Solve the nonhomogeneus wave equation with NO initial conditions Sounds strange, but I've came up with this wave equation:
$$
\begin{cases}
u_{tt} - u_{xx} = -\sin u, \\
u(+\infty, t) = 2\pi, \\
u(-\infty, t) = 0,\\ 
u_x \geq 0, \\
\dfrac{\partial^n u}{\partial x^n}(\pm \infty, t) = 0, \quad \forall n \in \mathbb{N}.
\end{cases}$$
Yes, there are no initial conditions (when only $t = 0$). So I can't see how d'Alembert's formula will work. Meanwhile, I know the solution to the homogeneous version of this equation should be
$$
u = f(x - t) + g(x + t),
$$
where $f$ and $g$ are twice differentiable (i.e. they belong to $C^2(\mathbb{R})$) functions.
I also found out that when $x = \pm \infty$, the equation reduces to the homogeneous version, and the solution becomes $f(\pm \infty) + g(\pm \infty)$ which offers little help.
Or, if I try $y = \sin u$, then the equation can be converted to $y_{tt} - y_{xx} = -y/(\sqrt{1 - y^2} - y)$. Would this form help?
 A: Lul maybe I asked too specifically, let me know if I solve the equation wrong.
If we let $u(x, t) = f(y) = f(x - ct)$, then $u_{tt} = c^2 f''(y), u_{xx} = f''(y)$, and the original equation becomes \begin{equation}
    \begin{split}
        u_{tt} - u_{xx} = (c^2 - 1) f'' &= - \sin f \\
\Rightarrow (1 - c^2) f'' - \sin f &= 0; \\
        (1 - c^2) f'' f' - (\sin f)(f') &= 0; \\
        \Rightarrow 0.5(1 - c^2)(f')^2 + (\cos f) &= A, A \in \mathbb{R}.
    \end{split}
\end{equation} Given $u_x(\pm \infty, t) = f'(\pm \infty) = 0, \cos u(\pm \infty, t) = \cos (0$ or $2\pi) = 0$, we get $A = 1$. Then \begin{equation}
    \begin{split}
        \dfrac{1 - c^2}{2} \left(\dfrac{df(y)}{dy}\right)^2 = 1 - \cos (f(y)) \Rightarrow \dfrac{df(y)}{dy} &= \pm \sqrt{\dfrac{2(1 - \cos (f(y)))}{1 - c^2}}; \\
        \dfrac{df(y)}{\sqrt{2(1 - \cos (f(y)))}} = \dfrac{df(y)}{\sin (0.5f(y))} &= \pm \dfrac{dy}{\sqrt{1 - c^2}}; \\
        \Rightarrow \ln \left(\csc \left[\dfrac{f(y)}{2}\right] + \cot\left[\dfrac{f(y)}{2}\right]\right) = \ln \left(\tan \left[\dfrac{f(y)}{4}\right]\right) &= \pm \dfrac{y}{\sqrt{1 - c^2}}; \text{ (separation of variables)} \\
        \Rightarrow u(x, t) = f(x - ct) &= 4 \arctan \left[\exp\left(\pm \dfrac{x - ct}{\sqrt{1 - c^2}}\right)\right]. \text{ } \Box
    \end{split}
\end{equation}
