Method of characteristics. Small question about initial conditions.

Okay, so we're given a PDE

$$x \frac {\partial u} {\partial x} + (x+y) \frac{\partial u} {\partial y} = 1$$

with initial condition: $u(x=1,y)=y$

So $a=x, b=x+y, c=1$

$\Rightarrow$ characteristic equations: $$\frac{dx}{dt}=x, \frac{dy}{dt}=x+y, \frac{du}{dt}=1$$

This next part is my trouble:

Initial Conditions: $$x_0(0,s)=1,$$ $$y_0(0,s)=s,$$ $$u_0(0,s)=y=s.$$

So I can see that the $u(0,s)=s$ is coming from the original IC, but where are $x_0,$ and $y_0$ coming from? Many thanks in advance!

The idea with method of characteristics is there is a initial condition variable $s$ that parametrizes the initial curve, and a characteristic variable $t$ that dictates the "flow" of the characteristics away from the initial curve.

In your problem, your initial condition is $u(x=1,y) = y$, so the initial curve is the curve given by $x = 1$, which can be parametrized by $\{(1,s): s \in \mathbb{R}\}$. So this gives us $x(0,s) = 1, y(0,s) = s, u(0,s) = s$.

Here is a technique you can follow. We have the equations

$$\frac{dx}{dt}=x, \frac{dy}{dt}=x+y, \frac{du}{dt}=1 \longrightarrow (*).$$

From the first two equations, we get

$$\frac{dy}{dx}=\frac{x+y}{x} \implies y = x\ln(x) + c x \implies c=\frac{y-x\ln(x)}{x}\longrightarrow (1) .$$

Now, the first and the third equations in $(*)$ give

$$\frac{du}{dx}=\frac{1}{x}\implies u(x,y)= \ln(x)+f(c) \longrightarrow (2).$$

Using $(1)$, $(2)$ becomes

$$\implies u(x,y)=\ln(x)+f\left(\frac{y-x\ln(x)}{x}\right) \longrightarrow (**).$$

Now, we exploit the initial condition in $(**)$ to find the function $f$

$$u(1,y) = y = 0+f(y) \implies f(y)=y.$$

Substituting back in $(**)$, we have

$$u(x,y)=\ln(x)+\left(\frac{y-x\ln(x)}{x}\right)$$

$$u(x,y) = \frac{y}{x}.$$

Solving the ODEs, $$x(t,s)=e^t,\;\; y(t,s)=e^t(t+s),\;\; z(t,s)=t+s$$ Then $$u(x,y)=t+s=\frac{e^t(t+s)}{e^t}=\frac{y}{x}$$

$${{\rm d}y \over {\rm d}x} = 1 + {y \over x} \tag{1}$$ With the scaling $\tilde{x} = \mu x$ and $\tilde{y} = \nu\, y$, Eq. $\left(1\right)$ does not change its form whenever $\mu = \nu$ which is equivalent to $\tilde{y}/\tilde{x} = y/x$. It means Ec. $\left(1\right)$ is simplified with the choice $y/x \equiv \phi\left(x\right)$: $$x\phi´\left(x\right) = 1 \quad\Longrightarrow\quad \phi\left(x\right) = \ln\left(x\right) + \overbrace{\alpha}^{\mbox{constant}} \quad\Longrightarrow\quad y = x\ln\left(x\right) + \alpha\, x \tag{2}$$ In addition, ${\rm d}{\rm u}\left(x,y\left(x\right)\right)/{\rm d}x = 1/x$ leads to ${\rm u}\left(x,y\left(x\right)\right) = \ln\left(x\right)\ +\ \overbrace{\beta}^{\mbox{constant}}$. It is reduced, with Eq. $\left(2\right)$, to

$${\rm u}\left(x,y\right) = \left({y \over x} - \alpha\right) + \beta$$

$${\rm u}\left(1, y\right) = y\,, \quad\Longrightarrow\quad -\alpha + \beta = 0. \quad\Longrightarrow\quad \color{#ff0000}{\large{\rm u}\left(x, y\right) \color{#000000}{\ =\ }{y \over x}}$$