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Advection equation for a gradient and higher order derivatives

0

$begingroup$

If a scalar function $phi$ satisfy the advection equation
$$
partial_t phi + v_k partial_k phi = 0,
$$
where $v_k$ is some vector field. Then the gradient $partial_i phi$ satisfy the following advection equation
$$
partial_t partial_i phi + v_k partial_i partial_k phi + (partial_i v_k) partial_k phi = 0.
$$

With the help of Lie derivative we can rewrite the above equations as
beginalign
partial_t phi + mathcalL_v phi
&=
0,
\
partial_t nabla phi + mathcalL_v nabla phi
&=
0.
endalign

What about the second derivative(Hessian) $nabla^2 phi$, what equation does it satisfy? I did the calculation in coordinates and I got
$$
partial_t nabla^2 phi + mathcalL_v nabla^2 phi
= - nabla phi cdot nabla^2 v,
$$
just to clarify, we have $(nabla phi cdot nabla^2 v)_ij = (partial_k phi)(partial_i partial_j v_k)$.

Is there a nice way how to derive the equation for $nabla^2 phi$ without doing the calculation in coordinates? Does it scale to higher derivatives?

differential-geometry pde lie-derivative

share|cite|improve this question

edited Apr 2 at 7:52

tom

asked Apr 1 at 14:10

tomtom

2,96711233

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How can you dot a vector with a scalar in $nabla phi cdot nabla^2 v$?
  $endgroup$
  – Mattos
  Apr 2 at 1:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was expecting that the notation might be confusing, that is why I wrote an explicit formula in coordinates: $(nabla phi cdot nabla^2 v)_ij = (partial_k phi)(partial_i partial_j v_k)$. Note, that $v$ is a vector, its second derivative $nabla^2 v$ is a rank three tensor and $nabla phi$ is a vector.
  $endgroup$
  – tom
  Apr 2 at 7:51
  

  
  
  
  

add a comment | 

0

$begingroup$

If a scalar function $phi$ satisfy the advection equation
$$
partial_t phi + v_k partial_k phi = 0,
$$
where $v_k$ is some vector field. Then the gradient $partial_i phi$ satisfy the following advection equation
$$
partial_t partial_i phi + v_k partial_i partial_k phi + (partial_i v_k) partial_k phi = 0.
$$

With the help of Lie derivative we can rewrite the above equations as
beginalign
partial_t phi + mathcalL_v phi
&=
0,
\
partial_t nabla phi + mathcalL_v nabla phi
&=
0.
endalign

What about the second derivative(Hessian) $nabla^2 phi$, what equation does it satisfy? I did the calculation in coordinates and I got
$$
partial_t nabla^2 phi + mathcalL_v nabla^2 phi
= - nabla phi cdot nabla^2 v,
$$
just to clarify, we have $(nabla phi cdot nabla^2 v)_ij = (partial_k phi)(partial_i partial_j v_k)$.

Is there a nice way how to derive the equation for $nabla^2 phi$ without doing the calculation in coordinates? Does it scale to higher derivatives?

differential-geometry pde lie-derivative

share|cite|improve this question

edited Apr 2 at 7:52

tom

asked Apr 1 at 14:10

tomtom

2,96711233

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How can you dot a vector with a scalar in $nabla phi cdot nabla^2 v$?
  $endgroup$
  – Mattos
  Apr 2 at 1:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was expecting that the notation might be confusing, that is why I wrote an explicit formula in coordinates: $(nabla phi cdot nabla^2 v)_ij = (partial_k phi)(partial_i partial_j v_k)$. Note, that $v$ is a vector, its second derivative $nabla^2 v$ is a rank three tensor and $nabla phi$ is a vector.
  $endgroup$
  – tom
  Apr 2 at 7:51
  

  
  
  
  

add a comment | 

$begingroup$

If a scalar function $phi$ satisfy the advection equation
$$
partial_t phi + v_k partial_k phi = 0,
$$
where $v_k$ is some vector field. Then the gradient $partial_i phi$ satisfy the following advection equation
$$
partial_t partial_i phi + v_k partial_i partial_k phi + (partial_i v_k) partial_k phi = 0.
$$

With the help of Lie derivative we can rewrite the above equations as
beginalign
partial_t phi + mathcalL_v phi
&=
0,
\
partial_t nabla phi + mathcalL_v nabla phi
&=
0.
endalign

What about the second derivative(Hessian) $nabla^2 phi$, what equation does it satisfy? I did the calculation in coordinates and I got
$$
partial_t nabla^2 phi + mathcalL_v nabla^2 phi
= - nabla phi cdot nabla^2 v,
$$
just to clarify, we have $(nabla phi cdot nabla^2 v)_ij = (partial_k phi)(partial_i partial_j v_k)$.

Is there a nice way how to derive the equation for $nabla^2 phi$ without doing the calculation in coordinates? Does it scale to higher derivatives?

differential-geometry pde lie-derivative

share|cite|improve this question

edited Apr 2 at 7:52

tom

asked Apr 1 at 14:10

tomtom

2,96711233

$endgroup$

share|cite|improve this question

edited Apr 2 at 7:52

tom

asked Apr 1 at 14:10

tomtom

2,96711233

share|cite|improve this question

edited Apr 2 at 7:52

tom

asked Apr 1 at 14:10

tomtom

2,96711233

asked Apr 1 at 14:10

tomtom

2,96711233

asked Apr 1 at 14:10

tomtom

2,96711233