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To show a function is in a Sobolev space, can we use weak spherical derivatives?

0

$begingroup$

For example, say $Omega=B(0,1)$ in $mathbbR^2$, and I have a function represented by
$$u(x)=begincaseslnln1/|x|& |x|leq e^-2\ln(2) &e^-2<|x|leq 1.endcases$$
There is a natural candidate for a representative of the weak partial derivatives. Namely
$$D^alphau(x)=begincasesD^alphalnln1/|x|&|x|leq e^-2\0&textotherwise.endcases$$
Since the circle has measure $0$ we need not even define the derivative there. Now my reasoning for spherical derivatives is that
$$fracpartiallnln1/partial x_1=fracx_1x,$$
which has a singularity at $(0,0).$ While
$$fracpartiallnln1/rpartial r=frac1rln r$$
may also have a singularity at $0,$ when we go to integrate it using spherical coordinates the singularity is resolved when we multiply by $r.$ Hence
$$int_0^2piint_0^e^-2frac1rln rr drdtheta$$
is finite as $1/ln r$ has an extension to the closure of $B(0,e^-2)$ and is therefore bounded.

partial-derivative sobolev-spaces weak-derivatives

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asked Apr 1 at 7:19

MelodyMelody

1,29012

$endgroup$

add a comment | 

0

$begingroup$

For example, say $Omega=B(0,1)$ in $mathbbR^2$, and I have a function represented by
$$u(x)=begincaseslnln1/|x|& |x|leq e^-2\ln(2) &e^-2<|x|leq 1.endcases$$
There is a natural candidate for a representative of the weak partial derivatives. Namely
$$D^alphau(x)=begincasesD^alphalnln1/|x|&|x|leq e^-2\0&textotherwise.endcases$$
Since the circle has measure $0$ we need not even define the derivative there. Now my reasoning for spherical derivatives is that
$$fracpartiallnln1/partial x_1=fracx_1x,$$
which has a singularity at $(0,0).$ While
$$fracpartiallnln1/rpartial r=frac1rln r$$
may also have a singularity at $0,$ when we go to integrate it using spherical coordinates the singularity is resolved when we multiply by $r.$ Hence
$$int_0^2piint_0^e^-2frac1rln rr drdtheta$$
is finite as $1/ln r$ has an extension to the closure of $B(0,e^-2)$ and is therefore bounded.

partial-derivative sobolev-spaces weak-derivatives

share|cite|improve this question

asked Apr 1 at 7:19

MelodyMelody

1,29012

$endgroup$

add a comment | 

$begingroup$

For example, say $Omega=B(0,1)$ in $mathbbR^2$, and I have a function represented by
$$u(x)=begincaseslnln1/|x|& |x|leq e^-2\ln(2) &e^-2<|x|leq 1.endcases$$
There is a natural candidate for a representative of the weak partial derivatives. Namely
$$D^alphau(x)=begincasesD^alphalnln1/|x|&|x|leq e^-2\0&textotherwise.endcases$$
Since the circle has measure $0$ we need not even define the derivative there. Now my reasoning for spherical derivatives is that
$$fracpartiallnln1/partial x_1=fracx_1x,$$
which has a singularity at $(0,0).$ While
$$fracpartiallnln1/rpartial r=frac1rln r$$
may also have a singularity at $0,$ when we go to integrate it using spherical coordinates the singularity is resolved when we multiply by $r.$ Hence
$$int_0^2piint_0^e^-2frac1rln rr drdtheta$$
is finite as $1/ln r$ has an extension to the closure of $B(0,e^-2)$ and is therefore bounded.

partial-derivative sobolev-spaces weak-derivatives

share|cite|improve this question

asked Apr 1 at 7:19

MelodyMelody

1,29012

$endgroup$

share|cite|improve this question

asked Apr 1 at 7:19

MelodyMelody

1,29012

share|cite|improve this question

asked Apr 1 at 7:19

MelodyMelody

1,29012

asked Apr 1 at 7:19

MelodyMelody

1,29012

asked Apr 1 at 7:19

MelodyMelody

1,29012