# LaPlace equation in 1D

1. Feb 22, 2012

### Jgoshorn1

1. The problem statement, all variables and given/known data

Solve the Laplace equation in one dimension (x, i.e. (∂^2h)/(∂x^2)= 0)
Boundary conditions are as follows:
h= 1m @ x=0m
h= 13m @ x=10m
For 0≤x≤5 K1= 6ms^-1
For 5≤x≤10 K2 = 3ms^-1

What is the head at x = 3, x = 5, and x = 8?

What is the Darcy velocity (specific discharge)?

NOTE: There are multiple steps that will need to be done. Realize that system is heterogeneous. In a multiple layer system with steady-state conditions, Darcy velocity in one layer must equal the Darcy velocity in the other layers

2. Relevant equations

h(x) = ho - [(h0 - hD )/D]*x

3. The attempt at a solution

I tried to use the equation above subbing in 3, 5, and 8 for the x and using 10m as D, 1m as h0, and 13m for hD

Then I used the specific specific discharge for the Darcy's velocity (q=K(dh/dL))

That was apparently all wrong. Apparently this needs to be broken into 2 systems, coupled. Each individual system can be treated as homogenous. So it need two separate LaPlace equations? I really don't know what to do with this problem, please help!

2. Feb 24, 2012

### LawrenceC

Since the discharge goes through both portions of the domain, you have continuity of the dependent variable h at the interface. In addition you also have continuity of discharge at the interface.

When you integrate the Laplace equation in 1-D, you have two constants of integration for each section of the domain. The matchings of discharge and h serve to eliminate some of the integration constants.

3. Feb 24, 2012

### Jgoshorn1

So, what I've done is set q(in) = q(out) => -K1(dh/dl)=K2(dh/dl) => 6((h-1)/5) = 3((13-h)/5) => h(@x=5) = 5m

Then I set the used set of boundary Conditions (BC) to interpret the solution to LaPlace in 1D (h=cx+D):
For BC @ x=0 and h=1 =>1=c(0)+D => 1=D
For BC @ x=5 and h=5 => 5=c(5)+1 => c=4/5
the I used h=cx+D again to solve for the head at x=3
=>h(@x=3)=(4/5)(3)+1=3.4

For the second system I used the point slope eq (y1-y)=m(x1-x) to get c
=> (5-13)/(5-10) = m = 8/5 = c
Then I used h=cx+D to solve for D
=> 5=(8/5)(5) + D => D=-3
Then to solve for h at x=8
h=cx+D => h(@x=8) => (8/5)(8)-3 = 9.8

To get Darcy's velocity I just used q=K(dh/dl) for each system
System 1 = 6(4/5) = 4.8 = q
System 2 = 4(8/5) = 4.8 = q

Yes?

4. Feb 24, 2012

### LawrenceC

I get the same answers as you but went about the problem a little differently.

I solve the ODE over two separate domains:

h1(x)=c1+c2*x
h2(x)=c3+c4*x

I used the Dirichlet boundary conditions along with continuity of head and continuity of Darcy velocity to solve for the 4 constants of integration. Equations worked out to be:

h(x) = 1 + .8x 0<x<5

h(x) = -3 + 1.6x 5<x<10

The < signs should be weak inequalities. I cannot make them on my computer.

Last edited: Feb 24, 2012