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So I've got: $\frac{1}{r}\frac{du}{dr}=\frac{1}{2\mu}\frac{dp}{dx}$ as an equation in a fluid mechanics book. The book stipulates that if we have a function that only depends on r ont he left side and that only depends on x on the right side ($\mu$ is constant) then both functions have to be constant (I'm supposing here that means $\frac{1}{r}\frac{du}{dr}$ and $\frac{dp}{dx}$ are constant).

Is it possible to prove this? My take on it is:

We suppose that one function isn't constant (which in turn means the other isn't constant either): and so there exist two couples ($x_1$,$x_2$) and ($r_1$,$r_2$) in the respective domains of definition of both functions, so that: $$ \frac{1}{r_1}\frac{du}{dr}(r_1)\neq\frac{1}{r_2}\frac{du}{dr}(r_2)\space and\space \frac{dp}{dx}(x_1)\neq\frac{dp}{dx}(x_2) $$ Yet, $\frac{1}{r}\frac{du}{dr}(r)=\frac{1}{2\mu}\frac{dp}{dx}(x)$ for every ($x$,$r$) couple in the domain of definition of the two functions respectively.

So, $\frac{1}{r_1}\frac{du}{dr}(r_1)=\frac{1}{2\mu}\frac{dp}{dx}(x_1)=\frac{1}{2\mu}\frac{dp}{dx}(x_2)$, which leads us to the contradiction.

However I am unsure that I can say that the equation means that for every ($x$,$r$) couple it is true and go onfrom there. If I cannot then is there another way to prove that the two functions are constant?

Harry Peter
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user104521
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We can only write $\frac{du}{dr}$ if we know that $u$ is a differentiable function with no variables other than (possibly) $r$. Consequently, $\frac{du}{dr}$ will also be a function with no variables other than (possibly) $r$, and so will $\frac1r\frac{du}{dr}$.

By the same token, we can conclude (since $\mu$ is a constant) that $\frac1{2\mu}\frac{dp}{dx}$ is a function with no variables other than (possibly) $x$. Since these two functions are equal, then they have no variables at all, and so are constant.

Cameron Buie
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