Okay, I have to find $u(x,t)$ for the string of length $L=\pi$ when $c^2=1$.
I know: $$\text{wave equation}: \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}$$ $$u(x,0)=\frac x{10}(\pi-x)$$ $$u_t(x,0)=0$$
Now I am fairly sure I set $u(x,0)=f(x)$ and $u_t(x,0)=g(x)$
My textbook starts using $\frac{\partial^2 u}{\partial t^2}=F\ddot{G}$ and $\frac{\partial^2 u}{\partial x^2}=F''G$, I am not sure, are these(F,G) the same as $f(x)$ and $g(x)$? Why are they captilised?
Are they transformed using Laplace or Fourier?
The method used there is separation of variables. We suppose that we can separate $u(x,t)$ as a product of two functions $G(t)$ and $F(x)$. Then we find the derivatives in the differential equation as $$\frac{\partial^2 u}{\partial t^2}=F\ddot{G} \\ \frac{\partial^2 u}{\partial x^2}=F''G$$ we then substitute in the given PDE $$F\ddot{G}=c^2 F''G\implies \frac{\ddot G}{c^2G}=\frac{F''}{F}=\lambda $$ Now we have two ordinary differential equations $$\ddot G-\lambda c^2 G=0\\F''-\lambda F=0$$ Now we have three cases($\lambda=0, \lambda\lt 0,\lambda\gt0$). The first case implies that $F$ is linear and this doesn't agree with the initial condition. The second case: put $\lambda=-\alpha^2$ then $$\ddot G+\alpha^2 c^2 G=0\\F''+\alpha^2 F=0$$ These equations say that $$G=A\cos(\alpha c t)+B\sin(\alpha c t) \\ F=C\cos(\alpha x)+D\sin(\alpha x)$$ The second condition $\dot G(0)=0$ implies that $B=0$. Thus the general solution is $$u(x,t)=A\cos(\alpha c t)\{C\cos(\alpha x)+D\sin(\alpha x)\} \\ u(x,t)=\cos(\alpha c t)\{a\cos(\alpha x)+b\sin(\alpha x)\}$$ Now you can it from here.
You can see that