Help solving the PDE $u_x^2 + u_y^2 = u$

Question

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?

Answer 1

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.

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

Answer 2

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).

Answer 3

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 $$ z=e^{2s}(1+r^2)=\frac{(x+3r)^2}{16}(1+r^2). $$