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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)

Dylan
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cauchyprobability
  • 159

2 Answers2

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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. $v=G\left(x\pm\dfrac{vy}{\sqrt{1-v^2}}\right)$

$v(x,0)=f_x(x)$ :

$G(x)=f_x(x)$

$\therefore v=f_x\left(x\to x\pm\dfrac{vy}{\sqrt{1-v^2}}\right)$

doraemonpaul
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0

The Eikonal equation $u_x^2+u_y^2=1$ in upper half-plane $y\geq 0$ for
$u(x,0)=u_0(x)$. In terms of $p=u_x$, $q=u_y$, the equation becomes $p^2+q^2=C$, we differentiate it in $x$ and obtain $pp_x+qq_x=0$. Since $p_y=q_x$, then $pp_x+qp_y=0$. Similarly, by differentiation in $y$, we have $pq_x+qq_y=0$. Combining the characteristic equations, we get $p$ and $q$ are constants on characteristics $\frac{dx}{p}=\frac{dy}{q}=\frac{du}{1}$. The last equality is thanks $p^2+q^2=1$. If the characteristic straight line goes through point $(x,y)$ then its equation is $y=\frac{q}{p}(x-x_0)$. Hemce, $x_0=x-\frac{p(x_0)}{q(x_0)}y$. Since $p(x_0)=u_0'(x_0)$, $q(x_0)=1-u_0'(x_0)$, then let us assume that the equation for $x_0$ is uniquely solvable and we know $x_0$.Then we integrate along the characteristics and obtain $$ u(x,y)=u_0(x_0)+\int_{(x_0,0)}^{(x,y)} ds=u_0(x_0)+\left((x-x_0)^2+y^2\right)^{1/2} $$. For example, the case $u_0(x)=ax$, $|a|<1$, yields $p=a$, $q=\pm\sqrt{1-a^2}$, $x_0=x-\frac{p}{q}y=x\mp\frac{a}{\sqrt{1-a^2}}y$, and we obtain $$ u(x,y)=a x_0+\left((x-x_0)^2+y^2\right)^{1/2}=ax\mp\frac{a^2}{\sqrt{1-a^2}}y+ \frac{y}{\sqrt{1-a^2}}=ax\pm\sqrt{1-a^2}y. $$

pabodu
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