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Calculating divergence of function containing mollyfier
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Let $(phi_alpha)_alpha>0$ be a familiy of mollyfiers, $phi_alpha:mathbbR^n rightarrow mathbbR$ defined as:
beginalign
phi_1(x)=left{beginarrayrcl
c cdot exp(frac-11-vert xvert^2) &,& vert x vert < 1\
0 &,& textotherwise
endarrayright.
endalign
with $c>0$ such that $int_R^n phi_1(x) dx=1$ and $phi_alpha(x)=alpha^-n phi_1(x/alpha)$
Consider the function $w(x)=(y_1-y_2) int_0^1 phi_epsilon(x-y_1+t(y_1-y_2)) ~dt$ with $y_1,y_2 in mathbbR^n $ and calculate its divergence
$div~ w(x)= sum_i=1^n frac partial w_i(x)partial x_i.$
The result should be
$div ~w(x) = phi_epsilon(x-y_2)-phi_epsilon(x-y_1)$.
How do I calculate the divergence in this case? So far I tried to use the fundamental theorem of calculus, however the result was not the same.
calculus divergence
asked Jul 18 at 8:32
akwa
305
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up vote
0
down vote
favorite
Let $(phi_alpha)_alpha>0$ be a familiy of mollyfiers, $phi_alpha:mathbbR^n rightarrow mathbbR$ defined as:
beginalign
phi_1(x)=left{beginarrayrcl
c cdot exp(frac-11-vert xvert^2) &,& vert x vert < 1\
0 &,& textotherwise
endarrayright.
endalign
with $c>0$ such that $int_R^n phi_1(x) dx=1$ and $phi_alpha(x)=alpha^-n phi_1(x/alpha)$
Consider the function $w(x)=(y_1-y_2) int_0^1 phi_epsilon(x-y_1+t(y_1-y_2)) ~dt$ with $y_1,y_2 in mathbbR^n $ and calculate its divergence
$div~ w(x)= sum_i=1^n frac partial w_i(x)partial x_i.$
The result should be
$div ~w(x) = phi_epsilon(x-y_2)-phi_epsilon(x-y_1)$.
How do I calculate the divergence in this case? So far I tried to use the fundamental theorem of calculus, however the result was not the same.
calculus divergence
asked Jul 18 at 8:32
akwa
305
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(phi_alpha)_alpha>0$ be a familiy of mollyfiers, $phi_alpha:mathbbR^n rightarrow mathbbR$ defined as:
beginalign
phi_1(x)=left{beginarrayrcl
c cdot exp(frac-11-vert xvert^2) &,& vert x vert < 1\
0 &,& textotherwise
endarrayright.
endalign
with $c>0$ such that $int_R^n phi_1(x) dx=1$ and $phi_alpha(x)=alpha^-n phi_1(x/alpha)$
Consider the function $w(x)=(y_1-y_2) int_0^1 phi_epsilon(x-y_1+t(y_1-y_2)) ~dt$ with $y_1,y_2 in mathbbR^n $ and calculate its divergence
$div~ w(x)= sum_i=1^n frac partial w_i(x)partial x_i.$
The result should be
$div ~w(x) = phi_epsilon(x-y_2)-phi_epsilon(x-y_1)$.
How do I calculate the divergence in this case? So far I tried to use the fundamental theorem of calculus, however the result was not the same.
calculus divergence
asked Jul 18 at 8:32
akwa
305
Let $(phi_alpha)_alpha>0$ be a familiy of mollyfiers, $phi_alpha:mathbbR^n rightarrow mathbbR$ defined as:
beginalign
phi_1(x)=left{beginarrayrcl
c cdot exp(frac-11-vert xvert^2) &,& vert x vert < 1\
0 &,& textotherwise
endarrayright.
endalign
with $c>0$ such that $int_R^n phi_1(x) dx=1$ and $phi_alpha(x)=alpha^-n phi_1(x/alpha)$
Consider the function $w(x)=(y_1-y_2) int_0^1 phi_epsilon(x-y_1+t(y_1-y_2)) ~dt$ with $y_1,y_2 in mathbbR^n $ and calculate its divergence
$div~ w(x)= sum_i=1^n frac partial w_i(x)partial x_i.$
The result should be
$div ~w(x) = phi_epsilon(x-y_2)-phi_epsilon(x-y_1)$.
How do I calculate the divergence in this case? So far I tried to use the fundamental theorem of calculus, however the result was not the same.
calculus divergence
asked Jul 18 at 8:32
akwa
305
asked Jul 18 at 8:32
akwa
305
asked Jul 18 at 8:32
akwa
305
asked Jul 18 at 8:32
akwa
305
305
add a comment |Â
add a comment |Â
1 Answer
1
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up vote
1
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accepted
If $v$ is a constant vector and $f$ a function, then $mathrmdiv ,(vf)=vcdot nabla f$. Applying this to $f=w$ and $v=y_1-y_2$, you have to differentiate under the integral sign and get
$$
mathrmdiv, w = int_0^1 (y_1-y_2)cdot nabla phi_epsilon(x-y_1+t(y_1-y_2)), dt
$$
Now you just need to note that the integrand is equal to
$$
fracddtphi_epsilon(x-y_1+t(y_1-y_2))
$$
and then your claim follows from the fundamental theorem of calculus.
answered Jul 18 at 9:03
Kusma
1,127112
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $v$ is a constant vector and $f$ a function, then $mathrmdiv ,(vf)=vcdot nabla f$. Applying this to $f=w$ and $v=y_1-y_2$, you have to differentiate under the integral sign and get
$$
mathrmdiv, w = int_0^1 (y_1-y_2)cdot nabla phi_epsilon(x-y_1+t(y_1-y_2)), dt
$$
Now you just need to note that the integrand is equal to
$$
fracddtphi_epsilon(x-y_1+t(y_1-y_2))
$$
and then your claim follows from the fundamental theorem of calculus.
answered Jul 18 at 9:03
Kusma
1,127112
add a comment |Â
up vote
1
down vote
accepted
If $v$ is a constant vector and $f$ a function, then $mathrmdiv ,(vf)=vcdot nabla f$. Applying this to $f=w$ and $v=y_1-y_2$, you have to differentiate under the integral sign and get
$$
mathrmdiv, w = int_0^1 (y_1-y_2)cdot nabla phi_epsilon(x-y_1+t(y_1-y_2)), dt
$$
Now you just need to note that the integrand is equal to
$$
fracddtphi_epsilon(x-y_1+t(y_1-y_2))
$$
and then your claim follows from the fundamental theorem of calculus.
answered Jul 18 at 9:03
Kusma
1,127112
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $v$ is a constant vector and $f$ a function, then $mathrmdiv ,(vf)=vcdot nabla f$. Applying this to $f=w$ and $v=y_1-y_2$, you have to differentiate under the integral sign and get
$$
mathrmdiv, w = int_0^1 (y_1-y_2)cdot nabla phi_epsilon(x-y_1+t(y_1-y_2)), dt
$$
Now you just need to note that the integrand is equal to
$$
fracddtphi_epsilon(x-y_1+t(y_1-y_2))
$$
and then your claim follows from the fundamental theorem of calculus.
answered Jul 18 at 9:03
Kusma
1,127112
If $v$ is a constant vector and $f$ a function, then $mathrmdiv ,(vf)=vcdot nabla f$. Applying this to $f=w$ and $v=y_1-y_2$, you have to differentiate under the integral sign and get
$$
mathrmdiv, w = int_0^1 (y_1-y_2)cdot nabla phi_epsilon(x-y_1+t(y_1-y_2)), dt
$$
Now you just need to note that the integrand is equal to
$$
fracddtphi_epsilon(x-y_1+t(y_1-y_2))
$$
and then your claim follows from the fundamental theorem of calculus.
answered Jul 18 at 9:03
Kusma
1,127112
answered Jul 18 at 9:03
Kusma
1,127112
answered Jul 18 at 9:03
Kusma
1,127112
answered Jul 18 at 9:03
Kusma
1,127112
1,127112
add a comment |Â
add a comment |Â
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