I am looking at the following example of solving an hyperbolic system of equations using the method of characteristics:
$$\left.\begin{matrix} \frac{\partial{u_1}}{\partial{t}}+a \frac{\partial{u_2}}{\partial{x}}=0\\ \frac{\partial{u_2}}{\partial{t}}+b \frac{\partial{u_1}}{\partial{x}}=0 \end{matrix}\right\}$$
$a,b>0 \ , \ \ \ \displaystyle{u=u(u_1, u_2)} \ , \ \ \ A=\begin{bmatrix} 0 & a\\ b & 0 \end{bmatrix}$
$\displaystyle{u_t+Au_x=0}$
To check if it is hyperbolic, we have to find the eigenvalues:$\displaystyle{det(A-\lambda I)=\lambda^2-ab=0 \Rightarrow \lambda= \pm \sqrt{ab}}$
$\gamma=(\gamma_1, \gamma_2)$
$\displaystyle{\gamma_1(\frac{\partial{u_1}}{\partial{t}}+a \frac{\partial{u_2}}{\partial{x}})+\gamma_2(\frac{\partial{u_2}}{\partial{t}}+b \frac{\partial{u_1}}{\partial{x}})=0}$
$\Rightarrow \displaystyle{\gamma_1(\frac{\partial{u_1}}{\partial{t}}+\frac{b \gamma_2}{\gamma_1} \frac{\partial{u_1}}{\partial{x}})+\gamma_2(\frac{\partial{u_2}}{\partial{t}}+\frac{a \gamma_1}{\gamma_2} \frac{\partial{u_2}}{\partial{x}})=0}$
Since it should be in the form: $\gamma^T(u_t+\lambda u_x)=\gamma^T b$, it should be: $\displaystyle{\frac{b \gamma_2}{\gamma_1}=\lambda=\frac{a \gamma_1}{\gamma_2}}$
- $\lambda=\sqrt{ab}:$
We set $\displaystyle{\gamma_1=1}$, so $\displaystyle{\gamma_2= \sqrt{\frac{a}{b}}}$
$\displaystyle{(\frac{\partial{u_1}}{\partial{t}}+\sqrt{ab} \frac{\partial{u_1}}{\partial{x}})+\sqrt{\frac{a}{b}}(\frac{\partial{u_2}}{\partial{t}}+\sqrt{ab} \frac{\partial{u_2}}{\partial{x}})=0}$
$\displaystyle{\frac{\partial}{\partial{t}}(u_1+\sqrt{\frac{a}{b}}u_2)+\sqrt{ab} \frac{\partial}{\partial{x}}(u_1+\sqrt{\frac{a}{b}}u_2)=0}$
We set $\displaystyle{v_+=u_1+\sqrt{\frac{a}{b}}u_2}$. So we have: $\displaystyle{\frac{\partial{v_+}}{\partial{t}}+\sqrt{ab} \frac{\partial{v_+}}{\partial{x}}=0}$
The characteristic system is: $$\frac{dt}{1}=\frac{dx}{\sqrt{ab}}=\frac{dv_+}{0}$$
$\displaystyle{v=\text{ constant }, v=u_1+\sqrt{\frac{a}{b}}u_2}$, when $\displaystyle{\frac{dx}{dt}=\sqrt{ab}}$
$\displaystyle{x=\sqrt{ab}t+c \Rightarrow c=x- \sqrt{ab}t}$
$\displaystyle{v}$ is constant when $\displaystyle{x-\sqrt{ab}t}$ is constant.
That means that $$u_1+\sqrt{\frac{a}{b}}u_2=\frac{f(x-\sqrt{ab}t)}{2}$$
- $\lambda=-\sqrt{ab}:$
We conclude that $\displaystyle{v_-=u_1-\sqrt{\frac{a}{b}}u_2= \text{ constant }}$ when $\displaystyle{x+\sqrt{ab}t}$ is constant.
$$u_1-\sqrt{\frac{a}{b}}u_2=\frac{g(x+\sqrt{ab}t)}{2}$$
$$\text{ So } u_1=f(x-\sqrt{ab}t)+g(x+\sqrt{ab}t) \text{ AND } u_2=\sqrt{\frac{b}{a}}[f(x-\sqrt{ab}t)-g(x+\sqrt{ab}t)]$$
Could you explain how we conclude that at the one case: $$u_1+\sqrt{\frac{a}{b}}u_2=\frac{f(x-\sqrt{ab}t)}{2}$$ and at the other: $$u_1-\sqrt{\frac{a}{b}}u_2=\frac{g(x+\sqrt{ab}t)}{2}$$ ?? What are these functions $f$ and $g$??
$f$ and $g$ are "free functions". That is to say, for any pairs of functions $f$ and $g$ of one real variable, the pair $$ \begin{align} u_1(t,x) &= \frac14 \left( f(x - \sqrt{ab}t) + g(x + \sqrt{ab}t)\right) \\ u_2(t,x) &= \frac14 \sqrt{\frac{b}{a}} \left( f(x - \sqrt{ab}t) - g(x+\sqrt{ab}t)\right) \end{align} $$ are solutions. (These two expressions are obtained by solving the ones you wrote near the end of your question.)
In actual applications, to determine the functions $f$ and $g$ you need to appeal to the initial conditions: you have $$ \frac12 f(x) = u_1(x,0) + \sqrt{\frac{a}{b}} u_2(x,0) $$ and similarly for $g(x)$, so the conditions given at $t = 0$ completely determines the functions $f$ and $g$.
As to how you arrive at the condition that $u_1 + \sqrt{\frac{a}b} u_2 = f/2$ etc.: this is the entire point of the method of characteristics. The method of characteristics tells you that, by examining the equation, you find $v_+$ to be a function that depends only on $c = x - \sqrt{ab}t$. Ditto for $v_-$. Now, naively an arbitrary function of $x$ and $t$ variables should have two degrees of freedom. The fact that $v_+$ depends only on $c = x-\sqrt{ab}t$ says that $v_+$ has only one degree of freedom, and so is really a "one dimensional" function in disguise.