a special condition about vibrating string problem

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I have the following problem

$$ \begin{align}\begin{cases} u_{tt} - 4 u_{xx} =0 & t > 0 , 0 < x < 3\pi \\ u(0,t) = -\pi, u(3\pi, t) = 2\pi & t> 0 \\ u(x,0) = 0 , u_{t}(x,0) = 0 & 0 < x < 3\pi \end{cases} \end{align} $$

In fact I do not know how to fix with $u_{t}(x,0) = 0$

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For the homogeneous wave equation with non-homogeneous boundary conditions like below

$$ \begin{align}\begin{cases} u_{tt} - c^{2} u_{xx} =0 & t > 0 , 0 < x < L \\ u(0,t) = \phi(t), u(L, t) = \psi(t) & t> 0 \\ u(x,0) = f(x) , u_{t}(x,0) = g(x) & 0 < x < L \end{cases} \end{align} $$

you introduce a reference function $r(x,t)$ such that $u(x,t) = v(x,t) +r(x,t)$ where $v(x,t)$ is a function satisifying the homogeneous solution and $r(x,t)$ satisfies the nonhomogeneous boundary conditions.

To do this we can make $r(x,t)$ the following function

$$ r(x,t) = \frac{x}{L}\left[ \psi(t) - \phi(t) \right] + \phi(t)$$

in this case, your $r(x,t)$ isn't time-dependent and we can write is as

$$ r(x) = \frac{x}{3\pi}\left[ 2\pi - (-\pi) \right] - \pi = \frac{3\pi x}{3\pi} - \pi = x - \pi$$

then your $v(x,t)$ is the homogeneous solution as usual which is

$$ v(x,t) = \sum_{n=1}^{\infty} \bigg( A_{n} \sin(\frac{n \pi x}{3\pi})\cos(\frac{2 n\pi t}{3\pi} )+ B_{n} \sin(\frac{n \pi x}{3\pi}) \sin(\frac{2 n \pi t}{3\pi})\bigg) $$

which can be reduced as

$$ v(x,t) = \sum_{n=1}^{\infty} \bigg( A_{n} \sin(\frac{n x}{3})\cos(\frac{2 n t}{3}) + B_{n} \sin(\frac{n x}{3}) \sin(\frac{2 n t}{3})\bigg) $$

$$ v(x,t) = \sum_{n=1}^{\infty} \bigg( A_{n} \cos(\frac{2 n t}{3}) + B_{n} \sin(\frac{2 n t}{3})\bigg)\sin(\frac{n x}{3}) $$

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If you let $$ v(x,t)=u(x,t)+(\pi-x), $$ then $v_{tt}-4v_{xx}=0$. However, unlike for $u$, $v$ satisfies homogeneous conditions at $x=0,3\pi$: $$v(0,t)=u(0,t)+\pi=0 \\ v(3\pi,t)=u(3\pi,t)+(\pi-3\pi)=0. $$ That makes separation of variables work for $v$. The initial conditions become $$ v(x,0)=u(x,0)+(\pi-x)=\pi-x\\ v_t(x,0)=u_t(x,0)=0. $$ The separated solution is $$ v(x,t)=\sum_{n=1}^{\infty}(A_n\cos(2n\pi t/3)+B_n\sin(2n\pi t/3))\sin(n\pi x/3). $$ The condition $v_t(x,0)=0$ forces all $B_n=0$. The condition $v(x,0)=\pi-x$ determines the $A_n$: $$ \pi-x = \sum_{n=1}^{\infty}A_n\sin(n\pi x/3). $$