Richards equation unique solution
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 at 20:40
Leonardo Sandoval
32
add a comment |
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 at 20:40
Leonardo Sandoval
32
Why are you subscripting $C(h)$ and $K(h)$? You do treat them as constants in your simulation?
– Lee David Chung Lin
Oct 11 at 21:51
No, in the simulation they should be computed with the actual value of $h$. Those are empirical functions of $h$.
– Leonardo Sandoval
Oct 12 at 20:10
So would you mind correcting the typo? It's very misleading.
– Lee David Chung Lin
Oct 12 at 20:13
Please specify all your variables. What is $z$?
– Hans
Oct 12 at 21:46
add a comment |
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 at 20:40
Leonardo Sandoval
32
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 at 20:40
Leonardo Sandoval
32
asked Oct 11 at 20:40
Leonardo Sandoval
32
edited Dec 1 at 19:13
asked Oct 11 at 20:40
Leonardo Sandoval
32
asked Oct 11 at 20:40
Leonardo Sandoval
32
asked Oct 11 at 20:40
Leonardo Sandoval
32
32
Why are you subscripting $C(h)$ and $K(h)$? You do treat them as constants in your simulation?
– Lee David Chung Lin
Oct 11 at 21:51
No, in the simulation they should be computed with the actual value of $h$. Those are empirical functions of $h$.
– Leonardo Sandoval
Oct 12 at 20:10
So would you mind correcting the typo? It's very misleading.
– Lee David Chung Lin
Oct 12 at 20:13
Please specify all your variables. What is $z$?
– Hans
Oct 12 at 21:46
add a comment |
Why are you subscripting $C(h)$ and $K(h)$? You do treat them as constants in your simulation?
– Lee David Chung Lin
Oct 11 at 21:51
No, in the simulation they should be computed with the actual value of $h$. Those are empirical functions of $h$.
– Leonardo Sandoval
Oct 12 at 20:10
So would you mind correcting the typo? It's very misleading.
– Lee David Chung Lin
Oct 12 at 20:13
Please specify all your variables. What is $z$?
– Hans
Oct 12 at 21:46
Why are you subscripting $C(h)$ and $K(h)$? You do treat them as constants in your simulation?
– Lee David Chung Lin
Oct 11 at 21:51
Why are you subscripting $C(h)$ and $K(h)$? You do treat them as constants in your simulation?
– Lee David Chung Lin
Oct 11 at 21:51
No, in the simulation they should be computed with the actual value of $h$. Those are empirical functions of $h$.
– Leonardo Sandoval
Oct 12 at 20:10
No, in the simulation they should be computed with the actual value of $h$. Those are empirical functions of $h$.
– Leonardo Sandoval
Oct 12 at 20:10
So would you mind correcting the typo? It's very misleading.
– Lee David Chung Lin
Oct 12 at 20:13
So would you mind correcting the typo? It's very misleading.
– Lee David Chung Lin
Oct 12 at 20:13
Please specify all your variables. What is $z$?
– Hans
Oct 12 at 21:46
Please specify all your variables. What is $z$?
– Hans
Oct 12 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 at 21:53
Hans
4,94521032
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 at 21:53
Hans
4,94521032
add a comment |
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 at 21:53
Hans
4,94521032
add a comment |
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 at 21:53
Hans
4,94521032
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 at 21:53
Hans
4,94521032
answered Oct 12 at 21:53
Hans
4,94521032
answered Oct 12 at 21:53
Hans
4,94521032
answered Oct 12 at 21:53
Hans
4,94521032
4,94521032
add a comment |
add a comment |
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Why are you subscripting $C(h)$ and $K(h)$? You do treat them as constants in your simulation?
– Lee David Chung Lin
Oct 11 at 21:51
No, in the simulation they should be computed with the actual value of $h$. Those are empirical functions of $h$.
– Leonardo Sandoval
Oct 12 at 20:10
So would you mind correcting the typo? It's very misleading.
– Lee David Chung Lin
Oct 12 at 20:13
Please specify all your variables. What is $z$?
– Hans
Oct 12 at 21:46