Greens first identity
$begingroup$
Good afternoon all.
Using the greens first identity I am finding weak formulation. Can someone check I did this correct with below mentioned greens first identity
$$int_U (nabla cdot nabla mathbf{u}) cdot , vec{omega} , dU = int_{partial U} frac{partial {mathbf{u}}}{partial n} cdot vec{omega}, dS - int_U nabla mathbf{u} : nabla vec{omega} , dU $$
My answer is :
finite-element-method greens-function greens-theorem
asked Feb 14 '16 at 14:28
NithishNithish
5110
$endgroup$
add a comment |
$begingroup$
Good afternoon all.
Using the greens first identity I am finding weak formulation. Can someone check I did this correct with below mentioned greens first identity
$$int_U (nabla cdot nabla mathbf{u}) cdot , vec{omega} , dU = int_{partial U} frac{partial {mathbf{u}}}{partial n} cdot vec{omega}, dS - int_U nabla mathbf{u} : nabla vec{omega} , dU $$
My answer is :
finite-element-method greens-function greens-theorem
asked Feb 14 '16 at 14:28
NithishNithish
5110
$endgroup$
$begingroup$
I think you need to use the scalar Green's first identity: en.wikipedia.org/wiki/Green%27s_identities
$endgroup$
– Svetoslav
Feb 14 '16 at 15:37
1
$begingroup$
It is correct, but if you want to get a symmetric variational problem, then you should split the term $alphafrac{partial^2u}{partial xpartial y}=frac{alpha}{2}frac{partial^2u}{partial xpartial y}+frac{alpha}{2}frac{partial^2u}{partial ypartial x}=frac{alpha}{2}frac{partial }{partial x}left(frac{partial u}{partial y}right)+frac{alpha}{2}frac{partial }{partial y}left(frac{partial u}{partial x}right)$
$endgroup$
– Svetoslav
Feb 14 '16 at 17:15
1
$begingroup$
This means to write your initial PDE in divergence form (because $alpha,beta inmathbb R$ and so you can insert them in the derivatives)
$endgroup$
– Svetoslav
Feb 14 '16 at 17:20
add a comment |
$begingroup$
Good afternoon all.
Using the greens first identity I am finding weak formulation. Can someone check I did this correct with below mentioned greens first identity
$$int_U (nabla cdot nabla mathbf{u}) cdot , vec{omega} , dU = int_{partial U} frac{partial {mathbf{u}}}{partial n} cdot vec{omega}, dS - int_U nabla mathbf{u} : nabla vec{omega} , dU $$
My answer is :
finite-element-method greens-function greens-theorem
asked Feb 14 '16 at 14:28
NithishNithish
5110
$endgroup$
Good afternoon all.
Using the greens first identity I am finding weak formulation. Can someone check I did this correct with below mentioned greens first identity
$$int_U (nabla cdot nabla mathbf{u}) cdot , vec{omega} , dU = int_{partial U} frac{partial {mathbf{u}}}{partial n} cdot vec{omega}, dS - int_U nabla mathbf{u} : nabla vec{omega} , dU $$
My answer is :
finite-element-method greens-function greens-theorem
finite-element-method greens-function greens-theorem
asked Feb 14 '16 at 14:28
NithishNithish
5110
asked Feb 14 '16 at 14:28
NithishNithish
5110
edited Feb 14 '16 at 16:37
Nithish
asked Feb 14 '16 at 14:28
NithishNithish
5110
asked Feb 14 '16 at 14:28
NithishNithish
5110
asked Feb 14 '16 at 14:28
NithishNithish
5110
5110
$begingroup$
I think you need to use the scalar Green's first identity: en.wikipedia.org/wiki/Green%27s_identities
$endgroup$
– Svetoslav
Feb 14 '16 at 15:37
1
$begingroup$
It is correct, but if you want to get a symmetric variational problem, then you should split the term $alphafrac{partial^2u}{partial xpartial y}=frac{alpha}{2}frac{partial^2u}{partial xpartial y}+frac{alpha}{2}frac{partial^2u}{partial ypartial x}=frac{alpha}{2}frac{partial }{partial x}left(frac{partial u}{partial y}right)+frac{alpha}{2}frac{partial }{partial y}left(frac{partial u}{partial x}right)$
$endgroup$
– Svetoslav
Feb 14 '16 at 17:15
1
$begingroup$
This means to write your initial PDE in divergence form (because $alpha,beta inmathbb R$ and so you can insert them in the derivatives)
$endgroup$
– Svetoslav
Feb 14 '16 at 17:20
add a comment |
$begingroup$
I think you need to use the scalar Green's first identity: en.wikipedia.org/wiki/Green%27s_identities
$endgroup$
– Svetoslav
Feb 14 '16 at 15:37
1
$begingroup$
It is correct, but if you want to get a symmetric variational problem, then you should split the term $alphafrac{partial^2u}{partial xpartial y}=frac{alpha}{2}frac{partial^2u}{partial xpartial y}+frac{alpha}{2}frac{partial^2u}{partial ypartial x}=frac{alpha}{2}frac{partial }{partial x}left(frac{partial u}{partial y}right)+frac{alpha}{2}frac{partial }{partial y}left(frac{partial u}{partial x}right)$
$endgroup$
– Svetoslav
Feb 14 '16 at 17:15
1
$begingroup$
This means to write your initial PDE in divergence form (because $alpha,beta inmathbb R$ and so you can insert them in the derivatives)
$endgroup$
– Svetoslav
Feb 14 '16 at 17:20
$begingroup$
I think you need to use the scalar Green's first identity: en.wikipedia.org/wiki/Green%27s_identities
$endgroup$
– Svetoslav
Feb 14 '16 at 15:37
$begingroup$
I think you need to use the scalar Green's first identity: en.wikipedia.org/wiki/Green%27s_identities
$endgroup$
– Svetoslav
Feb 14 '16 at 15:37
1
1
$begingroup$
It is correct, but if you want to get a symmetric variational problem, then you should split the term $alphafrac{partial^2u}{partial xpartial y}=frac{alpha}{2}frac{partial^2u}{partial xpartial y}+frac{alpha}{2}frac{partial^2u}{partial ypartial x}=frac{alpha}{2}frac{partial }{partial x}left(frac{partial u}{partial y}right)+frac{alpha}{2}frac{partial }{partial y}left(frac{partial u}{partial x}right)$
$endgroup$
– Svetoslav
Feb 14 '16 at 17:15
$begingroup$
It is correct, but if you want to get a symmetric variational problem, then you should split the term $alphafrac{partial^2u}{partial xpartial y}=frac{alpha}{2}frac{partial^2u}{partial xpartial y}+frac{alpha}{2}frac{partial^2u}{partial ypartial x}=frac{alpha}{2}frac{partial }{partial x}left(frac{partial u}{partial y}right)+frac{alpha}{2}frac{partial }{partial y}left(frac{partial u}{partial x}right)$
$endgroup$
– Svetoslav
Feb 14 '16 at 17:15
1
1
$begingroup$
This means to write your initial PDE in divergence form (because $alpha,beta inmathbb R$ and so you can insert them in the derivatives)
$endgroup$
– Svetoslav
Feb 14 '16 at 17:20
$begingroup$
This means to write your initial PDE in divergence form (because $alpha,beta inmathbb R$ and so you can insert them in the derivatives)
$endgroup$
– Svetoslav
Feb 14 '16 at 17:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have the equation:
$$-9u_{xx}-5u_{yy}=f$$
Now multiplying by a test function $vin H_0^1(Omega)$ both sides and integrating over $Omega$ you get.
$$-9intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}-5intlimits_{Omega}{frac{partial}{partial y}left(frac{partial u}{partial y}right)vdxdy}=intlimits_{Omega}{fvdxdy}$$
$$9intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}+5intlimits_{Omega}{frac{partial u}{partial y}frac{partial v}{partial y}dxdy}=intlimits_{Omega}{fvdxdy}quadquad (*)$$
which is the weak formulation that you should get.
We used $$-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}=-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}vright)dxdy}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=-underbrace{intlimits_{partialOmega}{frac{partial u}{partial x}vn_1ds}}_{=0}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}$$
Here, the tool that we used is the divergence theorem (with which is actually derived the Green's first identity). Note that the surface integral is $0$ because $v$ is zero on $partial Omega$ (to be more speciffic, it is zero in the trace sense). Or if you want to avoid the use of the divergence formula in the weak sense, then you can start by multiplying by a smooth test function $vin C_0^infty(Omega)$, derive the equation $(*)$ above, and then use density argument to show that $(*)$ is actually satisfied also for all $vin H_0^1(Omega)$. The integral with the $y$-derivative is done in similar way, just having in mind that in the surface integral there will be the second component $n_2$ of the normal vector $n$ to $partialOmega$ .
answered Feb 14 '16 at 16:05
SvetoslavSvetoslav
4,5672828
$endgroup$
$begingroup$
I understood the concept mentoined above, which is pretty much similar to integration by parts. I have confusion about the xy terms together if you can check with my updated question please
$endgroup$
– Nithish
Feb 14 '16 at 16:38
$begingroup$
Yes, it is actually just integration by parts. I will check your solution and write here possibly in several hours.
$endgroup$
– Svetoslav
Feb 14 '16 at 16:57
$begingroup$
I updated my solution
$endgroup$
– Nithish
Feb 14 '16 at 16:59
1
$begingroup$
Here is another weak PDE math.stackexchange.com/questions/1653737/…
$endgroup$
– Svetoslav
Feb 14 '16 at 17:36
add a comment |
Your Answer
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1 Answer
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$begingroup$
You have the equation:
$$-9u_{xx}-5u_{yy}=f$$
Now multiplying by a test function $vin H_0^1(Omega)$ both sides and integrating over $Omega$ you get.
$$-9intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}-5intlimits_{Omega}{frac{partial}{partial y}left(frac{partial u}{partial y}right)vdxdy}=intlimits_{Omega}{fvdxdy}$$
$$9intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}+5intlimits_{Omega}{frac{partial u}{partial y}frac{partial v}{partial y}dxdy}=intlimits_{Omega}{fvdxdy}quadquad (*)$$
which is the weak formulation that you should get.
We used $$-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}=-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}vright)dxdy}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=-underbrace{intlimits_{partialOmega}{frac{partial u}{partial x}vn_1ds}}_{=0}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}$$
Here, the tool that we used is the divergence theorem (with which is actually derived the Green's first identity). Note that the surface integral is $0$ because $v$ is zero on $partial Omega$ (to be more speciffic, it is zero in the trace sense). Or if you want to avoid the use of the divergence formula in the weak sense, then you can start by multiplying by a smooth test function $vin C_0^infty(Omega)$, derive the equation $(*)$ above, and then use density argument to show that $(*)$ is actually satisfied also for all $vin H_0^1(Omega)$. The integral with the $y$-derivative is done in similar way, just having in mind that in the surface integral there will be the second component $n_2$ of the normal vector $n$ to $partialOmega$ .
answered Feb 14 '16 at 16:05
SvetoslavSvetoslav
4,5672828
$endgroup$
$begingroup$
I understood the concept mentoined above, which is pretty much similar to integration by parts. I have confusion about the xy terms together if you can check with my updated question please
$endgroup$
– Nithish
Feb 14 '16 at 16:38
$begingroup$
Yes, it is actually just integration by parts. I will check your solution and write here possibly in several hours.
$endgroup$
– Svetoslav
Feb 14 '16 at 16:57
$begingroup$
I updated my solution
$endgroup$
– Nithish
Feb 14 '16 at 16:59
1
$begingroup$
Here is another weak PDE math.stackexchange.com/questions/1653737/…
$endgroup$
– Svetoslav
Feb 14 '16 at 17:36
add a comment |
$begingroup$
You have the equation:
$$-9u_{xx}-5u_{yy}=f$$
Now multiplying by a test function $vin H_0^1(Omega)$ both sides and integrating over $Omega$ you get.
$$-9intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}-5intlimits_{Omega}{frac{partial}{partial y}left(frac{partial u}{partial y}right)vdxdy}=intlimits_{Omega}{fvdxdy}$$
$$9intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}+5intlimits_{Omega}{frac{partial u}{partial y}frac{partial v}{partial y}dxdy}=intlimits_{Omega}{fvdxdy}quadquad (*)$$
which is the weak formulation that you should get.
We used $$-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}=-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}vright)dxdy}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=-underbrace{intlimits_{partialOmega}{frac{partial u}{partial x}vn_1ds}}_{=0}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}$$
Here, the tool that we used is the divergence theorem (with which is actually derived the Green's first identity). Note that the surface integral is $0$ because $v$ is zero on $partial Omega$ (to be more speciffic, it is zero in the trace sense). Or if you want to avoid the use of the divergence formula in the weak sense, then you can start by multiplying by a smooth test function $vin C_0^infty(Omega)$, derive the equation $(*)$ above, and then use density argument to show that $(*)$ is actually satisfied also for all $vin H_0^1(Omega)$. The integral with the $y$-derivative is done in similar way, just having in mind that in the surface integral there will be the second component $n_2$ of the normal vector $n$ to $partialOmega$ .
answered Feb 14 '16 at 16:05
SvetoslavSvetoslav
4,5672828
$endgroup$
$begingroup$
I understood the concept mentoined above, which is pretty much similar to integration by parts. I have confusion about the xy terms together if you can check with my updated question please
$endgroup$
– Nithish
Feb 14 '16 at 16:38
$begingroup$
Yes, it is actually just integration by parts. I will check your solution and write here possibly in several hours.
$endgroup$
– Svetoslav
Feb 14 '16 at 16:57
$begingroup$
I updated my solution
$endgroup$
– Nithish
Feb 14 '16 at 16:59
1
$begingroup$
Here is another weak PDE math.stackexchange.com/questions/1653737/…
$endgroup$
– Svetoslav
Feb 14 '16 at 17:36
add a comment |
$begingroup$
You have the equation:
$$-9u_{xx}-5u_{yy}=f$$
Now multiplying by a test function $vin H_0^1(Omega)$ both sides and integrating over $Omega$ you get.
$$-9intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}-5intlimits_{Omega}{frac{partial}{partial y}left(frac{partial u}{partial y}right)vdxdy}=intlimits_{Omega}{fvdxdy}$$
$$9intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}+5intlimits_{Omega}{frac{partial u}{partial y}frac{partial v}{partial y}dxdy}=intlimits_{Omega}{fvdxdy}quadquad (*)$$
which is the weak formulation that you should get.
We used $$-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}=-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}vright)dxdy}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=-underbrace{intlimits_{partialOmega}{frac{partial u}{partial x}vn_1ds}}_{=0}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}$$
Here, the tool that we used is the divergence theorem (with which is actually derived the Green's first identity). Note that the surface integral is $0$ because $v$ is zero on $partial Omega$ (to be more speciffic, it is zero in the trace sense). Or if you want to avoid the use of the divergence formula in the weak sense, then you can start by multiplying by a smooth test function $vin C_0^infty(Omega)$, derive the equation $(*)$ above, and then use density argument to show that $(*)$ is actually satisfied also for all $vin H_0^1(Omega)$. The integral with the $y$-derivative is done in similar way, just having in mind that in the surface integral there will be the second component $n_2$ of the normal vector $n$ to $partialOmega$ .
answered Feb 14 '16 at 16:05
SvetoslavSvetoslav
4,5672828
$endgroup$
You have the equation:
$$-9u_{xx}-5u_{yy}=f$$
Now multiplying by a test function $vin H_0^1(Omega)$ both sides and integrating over $Omega$ you get.
$$-9intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}-5intlimits_{Omega}{frac{partial}{partial y}left(frac{partial u}{partial y}right)vdxdy}=intlimits_{Omega}{fvdxdy}$$
$$9intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}+5intlimits_{Omega}{frac{partial u}{partial y}frac{partial v}{partial y}dxdy}=intlimits_{Omega}{fvdxdy}quadquad (*)$$
which is the weak formulation that you should get.
We used $$-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}=-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}vright)dxdy}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=-underbrace{intlimits_{partialOmega}{frac{partial u}{partial x}vn_1ds}}_{=0}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}$$
Here, the tool that we used is the divergence theorem (with which is actually derived the Green's first identity). Note that the surface integral is $0$ because $v$ is zero on $partial Omega$ (to be more speciffic, it is zero in the trace sense). Or if you want to avoid the use of the divergence formula in the weak sense, then you can start by multiplying by a smooth test function $vin C_0^infty(Omega)$, derive the equation $(*)$ above, and then use density argument to show that $(*)$ is actually satisfied also for all $vin H_0^1(Omega)$. The integral with the $y$-derivative is done in similar way, just having in mind that in the surface integral there will be the second component $n_2$ of the normal vector $n$ to $partialOmega$ .
answered Feb 14 '16 at 16:05
SvetoslavSvetoslav
4,5672828
answered Feb 14 '16 at 16:05
SvetoslavSvetoslav
4,5672828
answered Feb 14 '16 at 16:05
SvetoslavSvetoslav
4,5672828
answered Feb 14 '16 at 16:05
SvetoslavSvetoslav
4,5672828
4,5672828
$begingroup$
I understood the concept mentoined above, which is pretty much similar to integration by parts. I have confusion about the xy terms together if you can check with my updated question please
$endgroup$
– Nithish
Feb 14 '16 at 16:38
$begingroup$
Yes, it is actually just integration by parts. I will check your solution and write here possibly in several hours.
$endgroup$
– Svetoslav
Feb 14 '16 at 16:57
$begingroup$
I updated my solution
$endgroup$
– Nithish
Feb 14 '16 at 16:59
1
$begingroup$
Here is another weak PDE math.stackexchange.com/questions/1653737/…
$endgroup$
– Svetoslav
Feb 14 '16 at 17:36
add a comment |
$begingroup$
I understood the concept mentoined above, which is pretty much similar to integration by parts. I have confusion about the xy terms together if you can check with my updated question please
$endgroup$
– Nithish
Feb 14 '16 at 16:38
$begingroup$
Yes, it is actually just integration by parts. I will check your solution and write here possibly in several hours.
$endgroup$
– Svetoslav
Feb 14 '16 at 16:57
$begingroup$
I updated my solution
$endgroup$
– Nithish
Feb 14 '16 at 16:59
1
$begingroup$
Here is another weak PDE math.stackexchange.com/questions/1653737/…
$endgroup$
– Svetoslav
Feb 14 '16 at 17:36
$begingroup$
I understood the concept mentoined above, which is pretty much similar to integration by parts. I have confusion about the xy terms together if you can check with my updated question please
$endgroup$
– Nithish
Feb 14 '16 at 16:38
$begingroup$
I understood the concept mentoined above, which is pretty much similar to integration by parts. I have confusion about the xy terms together if you can check with my updated question please
$endgroup$
– Nithish
Feb 14 '16 at 16:38
$begingroup$
Yes, it is actually just integration by parts. I will check your solution and write here possibly in several hours.
$endgroup$
– Svetoslav
Feb 14 '16 at 16:57
$begingroup$
Yes, it is actually just integration by parts. I will check your solution and write here possibly in several hours.
$endgroup$
– Svetoslav
Feb 14 '16 at 16:57
$begingroup$
I updated my solution
$endgroup$
– Nithish
Feb 14 '16 at 16:59
$begingroup$
I updated my solution
$endgroup$
– Nithish
Feb 14 '16 at 16:59
1
1
$begingroup$
Here is another weak PDE math.stackexchange.com/questions/1653737/…
$endgroup$
– Svetoslav
Feb 14 '16 at 17:36
$begingroup$
Here is another weak PDE math.stackexchange.com/questions/1653737/…
$endgroup$
– Svetoslav
Feb 14 '16 at 17:36
add a comment |
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$begingroup$
I think you need to use the scalar Green's first identity: en.wikipedia.org/wiki/Green%27s_identities
$endgroup$
– Svetoslav
Feb 14 '16 at 15:37
1
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It is correct, but if you want to get a symmetric variational problem, then you should split the term $alphafrac{partial^2u}{partial xpartial y}=frac{alpha}{2}frac{partial^2u}{partial xpartial y}+frac{alpha}{2}frac{partial^2u}{partial ypartial x}=frac{alpha}{2}frac{partial }{partial x}left(frac{partial u}{partial y}right)+frac{alpha}{2}frac{partial }{partial y}left(frac{partial u}{partial x}right)$
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– Svetoslav
Feb 14 '16 at 17:15
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This means to write your initial PDE in divergence form (because $alpha,beta inmathbb R$ and so you can insert them in the derivatives)
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– Svetoslav
Feb 14 '16 at 17:20