Parallelogram identity in the wave equation Using the parallelogram identity, I need to solve the following initial boundary value problem
for a vibrating semi-infinite string with a nonhomogeneous boundary condition:


  $$ u_{tt} − u_{xx} = 0 , \ 0 &lt; x &lt; \infty, t &gt; 0 $$
  $$u(0,t) = h(t)$$
  $$u(x,0) = f(x), \ u_{t}(x,0) = g(x)$$
  where $f, g, h ∈ C_2\{[0, ∞)\}$


I really have try to solve it, be I still dont know how to use the parallelogram identity. Thanks for your help.

Edit: The parallelogram identity is

$u(x_0 − a, t_0 − b) + u(x_0 + a, t_0 + b) = u(x_0 − b, t_0 − a) + u(x_0 + b, t_0 + a).
$
I have no idea about "parallelogram identity" in PDE, I only know that this PDE problem with those types of I.C.s and B.C.s can be found the solution exactly in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf.

Case $1$: $\dfrac{f(t)+f(-t)}{2}+\dfrac{1}{2}\int_{-t}^t g(s)~ds=h(t)$

$u(x,t)=\dfrac{f(x+t)+f(x-t)}{2}+\dfrac{1}{2}\int_{x-t}^{x+t}g(s)~ds$

Case $2$: $\dfrac{f(t)+f(-t)}{2}+\dfrac{1}{2}\int_{-t}^t g(s)~ds\neq h(t)$

$u(x,t)=\begin{cases}\dfrac{f(x+t)+f(x-t)}{2}+\dfrac{1}{2}\int_{x-t}^{x+t}g(s)~ds&amp;\text{when}~x&gt;t\\\dfrac{f(x+t)-f(t-x)}{2}+\dfrac{1}{2}\int_{t-x}^{x+t}g(s)~ds+h(t-x)&amp;\text{when}~x&lt;t\end{cases}$
I have no idea about "parallelogram identity" in PDE, I only know that this PDE problem with those types of I.C.s and B.C.s can be found the solution exactly in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf.

Case $1$: $\dfrac{f(t)+f(-t)}{2}+\dfrac{1}{2}\int_{-t}^t g(s)~ds=h(t)$

$u(x,t)=\dfrac{f(x+t)+f(x-t)}{2}+\dfrac{1}{2}\int_{x-t}^{x+t}g(s)~ds$

Case $2$: $\dfrac{f(t)+f(-t)}{2}+\dfrac{1}{2}\int_{-t}^t g(s)~ds\neq h(t)$

$u(x,t)=\begin{cases}\dfrac{f(x+t)+f(x-t)}{2}+\dfrac{1}{2}\int_{x-t}^{x+t}g(s)~ds&amp;\text{when}~x&gt;t\\\dfrac{f(x+t)-f(t-x)}{2}+\dfrac{1}{2}\int_{t-x}^{x+t}g(s)~ds+h(t-x)&amp;\text{when}~x&lt;t\end{cases}$