- #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.

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- Thread starter andrey21
- Start date

- #1

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q(x.t) = -k

Show that:

Can be written as:

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

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

- #2

hunt_mat

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[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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- #4

hunt_mat

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check the signs...

- #5

- 466

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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)

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

- #10

hunt_mat

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Mat

- #11

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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

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[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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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?

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

so with that in mind:

u =h(t,x)

v = dh/dx

correct?

- #16

Char. Limit

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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?

You got it.

- #17

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Great, problem is Im nt sure how to differentiate either of those numbers??

- #18

Char. Limit

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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?

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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- #22

Char. Limit

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Just differentiate it again. You should get d^2 h / dx^2

- #23

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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

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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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