$ u\partial_x u + \partial_y u = 2$ using Method of Characteristics I'm looking at the Solution of Example 2.7 from https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf with the following PDE
$$ u\partial_x u + \partial_y u = 2$$ with $x,y > 0$ and Initial Condition $u(x,x) = \frac{x}{2}$.
The Method of Characteristics leads to the following equations:
\begin{align*} &\frac{dx}{ds} = u \\&\frac{dy}{ds} = 1 \\& \frac{du}{ds} = 2 \end{align*}
Then they write \begin{align*} x(0,\tau)=\tau ,\: y(0,\tau)=\tau ,\: u(0,\tau)=\tau \, \text{  with  } \, u(\tau,\tau) = \frac{\tau}{2}  \end{align*}
This  step is not clear to me. Where does this come from?
 A: Possibly this answer isn't exactly what you expect. However I hope that will help.
$$ u\partial_x u + \partial_y u = 2$$
$$\begin{cases} 
\frac{dx}{ds} = u \\
\frac{dy}{ds} = 1 \\
\frac{du}{ds} = 2 
\end{cases}\qquad\text{is equivalent to}\qquad ds=\frac{dx}{u}=\frac{dy}{1}=\frac{du}{2}$$
A first characteristic equation comes from solving $\quad \frac{dx}{u}=\frac{du}{2}$ :
$$u^2-4x=c_1$$
A second characteristic equation comes from solving $\quad \frac{dy}{1}=\frac{du}{2}$ :
$$u-2y=c_2$$
The general solution of the PDE expressed on the form of implicit equation $\Phi(c_1,c_2)=0$ is :
$$\Phi(u^2-4x\:,\:u-2y)=0$$
$\Phi$ is an arbitrary function of two variables. Or equivalently :
$$\boxed{u^2-4x=F(u-2y)}$$
$F$ is an arbitrary function of one variable. Of course they are many equivalent forms of implicit equations to express the general solution.
Condition : $\quad u(x,x)=\frac{x}{2}$
$$\left(\frac{x}{2}\right)^2-4x=F\left(\frac{x}{2}-2x\right)\quad\implies\quad F\left(-\frac{3x}{2}\right)=\frac{x^2}{4}-4x$$
Let $X=-\frac{3x}{2}\quad;\quad x=-\frac{2X}{3}$
$$F(X)=\frac{\left(-\frac{2X}{3}\right)^2}{4}-4\left(-\frac{2X}{3}\right)=\frac{X^2}{9}+\frac{8X}{3}$$
Now the function $F$ is known. We put it into the above general solution where $X=u-2y$ :
$$u^2-4x=\frac{(u-2y)^2}{9}+\frac{8(u-2y)}{3}$$
Solving for $u$ gives the particular solution of the PDE which satisfies the specified condition :
$$u(x,y)=-\frac14 y+\frac32\pm\frac34\sqrt{y^2-12y+8x+4}$$
Note : the sign of the square root has to be determined according to $u(x,x)=\frac{x}{2}$ .
