Find the function $u: \mathbb R ^2\to \mathbb R^2$ that holds the property $d^3u(x,y)=0$

Question

Find the function $u: \mathbb R ^2\to \mathbb R^2$ that holds the property $d^3u(x,y)=0$

I really would like some help with this problem because I don't even know where to start.

I thought taking the formula

$$d^3u=\frac{\partial ^3u(x,y)}{\partial x^3}dx^3+3\frac{\partial ^3u(x,y)}{\partial x^2 \partial y}dx^2dy+3\frac{\partial ^3u(x,y)}{\partial x \partial y^2}dxdy^2+\frac{\partial ^3u(x,y)}{\partial y^3}dy^3$$

but I don't know where to go from here. I would appreciate some help

Answer 1

By dividing $d^3 u(x,y)$ over $(dx)^m(dy)^n$ where $m+n=3$ and $m,n\in\{0,1,2,3\}$, we obtain$${d^3 u(x,y)\over dx^3}=0\\{d^3 u(x,y)\over dx^2dy}=0\\{d^3 u(x,y)\over dxdy^2}=0\\{d^3 u(x,y)\over dy^3}=0$$which yields to the general solution$$u(x,y)=a_1x^2+a_2xy+a_3y^2+a_4x+a_5y+a_6$$

Answer 2

Looking at your formula one can say the following: $d^3u(x,y)=0$ iff all third partial derivatives $$u_{xxx},\quad u_{xxy},\quad u_{xyy},\quad u_{yyy}$$ vanish identically. This is the case if all second partial derivatives $$u_{xx},\quad u_{xy},\quad u_{yy}$$ are constants, etcetera. It follows that the most general $u:\>{\mathbb R}^2\to{\mathbb R}^2$ with $d^3u=0$ has the form $$u(x,y)=\bigl(f(x,y),g(x,y)\bigr)$$ with $f$ and $g$ polynomials of total degree $\leq2$ in $x$ and $y$.