Richards equation unique solution
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Richards equation is used for describe flow in unsaturated porous media:
$$C{(h)}frac{partial h}{partial t}= nabla [K{(h)} nabla (H)]$$
$h=$ capillary pressure
$K$ and C properties in function of $h$
$H=$ energy of the fluid $= h+z$ with z being the vertical axis.
I performed some 2D simulations with neumann BC in all the boundaries with FVM. The point is that no mather the initial condition, when the simulation reach steady state, the solution is always the same. Could anybody explains me why this happen??
I suppose this is a parabolic PDE but reading in some books this kind of equations require at least one dirichlet BC to be well-posed. However, I suppose this only applies for linear equations and Richards equation is not linear.
pde simulation
asked Oct 11 '18 at 20:40
Leonardo SandovalLeonardo Sandoval
32
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add a comment |
$begingroup$
Richards equation is used for describe flow in unsaturated porous media:
$$C{(h)}frac{partial h}{partial t}= nabla [K{(h)} nabla (H)]$$
$h=$ capillary pressure
$K$ and C properties in function of $h$
$H=$ energy of the fluid $= h+z$ with z being the vertical axis.
I performed some 2D simulations with neumann BC in all the boundaries with FVM. The point is that no mather the initial condition, when the simulation reach steady state, the solution is always the same. Could anybody explains me why this happen??
I suppose this is a parabolic PDE but reading in some books this kind of equations require at least one dirichlet BC to be well-posed. However, I suppose this only applies for linear equations and Richards equation is not linear.
pde simulation
asked Oct 11 '18 at 20:40
Leonardo SandovalLeonardo Sandoval
32
$endgroup$
$begingroup$
Why are you subscripting $C(h)$ and $K(h)$? You do treat them as constants in your simulation?
$endgroup$
– Lee David Chung Lin
Oct 11 '18 at 21:51
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No, in the simulation they should be computed with the actual value of $h$. Those are empirical functions of $h$.
$endgroup$
– Leonardo Sandoval
Oct 12 '18 at 20:10
$begingroup$
So would you mind correcting the typo? It's very misleading.
$endgroup$
– Lee David Chung Lin
Oct 12 '18 at 20:13
$begingroup$
Please specify all your variables. What is $z$?
$endgroup$
– Hans
Oct 12 '18 at 21:46
add a comment |
$begingroup$
Richards equation is used for describe flow in unsaturated porous media:
$$C{(h)}frac{partial h}{partial t}= nabla [K{(h)} nabla (H)]$$
$h=$ capillary pressure
$K$ and C properties in function of $h$
$H=$ energy of the fluid $= h+z$ with z being the vertical axis.
I performed some 2D simulations with neumann BC in all the boundaries with FVM. The point is that no mather the initial condition, when the simulation reach steady state, the solution is always the same. Could anybody explains me why this happen??
I suppose this is a parabolic PDE but reading in some books this kind of equations require at least one dirichlet BC to be well-posed. However, I suppose this only applies for linear equations and Richards equation is not linear.
pde simulation
asked Oct 11 '18 at 20:40
Leonardo SandovalLeonardo Sandoval
32
$endgroup$
Richards equation is used for describe flow in unsaturated porous media:
$$C{(h)}frac{partial h}{partial t}= nabla [K{(h)} nabla (H)]$$
$h=$ capillary pressure
$K$ and C properties in function of $h$
$H=$ energy of the fluid $= h+z$ with z being the vertical axis.
I performed some 2D simulations with neumann BC in all the boundaries with FVM. The point is that no mather the initial condition, when the simulation reach steady state, the solution is always the same. Could anybody explains me why this happen??
I suppose this is a parabolic PDE but reading in some books this kind of equations require at least one dirichlet BC to be well-posed. However, I suppose this only applies for linear equations and Richards equation is not linear.
pde simulation
pde simulation
asked Oct 11 '18 at 20:40
Leonardo SandovalLeonardo Sandoval
32
asked Oct 11 '18 at 20:40
Leonardo SandovalLeonardo Sandoval
32
edited Dec 1 '18 at 19:13
Leonardo Sandoval
asked Oct 11 '18 at 20:40
Leonardo SandovalLeonardo Sandoval
32
asked Oct 11 '18 at 20:40
Leonardo SandovalLeonardo Sandoval
32
asked Oct 11 '18 at 20:40
Leonardo SandovalLeonardo Sandoval
32
32
$begingroup$
Why are you subscripting $C(h)$ and $K(h)$? You do treat them as constants in your simulation?
$endgroup$
– Lee David Chung Lin
Oct 11 '18 at 21:51
$begingroup$
No, in the simulation they should be computed with the actual value of $h$. Those are empirical functions of $h$.
$endgroup$
– Leonardo Sandoval
Oct 12 '18 at 20:10
$begingroup$
So would you mind correcting the typo? It's very misleading.
$endgroup$
– Lee David Chung Lin
Oct 12 '18 at 20:13
$begingroup$
Please specify all your variables. What is $z$?
$endgroup$
– Hans
Oct 12 '18 at 21:46
add a comment |
$begingroup$
Why are you subscripting $C(h)$ and $K(h)$? You do treat them as constants in your simulation?
$endgroup$
– Lee David Chung Lin
Oct 11 '18 at 21:51
$begingroup$
No, in the simulation they should be computed with the actual value of $h$. Those are empirical functions of $h$.
$endgroup$
– Leonardo Sandoval
Oct 12 '18 at 20:10
$begingroup$
So would you mind correcting the typo? It's very misleading.
$endgroup$
– Lee David Chung Lin
Oct 12 '18 at 20:13
$begingroup$
Please specify all your variables. What is $z$?
$endgroup$
– Hans
Oct 12 '18 at 21:46
$begingroup$
Why are you subscripting $C(h)$ and $K(h)$? You do treat them as constants in your simulation?
$endgroup$
– Lee David Chung Lin
Oct 11 '18 at 21:51
$begingroup$
Why are you subscripting $C(h)$ and $K(h)$? You do treat them as constants in your simulation?
$endgroup$
– Lee David Chung Lin
Oct 11 '18 at 21:51
$begingroup$
No, in the simulation they should be computed with the actual value of $h$. Those are empirical functions of $h$.
$endgroup$
– Leonardo Sandoval
Oct 12 '18 at 20:10
$begingroup$
No, in the simulation they should be computed with the actual value of $h$. Those are empirical functions of $h$.
$endgroup$
– Leonardo Sandoval
Oct 12 '18 at 20:10
$begingroup$
So would you mind correcting the typo? It's very misleading.
$endgroup$
– Lee David Chung Lin
Oct 12 '18 at 20:13
$begingroup$
So would you mind correcting the typo? It's very misleading.
$endgroup$
– Lee David Chung Lin
Oct 12 '18 at 20:13
$begingroup$
Please specify all your variables. What is $z$?
$endgroup$
– Hans
Oct 12 '18 at 21:46
$begingroup$
Please specify all your variables. What is $z$?
$endgroup$
– Hans
Oct 12 '18 at 21:46
add a comment |
1 Answer
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Steady state means $frac{partial h}{partial t}=0$ and the boundary condition is predetermined and time independent as well. So the steady state PDE is an elliptical equation. Its solution is uniquely determined by the boundary condition which is now time independent. Therefore the steady state solution is independent of the initial condition and unique.
answered Oct 12 '18 at 21:53
HansHans
4,98021032
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Steady state means $frac{partial h}{partial t}=0$ and the boundary condition is predetermined and time independent as well. So the steady state PDE is an elliptical equation. Its solution is uniquely determined by the boundary condition which is now time independent. Therefore the steady state solution is independent of the initial condition and unique.
answered Oct 12 '18 at 21:53
HansHans
4,98021032
$endgroup$
add a comment |
$begingroup$
Steady state means $frac{partial h}{partial t}=0$ and the boundary condition is predetermined and time independent as well. So the steady state PDE is an elliptical equation. Its solution is uniquely determined by the boundary condition which is now time independent. Therefore the steady state solution is independent of the initial condition and unique.
answered Oct 12 '18 at 21:53
HansHans
4,98021032
$endgroup$
add a comment |
$begingroup$
Steady state means $frac{partial h}{partial t}=0$ and the boundary condition is predetermined and time independent as well. So the steady state PDE is an elliptical equation. Its solution is uniquely determined by the boundary condition which is now time independent. Therefore the steady state solution is independent of the initial condition and unique.
answered Oct 12 '18 at 21:53
HansHans
4,98021032
$endgroup$
Steady state means $frac{partial h}{partial t}=0$ and the boundary condition is predetermined and time independent as well. So the steady state PDE is an elliptical equation. Its solution is uniquely determined by the boundary condition which is now time independent. Therefore the steady state solution is independent of the initial condition and unique.
answered Oct 12 '18 at 21:53
HansHans
4,98021032
answered Oct 12 '18 at 21:53
HansHans
4,98021032
answered Oct 12 '18 at 21:53
HansHans
4,98021032
answered Oct 12 '18 at 21:53
HansHans
4,98021032
4,98021032
add a comment |
add a comment |
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$begingroup$
Why are you subscripting $C(h)$ and $K(h)$? You do treat them as constants in your simulation?
$endgroup$
– Lee David Chung Lin
Oct 11 '18 at 21:51
$begingroup$
No, in the simulation they should be computed with the actual value of $h$. Those are empirical functions of $h$.
$endgroup$
– Leonardo Sandoval
Oct 12 '18 at 20:10
$begingroup$
So would you mind correcting the typo? It's very misleading.
$endgroup$
– Lee David Chung Lin
Oct 12 '18 at 20:13
$begingroup$
Please specify all your variables. What is $z$?
$endgroup$
– Hans
Oct 12 '18 at 21:46