A Charpit-type problem

Question

Solve in parametric form for $u(x,y)$: $$u + u_x^2 + u_y^2 - 2 = 0$$ with the data $u(0,y) = y$ for $0\leq y \leq 1$ and the restriction $u_x \geq 0$. Determine (and show on a sketch) the domain in which the solution is uniquely determined.

Progress: If $$F = u + u_x^2+ u_y^2 - 2,$$ $p = u_x$, $q = u_y$, then $$p_{\tau} = -p, ~ q_{\tau} = -q, ~ x_{\tau} = 2p, ~ y_{\tau} = 2q, ~ u_{\tau} = 2(p^2 + q^2) = 4-2u$$ so with the initial data (and $p = u_x \geq 0$) we get $x_0 = 0$, $y_0 = u_0 = s$ we have $$x = 2\sqrt{1-s}(1-e^{-\tau}), ~ y= s + 2(1-e^{-\tau}), \\ p=\sqrt{1-s}e^{-\tau}, ~ q=e^{-\tau}\\ u = 2 + (s-2)e^{-2\tau}.$$

Now to determine the domain note that $y = s + \frac{x}{\sqrt{1-s}}$ where $0 \leq s < 1$ (and $x = 0$ for $s=1$), that $x = 2\sqrt{1-s}(1-e^{-\tau}) \leq 2$ and that the envelope is determined by differentiating $y = s + \frac{x}{\sqrt{1-s}}$ with respect to $s$, i.e. $ 0 = 1 - \frac{x}{2}(1-s)^{-\frac{3}{2}}$, i.e. $s = 1 - (\frac{x}{2})^{\frac{2}{3}}$, so we obtain $$y = 1 - (\frac{x}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}}x^{\frac{2}{3}}$$ as the envelope curve. Moreover $y=x$ at $s=0$.

But I don't think I can just say that the domain is between $y=x$, $x\leq 2$ and $y = 1 - (\frac{x}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}}x^{\frac{2}{3}}$ as the lines $y = s + \frac{x}{\sqrt{1-s}}$, $0\leq s \leq 1$, do not cover all of it (this can be seen via a diagram).

So any advice?

Answer 1

Hint:

$u+u_x^2+u_y^2-2=0$ with $u(0,y)=y$

$u_x^2+u_y^2=2-u$ with $u(0,y)=y$

Let $v=2-u$ ,

Then $v_x=-u_x$

$v_y=-u_y$

$\therefore(-v_x)^2+(-v_y)^2=v$ with $v(0,y)=2-y$

$v_x^2+v_y^2=v$ with $v(0,y)=2-y$

Similar to Characteristics method applied to the PDE $u_x^2 + u_y^2=u$:

Let $v=w^2$ ,

Then $v_x=2ww_x$

$v_y=2ww_y$

$\therefore(2ww_x)^2+(2ww_y)^2=w^2$ with $w(0,y)=\pm\sqrt{2-y}$

$4w^2(w_x)^2+4w^2(w_y)^2=w^2$ with $w(0,y)=\pm\sqrt{2-y}$

$w_x^2+w_y^2=\dfrac{1}{4}$ with $w(0,y)=\pm\sqrt{2-y}$

$w_x^2=\dfrac{1}{4}-w_y^2$ with $w(0,y)=\pm\sqrt{2-y}$

$w_x=\pm\sqrt{\dfrac{1}{4}-w_y^2}$ with $w(0,y)=\pm\sqrt{2-y}$

$w_{xy}=\mp\dfrac{w_yw_{yy}}{\sqrt{\dfrac{1}{4}-w_y^2}}$ with $w(0,y)=\pm\sqrt{2-y}$

Let $z=w_y$ ,

Then $z_x=\mp\dfrac{zz_y}{\sqrt{\dfrac{1}{4}-z^2}}$ with $z(0,y)=\mp\dfrac{1}{2\sqrt{2-y}}$

$z_x\pm\dfrac{zz_y}{\sqrt{\dfrac{1}{4}-z^2}}=0$ with $z(0,y)=\mp\dfrac{1}{2\sqrt{2-y}}$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$

$\dfrac{dz}{dt}=0$ , letting $z(0)=z_0$ , we have $z=z_0$

$\dfrac{dy}{dt}=\pm\dfrac{z}{\sqrt{\dfrac{1}{4}-z^2}}=\pm\dfrac{z_0}{\sqrt{\dfrac{1}{4}-z_0^2}}$ , letting $y(0)=f(z_0)$ , we have $y=\pm\dfrac{z_0t}{\sqrt{\dfrac{1}{4}-z_0^2}}+f(z_0)=\pm\dfrac{2zx}{\sqrt{1-4z^2}}+f(z)$ , i.e. $z=F\left(y\mp\dfrac{2zx}{\sqrt{1-4z^2}}\right)$

$z(0,y)=\mp\dfrac{1}{2\sqrt{2-y}}$ :

$F(y)=\mp\dfrac{1}{2\sqrt{2-y}}$

$\therefore z=\mp\dfrac{1}{2\sqrt{2-y\pm\dfrac{2zx}{\sqrt{1-4z^2}}}}$

$\sqrt{2-y\pm\dfrac{2zx}{\sqrt{1-4z^2}}}=\mp\dfrac{1}{2z}$

$2-y\pm\dfrac{2zx}{\sqrt{1-4z^2}}=\dfrac{1}{4z^2}$

$\pm\sqrt{1-4z^2}=\dfrac{8xz^3}{4(y-2)z^2+1}$

$1-4z^2=\dfrac{64x^2z^6}{16(y-2)^2z^4+8(y-2)z^2+1}$

$64x^2z^6=16(y-2)^2z^4+8(y-2)z^2+1-64(y-2)^2z^6-32(y-2)z^4-4z^2$

$64(x^2+(y-2)^2)z^6-16(y^2-6y+8)z^4-4(2y-5)z^2-1=0$

Answer 2

First, for negative $x$ the characteristics intersect thus we can not have a unique solution

Thus, when searching for the envelope there is in fact no extremal point of $y(s)$. Therfore, the region is determined by the maximal $x$ values given by $\lim_{\tau\to\infty}$. Taking the limit in $x$ and $y$ and substituting for $s$ yields $x\leq y\leq3-(x/2)^2$.

Second since the maps are continuous I see no reason why the domain should not be covered fully.

Third, you also need to think about the uniqueness which is given by the injectivity of the $y(x,s)$ map.