Partial Differential Equation(NEED HELP)

  • Thread starter andrey21
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  • #1
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Using Darcy's Law given below:

q(x.t) = -k dh/dx

Show that:

-d/dx.(q(x,t)h(x,t)) = phi dh/dt

Can be written as:

K(h(d^(2)u/dx^(2)) +(du/dx)^2)) = phi dh/dt

Any help would be greatly appreciated, Im nt to sure where to begin.
 

Answers and Replies

  • #2
hunt_mat
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You have:
[tex]
\frac{\partial}{\partial x}(h(t,x)q(t,x))=\phi\frac{\partial h}{\partial t}
[/tex]
Insert Darcy's law to obtain:
[tex]
\frac{\partial}{\partial x}\left( -kh(t,x)\frac{\partial h}{\partial x}\right)=\phi\frac{\partial h}{\partial t}
[/tex]
 
  • #3
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Ok I have done what u suggested although think u missed a minus sign out at the front. Do I pull the k outside the brackets??
 
  • #4
hunt_mat
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check the signs...
 
  • #5
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Ah ok so what is the next step, take out k as I suggested??
 
  • #6
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Should I use the fact that:

h(x,t) can be represented by:

dh/dt = k d^(2) h/dx^(2)
 
  • #7
hunt_mat
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I was thinking, should tou have q(t,x)h(t,x)? Should it just be q(t,x)?
 
  • #8
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its q(t,x)h(t,x).
 
  • #9
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Sorry but the first post contained an error. I need to rewrite the PDE so it becomes:

K(h(d^(2)h/dx^(2)) +(dh/dx)^2)) = phi dh/dt
 
  • #10
hunt_mat
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The k is a constant so that can be taken out of the derivative and then it should all fall into place.

Mat
 
  • #11
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Ye ok so doing that I obtain:

k(d/dx h(t,x) dh/dx) = phi dh/dt

How do I get to the required answer from here.

Thanks in advance.
 
  • #12
hunt_mat
Homework Helper
1,745
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Just take k out and differentiate:
[tex]
-k\frac{\partial}{\partial x}\right( h(t,x)\frac{\partial h}{\partial x}\right)=-k\frac{\partial^{2}h}{\partial x^{2}}-k\left(\frac{\partial h}{\partial x}\right)^{2}
[/tex]
 
  • #13
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Rite so:

k(d/dx h(t,x) dh/dx) = phi dh/dt

now differentating:

d/dx h(x,t) = d^2h/dx^2

dh/dx = (dh/dx)^2

is this correct??

Also there is a h missing from the final answer.
 
  • #14
Char. Limit
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Rite so:

k(d/dx (h(t,x) dh/dx)) = phi dh/dt

now differentating:

d/dx h(x,t) = d^2h/dx^2

dh/dx = (dh/dx)^2

is this correct??

Also there is a h missing from the final answer.

Remember the product rule? You use it here. (parentheses bolded, because that's the part to concentrate on).
 
  • #15
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Ah yes:

d/dx(uv) = udv/dx + vdu/dx

so with that in mind:

u =h(t,x)
v = dh/dx

correct?
 
  • #16
Char. Limit
Gold Member
1,208
14
Ah yes:

d/dx(uv) = udv/dx + vdu/dx

so with that in mind:

u =h(t,x)
v = dh/dx

correct?

You got it.
 
  • #17
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Great, problem is Im nt sure how to differentiate either of those numbers??
 
  • #18
Char. Limit
Gold Member
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Great, problem is Im nt sure how to differentiate either of those numbers??

Well, what's the partial derivative wrt x of [itex]\frac{\partial h}{\partial x}[/itex]? How about [itex]h(x,t)[/itex]?
 
  • #19
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well the partial derivative of h(x,t) is

d^(2) h/ dxdt

is this correct?
 
  • #20
Char. Limit
Gold Member
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well the partial derivative of h(x,t) is

d^(2) h/ dxdt

is this correct?

Not quite. That is the "mixed second partial derivative" of h(x,t). The partial derivative of h(x,t) wrt x is just [tex]\partial h / \partial x[/tex].
 
  • #21
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Ah i see got a little confused there so I now have value for v du/dx as (dh/dx)^2. I just cant work out partial dervative for dh/dx
 
  • #22
Char. Limit
Gold Member
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14
Ah i see got a little confused there so I now have value for v du/dx as (dh/dx)^2. I just cant work out partial dervative for dh/dx

Just differentiate it again. You should get d^2 h / dx^2
 
  • #23
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Yes i see now so putting that together I obtain:

u dv/dx + vdu/dx

h(t,x).d^2h/dx^2 + (dh/dx)^2

In the final answer given the h(t,x) isnt there. Do this cancel out?
 
  • #24
Char. Limit
Gold Member
1,208
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Yes i see now so putting that together I obtain:

u dv/dx + vdu/dx

h(t,x).d^2h/dx^2 + (dh/dx)^2

In the final answer given the h(t,x) isnt there. Do this cancel out?

I see an h in your original post. I think that's the h(t,x) we're talking about, and so I'm pretty sure it is there.
 
  • #25
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Ah great thanks Char.limit for all ur hep.
 

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