Wave Equation on Semi-Infinite Domain : Dirichlet and Neumann Boundary Conditions

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Show each step in the solution process so that I can follow your method.
Thanks!

I have attached a different PDF file with the same problem restricted to
Dirichlet boundary conditions instead of Neumann boundary conditions. I
would like to use this same method if possible, making the necessary
changes due to the different boundary conditions.

Wave Equation on R+ with Dirichlet BC

We consider the initial value problem for the wave equation on0 < x <

utt (x, t) = c2 uxx (x, t), 0 < x < , t > 0 (1)
u(x, 0) = f (x), 0 < x < ,
ut (x, t) = g(x), 0 < x < ,
u(0, t) = 0 (2)

Just as for the whole line it can be shown that the solution can be expressed as

u(x, t) = (x - ct) + (x + ct) (3)

for some sufficiently smooth functions and . Recall that we obtained for x - ct > 0
x-ct
1 1 K
(x - ct) = f (x - ct) - g(s) ds +
2 2c x0 2

and x+ct
1 1 K
(x + ct) = f (x + ct) + g(s) ds +
2 2c x0 2

But for this problem we need to find

(x - ct) for - < x - ct <

and
(x + ct) for 0 < x + ct < .

For x - ct 0 (which also implies x - ct > 0 (since x > 0 and t > 0) we obtain the same
result as before. Namely,

x+ct
f (x - ct) + f (x + ct) 1
u(x, t) = + g(s) ds. (4)
2 2c x-ct


But for x - ct < 0 we need a different formula for (x - ct) since f is only defined for
values of its argument which are greater than or equal to zero. What bails us out here is the
boundary condition at x = 0.

Applying the boundary condition at x = 0 to u(x, t) = (x - ct) + (x + ct) we have
0 = u(0, t) = (-ct) + (+ct) which implies

(-ct) = -(ct).

That is, for any p < 0 we can define (p) = -(-p).


1
Thus for x - ct < 0 we define

(x - ct) = -(-(x - ct))
= -(ct - x)
ct-x
1 1 K
= - f (ct - x) - g(s) ds + .
2 2c x0 2

Next we add the terms together to obtain, for x - ct < 0,
x+ct
f (x - ct) - f (ct - x) 1
u(x, t) = + g(s) ds.
2 2c ct-x



Finally we can write the solution as
x+ct

f (x - ct) + f (x + ct) 1
+ g(s) ds, x > ct
2 2c
x-ct
u(x, t) = . (5)
f (x - ct) - f (ct - x) x+ct
1

+ g(s) ds, x < ct

2 2c ct-x

Remark 1. 1. Equation (5) says that for x > ct the solution is exactly the same as
D'Alembert's solution for an infinite wave, while for x - ct, the solution is modified as
a result of the wave reflecting from the boundary (notice also that the sign of the wave
is reflected, i.e. it becomes negative) (see Example 1 below).

2. The solution would, of course, change if we changed the boundary condition at zero.
For example we could impose a Neumann condition.

Example 1. We consider the problem (1) with the initial conditions

1 4 f (x) =
0 otherwise

and g(x) = 0. In this case the solution (5) becomes


1/2 1 4 + 1/2 t 0 otherwise 0 otherwise

u(x, t) =
1 4
- 1/2
1/2
x 0 otherwise 0 otherwise





2
1 0.5
0.5 0.5

0.8 0.4
0.4 0.4

0.6 0.3
0.3 0.3

0.4 0.2 0.2 0.2


0.2 0.1 0.1 0.1


0 0 0 0
0 2 4 6 8 10 12 14 0 2 4 6 8 10 12 14 0 2 4 6 8 10 12 14 0 2 4 6 8 10 12 14
x x x x

t=0 t=1 t=2 t=3



0.4 0.4
0.5

0.2 0.2
0.4


0.3 0 0
0 2 4 6 8 10 12 14 0 2 4 6 8 10 12 14
0.2 x x
-0.2 -0.2

0.1
-0.4 -0.4
0
0 2 4 6 8 10 12 14
x

t=5 t=6 t=7




3