Solve a mixed problem in the domain $t > 0, x > 0$ : $u_{t t}-4 u_{x x}=0, u(0, x)=e^x, u_t(0, x)=1, u(t, 0)=0$
Applying d'Alembert's method, the general solution to this diffetential equation it the following
$$ u(x,t)= f_1(x-at) +f_2(x+at). $$ where $$ f_1(x)= \dfrac{e^x}{2} -\frac{1}{4}\int_{0}^{x} 1 \,ds, $$ $$ f_2(x)= \dfrac{e^x}{2} +\frac{1}{4}\int_{0}^{x} 1 \,ds. $$
Because of the boundry condition $u(t, 0)=0$ the solution must be found (!) in the the form of
$$\Phi(x)=\left\{\begin{array}{ll}\varphi(x), & x>0, \\ -\varphi(-x), & x<0,\end{array} \quad \Psi(x)= \begin{cases}\psi(x), & x>0 \\ -\psi(-x), & x<0\end{cases}\right.$$
Introducing the new expressions for our initial conditions $$ u(0, x)=e^x=\varphi(x)\ \text{and} \ u_t(0, x)=1=\psi(x)$$
I suppoused that the following sytem is the an extension of the initial conditions on the whole plane (!!)
$$\Phi(x)=\left\{\begin{array}{ll} e^{x}, & x>0, \\ -e^{-x}, & x<0,\end{array} \quad \Psi(x)= \begin{cases}1, & x>0 \\ -1 & x<0\end{cases}\right.$$
Questions
(!) Why do we search the function on the whole plane if we are only interested by the conditions of the problem in $x>0, \ t>0$.
(!!) Can we say that this is the solution to Cauchy problem? $\left\{\begin{array}{l}u_{t t}=4^2 u_{x x}, t>0, x \in \mathbb{R} \\ \left.u\right|_{t=0}=\Phi(x),\left.u_t\right|_{t=0}=\Psi(x), x \in \mathbb{R}\end{array}\right.$