Evolution of a Free Surface in a Porous Medium - Instituto de FÃsica ...

Evolution of a Free Surface in a Porous Medium 


Roberto André Kraenkel 

Instituto de Física Teórica-UNESP 

São Paulo - Brasil 

August 2007 / PDE’s-IMPA-Rio


Overview 

Underground water


Overview 

Underground water 

Free Surface Evolution in Porous Medium: setting the problem


Overview 


Free Surface Evolution in Porous Medium: setting the problem 

Well-known results


Overview 



Well-known results 

Exact equation


Overview 




Exact equation 

Conclusions




◮ Water that lies below the surface of the Earth




◮ Water that lies below the surface of the Earth 

◮ The underground is modeled as being an aggregation of particles 

with pores between them;






with pores between them; 

◮ Water can flow through these pores.







◮ Water can flow through these pores. 

◮ This flow will be considered laminar.








◮ This flow will be considered laminar. 

◮ The medium is saturated of water( the pore space is filled with 

water)










water) up to a certain constant-pressure surface,










water) up to a certain constant-pressure surface, 

◮ This is called a piezometric surface;











◮ This is called a piezometric surface; 

◮ Above this surface, pores are partially filled with water.












◮ Above this surface, pores are partially filled with water. 

◮ We will only consider a medium that is:













◮ We will only consider a medium that is: 

◮ “constant”, that is, there is no movement of the material that 

composes it;















composes it; 

◮ homogeneous















composes it; 

◮ homogeneous 

◮ We will consider a macroscopic description of the flow.















composes it; 


◮ We will consider a macroscopic description of the flow. 

◮ Our porous medium is bounded by below by an impermeable 

bottom.















composes it; 


◮ We will consider a macroscopic description of the flow. 

◮ Our porous medium is bounded by below by an impermeable 

bottom.



Setting the problem



Setting the problem 

◮ The problem is to study the propagation of a disturbance on the 

surface of a fluid totally immersed in a porous medium.





surface of a fluid totally immersed in a porous medium. 

◮ We want to get the equation governing the evolution of this 

disturbance.







disturbance. 

◮ Main Application:







disturbance. 

◮ Main Application: 

◮ Groundwater







disturbance. 


◮ Groundwater ⇒ Water table fluctuations







disturbance. 


◮ Groundwater ⇒ Water table fluctuations 

◮ Unconfined Aquifers







disturbance. 



◮ Unconfined Aquifers 

◮ Dam seepage problem







disturbance. 




◮ 

Dam seepage problem 

◮ We will consider 2 dimensional problems.







disturbance. 




◮ 

Dam seepage problem 

◮ We will consider 2 dimensional problems.



The equations 

◮ The very basics:



The equations 

◮ The very basics: the Darcy’s law.



The equations 

◮ The very basics: the Darcy’s law. 

◮ “the volume flux per unit area is proportional to an applied 

pressure”



The equations 



pressure” 

u = −K∇p



The equations 



pressure” 

u = −K∇p 

◮ 

where u has dimensions of velocity,



The equations 



pressure” 

u = −K∇p 

◮ 

where u has dimensions of velocity, and is called discharge.



The equations 



pressure” 

u = −K∇p 

◮ 

where u has dimensions of velocity, and is called discharge. 

◮ We also impose the conservation of the fluid mass



The equations 



pressure” 

u = −K∇p 

◮ 


◮ We also impose the conservation of the fluid mass 

◮ 

∇ · u = 0



The equations 



pressure” 

u = −K∇p 

◮ 



◮ 

∇ · u = 0 

◮ Leading to 

∇ 2 p = 0



The equations 



pressure” 

u = −K∇p 

◮ 



◮ 

∇ · u = 0 

◮ Leading to 

Nice! 

∇ 2 p = 0



More on equations 

◮ It is usual to consider instead of p, the piezometric head




◮ It is usual to consider instead of p, the piezometric head 

Φ = C · p + y





Φ = C · p + y 

where y is the coordinate in the direction of gravity, C is a 

constant





Φ = C · p + y 


constant 

◮ and obviously,





Φ = C · p + y 


constant 

◮ and obviously, 

◮ 

∇ 2 Φ = 0



The non-dimensional equations and boundary conditions



The non-dimensional equations and boundary conditions 

∇ 2 Φ = 0 0 < y < µh(x, t), (1) 

Φ = h − 1 y = µh(x, t), (2) 

0 = h t − Φ x h x + 1 µ Φ y y = µh(x, t), (3) 

Φ y = 0 y = 0 , (4) 

In nondimensional variables where




∇ 2 Φ = 0 0 < y < µh(x, t), (1) 

Φ = h − 1 y = µh(x, t), (2) 


Φ y = 0 y = 0 , (4) 

In nondimensional variables where 

◮ h 0 (x, t) is the free surface,




∇ 2 Φ = 0 0 < y < µh(x, t), (1) 

Φ = h − 1 y = µh(x, t), (2) 


Φ y = 0 y = 0 , (4) 


◮ h 0 (x, t) is the free surface, 

◮ µ = λ/h 0




∇ 2 Φ = 0 0 < y < µh(x, t), (1) 

Φ = h − 1 y = µh(x, t), (2) 


Φ y = 0 y = 0 , (4) 



◮ µ = λ/h 0 

◮ λ is the the typical length of the surface disturbance




∇ 2 Φ = 0 0 < y < µh(x, t), (1) 

Φ = h − 1 y = µh(x, t), (2) 


Φ y = 0 y = 0 , (4) 



◮ µ = λ/h 0 

◮ λ is the the typical length of the surface disturbance 

◮ h 0 is the undisturbed depth.




∇ 2 Φ = 0 0 < y < µh(x, t), (1) 

Φ = h − 1 y = µh(x, t), (2) 


Φ y = 0 y = 0 , (4) 



◮ µ = λ/h 0 


◮ h 0 is the undisturbed depth. 

◮ N.B.: we consider a flat bottom.




∇ 2 Φ = 0 0 < y < µh(x, t), (1) 

Φ = h − 1 y = µh(x, t), (2) 


Φ y = 0 y = 0 , (4) 



◮ µ = λ/h 0 



◮ N.B.: we consider a flat bottom. 

◮ Looks like usual hydrodynamics,




∇ 2 Φ = 0 0 < y < µh(x, t), (1) 

Φ = h − 1 y = µh(x, t), (2) 


Φ y = 0 y = 0 , (4) 



◮ µ = λ/h 0 




◮ Looks like usual hydrodynamics,but it is simpler!.




∇ 2 Φ = 0 0 < y < µh(x, t), (1) 

Φ = h − 1 y = µh(x, t), (2) 


Φ y = 0 y = 0 , (4) 



◮ µ = λ/h 0 




◮ Looks like usual hydrodynamics,but it is simpler!.




◮ The Dupuit-Forchheimer approximation




◮ The Dupuit-Forchheimer approximation 

◮ µ ≪ 1





◮ µ ≪ 1 

◮ that is, we look for a longwave limit.





◮ µ ≪ 1 

◮ that is, we look for a longwave limit. 

◮ for instance, if you are interested in a long and thin dam, this is 

OK.





◮ µ ≪ 1 

◮ that is, we look for a longwave limit. 

◮ for instance, if you are interested in a long and thin dam, this is 

OK. 

◮



The equations for the surface in the Dupuit approximation 

◮ Having a perturbative parameter




◮ Having a perturbative parameter , let us take advange of it.




◮ Having a perturbative parameter , let us take advange of it. 

◮ A long-wave expansion based on µ ≪ 1 gives us an equation for 

h(x, t)






h(x, t)correct to order µ:






h(x, t)correct to order µ: 

◮ 

h t = (hh x ) x







◮ 

h t = (hh x ) x 

◮ This equation is called the Boussinesq or







◮ 

h t = (hh x ) x 

◮ This equation is called the Boussinesq or Dupuit-Boussinesq 

equation



Dupuit-Boussinesq equation 

◮ A lot is known about this equation:




◮ A lot is known about this equation: 

◮ It possesses self-similar solutions.





◮ It possesses self-similar solutions. 

◮ For instance: h = t −1/3 g(ξ), with ξ = x/t 1/3






◮ For instance: h = t −1/3 g(ξ), with ξ = x/t 1/3 and 

g = 0 if ξ > ξ 0 , and g = 1 6 (ξ2 0 − ξ2 ) if ξ < ξ 0 .







g = 0 if ξ > ξ 0 , and g = 1 6 (ξ2 0 − ξ2 ) if ξ < ξ 0 . 

◮ This is an opening parabola with a shock at ξ0 .








◮ This is an opening parabola with a shock at ξ0 . 

◮ Self-similar solutions solutions dominate large time dynamics.





















◮ You can go beyond the first order approximation:



◮ You can go beyond the first order approximation: 

h t = ( 1 2 h2 + µ2 

3 h3 h xx ) xx




h t = ( 1 2 h2 + µ2 

3 h3 h xx ) xx 

◮ Which also desplays a – higher-order – shock.




h t = ( 1 2 h2 + µ2 


◮ Which also desplays a – higher-order – shock. 

◮ So, what if µ is not small.




h t = ( 1 2 h2 + µ2 


◮ Which also desplays a – higher-order – shock. 

◮ So, what if µ is not small.



Less well-known results 

◮ So, let us say that µ is just a parameter.




◮ So, let us say that µ is just a parameter. 

◮ No perturbation theory





◮ No perturbation theory. 

◮ New strategy:






◮ New strategy: 

◮ define a conformal transformation from the strip R × (0, h(x)) to 

R × (0, µ)








R × (0, µ) 

◮ this transformation eliminates the free-boundary problem, 

replacing it by a Laplace equation with new boundary conditions, 

envolving the Jacobian of the transformation








R × (0, µ) 



envolving the Jacobian of the transformation 

◮ then solve the Laplace equation, with two of the boundary 

conditions








R × (0, µ) 





conditions 

◮ this solution depends on h(x) , which is yet undetermined








R × (0, µ) 





conditions 

◮ this solution depends on h(x) , which is yet undetermined 

◮ use the third boundary solution to get the equation for h(x).








R × (0, µ) 





conditions 

◮ this solution depends on h(x) , which is yet undetermined 

◮ use the third boundary solution to get the equation for h(x). 

◮ Seems nice, let’s do it!.



Conformal transformation



Conformal transformation



Conformal transformation 

◮ We know that the transformation exists




◮ We know that the transformation exists 

◮ preserves the Laplacian





◮ preserves the Laplacian 

◮ preserves angles






◮ preserves angles 

◮ But the boundary conditions are not invariant.






◮ preserves angles 

◮ But the boundary conditions are not invariant.



The transformed equations



The transformed equations 

∇ 2 Φ = 0 0 < ζ < µ, (5) 

Φ = h − 1 ζ = µ, (6) 

0 = h t + Φ ζ 

µx ξ 

ζ = µ, (7) 

Φ ζ = 0 ζ = 0. (8)




∇ 2 Φ = 0 0 < ζ < µ, (5) 

Our free-boundary problem 

Φ = h − 1 ζ = µ, (6) 

0 = h t + Φ ζ 

µx ξ 

ζ = µ, (7) 

Φ ζ = 0 ζ = 0. (8)




∇ 2 Φ = 0 0 < ζ < µ, (5) 

Φ = h − 1 ζ = µ, (6) 

0 = h t + Φ ζ 

µx ξ 

ζ = µ, (7) 

Φ ζ = 0 ζ = 0. (8) 

Our free-boundary problemhas been transformed to a fixed-boundary 

problem




∇ 2 Φ = 0 0 < ζ < µ, (5) 

Φ = h − 1 ζ = µ, (6) 

0 = h t + Φ ζ 

µx ξ 

ζ = µ, (7) 

Φ ζ = 0 ζ = 0. (8) 


problem with auxiliary conditions involving an unknown function, 

h(x, t).




∇ 2 Φ = 0 0 < ζ < µ, (5) 

Φ = h − 1 ζ = µ, (6) 

0 = h t + Φ ζ 

µx ξ 

ζ = µ, (7) 

Φ ζ = 0 ζ = 0. (8) 



h(x, t). So, let’s just solve Laplace equation!




∇ 2 Φ = 0 0 < ζ < µ, (5) 

Φ = h − 1 ζ = µ, (6) 

0 = h t + Φ ζ 

µx ξ 

ζ = µ, (7) 

Φ ζ = 0 ζ = 0. (8) 



h(x, t). So, let’s just solve Laplace equation!



Solving Laplace’s equation 

The following expression solves Laplace equation,




The following expression solves Laplace equation, 

◮ with Φ = h − 1 at ζ = µ and





◮ with Φ = h − 1 at ζ = µ and 

◮ Φ ζ = 0 at ζ = 0:






◮ Φ ζ = 0 at ζ = 0: 

Φ(ξ, ζ, t) = 

∫ ∞ 

−∞ 

F[Φ(ξ, µ, t)] cosh[2πκζ] 

cosh[2πκµ] e2πiκξ dκ, (9)






◮ Φ ζ = 0 at ζ = 0: 

Φ(ξ, ζ, t) = 

∫ ∞ 

−∞ 


cosh[2πκµ] e2πiκξ dκ, (9) 

where F[Φ](κ, µ, t) is the Fourier Transform of the piezometric head 

Φ at the surface ζ = µ, given in terms of h − 1.






◮ Φ ζ = 0 at ζ = 0: 

Φ(ξ, ζ, t) = 

∫ ∞ 

−∞ 


cosh[2πκµ] e2πiκξ dκ, (9) 

where F[Φ](κ, µ, t) is the Fourier Transform of the piezometric head 

Φ at the surface ζ = µ, given in terms of h − 1. 

As noted before, everything is given in terms of h(x(ξ, ζ), t).



The EQUATION



The EQUATION 

There remains the dynamics boundary condition:



The EQUATION 

There remains the dynamics boundary condition: 

h t + Φ ζ 

µx ξ 

= 0 at ζ = µ



The EQUATION 


We already calculated Φ(ξ, ζ, t). 

h t + Φ ζ 

µx ξ 

= 0 at ζ = µ



The EQUATION 


h t + Φ ζ 

µx ξ 

= 0 at ζ = µ 

We already calculated Φ(ξ, ζ, t).Just calculate the derivative and plug 

it into this equation:



The EQUATION 


h t + Φ ζ 

µx ξ 

= 0 at ζ = µ 


it into this equation:The derivative:



The EQUATION 


h t + Φ ζ 

µx ξ 

= 0 at ζ = µ 


it into this equation:The derivative: 

Φ ζ (ξ, µ, t) = 

= 

∫ ∞ 

−∞ 

∫ ∞ 

−∞ 

2πκ tanh[2πκµ]F[Φ] e 2πiκξ dκ (10) 

−i tanh[2πκµ]F[Φ ξ ] e 2πiκξ dκ ≡ T[Φ ξ ] , (11)



The EQUATION 


h t + Φ ζ 

µx ξ 

= 0 at ζ = µ 



Φ ζ (ξ, µ, t) = 

= 

∫ ∞ 

−∞ 

∫ ∞ 

−∞ 


−i tanh[2πκµ]F[Φ ξ ] e 2πiκξ dκ ≡ T[Φ ξ ] , (11) 

where T[−] is an integral operator defined by the above equation.



The EQUATION 


h t + Φ ζ 

µx ξ 

= 0 at ζ = µ 



Φ ζ (ξ, µ, t) = 

= 

∫ ∞ 

−∞ 

∫ ∞ 

−∞ 



where T[−] is an integral operator defined by the above equation. So, 

now we have the equation:



The EQUATION 


h t + Φ ζ 

µx ξ 

= 0 at ζ = µ 



Φ ζ (ξ, µ, t) = 

= 

∫ ∞ 

−∞ 

∫ ∞ 

−∞ 




now we have the equation: 

h t + 1 

µx ξ 

T[h ξ ] = 0 , (12)



The EQUATION 


h t + Φ ζ 

µx ξ 

= 0 at ζ = µ 



Φ ζ (ξ, µ, t) = 

= 

∫ ∞ 

−∞ 

∫ ∞ 

−∞ 




now we have the equation: 

h t + 1 

µx ξ 

T[h ξ ] = 0 , (12)



Appraisal of the EQUATION



Appraisal of the EQUATION 

◮ It is an exact equation




◮ It is an exact equation 

◮ It is an elegant equation





◮ It is an elegant equation 

◮ It is (1+1)-dimensional equation






◮ It is (1+1)-dimensional equation 

◮ Although it looks complicated,it is amenable to numerics with 

FFT implementation.






◮ It is (1+1)-dimensional equation 

◮ Although it looks complicated,it is amenable to numerics with 

FFT implementation.



BUT,...



BUT,... 

◮ Isn’t that a linear equation?



BUT,... 

◮ Isn’t that a linear equation? 

◮ NO!.



BUT,... 


◮ NO!. 

◮ The EQUATION gives the time evolution of the free surface in 

the conformal coordinates (ξ, ζ).



BUT,... 


◮ NO!. 


the conformal coordinates (ξ, ζ). 

◮ x ξ , depends on h(x, t), making Eq.(12) nonlinear.



BUT,... 


◮ NO!. 



◮ x ξ , depends on h(x, t), making Eq.(12) nonlinear. 

◮ But still...



BUT,... 


◮ NO!. 




◮ But still... All of this is in the conformal coordinates, which have 

not been shown.



BUT,... 


◮ NO!. 





not been shown. 

◮ Can one obtain them?



BUT,... 


◮ NO!. 






◮ Can one obtain them? Sure.



BUT,... 


◮ NO!. 






◮ Can one obtain them? Sure.



Transformed coordinates 

∫ 

y(ξ, ζ) = µ 

∫ 

= µ 

R 

R 

sinh[2πkζ] 

sinh[2πkµ] F[h]e2πiκξ dκ 


sinh[2πkµ] F[h − 1]e2πiκξ dκ + ζ. 

(13)



Transformed coordinates 

∫ 

y(ξ, ζ) = µ 

∫ 

= µ 

R 

R 

∫ ∞ 

x(ξ, ζ) = −iµ 

= −iµ 

−∞ 

∫ ∞ 

−∞ 


sinh[2πkµ] F[h]e2πiκξ dκ 


sinh[2πkµ] F[h − 1]e2πiκξ dκ + ζ. 

cosh[2πκζ] 

sinh[2πκµ] F[h]e2πiκξ dκ 

cosh[2πκζ] 

sinh[2πκµ] F[h − 1]e2πiκξ dκ + ξ, 

(13) 

(14)



More.. 

◮ You can do asymptotics for small µ and recover previous results.



More.. 

◮ You can do asymptotics for small µ and recover previous results. 

◮ You can also study the case linear case:



More.. 


◮ You can also study the case linear case: 

◮ Take h = 1 + η, where η is the displacement of the free surface 

from its undisturbed position.



More.. 




from its undisturbed position. 

◮ In the case where η ≪ 1, as a first approximation we can obtain 

a differential equation for η by noting that:



More.. 






a differential equation for η by noting that: 

η t = − 1 T[η ξ ] 

µ 1 − µT −1 [η ξ ] 

= − 1 µ T[η ξ] − T[η ξ ]T −1 [η ξ ] + . . . , 

(15)



More.. 







η t = − 1 T[η ξ ] 

µ 1 − µT −1 [η ξ ] 

= − 1 µ T[η ξ] − T[η ξ ]T −1 [η ξ ] + . . . , 

◮ and thus the lowest order, linear, equation reads 

(15)



More.. 







η t = − 1 T[η ξ ] 

µ 1 − µT −1 [η ξ ] 

= − 1 µ T[η ξ] − T[η ξ ]T −1 [η ξ ] + . . . , 


(15) 

η t = − 1 µ T[η ξ] (16)



More.. 







η t = − 1 T[η ξ ] 

µ 1 − µT −1 [η ξ ] 

= − 1 µ T[η ξ] − T[η ξ ]T −1 [η ξ ] + . . . , 


(15) 

η t = − 1 µ T[η ξ] (16)


Conclusions 

Conclusions


Conclusions 

Conclusions 

◮ We have introduced a new integro-differential equation 

describing exactly the evolution of a free surface of a fluid totally 

imersed in a saturated porous medium and bounded by below by 

an impermeable bottom.


Conclusions 

Conclusions 




an impermeable bottom. 

◮ Our equation is a porous-media analog of the exact equation 

found for water waves by Zakharov.


Conclusions 

Conclusions 






found for water waves by Zakharov. 

◮ We have also shown that the asymptotic long-wave expansion for 

this equations leads to known equations.


Conclusions 

Conclusions 








this equations leads to known equations. 

◮ We obtained the linear limit and ( although not shown here) 

obtain and exact solution.


Conclusions 

Conclusions 










obtain and exact solution. 

◮ Reference: W. Artiles and R.A. Kraenkel, SIAM J. Appl. 

Math,67, 619, (2006)


Conclusions 

Conclusions 










obtain and exact solution. 

◮ Reference: W. Artiles and R.A. Kraenkel, SIAM J. Appl. 

Math,67, 619, (2006) 

◮ Obrigado = Thanks

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