Solution of partial differential equation

Question

Solve the differential equation, $$ z=\frac{\partial z}{\partial x}x + \frac{\partial z}{\partial y}y+ (\frac{\partial z}{\partial x})^2 + \frac{\partial z}{\partial x}\frac{\partial z}{\partial y}+ (\frac{\partial z}{\partial y})^2.$$

Can you please solve and explain the solution assuming that I don't have any idea.

Answer 1

For any function $\phi$ and coordinate variable $w$, we will use $\partial_w$ and $\phi_w$ as a shorthand of the differential operator $\frac{\partial}{\partial w}$ and the partial derivative $\partial_w \phi = \frac{\partial\phi}{\partial w}$.

Consider following change of variables

$$(x,y) = \left(\frac{\sqrt{3}u+v}{2}, \frac{\sqrt{3}u-v}{2}\right)\quad\iff\quad (u,v) = \left(\frac{x+y}{\sqrt{3}},x-y\right)$$

We have $$ \begin{cases} \partial_u &= \frac{\sqrt{3}}{2}(\partial_x + \partial_y)\\ \partial_v &= \frac12(\partial_x - \partial_y) \end{cases} \;\;\implies\;\; \begin{cases} u z_u + v z_v &= x z_x + y z_y\\ z_u^2 + z_v^2 &= \frac34(z_x + z_y)^2 + \frac14(z_x - z_y)^2 = z_x^2 + z_xz_y + z_y^2 \end{cases} $$

Let $\rho^2 = u^2 + v^2$ and $\varphi = z + \frac{\rho^2}{4}$. The PDE at hand can be transformed to

$$\begin{array}{rrcl} & z &=& uz_u + v z_v + z_u^2 + z_v^2\\ \iff & z + \frac{\rho^2}{4} &=& \left(z_u + \frac{u}{2}\right)^2 + \left(z_v + \frac{v}{2}\right)^2\\ \iff & \varphi &=& \varphi_u^2 + \varphi_v^2\tag{*1} \end{array} $$ The last equation in $\varphi$ has a trivial solution $\varphi \equiv 0$. This leads to one solution for $z$:

$$z = -\frac{\rho^2}{4} = -\frac{u^2+v^2}{4} = -\frac{x^2 - xy + y^2}{3}\tag{S.1}$$

On those places where $\varphi \ne 0$, we have $\varphi > 0$. Introduce another function $\psi$ such that $$\psi = \sqrt{4\varphi}\quad\iff\quad \varphi = \frac{\psi^2}{4}$$

the function $\psi$ will satisfy

$$|\nabla \psi|^2 = \psi_u^2 + \psi_v^2 = 1\tag{*2}$$

Conversely, given any solution of $(*2)$, the PDE in $z$ will have a solution of the from

$$z = \frac14(\psi^2 - \rho^2)\tag{*3}$$

For example,

• Set $\psi$ to the distance between $(u,v)$ and some fixed point $(-\sqrt{3}(a+b),-(a-b))$, one obtain following solution found by JJacquelin:

$$\begin{align} z &= \frac14\left((u+\sqrt{3}(a+b))^2 + (v+(a-b))^2 - u^2 - v^2\right)\\ &= a x + b y + a^2 + ab +b^2 \end{align}\tag{S.2} $$

• Set $\psi$ to any affine function with slope $1$ along direction $\theta$ in $uv$-plane, one find a family of quadratic surfaces as solutions of $z$: $$\begin{align} z &= \frac14\left[ \left(\cos\theta u + \sin\theta v + m\right)^2 - (u^2+v^2)\right]\\ &= -\frac14\left[ \left(-\sin\theta u + \cos\theta v\right)^2 - 2m \left(\cos\theta u + \sin\theta v\right) - m^2 \right]\tag{S.3} \end{align} $$ where $\theta, m$ are constants.

These $3$ solutions $(S.1)$, $(S.2)$ and $(S.3)$ are all the global solutions I can find. If one drop the requirement that $z$ is defined for all $(x,y)$, there are other ways to construct more solutions.

Let

• $K$ be any convex body in the plane bounded by some smooth curve.
• Let $d_K(p)$ be the distance of any point $p$ to $K$. i.e. $$d_K(p) = \inf\{ | p - q | : q \in K \}$$

Outside $K$, $d_K(p)$ is smooth, positive and $|\nabla d_K| = 1$. For any constant $\psi_0 \ge 0$, we can set $\psi$ to $d_k + \psi_0$ and use $(*3)$ to construct a solution for the PDE over there.

Answer 2

HINT (partial answer) :

Search for a particular linear solution on the form $z=ax+by+c$

Puting into the PDE leads to $$z=ax+by+a^2+ab+b^2 \quad\text{any }a\:,\: b.$$