Non-linear partial differential equation of order 1

Question

Solve the PDE $$2(pq+py+qx)+x^2+y^2=0$$

where $\displaystyle p = \frac{\partial z}{\partial x}$, $\,\displaystyle q = \frac{\partial z}{\partial y}$

$f_p=2q+2x\\ f_q=2p+2x\\ f_z=0\\ f_x=2q+2x\\ f_y=2p+2y$

Using Charpit's method

$$\frac{dp}{2q+2x}=\frac{dq}{2p+2y}=\frac{dz}{-4pq-2py-2qx}=\frac{dx}{-2q-2y}=\frac{dy}{-2p-2x}$$

I deduce $$p+q+x+y=a \\ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$

I am not sure how to proceed from here. Kinda stuck.

Answer 1

$$2(pq+py+qx)+x^2+y^2=0$$ You got to : $$p+q+x+y=a \\ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$ I checked and agree.

HINT :

Solve this system on two equations for the two unknown $p(x,y)$ and $q(x,y)$.

For example put $p=a-q-x-y$ into the second equation. It becomes a quadratic equation to be solved for $q$.

Now you have explicitly $z_x=p(x,y)$ and $z_y=q(x,y)$. Integrate.

$z=\int p(x,y)dx+F(y)$ and $z=\int q(x,y)dy+G(x)$

Determine $F(y)$ and $G(x)$ to make them consistent.

The integrals are complicated. This seems a boring calculus. I prefer to simplify the original PDE at first place. Let :

$$z(x,y)=u(x,y)-\frac12 x^2-\frac12 y^2$$

$p=z_x=u_x-x=P-x$

$q=z_y=u_y-y=Q-y$

$$2\left((P-x)(Q-y)+(P-x)y+(Q-y)x\right)+x^2+y^2=0$$ After simplification : $$PQ+\frac12(x-y)^2=0$$

$$u_xu_y+\frac12(x-y)^2=0$$ $$\frac{dx}{Q}=\frac{dy}{P}=\frac{du}{2PQ}=\frac{dP}{y-x}=\frac{dQ}{x-y}=ds$$ $$\begin{cases} P+Q=c_1 \\ PQ=-\frac12(x-y)^2 \end{cases}$$ etc.

Note: $P+Q=c_1$ is equivalent to your equation $p+q+x+y=a$ .

Note: It is a pity that no boundary condition be specified.

Answer 2

Using Charpit's method $$\frac{dp}{2q+2x}=\frac{dq}{2p+2y}=\frac{dz}{-4pq-2py-2qx}=\frac{dx}{-2q-2y}=\frac{dy}{-2p-2x}$$ Adding first two and last two fractions $$\frac{dp+dq}{2p+2y+2q+2x}=\frac{dx+dy}{-2q-2y-2p-2x}$$ This implies, $a$ being an arbitrary constant $$p+x+q+y=a \tag{1}$$ Now the equation can be rewritten as $$2(p+x)(q+y)+(x-y)^2=0\tag{2}$$ Using $1,2$ $$(p+x)-(q+y)=\sqrt{a^2-2(x-y)^2}$$ Solving for $p+x$ and $q+y$ , $$\begin{align} p+x &= \frac12(a+\sqrt{a^2+2(x-y)^2}) \\ q+y &= \frac12(a-\sqrt{a^2+2(x-y)^2}) \end{align}$$ This gives $$\begin{align} p&= -x+ \frac12(a+\sqrt{a^2+2(x-y)^2}) \\ q &= -y+ \frac12(a-\sqrt{a^2+2(x-y)^2}) \end{align}$$

$$dz=pdx+qdy$$ Substituting and integrating gives $$z=-\frac{x^2+y^2}{2}+\frac{a}{2}(x+y) +\frac12\int\sqrt{a^2-2(x-y)^2}d(x-y)$$

So the solution is $$z=-\frac{x^2+y^2}{2}+\frac{a}{2}(x+y)+\dfrac{a^2\ln\left(\left|a\sqrt{2(x-y)^2+a^2}+\sqrt{2}\left|a\right|(x-y)\right|\right)}{2^\frac{5}{2}}+\dfrac{a(x-y)\sqrt{\frac{2(x-y)^2}{a^2}+1}}{2^2}+b$$