# Partial Differential Equation(NEED HELP)

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.

hunt_mat
Homework Helper
You have:
$$\frac{\partial}{\partial x}(h(t,x)q(t,x))=\phi\frac{\partial h}{\partial t}$$
Insert Darcy's law to obtain:
$$\frac{\partial}{\partial x}\left( -kh(t,x)\frac{\partial h}{\partial x}\right)=\phi\frac{\partial h}{\partial t}$$

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??

hunt_mat
Homework Helper
check the signs...

Ah ok so what is the next step, take out k as I suggested??

Should I use the fact that:

h(x,t) can be represented by:

dh/dt = k d^(2) h/dx^(2)

hunt_mat
Homework Helper
I was thinking, should tou have q(t,x)h(t,x)? Should it just be q(t,x)?

its q(t,x)h(t,x).

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

hunt_mat
Homework Helper
The k is a constant so that can be taken out of the derivative and then it should all fall into place.

Mat

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.

hunt_mat
Homework Helper
Just take k out and differentiate:
$$-k\frac{\partial}{\partial x}\right( h(t,x)\frac{\partial h}{\partial x}\right)=-k\frac{\partial^{2}h}{\partial x^{2}}-k\left(\frac{\partial h}{\partial x}\right)^{2}$$

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.

Char. Limit
Gold Member
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).

Ah yes:

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

so with that in mind:

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

correct?

Char. Limit
Gold Member
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.

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

Char. Limit
Gold Member
Great, problem is Im nt sure how to differentiate either of those numbers??

Well, what's the partial derivative wrt x of $\frac{\partial h}{\partial x}$? How about $h(x,t)$?

well the partial derivative of h(x,t) is

d^(2) h/ dxdt

is this correct?

Char. Limit
Gold Member
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 $$\partial h / \partial x$$.

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

Char. Limit
Gold Member
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

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?

Char. Limit
Gold Member
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.

Ah great thanks Char.limit for all ur hep.