Wave equation is a well posed problem? Let $g:\mathbb{R} \rightarrow \mathbb{R}$ and $h:\mathbb{R} \rightarrow \mathbb{R}$ be smooth functions.
Prove that the following problem is well-posed:
$$
u_{tt}(t,x)=u_{xx}(t,x) \:\:\: t>0, x\in \mathbb{R}\\
u(0,x)=g(x) \:\:\: x\in \mathbb{R}\\
u_t(0,x)=h(x) \:\:\: x\in \mathbb{R}
$$
I know that I need to prove the following:
1) Existence of solution
2) Uniqueness of that solution
3) Continuous dependence on the IC
$$
\\
$$
ad. 1) I know  how to derive d'Alemebert formula and check that it solves our problem.
ad. 2) Do I really need to check it if I've already derived d'Alemebert formula (any solution must be in form of d'Alemebert formula?)
ad. 3) I have no idea. Please help me here.
 A: For 2) no, since you have derived the solution, there is really no need to do more. However, be aware that you must have made some assumptions on the solution while deriving d'Alembert's formula, such as continuity, differentiability, and what not. As a consequence, you have only proved uniqueness for solutions in that particular class of functions. You should be careful to specify that class.
For 3) you notice that since you have an explicit formula for the solution, you basically just have to show that this formula defines a continuous mapping. Since it is linear, you only have to prove that «small» initial data results in a «small» solution. The trick is in finding out what measure of «smallness» you need to use: Do you use the supremum norm, or something more fancy? Basically, it's a question of investigating the formula and seeing what kind of smallness results you are able to prove, and what kind of smallness of the initial data is required to get it. I think I'll leave it at that, for you to ponder.
A: *

*Once you have a formula of a function which satisfies the initial value problem (IVP), then (regardless how you came up with it) you DO have the existence part established.

*Uniqueness can be established in various ways. The simplest one is the following: 
Suppose that I have the following IVP: $\mathcal L u=0$, ${\mathcal B}= u=f$, and $u$ is a solution of it. If I manage to obtain the expression of $u$ using "$\Longrightarrow$" (straight implications), then uniqueness is established. Why? Because, it $w$ were another solution, then using the same "$\Longrightarrow$" (straight implications),, we would get the same value for $w$. 
Smooth dependence on Initial Data DEPENDS on the norm.
However, if $u=u(x,t;f,g)$ the solution of your IVP, then
$$
w=u_1-u_2=u(x,t;f_1,g_1)-u(x,t;f_2,g_2)=u(x,t;f_1-f_2,g_1-g_2),
$$ 
and $w_{tt}=c^2w_{xx}$, but, assuming that $w_t(\cdot,t),w_x(\cdot,t),\in L^2(\mathbb R)$,
$$
\frac{d}{dt}
\int_{-\infty}^\infty 
\big(w^2_t(x,t)+w^2_x(x,t)\big)\,dx=2\int_{-\infty}^\infty (w_tw_{tt}+w_xw_{xt})
=2\int_{-\infty}^\infty (w_tw_{tt}-w_{xx}w_{t})=0.
$$
This is a "smooth dependence result" as a certain norm of $w$ remains small, if $f_1-f_2$
and $g_1-g_2$ are small, as for $w=u_1-u_2$, we have 
$$\int w_t^2(x,t)\,dx, \,\,\!\!\int w_x^2(x,t)\,dx \le\int \big((g'_1-g_2')^2(x)+(h_1-h_2)^2(x)\big)\,dx,
$$
for all $t$. Thus, small $L^2$-change in $h$ and $g'$ results in small $L^2$-change in $u_x$, $u_t$.
A: I may be wrong, but on $\mathbb{R}$, is Suppose $u$ and $v$ are $C^{2,2}$ solutions. Let $w\equiv u-v$ and note that
\begin{align*}
w_{tt}=w_{xx} & \text{on }\left(0,\infty\right)\times\mathbb{R}\\
w\left(0,x\right)=0 & \text{on }\mathbb{R}\\
w_{t}\left(0,x\right)=0 & \text{on }\mathbb{R}.
\end{align*}
Now we would like to show that $w=0$ everywhere (and hence $u=v$ everywhere). Using the energy method approach,
\begin{align*}
e\left(t\right) & \equiv\frac{1}{2}\int_{\mathbb{R}}w_{t}^{2}+w_{x}^{2}dx\\
\frac{\partial e}{\partial t}\left(t\right) & =\int_{\mathbb{R}}w_{t}w_{tt}+\int_{\mathbb{R}}w_{x}w_{tx}dx\\
 & =\int_{\mathbb{R}}w_{t}w_{tt}+w_{x}w_{t}\mid_{\mathbb{R}}-\int_{\mathbb{R}}w_{xx}w_{t}dx\text{ (integration by parts)}\\
 & =\int_{\mathbb{R}}w_{t}\left(w_{tt}-w_{xx}\right)dx+w_{x}w_{t}\mid_{\mathbb{R}}.
\end{align*}
The first term here is zero, but the second term is troublesome. This is not an issue on a domain with nice boundary data. The usual energy method fails here.
(Maybe you can construct other solutions in this case)
