General Solution to wave equation

[renegade05]

So the problem reads:

Use the general solution to solve the homogeneous wave equation on the half-line, with homogeneous initial conditions, and a non-homogeneous Neumann boundary condition.

\(\displaystyle u_{tt} - c^2 u_{xx} = 0, \quad\quad \quad \quad 0 < x < \infty,\quad t>0\)
\(\displaystyle u(x,0)=0, u_t (x,0)=0 \quad (For \quad All \quad 0 < x < \infty)\)
\(\displaystyle u_x(0,t)=h(t) \quad \quad \quad \quad \quad(For \quad All \quad t > 0)\)

I know the general solution has form:

\(\displaystyle u(x,t) = F(x-ct) + G(x-ct)\)

From the initial conditions, boundary conditions I get that:

\(\displaystyle
(1) F(x) + G(x) = 0\)
\(\displaystyle (2) -cF'(x) + cG'(x) = 0\)
\(\displaystyle (3) F'(-ct)+G'(ct) = h(t)
\)

From the first two it is clear that F(x) = G(x) = 0 ??

The third one is proving difficult. I tried to let z = -ct to get:

F'(z)+G'(-z) = h (-z/c)

Then:

\(\displaystyle F(z) - G(-z) = \int_{0}^{z} h(-n/c) dn\)

but if F(x) = G(x) = 0 then so does this integral which is trivial. Where am I going wrong? What is the answer to this one? Thanks!

 

[Ishuda]

So the problem reads:

Use the general solution to solve the homogeneous wave equation on the half-line, with homogeneous initial conditions, and a non-homogeneous Neumann boundary condition.

\(\displaystyle u_{tt} - c^2 u_{xx} = 0, \quad\quad \quad \quad 0 < x < \infty,\quad t>0\)
\(\displaystyle u(x,0)=0, u_t (x,0)=0 \quad (For \quad All \quad 0 < x < \infty)\)
\(\displaystyle u_x(0,t)=h(t) \quad \quad \quad \quad \quad(For \quad All \quad t > 0)\)

I know the general solution has form:

\(\displaystyle u(x,t) = F(x-ct) + G(x-ct)\) <====wrong

...

See above: The general solution is
u(x,t) = F(x-ct) + G(x+ct)
where the - represent a wave traveling in the +x direction [as time increases, x must increase to keep the wave front at x - c t stationary] and the + represents a wave traveling in the -x direction. The variable c is the speed the wave travels.

 

[renegade05]

See above: The general solution is
u(x,t) = F(x-ct) + G(x+ct)
where the - represent a wave traveling in the +x direction [as time increases, x must increase to keep the wave front at x - c t stationary] and the + represents a wave traveling in the -x direction. The variable c is the speed the wave travels.

Whoops, ya that was just a typo as my further work would indicate.

Still stuck though. Please help.

 

[Ishuda]

Whoops, ya that was just a typo as my further work would indicate.

Still stuck though. Please help.

It appears something is wrong. Another way to approach problems like this is separation of variables. That is let u(x,t) = X(x) T(t). Boundary conditions generally lead to a set (possibly infinite) of eigenvalues \(\displaystyle \lambda_n\) and the solution is the sum of the X_n(x) T_n(t).

So u_tt = c²u_xx implies
X T'' = c² T X''
or
\(\displaystyle \frac{T''}{c^2 T} = \frac{X''}{X} = \lambda^2\)
[\(\displaystyle \lambda^2\) for convience] That is, if we change x but not t, the t side remains the same and the other way around so both must be a constant parameter \(\displaystyle \lambda^2\). The solution to the equations are
T = \(\displaystyle A\, e^{c\, \lambda\, t} + B\, e^{-c\, \lambda\, t}\)
X = \(\displaystyle C\, e^{\lambda\, x} + D\, e^{-\lambda\, x}\)

u(x,0) = 0 implies T(0;\(\displaystyle \lambda\)) = 0
since we don't want X to be identically zero for all \(\displaystyle \lambda\). Thus
T(0;\(\displaystyle \lambda\)) = A + B = 0
and A = -B

u_t(x,0) = 0 implies T'(0;\(\displaystyle \lambda\)) = 0
since we don't want X to be identically zero for all \(\displaystyle \lambda\). Thus
T'(0;\(\displaystyle \lambda\)) = \(\displaystyle c\, \lambda\) (A - B) = 0
So, since we do not want T to be identically zero, we must have \(\displaystyle \lambda = 0\)
But, that means
T(t) = A + B = 0
and u must be zero.

Note also the general solution by d'Alembert's formula from
http://en.wikipedia.org/wiki/Wave_equation
gives u=0

 

[renegade05]

The problem indicates that u should be continuous though. I am not sure if u=0 is the solution? thoughts?

 

[renegade05]

View attachment 4982

The problem indicates that u should be continuous though. I am not sure if u=0 is the solution? thoughts?

Stop me if I'm wrong but don't we have two regions we must deal with? x>ct and o<x<ct.

For x>ct the answer is u(x,t) = 0

But for 0<x<ct we have:

General solution:
\(\displaystyle u(x,t) = F(x-ct) + G(x+ct)\)

Using B.C. :
\(\displaystyle u_x(0,t) = h(t) \)
\(\displaystyle F'(-ct) + G'(ct) = h(t)\)
\(\displaystyle F'(z) + G'(-z) = h(-z/c)\)
\(\displaystyle F(z) - G(-z) = \int_{0}^{z} h(-s/c)ds\)
\(\displaystyle F(x-ct) - G(ct-x) = \int_{0}^{x-ct} h(-s/c)ds\)
\(\displaystyle F(x-ct) = \int_{0}^{x-ct} h(-s/c)ds + G(ct-x)\)

SO:

\(\displaystyle u(x,t) = \int_{0}^{x-ct} h(-s/c)ds + G(ct-x) + G(x+ct)\)

But we know that \(\displaystyle G(ct-x), G(x+ct) =0\)

So we get:

\(\displaystyle u(x,t) = \int_{0}^{x-ct} h(-s/c)ds\)

Dos this solution satisfy BC: \(\displaystyle u_x(0,t) = h(t) \) ??

\(\displaystyle u_x(x,t) = h(\frac{ct-x}{c})\)
\(\displaystyle u_x(0,t) = h(t)\)

So this satisfies the BC,but I don't think this satisfies the IC - does it have to for this region?

Namely:

IC: \(\displaystyle u(x,0) =0\)

\(\displaystyle u(x,0) = \int_{0}^{x} h(-s/c)ds \stackrel{?}{=} 0\)

Lots going here, please someone help me out

 

[Ishuda]

View attachment 4982

The problem indicates that u should be continuous though. I am not sure if u=0 is the solution? thoughts?

u(x,t) = 0 is infinitely partial differentiable. All derivatives are zero. A function equal to zero is called the trivial solution and is the solution all homogeneous differential equations including partial differential equations. It's just that it is not generally the only solution.

BTW: Note that we can not satisfy u_x(0,t)=h(t) unless h(t) is also zero.

 

[renegade05]

u(x,t) = 0 is infinitely partial differentiable. All derivatives are zero. A function equal to zero is called the trivial solution and is the solution all homogeneous differential equations including partial differential equations. It's just that it is not generally the only solution.

BTW: Note that we can not satisfy u_x(0,t)=h(t) unless h(t) is also zero.

So so my answer with the integral is no good ?

Im still not convinced the answer is u=0

mainly because I would think my prof wouldn't give us a question with a trivial answer.

 

[Ishuda]

...
Namely:

IC: \(\displaystyle u(x,0) =0\)

\(\displaystyle u(x,0) = \int_{0}^{x} h(-s/c)ds \stackrel{?}{=} 0\)

Lots going here, please someone help me out

If
\(\displaystyle u(x,0) = \int_{0}^{x} h(-s/c)ds = 0\)
for all x, doesn't that imply that h is identically zero? As I mentioned before, it appears something is wrong. I would suggest you ask your prof.