D alembert solution proof I've seen quite a handful of "proofs" of the solution to the wave equation. However, all of them use some technique which I feel could use some more explanation. 
To be particular, I'm referring to he "proof" in Strauss textbook.
I have attached are the screenshots of the proof from the book below including one in the comments. I don't get any of the assertions made in the proof.
The author mentions that because the equation is linear g(x - ct) has to be a solution. Also, in the alternative proof, there is a change in coordinates which is unexplained. Is there any formal proof for these assertions?
1
2
Third image in the comment below
 A: Maybe you do know the brilliant book by Evans about PDE's. I find his approach very clear and formal (also quite long and explanatory). I also appreciate a good formalism. So I try to enlight the proof given by him a bit more. First we have to consider a simpler problem to solve the wave equation. 
The Transport Equation. (see p. 18) Consider the PDE $$u_t + bu_x = 0 \qquad \text{in } \mathbb{R} \times (0,\infty)$$ 
where $b \in \mathbb{R}$ and $u: \mathbb{R} \times [0,\infty) \rightarrow \mathbb{R}$ is the unknown, $u = u(x,t)$. Assume we have found a smooth function which solves above PDE, fix $(x,t) \in \mathbb{R} \times (0,\infty)$ and define $$z(s) := u(x + sb,t+s)$$ for $s \in \mathbb{R}$. Then $$z'(s) = \nabla u(x + sb,xt + s)\begin{pmatrix} b\\1\end{pmatrix} = u_x(x + sb,t + s)b + u_t(x + sb,t + s) = 0$$ Thus $z$ is a constant function of $s$ and therefore, $u$ is constant on the line through $(x,t)$ with direction $(b,1)$. Hence if we know the value of $u$ at any point on each such line, we know its value everywhere in $\mathbb{R} \times (0,\infty)$. Consider the IVP $$\begin{aligned}u_t + bu_x = 0 \qquad &\text{in } \mathbb{R} \times (0,\infty)\\u = g \qquad &\text{on } \mathbb{R} \times \{0\}\end{aligned}$$ where $g: \mathbb{R} \to \mathbb{R}$ is known. Let again $(x,t)$ be fixed. Then $$u(x + bs,t + s)$$ is constant for $s \in \mathbb{R}$ and from the initial condition we get by setting $s = -t$ $$u(x - tb,0) = g(x - tb)$$ Letting $s = 0$ we deduce $$u(x,t) = g(x - tb) \qquad (x,t) \in \mathbb{R} \times [0,\infty)$$ Remark. There is also an inhomogenous version for $$\begin{aligned}u_t + bu_x = f \qquad &\text{in } \mathbb{R} \times (0,\infty)\\u = g \qquad &\text{on } \mathbb{R} \times \{0\}\end{aligned}$$ which solution is given by $$u(x,t) = g(x - tb) + \int_0^tf(x + (s - t)b,s)ds \qquad (x,t) \in \mathbb{R} \times [0,\infty)$$ See p. 19.
The Wave Equation. Consider the so-called wave equation IVP $$\begin{aligned}u_{tt} + b^2u_{xx} = 0 \qquad &\text{in } \mathbb{R} \times (0,\infty)\\u = g, u_t = h \qquad &\text{on } \mathbb{R} \times \{0\}\end{aligned}$$ The PDE can be factorized as $$\left( \frac{\partial}{\partial t} + b\frac{\partial}{\partial x}\right)\left( \frac{\partial}{\partial t} + b\frac{\partial}{\partial x}\right)u = u_{tt} - b^2u_{xx} = 0$$ Stipulate $$v(x,t) := \left( \frac{\partial}{\partial t} + b\frac{\partial}{\partial x}\right)u(x,t)$$ Then $$v_{t}+bv_{x} = 0$$ Applying the homogenous version of the transport equation yields $$v(x,t) = f(x - bt)\qquad (x,t) \in \mathbb{R} \times [0,\infty)$$ Hence $$u_t - bu_x = f(x - bt)$$ Now you can apply the inhomogenous solution to the transport equation, applying initial conditions and so on. I leave this to you, since it is already a very long answer. But I think the crucial part is to understand the solution to the transport equation properly.  
