Solution of the initial value problem

Consider now an initial value problem for Eq. (2.2):

$$u_{tt} = c^2u_{xx}, \quad t\geq 0$$

$$u(x,0) = f(x),$$ (2.7)

$$u_t(x,0) = g(x).$$

To write down the general solution of the initial value problem for (2.2) one needs to exspress the arbitrary function $F$ and $G$ in terms of initial data $f$ and $g$ . Using the relation

$$\frac{\partial}{\partial t} F(x-ct)=-cF'(x-ct),$$   where$$\quad F'(x-ct):=\frac{\partial}{\partial \xi} F(\xi)$$

one becomes:

$$u(x,0) = F(x)+G(x)=f(x);$$

$$u_t(x,0) = c(-F'(x)+G'(x))=g(x).$$

After differentiation of the first equation with respect to $x$ one can solve the system in terms of $F'(x)$ and $G'(x)$ , i.e.,

$$F'(x)=\frac{1}{2} \bigl(f'(x)-\frac{1}{c}g(x)\bigr),\qquad G'(x)=\frac{1}{2} \bigl(f'(x)+\frac{1}{c}g(x)\bigr).$$

Hence

$$F(x)=\frac{1}{2}f(x)-\frac{1}{2c}\int_0^x g(y)dy+C,\quad G(x)=\frac{1}{2}f(x)+\frac{1}{2c}\int_0^x g(y)dy-C,$$

where the integration constant $C$ is chosen in such a way that the initial condition $F(x)+G(x)=f(x)$ is fullfield. Alltogether one obtains:

$$\boxed{ u(x,t)=\frac{1}{2}\bigl(f(x-ct)+f(x+ct)\bigr)+\frac{1}{2c}\int_{x-ct}^{x+ct} g(y)dy}$$ (2.8)