Wave Equation with initial conditions, boundary condtions

[mmmboh]

So, I do not think I did this properly, but if f(-x)=-f(x), then u(-x,0)=-u(x,0), and if g(-x)=-g(x), then u_t(-x,0)=-u_t(x,0).

According to D`Alambert`s formula,

u(x,t)=[f(x+t)+f(x-t)]/2 + 0.5∫g(s)ds (from x-t to x+t)
so, u(0,t)=[f(t)+f(-t)]/2 + 0.5∫g(s)ds (from -t to t)

f is odd, and so is g, so the equation ends up giving zero, as required. But I don`t think that`s what we have to do. How do do I extend the initial conditions so that f(-x)=-f(x)? and the same for g(x). I know how to create an odd function, I can just let h(x)=xf(x²), then h(-x)=-h(x), but I`m not sure what I`m suppose to do.

 

[mmmboh]

anyone?

 

[hunt_mat]

Write doen D'Almberts solution, what do you get?

 

[mmmboh]

What do you mean? I thought I did write it down,
$$u(x,t)=\frac{1}{2}[f(x-t)+f(x+t)]+\frac{1}{2}\int_{x-t}^{x+t} g(s)ds$$

 

[mmmboh]

Hm, I think I got it, I`ll try to post it soon.

 

[hunt_mat]

If you've done it, post it otherwise we can go from there. My first thought would be to define two new functions a,b, such that:
$$a(x)=\left\{\begin{array}{cc} f(x) & x\geqslant 0 \\ -f(x) & x<0 \end{array}\right.$$
Likewise fot b and work from there.