Cfrgtkky

Solve the PDE:

$begin{equation}
frac{partial^2u}{partial t^2} - frac{partial ^2 u}{partial x^2}=0, qquad 0<x<L,quad t>0 \
u(x,0)=f(x) qquad frac{partial u}{partial t }(x,0)=g(x)
end{equation}$

Find $u(x,t)$ for the values $f(x)=sin x$, $quad g(x)=0$ when $u(0,t)= 0$ and $frac{partial u}{partial x}(pi /2,t)=0$

My attempt

We know the PDE is only a wave equation, then applying D'Alembert formula we have:

$u(x,t)=frac{1}{2}[f(x-ct)+f(x+ct)]+frac{1}{2c}int^{x+ct}_{x-ct}g(s)ds. tag 1$

As $g(x)=$0 then $(1)$ is:

$u(x,t)=frac{1}{2}[f(x-t)+f(x+t)] tag 1$

Note we have a Neumman condition the we need make even periodic extension

Then
begin{equation}
f(x) = left{
begin{array}{ll}
sin(x) & mathrm{if } 0<x<frac{pi}{2} \
-sin(x) & mathrm{if } -frac{pi}{2}<x<0 \
end{array}
right.
end{equation}

such that $f(x pm pi/2)=f(x)$

This implies:

$
begin{equation}
f(x-t) = left{
begin{array}{ll}
sin(x-t) & mathrm{if } 0<x<frac{pi}{2} \
-sin(x-t) & mathrm{if } -frac{pi}{2}<x<0 \
end{array}
right.
end{equation}
$

and

$
begin{equation}
f(x+t) = left{
begin{array}{ll}
sin(x+t) & mathrm{if } 0<x<frac{pi}{2} \
-sin(x+t) & mathrm{if } -frac{pi}{2}<x<0 \
end{array}
right.
end{equation}
$

Is correct this?