The core is to invent a initial function that satisfies the boundary conditions as well as defined on the whole line.

\begin{equation} \end{equation}

a) need to think of an initial function, consider:

$$

f(x) = \left\{\begin{aligned}

&-g(-x) && -L<x<0 \\

&0 && x=...-2L, -L, 0, L, 2L ... \\

&g(x) &&0<x<L \\

&extended\ to\ be\ 2L-periodic \\

\end{aligned}

\right.$$

This function satisfy Dirichlet boundary conditions on the whole line.

Thus our solution is now:

\begin{equation} v(x,t) = \int_{-\infty}^{\infty} G(x,y,t)f(y)dy \end{equation}

Depends on the relative position of (x,t), the integral can be valued piecewisely.

b) by similar fashion, consider:

$$

f(x) = \left\{\begin{aligned}

&g(-x) && -L<x\le 0 \\

&g(x) &&0\le x<L \\

&extended\ to\ be\ 2L-periodic \\

\end{aligned}

\right.$$

This function is even, thus its derivative is an odd function, which satisfies Neumann condition.

\begin{equation} v(x,t) = \int_{-\infty}^{\infty} G(x,y,t)f(y)dy \end{equation} can be valued based on (x,t)'s position.

c) the condition ask for a function f(x) that has odd continuation at 0 + 2kL, but even continuation at L+2kL. Thus its period is 4L. consider:

$$

f(x) = \left\{\begin{aligned}

&-g(x-2L) && -2L<x\le -L\\

&-g(-x) && -L\le x<0 \\

&0 && x=...-4L,-2L , 0, 2L, 4L... \\

&g(x) &&0<x\le L \\

&g(2L-x) && L\le x<2L\\

&extended\ to\ be\ 4L-periodic \\

\end{aligned}

\right.$$

Solution \begin{equation} v(x,t) = \int_{-\infty}^{\infty} G(x,y,t)f(y)dy \end{equation} then can be valued based on (x,t)'s position.