Waves on the Half Line - difference between Dirichlet and Neumann problems? I'm having trouble finding the solution of the one-dimensional wave equation for $x \in (0,\infty) $ and $ t > 0 $ with initial conditions:
$$u(x,0) = f(x), \space u_t(x,0) = g(x), \space x > 0 $$
We are asked to examine the Dirichlet and Neumann problems separately.
$$ u(0,t) = 0 $$
$$ \frac{\partial u}{\partial x} (0,t) = 0$$ 
The question gives a hint to consider the 'method of images', but the only time I've encountered that is solving problems in electrostatics by the uniqueness of Poisson's equation, does that mean that if we extend the problem to the whole line satisfying the boundary conditions we are guaranteed to have the correct solution to the half line problem by uniqueness? 
How do we extend this problem to the whole line? Using an odd extension for the Dirichlet case and the even extension for the Neumann case, I can grind out an answer, but am confused as to why this works, but I'm not sure why this is the case. Why can't we use an even extension for a Dirichlet problem such that
$$ \phi_{even} = 0 \mathrm{\space for \space}x = 0 $$
and an odd extension for the Neumann problem satisfying the boundary conditions? If possible, could I get a physical as well as mathematical explanation? 
 A: This being a math site, I only give mathematical explanation. 


*

*Any odd function takes value $0$ at $0$. Indeed, if $f(-x)=-f(x)$, then $f(-0)=-f(0)$, which implies $f(0)=0$. 

*Any even differentiable function has derivative $0$ at $0$. Indeed, if $f(-x)=f(x)$, then  taking derivatives of both sides we get $-f'(-x) = f'(x)$, so $f'$ is odd. See item 1. 


The above suggests the idea of forcing boundary conditions via symmetry: if we can get the solution to be odd [or even] then the Dirichlet [or Neumann] condition at $x=0$ will hold. 
What makes this idea practical is that the wave equation on $\mathbb R$  preserves the even-ness or odd-ness of initial conditions. (So does the heat equation, by the way.) Indeed, if $u$ satisfies the PDE, then the even reflection $u_e(x,t)= u(-x,t)$ and odd reflection $u_o(x,t) = -u(-x,t)$ also satisfy the PDE.  If the initial condition is odd [or even], then the odd [or even] reflection of $u$ solve the same IVP, and therefore is equal to $u$ by a uniqueness theorem. 
Conclusion: if we arrange the initial condition on $\mathbb R$ to be odd [or even], then the solution will inherit this property, and therefore will satisfy the Dirichlet [or Neumann] condition at $x=0$. 

Why can't we use an even extension for a Dirichlet problem 

Because even functions don't have to be $0$ at $x=0$. Like $\cos x$, for example.
