# 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:

$u_{tt} − u_{xx} = 0$
$0 < x < ∞, t > 0,$

$u(0, t) = h(t)$

$u(x, 0) = f (x)$

$u_{t} (x, 0) = g(x)$

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.

The parallelogram identity is(edit):

$u(x0 − a, t0 − b) + u(x0 + a, t0 + b) = u(x0 − b, t0 − a) + u(x0 + b, t0 + a).$

-

## 1 Answer

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

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

-