial B(0,epsilon)$ and not $partial(U - B(0, epsilon))$, is this because of the compact support that the test function has?
$endgroup$
– DaveNine
Dec 30 '18 at 7:15
$begingroup$
Because $phi$ has compact support in $U$ and vanishes on the boundary of $U$.
$endgroup$
– Julián Aguirre
Dec 30 '18 at 17:09
$begingroup$
Thank you for the answer! One more question: why do we want $lvert Durvert$ in $L^1$ when we calculate the integral of the boundary therm?
$endgroup$
– math_novice
Dec 30 '18 at 23:02
$begingroup$
In $L^p$, not in $L^1$. Is the definition of $W^{1,p}$.
$endgroup$
– Julián Aguirre
Dec 31 '18 at 4:03
$begingroup$
Not to piggyback on this question, but why is the boundary in the second integral after performing integration by parts $partial B(0,epsilon)$ and not $partial(U - B(0, epsilon))$, is this because of the compact support that the test function has?
$endgroup$
– DaveNine
Dec 30 '18 at 7:15
$begingroup$
Not to piggyback on this question, but why is the boundary in the second integral after performing integration by parts $partial B(0,epsilon)$ and not $partial(U - B(0, epsilon))$, is this because of the compact support that the test function has?
$endgroup$
– DaveNine
Dec 30 '18 at 7:15
$begingroup$
Because $phi$ has compact support in $U$ and vanishes on the boundary of $U$.
$endgroup$
– Julián Aguirre
Dec 30 '18 at 17:09
$begingroup$
Because $phi$ has compact support in $U$ and vanishes on the boundary of $U$.
$endgroup$
– Julián Aguirre
Dec 30 '18 at 17:09
$begingroup$
Thank you for the answer! One more question: why do we want $lvert Durvert$ in $L^1$ when we calculate the integral of the boundary therm?
$endgroup$
– math_novice
Dec 30 '18 at 23:02
$begingroup$
Thank you for the answer! One more question: why do we want $lvert Durvert$ in $L^1$ when we calculate the integral of the boundary therm?
$endgroup$
– math_novice
Dec 30 '18 at 23:02
$begingroup$
In $L^p$, not in $L^1$. Is the definition of $W^{1,p}$.
$endgroup$
– Julián Aguirre
Dec 31 '18 at 4:03
$begingroup$
In $L^p$, not in $L^1$. Is the definition of $W^{1,p}$.
$endgroup$
– Julián Aguirre
Dec 31 '18 at 4:03
add a comment |
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