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?
Third image in the comment below
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.