# D'Alembert solution of wave equation on semi infinite domain

Tags:
1. Jan 8, 2016

### Name15

1. The problem statement, all variables and given/known data
Wave equation: ytt=yxx
Initial conditions: Y(x,0) =f(x) =
x (0 ≤ x < 1)
2.5(5-x) (1 ≤ x < 3)
0 (Otherwise)
and yt(x,0) = 0

Boundary condition: y(0,t) =0

Semi infinite domain: 0 ≤ x < infinity
2. Relevant equations
d'Alembert solution

3. The attempt at a solution
I can plot the solution of a wave equation on an infinite domain. However I don't understand how to change f(x) in order to plot the solution on a semi infinite domain.

For the above example, would the initial conditions for a semi-infinite domain be:
Y(x,0) =f(x) =
x (0 ≤ x < 1)
2.5(5-x) (1 ≤ x < 3)
-x (-1 <x ≤ 0)
2.5(x-5) (-3 <x ≤ -1)
0 (Otherwise)

2. Jan 9, 2016

### Brian T

In order to ensure that the Dirichlet boundary condition (y(0,t)=0) is satisfied, you want to perform an odd reflection of the initial data y(x,0), not an even reflection, since odd functions have to be zero at x=0. Once you have extended the problem to the whole line -∞<x<∞, you use the standard d'Alambert solution with your initial data on the half line replaced by the odd-extension of the data on the whole line. The solution you get from the d'Alambert solution will (i) satisfy the initial conditions since you constructed it using the d'Alambert solution, and (ii) meet the B.C.'s since you performed an odd-extension. By the uniqueness theorem, this is the desired solution to your DE. Lastly, to complete the problem, you should note that your solution holds on the original half line (Essentially, you are making up the wave on the other half line -∞<x<0 to help you make your solution, but after using it, you have to discard of it).

Given initial data $y(x,0) = φ(x)$, you can extend it to an odd function as follows:
$y_{odd}(x,0) = \begin{cases} φ(x) \text{ for } x > 0 \\ -φ(-x) \text{ for } x < 0 \\ 0 \text{ for } x = 0 \end{cases}$

Your attempt at a solution makes use of an even extension (that is what you would use if you had the Neumann boundary condition, the spatial derivative at x=0 vanishes).

Last edited: Jan 9, 2016
3. Jan 9, 2016

### Name15

Thanks for the response. I'm still a little confused, I thought I these were the odd functions for x is less than 0:
-x (-1 <x ≤ 0)
2.5(x-5) (-3 <x ≤ -1)

4. Jan 9, 2016

### Brian T

Well let's just look at the interval (0,1), we have the function x. If we want to extend it, we should take -f(-x) which is still just x. If you think about it, x is already an odd function so it should stay as x. If you use - x on the negative part, you get a V shape which is an even function not an odd function (you essentially have the function abs(x))

Since x takes on positive values ranging from 0 to 1 on the interval 0 to 1, we want our odd extension to take the opposite values 0 to - 1 on the interval 0 to - 1, which would require we still use x

5. Jan 9, 2016

### Name15

Ok I think i'm getting there. So i only want solutions in the negative axis, and -x wouldn't work since it'll give a positive solution for negative x values.
However, 2.5(x-5) (-3 <x ≤ -1) would work as this gives a negative solutions?

6. Jan 9, 2016

### Brian T

Looks like you're starting to get it, you got it down for the x part, but it's not sufficient that its just negative for the other function. You need the exact opposite range of values. (Also, note that if your initial data was negative, the odd extension would then be positive)

Look at the original function on (1,3), we have 2.5(5-x). Plugging in 1 and 3,we see this ranges from 2.5(4) to 2.5(2). So, we want our odd extension to range from 2.5(-2) to 2. 5(-4) when we plug in endpoints of (-3,-1). If you look at 2.5(x-5) and plug in (-3,-1), you get 2.5(-8 ) and 2.5(-6). This tells you that you don't have the proper odd extension.

The general procedure for extending to an odd function is, starting from f(x),

f(x) -> f(-x) -> -f(-x)

Try that for 2.5(5-x), and see if when you plug in the values -3 and -1, you get the right range. You want your data to look similar to this

7. Jan 9, 2016

### Name15

So to get a range from 2.5(-2) to 2. 5(-4) , I should use -f(-x), which gives -2.5(5+x)? If i plug in -1 and -3 into this i get the range you mentioned.

8. Jan 9, 2016

### Brian T

You got it

9. Jan 10, 2016

### Name15

thank you so much for your help!

10. Jan 10, 2016

### Name15

Quick question regarding d'Alembert, for a function f(x)=x at time t=1 and c=1, should I get a translation of 1 or 0.5? I thought the ct should be multiplied by the 0.5 from the d'Alembert solution giving a translation of 0.5, but something I read tells me otherwise..

11. Jan 10, 2016

### Brian T

You don't multiply the translation by 1/2, you multiply the function at the end by that amount. Recall the formula is 1/2(f(x-ct) + f(x+ct)) so the translation is just +/- ct which in your case is +/- 1. Then you just half the amplitude of the wave. Physically, the wave splits in half from its original position and propagates in both directions at a speed c (characterized by the x-ct). If it propagated at 1/2c, it would no longer be a solution to the given eqn

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted