Integrate Vector Laplacian By Parts For FEM
$begingroup$
I am currently trying to solve the following, vectorial Poisson equation using the FEM technique:
$$-nabla^2 vec{A}=vec{J} quad forall xinOmega$$
Now I know in the case of the scalar Poisson equation $-nabla^2 varphi=rho$ one can derive the weak form of the PDE by introducing a scalar test function $v$ which is multiplied with the PDE and integrated over $Omega$ using integration by parts:
$$int_{Omega} (nabla^2varphi) v mathrm{d}x=int_{partialOmega} (vec{nabla}varphi) v mathrm{d}vec{omega}-int_{Omega} (vec{nabla}varphi) cdot(vec{nabla}v) mathrm{d}x$$
My question is how this approach can be applied to the vectorial Poisson equation? In order to do so, one would have to integrate the following expression by parts (where $vec{v}$ is a vectorial test function):
$$int_{Omega} (nabla^2vec{A})cdotvec{v} mathrm{d}x$$
However, I am struggling to do this integration. I guess one would have to generalize the following identity for scalar functions $varphi$
$$vec{nabla}cdot(vvec{nabla}varphi)=(nabla^2varphi) v+(vec{nabla}varphi)cdot(vec{nabla}v)$$
into an identity of the following form for vectorial functions $vec{A}$
$$vec{nabla}cdot(...)=(nabla^2vec{A})vec{v}+(...)$$
integration ordinary-differential-equations multivariable-calculus partial-derivative
asked Dec 29 '18 at 13:32
MantabitMantabit
164
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add a comment |
$begingroup$
I am currently trying to solve the following, vectorial Poisson equation using the FEM technique:
$$-nabla^2 vec{A}=vec{J} quad forall xinOmega$$
Now I know in the case of the scalar Poisson equation $-nabla^2 varphi=rho$ one can derive the weak form of the PDE by introducing a scalar test function $v$ which is multiplied with the PDE and integrated over $Omega$ using integration by parts:
$$int_{Omega} (nabla^2varphi) v mathrm{d}x=int_{partialOmega} (vec{nabla}varphi) v mathrm{d}vec{omega}-int_{Omega} (vec{nabla}varphi) cdot(vec{nabla}v) mathrm{d}x$$
My question is how this approach can be applied to the vectorial Poisson equation? In order to do so, one would have to integrate the following expression by parts (where $vec{v}$ is a vectorial test function):
$$int_{Omega} (nabla^2vec{A})cdotvec{v} mathrm{d}x$$
However, I am struggling to do this integration. I guess one would have to generalize the following identity for scalar functions $varphi$
$$vec{nabla}cdot(vvec{nabla}varphi)=(nabla^2varphi) v+(vec{nabla}varphi)cdot(vec{nabla}v)$$
into an identity of the following form for vectorial functions $vec{A}$
$$vec{nabla}cdot(...)=(nabla^2vec{A})vec{v}+(...)$$
integration ordinary-differential-equations multivariable-calculus partial-derivative
asked Dec 29 '18 at 13:32
MantabitMantabit
164
$endgroup$
$begingroup$
Perhaps you'll find my old post useful: math.stackexchange.com/questions/745000/greens-first-identity. Best, Daniel.
$endgroup$
– Dmoreno
Dec 29 '18 at 15:21
$begingroup$
Thank you for your response. It seems like the vector calculus identity $nabla cdot(mathbf{T} cdot vec{omega} ) = T:nablavec{omega} + (nabla cdot mathbf{T}) cdotvec{omega}$ which you linked solves the problem. I was able to come to the same solution with some different reasoning. I'll check the identity with some symbolic math software and then formulate an answer to my own question.
$endgroup$
– Mantabit
Dec 29 '18 at 18:39
$begingroup$
Glad to help @Mantabit. Best, Daniel.
$endgroup$
– Dmoreno
Dec 30 '18 at 18:02
add a comment |
$begingroup$
I am currently trying to solve the following, vectorial Poisson equation using the FEM technique:
$$-nabla^2 vec{A}=vec{J} quad forall xinOmega$$
Now I know in the case of the scalar Poisson equation $-nabla^2 varphi=rho$ one can derive the weak form of the PDE by introducing a scalar test function $v$ which is multiplied with the PDE and integrated over $Omega$ using integration by parts:
$$int_{Omega} (nabla^2varphi) v mathrm{d}x=int_{partialOmega} (vec{nabla}varphi) v mathrm{d}vec{omega}-int_{Omega} (vec{nabla}varphi) cdot(vec{nabla}v) mathrm{d}x$$
My question is how this approach can be applied to the vectorial Poisson equation? In order to do so, one would have to integrate the following expression by parts (where $vec{v}$ is a vectorial test function):
$$int_{Omega} (nabla^2vec{A})cdotvec{v} mathrm{d}x$$
However, I am struggling to do this integration. I guess one would have to generalize the following identity for scalar functions $varphi$
$$vec{nabla}cdot(vvec{nabla}varphi)=(nabla^2varphi) v+(vec{nabla}varphi)cdot(vec{nabla}v)$$
into an identity of the following form for vectorial functions $vec{A}$
$$vec{nabla}cdot(...)=(nabla^2vec{A})vec{v}+(...)$$
integration ordinary-differential-equations multivariable-calculus partial-derivative
asked Dec 29 '18 at 13:32
MantabitMantabit
164
$endgroup$
I am currently trying to solve the following, vectorial Poisson equation using the FEM technique:
$$-nabla^2 vec{A}=vec{J} quad forall xinOmega$$
Now I know in the case of the scalar Poisson equation $-nabla^2 varphi=rho$ one can derive the weak form of the PDE by introducing a scalar test function $v$ which is multiplied with the PDE and integrated over $Omega$ using integration by parts:
$$int_{Omega} (nabla^2varphi) v mathrm{d}x=int_{partialOmega} (vec{nabla}varphi) v mathrm{d}vec{omega}-int_{Omega} (vec{nabla}varphi) cdot(vec{nabla}v) mathrm{d}x$$
My question is how this approach can be applied to the vectorial Poisson equation? In order to do so, one would have to integrate the following expression by parts (where $vec{v}$ is a vectorial test function):
$$int_{Omega} (nabla^2vec{A})cdotvec{v} mathrm{d}x$$
However, I am struggling to do this integration. I guess one would have to generalize the following identity for scalar functions $varphi$
$$vec{nabla}cdot(vvec{nabla}varphi)=(nabla^2varphi) v+(vec{nabla}varphi)cdot(vec{nabla}v)$$
into an identity of the following form for vectorial functions $vec{A}$
$$vec{nabla}cdot(...)=(nabla^2vec{A})vec{v}+(...)$$
integration ordinary-differential-equations multivariable-calculus partial-derivative
integration ordinary-differential-equations multivariable-calculus partial-derivative
asked Dec 29 '18 at 13:32
MantabitMantabit
164
asked Dec 29 '18 at 13:32
MantabitMantabit
164
edited Dec 29 '18 at 18:36
Mantabit
asked Dec 29 '18 at 13:32
MantabitMantabit
164
asked Dec 29 '18 at 13:32
MantabitMantabit
164
asked Dec 29 '18 at 13:32
MantabitMantabit
164
164
$begingroup$
Perhaps you'll find my old post useful: math.stackexchange.com/questions/745000/greens-first-identity. Best, Daniel.
$endgroup$
– Dmoreno
Dec 29 '18 at 15:21
$begingroup$
Thank you for your response. It seems like the vector calculus identity $nabla cdot(mathbf{T} cdot vec{omega} ) = T:nablavec{omega} + (nabla cdot mathbf{T}) cdotvec{omega}$ which you linked solves the problem. I was able to come to the same solution with some different reasoning. I'll check the identity with some symbolic math software and then formulate an answer to my own question.
$endgroup$
– Mantabit
Dec 29 '18 at 18:39
$begingroup$
Glad to help @Mantabit. Best, Daniel.
$endgroup$
– Dmoreno
Dec 30 '18 at 18:02
add a comment |
$begingroup$
Perhaps you'll find my old post useful: math.stackexchange.com/questions/745000/greens-first-identity. Best, Daniel.
$endgroup$
– Dmoreno
Dec 29 '18 at 15:21
$begingroup$
Thank you for your response. It seems like the vector calculus identity $nabla cdot(mathbf{T} cdot vec{omega} ) = T:nablavec{omega} + (nabla cdot mathbf{T}) cdotvec{omega}$ which you linked solves the problem. I was able to come to the same solution with some different reasoning. I'll check the identity with some symbolic math software and then formulate an answer to my own question.
$endgroup$
– Mantabit
Dec 29 '18 at 18:39
$begingroup$
Glad to help @Mantabit. Best, Daniel.
$endgroup$
– Dmoreno
Dec 30 '18 at 18:02
$begingroup$
Perhaps you'll find my old post useful: math.stackexchange.com/questions/745000/greens-first-identity. Best, Daniel.
$endgroup$
– Dmoreno
Dec 29 '18 at 15:21
$begingroup$
Perhaps you'll find my old post useful: math.stackexchange.com/questions/745000/greens-first-identity. Best, Daniel.
$endgroup$
– Dmoreno
Dec 29 '18 at 15:21
$begingroup$
Thank you for your response. It seems like the vector calculus identity $nabla cdot(mathbf{T} cdot vec{omega} ) = T:nablavec{omega} + (nabla cdot mathbf{T}) cdotvec{omega}$ which you linked solves the problem. I was able to come to the same solution with some different reasoning. I'll check the identity with some symbolic math software and then formulate an answer to my own question.
$endgroup$
– Mantabit
Dec 29 '18 at 18:39
$begingroup$
Thank you for your response. It seems like the vector calculus identity $nabla cdot(mathbf{T} cdot vec{omega} ) = T:nablavec{omega} + (nabla cdot mathbf{T}) cdotvec{omega}$ which you linked solves the problem. I was able to come to the same solution with some different reasoning. I'll check the identity with some symbolic math software and then formulate an answer to my own question.
$endgroup$
– Mantabit
Dec 29 '18 at 18:39
$begingroup$
Glad to help @Mantabit. Best, Daniel.
$endgroup$
– Dmoreno
Dec 30 '18 at 18:02
$begingroup$
Glad to help @Mantabit. Best, Daniel.
$endgroup$
– Dmoreno
Dec 30 '18 at 18:02
add a comment |
1 Answer
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Not sure if this is useful, but have a look at this identity for the divergence of a matrix $mathbf{A}$ acting on a vector $vec{v}$:
$$vecnabla cdot (mathbf{A}vec{v}) = (vecnabla cdot mathbf{A}) vec{v} + operatorname{Tr}left(mathbf{A}(vecnablavec{v})right)$$
Now if you take $mathbf{A}$ to be the Jacobi matrix of your vector field $vec{A}$, i.e. $mathbf{A}= vec{nabla} vec{A}$, you get
$$vecnabla cdot big((vec{nabla} vec{A})vec{v}big) = left(vecnabla cdot (vec{nabla} vec{A})right) vec{v} + operatorname{Tr}left((vec{nabla} vec{A})(vecnablavec{v})right) = left(nabla^2vec{A}right) vec{v} + operatorname{Tr}left((vec{nabla} vec{A})(vecnablavec{v})right)$$
answered Dec 29 '18 at 22:02
mechanodroidmechanodroid
28.9k62548
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1 Answer
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$begingroup$
Not sure if this is useful, but have a look at this identity for the divergence of a matrix $mathbf{A}$ acting on a vector $vec{v}$:
$$vecnabla cdot (mathbf{A}vec{v}) = (vecnabla cdot mathbf{A}) vec{v} + operatorname{Tr}left(mathbf{A}(vecnablavec{v})right)$$
Now if you take $mathbf{A}$ to be the Jacobi matrix of your vector field $vec{A}$, i.e. $mathbf{A}= vec{nabla} vec{A}$, you get
$$vecnabla cdot big((vec{nabla} vec{A})vec{v}big) = left(vecnabla cdot (vec{nabla} vec{A})right) vec{v} + operatorname{Tr}left((vec{nabla} vec{A})(vecnablavec{v})right) = left(nabla^2vec{A}right) vec{v} + operatorname{Tr}left((vec{nabla} vec{A})(vecnablavec{v})right)$$
answered Dec 29 '18 at 22:02
mechanodroidmechanodroid
28.9k62548
$endgroup$
add a comment |
$begingroup$
Not sure if this is useful, but have a look at this identity for the divergence of a matrix $mathbf{A}$ acting on a vector $vec{v}$:
$$vecnabla cdot (mathbf{A}vec{v}) = (vecnabla cdot mathbf{A}) vec{v} + operatorname{Tr}left(mathbf{A}(vecnablavec{v})right)$$
Now if you take $mathbf{A}$ to be the Jacobi matrix of your vector field $vec{A}$, i.e. $mathbf{A}= vec{nabla} vec{A}$, you get
$$vecnabla cdot big((vec{nabla} vec{A})vec{v}big) = left(vecnabla cdot (vec{nabla} vec{A})right) vec{v} + operatorname{Tr}left((vec{nabla} vec{A})(vecnablavec{v})right) = left(nabla^2vec{A}right) vec{v} + operatorname{Tr}left((vec{nabla} vec{A})(vecnablavec{v})right)$$
answered Dec 29 '18 at 22:02
mechanodroidmechanodroid
28.9k62548
$endgroup$
add a comment |
$begingroup$
Not sure if this is useful, but have a look at this identity for the divergence of a matrix $mathbf{A}$ acting on a vector $vec{v}$:
$$vecnabla cdot (mathbf{A}vec{v}) = (vecnabla cdot mathbf{A}) vec{v} + operatorname{Tr}left(mathbf{A}(vecnablavec{v})right)$$
Now if you take $mathbf{A}$ to be the Jacobi matrix of your vector field $vec{A}$, i.e. $mathbf{A}= vec{nabla} vec{A}$, you get
$$vecnabla cdot big((vec{nabla} vec{A})vec{v}big) = left(vecnabla cdot (vec{nabla} vec{A})right) vec{v} + operatorname{Tr}left((vec{nabla} vec{A})(vecnablavec{v})right) = left(nabla^2vec{A}right) vec{v} + operatorname{Tr}left((vec{nabla} vec{A})(vecnablavec{v})right)$$
answered Dec 29 '18 at 22:02
mechanodroidmechanodroid
28.9k62548
$endgroup$
Not sure if this is useful, but have a look at this identity for the divergence of a matrix $mathbf{A}$ acting on a vector $vec{v}$:
$$vecnabla cdot (mathbf{A}vec{v}) = (vecnabla cdot mathbf{A}) vec{v} + operatorname{Tr}left(mathbf{A}(vecnablavec{v})right)$$
Now if you take $mathbf{A}$ to be the Jacobi matrix of your vector field $vec{A}$, i.e. $mathbf{A}= vec{nabla} vec{A}$, you get
$$vecnabla cdot big((vec{nabla} vec{A})vec{v}big) = left(vecnabla cdot (vec{nabla} vec{A})right) vec{v} + operatorname{Tr}left((vec{nabla} vec{A})(vecnablavec{v})right) = left(nabla^2vec{A}right) vec{v} + operatorname{Tr}left((vec{nabla} vec{A})(vecnablavec{v})right)$$
answered Dec 29 '18 at 22:02
mechanodroidmechanodroid
28.9k62548
answered Dec 29 '18 at 22:02
mechanodroidmechanodroid
28.9k62548
answered Dec 29 '18 at 22:02
mechanodroidmechanodroid
28.9k62548
answered Dec 29 '18 at 22:02
mechanodroidmechanodroid
28.9k62548
28.9k62548
add a comment |
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$begingroup$
Perhaps you'll find my old post useful: math.stackexchange.com/questions/745000/greens-first-identity. Best, Daniel.
$endgroup$
– Dmoreno
Dec 29 '18 at 15:21
$begingroup$
Thank you for your response. It seems like the vector calculus identity $nabla cdot(mathbf{T} cdot vec{omega} ) = T:nablavec{omega} + (nabla cdot mathbf{T}) cdotvec{omega}$ which you linked solves the problem. I was able to come to the same solution with some different reasoning. I'll check the identity with some symbolic math software and then formulate an answer to my own question.
$endgroup$
– Mantabit
Dec 29 '18 at 18:39
$begingroup$
Glad to help @Mantabit. Best, Daniel.
$endgroup$
– Dmoreno
Dec 30 '18 at 18:02