Partial Differential Equation(NEED HELP)

[andrey21]

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, I am nt to sure where to begin.

 

[hunt_mat]

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

 

[andrey21]

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]

check the signs...

 

[andrey21]

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

 

[andrey21]

Should I use the fact that:

h(x,t) can be represented by:

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

 

[hunt_mat]

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

 

[andrey21]

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

 

[andrey21]

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]

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

Mat

 

[andrey21]

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.

 

[hunt_mat]

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

 

[andrey21]

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]

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).

 

[andrey21]

Ah yes:

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

so with that in mind:

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

correct?

 

[Char. Limit]

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.

 

[andrey21]

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

 

[Char. Limit]

Great, problem is I am 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 $h(x,t)$?

 

[andrey21]

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

d^(2) h/ dxdt

is this correct?

 

[Char. Limit]

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 $$\partial h / \partial x$$.

 

[andrey21]

Ah i see got a little confused there so I now have value for v du/dx as (dh/dx)^2. I just can't work out partial dervative for dh/dx

 

[Char. Limit]

Ah i see got a little confused there so I now have value for v du/dx as (dh/dx)^2. I just can't work out partial dervative for dh/dx

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

 

[andrey21]

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) isn't there. Do this cancel out?

 

[Char. Limit]

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) isn't 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.

 

[andrey21]

Ah great thanks Char.limit for all ur hep.

 

[Char. Limit]

No problem at all.

 

[andrey21]

Hey char.limit I have another question which is posing a few problems.

Given the following pde:

du/dt = y^2 d^2u/dx^2

where t> 0 0<x<L

where L is 4000 and Y^2 = HK/phi

if t=0 is the time when river flow is altered, write down an initial condition for u.

Again bit unsure where to start on this one.