Let $P$ and $Q$ be functions of $r$ and $r$ be a function of $(x,y,z)$. Also let $f$ be a function of $(x,y)$.

If: $$P(x,y,z) + f (x,y)= Q(x,y,z) \tag{1}$$ By $(1)$ $$\dfrac{\partial P}{\partial x} \neq \dfrac{\partial Q}{\partial x} \Rightarrow \dfrac{dP}{dr} \dfrac{\partial r}{\partial x} \neq \dfrac{dQ}{dr} \dfrac{\partial r}{\partial x}\Rightarrow \dfrac{dP}{dr} \neq \dfrac{dQ}{dr} \tag{2}$$ Also by $(1)$ $$\dfrac{\partial P}{\partial z} = \dfrac{\partial Q}{\partial z}\Rightarrow \dfrac{dP}{dr} \dfrac{\partial r}{\partial z} = \dfrac{dQ}{dr} \dfrac{\partial r}{\partial z} \Rightarrow \dfrac{dP}{dr} = \dfrac{dQ}{dr} \tag{3}$$

$(2)$ and $(3)$ contradict. Why is this so?

• The formula (1) seems wrong, missing an the "r" cc @Robert
– quid
Mar 3, 2018 at 12:46
• If (1) is true in the sense that $P$ and $Q$ are functions of $r$, then what does that say about $f(x,y)$? Or else, what does that say about $r$? Or $P$ and $Q$? Mar 3, 2018 at 17:01

You say that: $\qquad P(r(x,y,z))+f(x,y)=Q(r(x,y,z))$

Which is to say: $\quad f(x,y)=[Q-P]\circ r(x,y,z)$

Therefore $\qquad\quad~~ 0~{= \dfrac{\partial f(x,y)}{\partial z}\\=\left[\dfrac{\mathsf d [Q-P](r)}{\mathsf dr\qquad\qquad}\right]\!\!(x,y,z)\cdot\dfrac{\partial r(x,y,z)}{\partial z\qquad\quad}}$

So either $\dfrac{\mathsf d [Q-P](r)}{\mathsf dr\qquad\qquad}=0$ or $\dfrac{\partial r(x,y,z)}{\partial z\qquad\quad} =0$

In the first case, $f(x,y)$ must be a constant, and in the second $r$ is invariant with respect to $z$.   In either case, there is no contradiction involved.

If $f(x,y)$ is constant, then $\dfrac{\partial P}{\partial x}+0=\dfrac{\partial Q}{\partial x}$.

If $\dfrac{\partial r(x,y,z)}{\partial z\qquad\quad}=0$ then $\dfrac{\partial P}{\partial z}=\dfrac{\partial Q}{\partial z}=0$.

If $$P(r(x,y,z)) + f(x,y)= Q(r(x,y,z)) \tag{1}$$ then $$\dfrac{d P(r)}{d r}\dfrac{\partial r}{\partial x}+\dfrac{\partial f}{\partial x}=\dfrac{d Q(r)}{d r}\dfrac{\partial r}{\partial x} \tag{2}$$ and $$\dfrac{d P(r)}{d r}\dfrac{\partial r}{\partial z}=\dfrac{d Q(r)}{d r}\dfrac{\partial r}{\partial z} \tag{3}.$$ In my opinion (2) does not contradict (3): if $\dfrac{\partial r}{\partial z}\not=0$ then, from (3), $\dfrac{d P(r)}{d r}=\dfrac{d Q(r)}{d r}$ and, by (2), it follows that $\dfrac{\partial f}{\partial x}=0$. In the same way $\dfrac{\partial f}{\partial y}=0$. The difference $P(r)-Q(r)=f$ could be identically constant.