Solve the PDE $u_x + u_y = 0$ in the domain $y > \phi(x)$, $x \in \mathbb{R}$, given that $u = g(x)$ on the curve $y = \phi(x)$, where $\phi(x) = \dfrac{x}{1 + |x|}$.
The characteristic equations are
$$\frac{dx}{dt} = 1, \frac{dy}{dt} = 1$$
with solution
$$x = t + C_1, y = t + C_2$$
So $x – y$ is a constant, which implies that a general solution is of the form
$$u(x, y) = F(x – y)$$
where $F$ is an arbitrary function.
Question 1: Why does $x – y$ being a constant imply that a general solution is of the form $u(x, y) = F(x – y)?$
Given the general solution
$$u(x, y) = F(x – y)$$
with initial curve
$$\Gamma = (s, \phi(s), g(s))$$
we have the solution
$$u(s, \phi(s)) = F(s – \phi(s)) = g(s)$$
It just remains to write the solution in terms of $x$ and $y$.
Writing the solution in terms of $x$ and $y$ requires that $x = s – \phi(s)$ be solved for $s$ in terms of $x$ so as to give $A(x) = g(s(x))$.
When $\phi(x) = \dfrac{x}{1 + |x|}$, this is accomplished by treating the cases $x \ge 0$ and $x < 0$ separately.
$s(x) =
\begin{cases}
\frac{1}{2}(x + \sqrt{x^2 + 4x})&\text{if}\, x\ge 0\\
\frac{1}{2}(x – \sqrt{x^2 – 4x})&\text{if}\, x < 0\\
\end{cases}$
Question 2: How did the author get these values for $s(x)$? What is the process?
The solution can then be written as $u(x, y) = g(s(x – y))$ for $y > \phi(x)$.
I would be very thankful if someone could please help me with this.
Best Answer
For the second question. Simply isolate $s$ in $x=s-\phi(s)$ or in
$x=s-\dfrac{s}{1 + |s|}$. For the $s\ge 0$ part: $x(1+s)=s+s^2-s$ or
$s^2-xs-x=0\implies s=\frac{1}{2}(x\pm\sqrt{x^2 + 4x})$ We drop the solution with the minus sign because it entails that $s<0$, aganist the hypothesis for the posing of the equation. Because the same reason ($s<0$) we drop values with $x<0$. So $s=\frac{1}{2}(x + \sqrt{x^2 + 4x})\;\text{if}\, x\ge 0$. For $s<0$ we solve $x(1-s)=s-s^2-s$ or ($s^2-xs+x=0$) instead. Then, $s=\frac{1}{2}(x\pm\sqrt{x^2 -4x})$ Now, we drop the plus sign because on the contrary $s\ge0$ and, because the same reason, it must be $x<0$. This drives to the given solution:
$$s(x) = \begin{cases} \frac{1}{2}(x + \sqrt{x^2 + 4x})&\text{if}\, x\ge 0\\ \frac{1}{2}(x - \sqrt{x^2 - 4x})&\text{if}\, x < 0\\ \end{cases}$$
For the first question you have to consider that $u$ is constant along the characteristic curves, so is, along curves $x-y=c_1$ with $c_1$ some constant($u(x,x+c_1)=k$, $k$ constant). A solution must specify the value of $u$ for each of the values of $c_1$: $u(x,x+c_1)=F(c_1)$. So is, a particular choice of $F$ says the value of $u$ for each member of the family of characteristic curves: $u(x,y)=F(x-y)$. The function $F$ is arbitrary in the sense that by itself doesn't impose any condition on the solutions (taken apart that it must enjoy of some regularity)