# Wave equation how to derive the form $u(x,t)=f(x+ct)+g(x+ct)$

I am referring Walter A. Strauss PDE book. There in solving the wave equation I have several parts which I don't understand.
$U_{tt}=c^2U_{xx}$=(${\partial \over \partial t}$-$c \partial \over \partial x$ )(${\partial \over \partial t}$+$c \partial \over \partial x$ )$u$=0

Is this factorization done by simply considering $\partial$ as a operator and then factoring as $(a^2-b^2)=(a-b)(a+b)$

Then in the book it is said that let $v=u_t+cu_x$, then $v_t-cv_x=0$. When $v_t-cv_x=0$ is solved $v=h(x+ct)$.I understand upto this point.

But I don't understand from the point:
When solving $u_t+cu_x=v=h(x+ct)$ in the book it says " It is easy to check directly by differentiation that one solution is $u(x,t)= f(x+ct)$, where $f'(s)=h(s)/2c.$.To the solution $f(x+ct)$ we can add $g(x-ct )$ " to get another solution.

Please can someone explain to me how this solution $u(x,t)=f(x+ct)+g(x+ct)$ is obtained

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ Lets $\ds{\xi \equiv x - ct\,,\quad \eta \equiv x + ct}$. Then, \begin{align} \partiald{}{x}& =\partiald{\xi}{x}\,\partiald{}{\xi} + \partiald{\eta}{x}\,\partiald{}{\eta} =\partiald{}{\xi} + \partiald{}{\eta} \\[3mm] \partiald{}{t}& =\partiald{\xi}{t}\,\partiald{}{\xi} + \partiald{\eta}{t}\,\partiald{}{\eta} =-c\,\partiald{}{\xi} + c\,\partiald{}{\eta} \end{align}

\begin{align} \partiald[2]{}{x}&=\partiald[2]{}{\xi} + 2\,{\partial^{2} \over \partial\xi\,\partial\eta} + \partiald[2]{}{\eta} \\[3mm] \partiald[2]{}{t}&=c^{2}\pars{% \partiald[2]{}{\xi} - 2\,{\partial^{2} \over \partial\xi\,\partial\eta} + \partiald[2]{}{\eta}} \end{align}

$$\partiald[2]{}{x} - {1 \over c^{2}}\,\partiald[2]{}{t} =4\,{\partial^{2} \over \partial\xi\,\partial\eta}$$

$$0=\partiald[2]{u}{x} - {1 \over c^{2}}\,\partiald[2]{u}{t} =4\,{\partial^{2}u \over \partial\xi\,\partial\eta} \quad\imp\quad\partiald{u}{\eta} = {\rm F}\pars{\eta} \quad\imp\quad u = \overbrace{\int{\rm F}\pars{\eta}\,\dd\eta}^{\ds{\equiv\fermi\pars{\eta}}} + {\rm g}\pars{\xi}$$

$$\color{#44f}{\large% u = \fermi\pars{x + ct} + {\rm g}\pars{x - ct}}$$

I'm not too sure what the books doing, so I'm just going to put my approaches in case you like them.

Lazy way: Assume solution of the form $f(x+\alpha t)$. Subbing into the pde we get $\alpha^2 f''-c^2f''=0$ where the primes are differentiation with respect to argument. To avoid trivial solutions, we assume $f''\not=0$ and we get $\alpha=\pm c$. Hence $u=f(x+ct)+g(x-ct)$, where $f,g$ are arbitrary functions of variables corresponding to the possible values for alpha.

More involved (but basically the same) way: For an equation of the form $a(x,t)u_{xx}+2b(x,t)u_{xt}+c(x,t)u_{tt}=f(u_x, u_t, x, t)$ the ODE $a(\frac{dx}{dt})^2-2b\frac{dx}{dt}+c=0$ defines the gradient of the characteristic coordinates. Solving this for the wave equation gives the characteristic coordinates $\xi = x+ct$ and $\eta = x-ct$.

Then, with lots of chain rule (easy, but tedious), this change of coordinates will eventually lead you to the canonical form $\dfrac{\partial^2 u}{\partial\xi\partial\eta}=0$. Integrating twice yields $u=\displaystyle\int f(\xi)\ d\xi+g(\eta)=F(x+ct)+g(x-ct)$.