Method of characteristics - eliminating variables

Question

I am trying to follow a guide for the method of characteristics; quoting the first example:

We use the method of characteristics to solve the problem

$ 2u_x - u_y = 0, \;\; u(x, 0) = f(x) $

(...) we can easily solve [the characteristic equations] to get:

$x = 2s + x_0 \;\;\; y= -s + y_0 \;\;\; u = u_0$

Up to here I have no problem. However, the guide then substitutes in the initial condition and says this:

we now eliminate $x_0$ and $s$ to find that:

$ y = -s \;\;\; \Rightarrow \;\;\; u = f(x_0) = f(x - 2s) = f(x + 2y)$

My question is: where did $y_0$ and $u_0$ go?

Answer 1

You have characteristics given as functions $x(s)$ and $y(s)$ which depend on parameter $s$:

\begin{align} x = x(s) &= x_0 + 2s && \implies& x_0 &= x - 2s \label{1}\tag{1} \\ y = y(s) &= y_0 - s && \implies& y_0 &= y + s \label{2}\tag{2} \end{align}

The initial condition $u\left(x_0, 0\right) = f\left(x_0\right)$ is given for $y_0=0$, so in equation $\eqref{2}$ we set $y_0$ to be equal $0$:

$$0 = y_0 = y + s \qquad\;\implies \quad\bbox[5pt, border: 1.5pt solid #DD0712]{y = -s^{\,\!}} \label{3}\tag{3}$$

But then we can write general solutions as $$ u\left(x_0, y_0\right) = u\left(x_0, 0\right) = f\left(x_0\right) = f\left(x-2s\right) \overset{\eqref{3}}{=} f\left(x+2s\right) $$

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Now, one can ask why do we say that $u\left(x_0, y_0\right)$ is the general solution? Well, if you recall the definition of characteristics, which are the

$\;\big(\cdots\big)\;$ parametric curves along which solution $u\left(x, y\right)$ remains constant,

you will see that $u\left(x, y\right) = u\left(x_0, y_0\right)$ as long as $(x,y)$ and $\left(x_0,y_0\right)$ lie on the same curve $\eqref{1}$–$\eqref{2}$. Then we choose $y_0$ to be equal $0$ in order to incorporate initial condition $u(x,0) = f(x)$.

Answer 2

The characteristic equations are : $$\frac{dx}{2}=\frac{dy}{-1}=\frac{du}{0}$$ which leads to the equations of chararacteristics : $$\begin{cases}u=c_1\\x+2y=c_2 \end{cases}$$ The general solution on implicit form with is : $$\Phi\left(u\:,\:x+2y\right)=0$$ where $\Phi$ is any derivable function of two variables.

This is equivalent to : $$u=F(x+2y)$$ where $F$ is any derivable function.

The bounding condition leads to : $$u(x,0)=f(x)=F(x+0)=F(x)$$ Hense $F=f$ is now determined and the solution is : $$u(x,y)=F(x+2y)=f(x+2y)$$