# Method of characteristics - eliminating variables

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?

• More simply, the method shows that for every $(x_0,y_0,s)$, $$u(x_0+2s,y_0-s)=u(x_0,y_0).$$ For every $(x,y)$, choosing $(x_0,y_0,s)=(x+2y,0,-y)$ above, one gets $$u(x,y)=u(x+2y,0).$$ Rename $u(\ ,0)$ as $f$ and you are done. – Did Feb 5 '16 at 7:35
• @Did I.. don't quite follow. How does the method show that? – lvc Feb 5 '16 at 7:46
• First line simply rephrases the fact that "$u$ remains constant on the curve $(x(s),y(s))$", for $x(s)=2s+x_0$, $y(s)=-s+y_0$, that is, that $u(x(s),y(s))=u(x(0),y(0))$ for every $s$. Second line is self-explanatory. What remains? – Did Feb 5 '16 at 7:49
• @Did I think there's some deep background here that I've missed - "$u$ remains constant on the curve $(x(s), y(s))$" isn't a fact that I recognise. Where does it come from? – lvc Feb 5 '16 at 7:57
• Lines 3-5 of the text you linked to. – Did Feb 5 '16 at 7:59

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 $$u\left(x_0, y_0\right) = u\left(x_0, 0\right) = f\left(x_0\right) = f\left(x-2s\right) \overset{\eqref{3}}{=} f\left(x+2s\right)$$

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 $u(x,0) = f(x)$.

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 : $$u(x,y)=F(x+2y)=f(x+2y)$$