Semi-infinite wave equation

[the_godfather]

Homework Statement

solve the wave equation (dx^2 -dt^2)$$\Phi$$ = 0 on the semi-infinite line x<=0 with boundary conditions $$\Phi$$ at x=0 = 0 and initial conditions $$\Phi$$ at t=0 = tanh(x)

Homework Equations

solution of the wave equation is of the form $$\Phi$$ = f(x-t) + g(x+t).

$$\Phi$$ at t=0 = tanh(x)
tanh(x) is an odd function

The Attempt at a Solution

i know that $$\Phi$$ at t=0 = tanh(x) which i will call A
i know that $$\Phi$$ at x=0 is 0, which i will call B

am i correct in thinking that i can simply add the two equations together to get 2$$\Phi$$ = f(x-t) + g(x+t) = A + B
therefore giving $$\Phi$$ = [tanh(x+t) tanh(x-t)]/2
also what happens if tanh(x) is not an odd function? how can i turn it into an odd function?

 

[anathema]

Your solution is partially correct. You correctly identified that the general solution to the wave equation is of the form \Phi = f(x-t) + g(x+t). However, when solving for the boundary conditions, you should be using the initial conditions at t=0 and the boundary condition at x=0, not the other way around.

So, using the initial condition \Phi at t=0 = tanh(x), we can set f(x) = tanh(x) and g(x) = 0. Then, using the boundary condition \Phi at x=0 = 0, we can set f(0) = 0 and g(0) = 0.

Therefore, our solution becomes \Phi = tanh(x-t), which satisfies both the initial and boundary conditions.

If tanh(x) is not an odd function, we can still use the same method to solve for the solution. However, we would need to adjust the initial condition to make it an odd function. For example, if we have the initial condition \Phi at t=0 = x^2, we can redefine it as \Phi at t=0 = x^2 - 1, which is an odd function. Then, our solution would be \Phi = (x^2 - 1)(x-t) + g(x+t), where g(x) is determined by the boundary condition.