# Equal partial derivatives

Suppose $\frac{\partial}{\partial x}F(x,y,z)=\frac{\partial}{\partial y}F(x,y,z)=\frac{\partial}{\partial z}F(x,y,z)$. Does this imply $F(x,y,z)=G(x+y+z)$ for some function G?

• No. Not without additional hypotheses. – Potato Sep 16 '13 at 12:26
• Could you give a counter example? I'm having trouble finding one. Also can you say something more about these additional hypotheses? – MthQ Sep 16 '13 at 12:31
• Consider the function that is $0$ except for for some spiral that leads to the origin. And I don't know what additional hypotheses would be needed to make this true. – Potato Sep 16 '13 at 12:32

## 2 Answers

The statement is true if all 3 partial derivatives exist and continuous over all $\mathbb{R}^3$.

The proof relies on a multivariable differentiability theorem:

Let $\varphi : U \to \mathbb{R}\;$ be a function defined on some open subset $U$ of $\mathbb{R}^m$. If all $m$ partial derivatives of $\varphi$ exists and continuous over $U$, then $\varphi$ is differentiable over $U$.

If all 3 partial derivatives $\frac{\partial F}{\partial x}$, $\frac{\partial F}{\partial y}$, $\frac{\partial F}{\partial z}$ exists and continuous over some open convex set K, then by above theorem, $F$ is differentiable over $K$.

Now for any two point $\vec{v}_1 = (x_1,y_1,z_1),\;\;\vec{v}_2 = (x_2, y_2, z_2) \in K$ such that $$x_1+y_1+z_1 = x_2 + y_2+z_2,$$ the line segment between $\vec{v}_1$ and $\vec{v}_2$:

$$[0,1] \in t\quad\mapsto\quad\vec{v}(t) = \vec{v}_1 + (\vec{v}_2 - \vec{v}_1) t$$

lies completely in $K$. Consider following two functions defined on $[0,1]$:

$$\begin{cases} f(t) = F(\vec{v}(t))\\ \\ g(t) = \frac{\partial F}{\partial x}(\vec{v}(t)) = \frac{\partial F}{\partial y}(\vec{v}(t)) = \frac{\partial F}{\partial z}(\vec{v}(t)) \end{cases}$$

Since $F$ is differentiable over $K$, we can express the derivative of $f$ in terms of partial derivatives of $F$ and hence in $g$:

\begin{align}\frac{df(t)}{dt} = & (x_2-x_1) \frac{\partial F}{\partial x}(\vec{v}(t)) + (y_2-y_1) \frac{\partial F}{\partial y}(\vec{v}(t)) + (z_2-z_1) \frac{\partial F}{\partial z}(\vec{v}(t))\\ = & ((x_2+y_2+z_2) - (x_1+y_1+z_1)) g(t)\\ = & 0 \end{align}

From this, we can conclude $$f(0) = f(1) \quad\iff\quad F(\vec{v}_1) = F(\vec{v}_2)$$ i.e. $F(\vec{v})$ is constant on the intersection of $K$ with the planes $x+y+z = \text{constant}$.

If $K$ is the whole $\mathbb{R}^3$, then

$$F(x,y,z) = F(x+y+z,0,0)$$

and the statement follows immediately.

• Thanks, very clear and helpful. Also easy to extend to more dimensions. – MthQ Sep 16 '13 at 13:26

Partial Answer:

I tend to think that it is true, but not sure. For polynomials it is true:

Prove by induction on the grade $n$.

For $n=0$ it is true obiously.

Suppose $P_n$ is a polynomial of grade $n$ and has the property: $$\partial_xP_n=\partial_yP_n=\partial_zP_n=Q_{n-1}$$ The vector field $(Q_{n-1},Q_{n-1},Q_{n-1})$ is a gradient and then its rotor is null. This implies $$\partial_xQ_{n-1}=\partial_yQ_{n-1}=\partial_zQ_{n-1}$$ which implies by induction that $Q_{n-1}=q(x+y+z)$. $\;\square$

Update: In dimmension 2 it is true (and I suppose in more dimmensions also). Suppose $f$ is such that $f_x=f_y$. Then $f$ is the solution of the following differential problem: $$u_x(x,y)-u_y(x,y)=0 \\ u(x,0)=f(x,0)=g(x)$$ We can solve this problem with the method of the caractheristic. We obtain that $u(x,y)=g(x+y)$.

Update 2:

Fix z. Then using the previous result we conclude: $f(x,y,z)=g(x+y,z)=g(u,z)$. Now fix $y=0$. Then $f_x=f_z$ implies that $g_u=g_z$ and we conclude that $g(u,z)=h(u+z)$ and we are done.