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I'm not sure how to start this PDE since the initial and boundary conditions are nonzero. May someone point me in the right direction?
This is the problem:
$$u_{tt} = u_{xx}$$ $$u(x,0) = \frac{1}{2+ \sin(x)}$$ $$u_t(x,0) = -\frac{\cos(x)}{(2+ \sin(x))^2}$$ $$u(0,t) = u(2\pi,t)= \frac{1}{2+ \sin(t)}$$
Hint: The general solution of the 1D wave equation is $u(x,t)=F(x+t)+G(x-t)$.
Firstly, you should get homogenous boundary conditions. Let $v(x, t) = \frac{1}{2+ \sin(t)}$. We will seek the solution in the form $u(x, t) = v(x, t) + w(x, t)$, so $w(x, t) = u(x, t) - v(x, t)$. After substitution $u$ into equation, we get $\partial_t^2 w - \partial_x^2 w = -\frac{4 \sin (t)+\cos (2 t)+3}{2 (\sin (t)+2)^3}$. In this case, the initial conditions will take the form $w(x, 0) = u(x, 0) - v(x, 0) = \frac{1}{2+ \sin(x)} - \frac{1}{2}$ and $\partial_t w(x, 0) = \partial_t u(x, 0) - \partial_t v(x, 0) = -\frac{\cos(x)}{(2+ \sin(x))^2} + \frac{1}{4}$, and the boundary conditions $w(0, t) = w(2 \pi, t) =0$. We will find $w$ as $w = y + z$, where
$$\partial_t^2 y - \partial_x^2 y = 0, y(x, 0) = \frac{1}{2+ \sin(x)} - \frac{1}{2}, \partial_t y(x, 0) = -\frac{\cos(x)}{(2+ \sin(x))^2} + \frac{1}{4}, y(0, t) = y(2 \pi, t) =0$$ $$\partial_t^2 z - \partial_x^2 z= -\frac{4 \sin (t)+\cos (2 t)+3}{2 (\sin (t)+2)^3}, z(x, 0) = \partial_t z(x, 0) = z(0, t) = z(2 \pi, t) =0$$
These functions can be found using separation of variables method.
