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In our problem sheet on Real Analysis, there is a problem that seems to be unprecise (perhaps it's just me - I'm sorry if that's the case.). I'm asking you to please only answer on what the notation should say (and not spoil anything on the problem).

Let $f:\mathbb{R}^2 \to \mathbb{R}$ be differentiable. Show that a differentiable function $y: \mathbb{R} \to \mathbb{R}$ solves the differential equation $$ \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}y' = 0$$ if and only if there exists a $c \in \mathbb{R}$ with $$ f(x,y(x)) = c$$ for all $x \in \mathbb{R}$.

So should that differential equation stand for $$ \frac{\partial f}{\partial x_1} (x_1,x_2) + \frac{\partial f}{\partial x_2}(x_1,x_2)y(x_1) = 0 $$ for all $(x_1,x_2) \in \mathbb{R}^2$ or what exactly should it say? To which variables do my instructors intend to refer these functions?

Qi Zhu
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1 Answers1

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It's a bit awkward to not touch on the content of the question but just the notation ...

That differential equation is explicitly writing out the total derivative $\displaystyle \frac{ \mathrm{d}f }{ \mathrm{d}x_1 }$ when $f$ is expressed as a function of both $x_1$ and $x_2$ but while $x_2$ is "secretly" a function of $x_1$. $$\frac{\partial f}{\partial x_1} (x_1,x_2) + \frac{\partial f}{\partial x_2}(x_1,x_2) \frac{ \mathrm{d} x_2(x_1) }{ \mathrm{d} x_1} = 0$$ for all points along the trajectory $\left(x_1, x_2(x_1) \right)$ for all $x_1 \in \mathbb{R}$.

Seriously, this is just replacing the notation $(x,y)$ in the given question with your $(x_1, x_2)$.

Lee David Chung Lin
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  • I'm a bit embarrassed but it's still not clear to me. My problem is $\frac{\partial f}{\partial x_2}$ or in the original $\frac{\partial f}{\partial y}$ when $x_2$ or $y$ is a function. I'm only used to it being a variable of the function. (perhaps again I'm being stupid here but only asking helps in this case I guess...) E.g. in your case you too are using $x_2$ twice, once with $x_2(x_1)$ and once only as $x_2$. That's what confuses me. – Qi Zhu Apr 28 '18 at 10:02
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    @Kezer I added one sentence to see if that helps. There are certain things to say that I know would clarify the situation, but clearly you want to stay "clean" and I respect that. – Lee David Chung Lin Apr 28 '18 at 10:17
  • Thanks for the effort. I did actually suspect that it'd just be just that total derivative but I guess I'm still confused by the notation. Is it possible to give a concise definition of $\frac{\partial f}{\partial x_2} (x_1, x_2)$ with $x_2$ being a function? Thank you for respecting my wishes! – Qi Zhu Apr 28 '18 at 10:29
  • Let me call the function $y$ with the name $g$, then by chain rule on $f(x,g(x))$ I'll get $$ \frac{\partial f}{\partial x}(x,g(x)) + \frac{\partial f}{\partial y}(x,g(x)) \cdot g'(x) = 0.$$ That should be how it should be interpreted, right?

    For me it just seemed like $\frac{\partial f}{\partial x} + \dots = 0$ refers to $\frac{\partial f}{\partial x}(x,y)$ for all $(x,y) \in \mathbb{R}^2$.

     – Qi Zhu Apr 28 '18 at 10:48
  • I guess with this interpretation the problem becomes easy - and apparently it is related to implicit functions? We haven't introduced those yet but I guess this problem should serve as a quick intro. Thanks for your help. :) (sorry for the spam) – Qi Zhu Apr 28 '18 at 10:55
  • In fact, it's exactly the idea of an implicit function. I can see why the first notation was taken, it is still not good notation in my opinion - the one I proposed should be about the cleanest possible if I'm not mistaken. – Qi Zhu Apr 28 '18 at 11:51
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    @Kezer Hey I went away for a while not realizing the conversation was still on-going. The concepts and notations of partial derivatives can be confusing one way or another for new learners as well as veterans. My advice for you at this point is to familiarize yourself with the formal operations (manipulation in forms) with basic and concrete examples, don't aim too high or too abstract. Those examples you can always go back to and be sure of the "rules". Don't worry too much about the "best" interpretation or notations. Most likely as you progress your view point and taste will change. – Lee David Chung Lin Apr 28 '18 at 13:29
  • Yes, that's true, that's always the right thing to do. The first thing I did was to try examples on some easy functions to try this idea of "implicit functions" and their relationships with this problem! Thanks a lot for your replies, I appreciate it! – Qi Zhu Apr 28 '18 at 17:27