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I'm trying to solve this Cauchy problem:

$$ u_x^2 + u_y^2 = u \\ u(x,0) = x^2+1$$

Here's what I've tried so far:

Letting $p(r,s) = u_x$, $q(r,s) = u_y$ and $z(r,s) = u$, we have:

$$ F = p^2 + q^2 - z = 0 $$

and thus:

$$ \begin{align} \frac{dx}{ds} &= F_p = 2p \\ \frac{dy}{ds} &= F_q = 2q \\ \frac{dz}{ds} &= pF_p + qF_q = 2p^2 + 2q^2 = 2z\\ \frac{dp}{ds} &= -F_x -pF_z = -(0) -p(-1) = p \\ \frac{dq}{ds} &= -F_y -qF_z = -(0) -q(-1) = q \\ \end{align}$$

Also, $\Gamma(r,0) = r^2 +1 $, and

$$ \begin{align} (r^2+1)' &= \phi_1 \cdot (r)' + \phi_2 \cdot (0)' \\ \implies 2r &= \phi_1 \end{align}$$

so that:

$$ \begin{align} (2r)^2 + \phi_2^2 &= r^2+1 \\ \implies 4r^2 + \phi_2^2 &= r^2+1 \\ \implies \phi_2^2 &= -3r^2 + 1 \\ \implies \phi_2 &= \pm \sqrt{ -3r^2 +1 } \end{align} $$

Which leads to:

$$ \begin{align} p &= e^s 2r \\ q &= e^s \sqrt{ -3r^2 +1 } \\ \implies x &= 4e^s r - 3r \\ \text{and } y & = 2e^s \cdot ( \pm \sqrt{ -3r^2 +1 } ) \mp \sqrt{ -3r^2 +1 }\end{align} $$

And I'm pretty sure by now it's gone way off the rails, because I can't solve for for $s$ and $r$ in terms of $x$ and $y$, but I can't figure out where I went wrong.

Could anybody please shed some light?

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  • $\begingroup$ Similar to math.stackexchange.com/questions/1033906 $\endgroup$ – doraemonpaul May 16 '17 at 3:21
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    $\begingroup$ Are you sure your PDE and initial condition is correct? $\endgroup$ – mattos May 16 '17 at 14:33
  • $\begingroup$ An initial condition like $u = 1 + \frac{x^2}{4}$ would make it so much easier. Then taking an ansatz on the form $u = a + bx + cy + dxy + ex^2 + fy^2$ would work. In this case it does not. $\endgroup$ – Winther May 16 '17 at 23:40
  • $\begingroup$ @mlaci : The difficulty appears just after the calculus of $p$ and $q$ (about where you stopped). The solving involves a four degree polynomial equation. See an explanation in my answer. $\endgroup$ – JJacquelin May 17 '17 at 10:02
  • $\begingroup$ @doraemonpaul , Very similar, indeed! I had found that one using Google, but this problem was presented in the midst of other ones that were supposed to be solved using the method I started with above, so I figured the $x^2+1$ would have made a difference, and that I had just made a mistake somewhere in using the method. $\endgroup$ – mlaci May 18 '17 at 1:51
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Obviously, the main difficulty comes from the boundary condition $u(x,0)=x^2+1$ because it involves to solve a polynomial equation leading to huge formula. Even if the solving is theoretically possible, one have to merely accept a result on implicit form.

To make more clear where the difficulty arises, we will avoid the profusion of symbols introduced into the usual method of characteristics, but in following the same approach in fact. $$u_x^2+u_y^2=u \quad;\quad u(x,0)=x^2+1$$ First change of function :$\quad u=v^2\quad\begin{cases}u_x=2vv_x \\ u_y=2vv_y\end{cases}\quad\to\quad v_x^2+v_y^2=\frac{1}{4}\quad;\quad v(x,0)=\sqrt{x^2+1}$

$$v_y=\sqrt{\frac{1}{4}-v_x^2} \quad\to\quad v_{xy}=\frac{v_xv_{xx}}{\sqrt{\frac{1}{4}-v_x^2}}$$ Second change of function : $\quad w=v_x\quad\to\quad w_{y}=\frac{ww_{x}}{\sqrt{\frac{1}{4}-w^2}}$ $$\sqrt{\frac{1}{4}-w^2}\:w_y-2w\:w_x=0 \quad;\quad w(x,0)=v_x(x,0)=\frac{x}{\sqrt{x^2+1}}$$

This is a first order PDE. The set of characteristic ODEs is : $\quad \frac{dy}{\sqrt{\frac{1}{4}-w^2}}=\frac{dx}{-2w}=\frac{dw}{0}$

First family of characteristic curves, coming from $dw=0 \quad\to\quad w=c_1$

Second family of characteristic curves, from $\frac{dy}{\sqrt{\frac{1}{4}-c_1^2}}=\frac{dx}{-2c_1} \quad\to\quad 2c_1y+\sqrt{\frac{1}{4}-c_1^2}x=c_2$

The general solution can be presented on various forms :$\begin{cases} \Phi\left(w\:,\: 2wy+\sqrt{\frac{1}{4}-w^2}\:x\right)=0\\ w=f\left(2wy+\sqrt{\frac{1}{4}-w^2}\:x\right)\\ 2wy+\sqrt{\frac{1}{4}-w^2}\:x=F(w)\end{cases}$

where $\Phi$ , $f$ , $F$ are any differentiable functions. Any one of these functions has to be determined according to the boundary condition.

$$w(x,0)=\frac{x}{\sqrt{x^2+1}}\quad\to\quad 2\frac{x}{\sqrt{x^2+1}}0+\sqrt{\frac{1}{4}-\left(\frac{x}{\sqrt{x^2+1}}\right)^2}\:x=F\left(\frac{x}{\sqrt{x^2+1}}\right)$$

Let $t=\frac{x}{\sqrt{x^2+1}} \quad\to\quad x=\frac{t}{\sqrt{1-t^2}} \quad\to\quad \sqrt{\frac{1}{4}-t^2}\:\frac{t}{\sqrt{1-t^2}} =F\left(t\right)$

Now, the function $F$ is determined. We put it into the above general solution :

$$2wy+\sqrt{\frac{1}{4}-w^2}\:x=\sqrt{\frac{1}{4}-w^2}\:\frac{w}{\sqrt{1-w^2}}$$

$$\frac{4}{\sqrt{1-4w^2}}\:y+\frac{x}{w}=\frac{1}{\sqrt{1-w^2}}$$

Solving this equation for $w$ leads to $\quad w(x,y)$

In fact, this is a four degree polynomial equation. One can solve it analytically, but this involves huge formulas. That is the hitch.

So, we let $w$ on the implicit form of the above equation and, from it, we consider that $w(x,y)$ is known.

$w=\frac{u_x}{2\sqrt{u}}\quad\to\quad \int \frac{u_x}{2\sqrt{u}}=\sqrt{u}=\int w(x,y)dx$ $$u(x,y)=\left(\int w(x,y)dx \right)^2$$

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  • $\begingroup$ Thank you very much for that very detailed answer! Although I think that this wasn't what the teacher had in mind (I could be wrong), I was able to follow almost everything you did, so I'm very grateful. Just two questions though: 1) When you write $\Phi\left(w\:,\: 2c_1y+\sqrt{\frac{1}{4}-w^2}\:x\right)=0$ and $w=\frac{u_x}{\sqrt{u}}\quad\to\quad \int \frac{u_x}{\sqrt{u}}=2\sqrt{u}=\int w(x,y)dx$, is this $c_1$ is not equal to $w$ in this instance? 2) When you write $w=\frac{u_x}{\sqrt{u}}$, should there not be a 2 in the denominator in the right-hand side? $\endgroup$ – mlaci May 18 '17 at 2:13
  • $\begingroup$ Oh, sorry, third question: 3) It's possible to write the equations in the first question only because both $w$ and that whole $2c_1 y + ...x$ equal constants, right? Or would it be possible regardless? $\endgroup$ – mlaci May 18 '17 at 2:13
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    $\begingroup$ 1) $w$ is not equal to $c_1$ in general. $w=c_1$ only on the characteristics curves. 2) You are right, it's a typo, now corrected. 3) The two family of characteristic curves define two surfaces which intersection gives a specific value for $c_1$ and $c_2$ , that is one of the characteristic curves. In return, to each arbitrary relationship of the general solution corresponds a particular characteristic curve (with a particular couple $c_1,c_2$) on which the PDE is satisfied. $\endgroup$ – JJacquelin May 18 '17 at 7:10
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    $\begingroup$ For examples, see : math.ualberta.ca/~xinweiyu/436.A1.12f/… $\endgroup$ – JJacquelin May 18 '17 at 7:18
  • $\begingroup$ Once again, thank you very much! $\endgroup$ – mlaci May 19 '17 at 0:40
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Partial solution

Using the Lagrange Charpit method you find

$$ \mathrm{d}x/2p = \mathrm{d}y/2q = \mathrm{d}u/2u = \mathrm{d}p/p = \mathrm{d}q/q $$

from which $p = x/2 +c_1$ and $q = y/2 + c_2$, where $c_i$ are constants of integration. The Pfaff equation $p \, \mathrm{d} x + q \, \mathrm{d}y = \mathrm{d}u$ is thus integrable, providing the general integral of your equation

$$ u = (x/2 + c_1)^2 + (y/2 + c_2)^2 + c_3 $$

Inserting this into your original PDE yields $c_3 = 0$. However, I don't see any suitable values of $c_{1,2}$ for $u$ to satisfy the condition at $y = 0$ (note the factor $1/2$ affecting the $x$-term).

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  • $\begingroup$ I'll have to think about your answer a little more, as I haven't quite understood it yet, but thank you very much for posting it! I've marked JJacqueline's answer as accepted, though, as she gave a full answer to the problem. Thanks again. $\endgroup$ – mlaci May 18 '17 at 2:05
  • $\begingroup$ Glad to help, @mlaci. That's the way I learnt to solve this kind of problems. $\endgroup$ – Dmoreno May 18 '17 at 18:11
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You lost your way at the integration of $y$, all before looks good. You should have got \begin{align} x&=(4e^s-3)r&&=4(e^s-1)r+r\\ y&=\pm 2(e^s-1)\sqrt{1-3r^2} \end{align} Then $$ \pm2yr=(x-r)\sqrt{1-3r^2}\\ 4y^2r^2=(x-r)^2(1-3r^2) $$ and at this point you have to solve this degree 4 polynomial in $r$. After selecting the correct branch ($x>r$ is one condition) you obtain the value of $z$ as $$ z=e^{2s}(1+r^2)=\frac{(x+3r)^2}{16}(1+r^2). $$

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  • $\begingroup$ Yes, there's definitely a missing $2$ in that $y$ (and also a missing $\pm$ in the $q$ two lines above that). Thank you very much for your answer. Two further questions though: 1) Can I always require $x>r$ even if my parametrization for the prescribed data was $x=r$?; 2) In the last equality in the last line of your answer, are those supposed to be $r$'s or $y$'s? $\endgroup$ – mlaci Jun 24 '17 at 18:18
  • $\begingroup$ Also, I hand't come back to this question to add this bit, but afterward my teacher basically said, IIRC, that we weren't supposed to solve it for $x$ and $y$ (like we had every previous time), but rather that we were just supposed to know that the implicit function theorem guaranteed us that such a solution existed. :( $\endgroup$ – mlaci Jun 24 '17 at 18:21
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    $\begingroup$ No, $x>r$ is just a consequence of the solution formula. It might even be wrong as $s<0$ is probably not excluded. In the last equation I used the solution of $x$ to replace $e^s$, one could equally well also use the parametrization of $y$. And yes, one strives to simplify the problem as much as possible, and having a scalar polynomial at the end can be seen as a very favorable outcome. $\endgroup$ – Dr. Lutz Lehmann Jun 24 '17 at 18:29

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