# A Charpit-type problem

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).

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.

• What kind of error do I have - the one about the envelope or something else? And how do we get $x\geq0$? $\tau$ can be negative, I think it does not represent physical time, so it would give negative values for $x$. – DesmondMiles Jun 4 at 14:51
• If $x\geq 0$ is indeed true,I agree that the domain would be covered fully. Also, where does $y\leq 3 - (x/2)^2$ come from? – DesmondMiles Jun 4 at 14:57
• Sry, my bad, I edited my reply. Still, the domain is larger than the one you gave. – maxmilgram Jun 4 at 15:12
• What do you mean by "Motivated by the BC"? Nothing from it claims that $x\geq 0$ directly, I think. – DesmondMiles Jun 4 at 15:21
• No, not directly. If you are indeed interested in the case for negative $x$ I'll have a look at it in the evening! – maxmilgram Jun 4 at 15:26

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 $$v(0,y)=2-y$$

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

$$\dfrac{dx}{dt}=1$$ , 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. $$z=F\left(y\mp\dfrac{2zx}{\sqrt{1-4z^2}}\right)$$

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

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

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

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

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

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

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

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

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

• Sorry, but you just try to attack the problem in a completely different (and more complicated, since how are you going to go back to the main function?!) way. I have already given an explicit parametric form. – DesmondMiles Jun 4 at 12:16