For $$ u_x^2+u_y^2=1 $$ and initial condition $$u(x,0) = f(x)$$ Is there a reference I can look up to better understand how to go about this solution? If I can visualize the results, that'd be great.
I am really just looking to visualize this. If there is a video online or some pdf which is simple and has charts/graphs that would help to go about the solution.
I got this:
$$u_y = \pm \sqrt {1 - u_x^2}$$
and the diff w.r.t $x$, I get
$$ u_{xy} =\pm \frac{uu_x}{\sqrt{1-{u_x}^2}} $$
and then I substitute $v = u_x$ and end up with a linear equation like this:
$$v_y = \pm \frac{vv_x}{\sqrt{1-v^2}}$$
This gives 2 equations
$\sqrt{1-v^2}v_y + v v_x = 0$ and $\sqrt{1-v^2}v_y - v v_x = 0$
How do I go about solving these (let's say the first one)
Hint:
$u_x^2+u_y^2=1$ with $u(x,0)=f(x)$
$u_y=\mp\sqrt{1-u_x^2}$ with $u(x,0)=f(x)$
$u_{xy}=\pm\dfrac{u_xu_{xx}}{\sqrt{1-u_x^2}}$ with $u(x,0)=f(x)$
Let $v=u_x$ ,
Then $v_y=\pm\dfrac{vv_x}{\sqrt{1-v^2}}$ with $v(x,0)=f_x(x)$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$
$\dfrac{dv}{dt}=0$ , letting $v(0)=v_0$ , we have $v=v_0$
$\dfrac{dx}{dt}=\mp\dfrac{v}{\sqrt{1-v^2}}=\mp\dfrac{v_0}{\sqrt{1-v_0^2}}$ , letting $x(0)=g(v_0)$ , we have $x=g(v_0)\mp\dfrac{v_0t}{\sqrt{1-v_0^2}}=g(v)\mp\dfrac{vy}{\sqrt{1-v^2}}$ i.e. $v=G\left(x\pm\dfrac{vy}{\sqrt{1-v^2}}\right)$
$v(x,0)=f_x(x)$ :
$G(x)=f_x(x)$
$\therefore v=f_x\left(x\to x\pm\dfrac{vy}{\sqrt{1-v^2}}\right)$