Hello dear mathematical community.
I was wondering if anybody could help me, please, with the following question. (I apologize in advance in case my question reveals unawareness of mathematical basics).
I have a first order partial differential equation: $$\frac{\partial (A·V)}{\partial x} = -\frac{\partial A}{\partial t}$$ Here:
$A=f(x,t)$ - field of property $A$ as a function of spatial coordinate $x$ and time coordinate $t$
$V=f(x,t)$ - field of property $V$ as a function of spatial coordinate $x$ and time coordinate $t$
I did the following mathematical manipulations with it: $$\frac{\partial (A·V)}{\partial A} = -\frac{\partial x}{\partial t}$$
MY QUESTION IS: whether I can do such mathematical manipulation with partials or not? 1) If not then why? 2) If conditional yes, then what are the conditions? 3) If yes, how can it be physically/mathematically explained and what theorem/rule does allow me to do it?
Thank you in advance.
CLARIFICATIONS REQUIRED BY @Raskolnikov, @user7530, @MarsPlastic, @Shogun, @ParclyTaxel:
My question is pretty obvious: can I work with partial derivatives as if they were mere division operations. Like in my example: can I bring the numerator of the right hand side to the denominator of the left hand side. And in turn, denominator of the left hand side to the numerator of the right hand side.
So, I am interested if there exist mathematically proven rules prescribing allowed manipulations with derivatives. Basically, that's it - can I work with derivatives as with regular division operations? If yes - what theorem allows me to do it. If not - what theorem doesn't allow me to do it.
The answer of Hans Lundmark doesn't satisfy me - I explained the reason in my answer. Briefly, he states that $\frac{\partial x}{\partial t}$ must be zero. I explained why it is not necessarily zero. He also states that $\partial/\partial A$ doesn't mean anything. I explained its meaning.
CLARIFICATION OF THE REASON OF MY QUESTION:
Obviously, my equation is one equation with two unknowns. I want an extremely simple thing - I want to express one unknown through the other unknown. That's it. This is what we do without hesitation with any algebraic equations, e.g.: $x + y = 5 \Rightarrow x = 5 - y$. Why can I do it? Because rules of work with addition and subtraction are determined. How can I do the same thing with derivatives? Literature is full of solution methods for PDEs. But what if I don't need to solve it - instead, I want to express one variable through another one? What should I do in this case?
SUGGESTED CORRECTIONS:
1) $x$ and $t$ are independent variables, since I am considering two fields.
2) I can't tell what becomes a constant and what becomes a variable when I move $\partial A$ to the left hand side. This can be treated as a part of my question: whether or not I can make such manipulation and what does it mean.
3) One may treat $V$ as a one dimensional velocity field of a pipe fluid flow. Since it is one dimensional, then its $x$ component equals to its magnitude at every point. Therefore, right hand side of the second expression might be considered equal to $-V$ at every point.
4) In the first expression, $A$ and $V$ are allowed to vary with time and space. In the first expression on the right hand side, $x$ is held constant, on the left hand side $t$ is held constant. In the second expression on the left hand side - it seems that I begin considering $A$ and $V$ as the functions of each other. But I am not sure about it. It's my question. Right hand side of the second equation seems strange, since $x$ and $t$ are independent variables. But to my opinion, right hand side can be treated as an $x$ component of velocity field by analogy to the way we derive acceleration expression for a flow field.
5) One's first thought can apparently be that the solution of the equation should be $A = f(x,t)$ and $V = f(x,t)$. But I don't imply such limitation when I am asking the question. Yes, there are two functions in my equation and, therefore, it is obvious that my equation is a part of a system of PDEs. But it's not important now. I would be satisfied if I could only use my equation to express one function through another, for instance: $A$ through $V$, or vice versa.
6) Apart from all, I am interested in pure math: what mathematical manipulations are allowed for numerators and denominators of partial derivatives. In other words: am I allowed to freely manipulate with numerators and denominators of a partial derivative as if it were not a derivative but a division defined at a point.
RESPONSE TO THE @HansLundmark ANSWER AND @user7530 COMMENTS:
Dear @user7530 and @HansLundmark.
Thank you for the responds. You hinted me where to look at.
Regarding your comments I have only to say the following so far.
1) I already thought about $-\frac{\partial x}{\partial t}$. It can have some value other then zero. To my mind (I might be absolutely wrong), it is similar to the way, we derive an acceleration equation for fluid flows. Let me introduce this example for better understanding of what I mean.
\EXAMPLE
Assume, we have a velocity field of some transient 3D flow: $$v = f(x,y,z,t)$$ Let's take its differential (v is obviously a vector): $$dv = \frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial y}dy+\frac{\partial v}{\partial z}dz+\frac{\partial v}{\partial t}dt$$ Now, let's divide both sides by $dt$. We'll get: $$\frac{dv}{dt} = \frac{\partial v}{\partial x} \frac{dx}{dt}+\frac{\partial v}{\partial y} \frac{dy}{dt}+\frac{\partial v}{\partial z} \frac{dz}{dt}+\frac{\partial v}{\partial t}$$ Then we say that:
$\frac{dv}{dt} = a$ - is acceleration
$\frac{dx}{dt} = u$ - is x component of velocity
$\frac{dy}{dt} = v$ - is y component of velocity
$\frac{dz}{dt} = w$ - is z component of velocity
And we get that flow acceleration at every point in the flow field is:
$$a = u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} + w\frac{\partial v}{\partial z} + \frac{\partial v}{\partial t}$$
Now why do people say that $\frac{dx}{dt} = u$, $\frac{dy}{dt} = v$, $\frac{dz}{dt} = w$? We know that $x$, $y$, $z$, $t$ are independent variables. Therefore, these derivatives should equal zero. But they are not. They equal to the corresponding velocity components. Why? To my opinion, because, if for a moment we cease considering field where coordinates are fixed and instead consider a real fluid particle at some point in the flow field, then we'll see that magnitudes of its velocity components can be written as $\frac{dx}{dt} = u$, $\frac{dy}{dt} = v$, $\frac{dz}{dt} = w$. But in this case, $x$, $y$, $z$ are corresponding components of its radius-vector. Hence in the field equation for acceleration, the terms $\frac{dx}{dt} = u$, $\frac{dy}{dt} = v$, $\frac{dz}{dt} = w$ mean nothing. But they mathematically equal to the velocity components of a real fluid particle. Therefore, we plug the velocity components into the equation.
\END OF THE EXAMPLE
That is why, I think that in my equation the term $-\frac{\partial x}{\partial t}$ can be treated the same way and can be substituted with an $x$ component of velocity of the flow (because my equation is written for a 1D flow originally).
I apologize that I didn't mention it at once.
2) Now about what $\partial/\partial A$ means. I though of it. It apparently seems to me, that when I do my manipulation I automatically begin considering $A$ and $V$ as functions of $A$ on the left hand side (physically, it is possible due to the nature of my equation). Since, $A$ cannot be a function of itself, then I am left with only $V$ as a function of $A$ on the left hand side. Which allows me to move from partial derivatives to the total ones and solve the equation getting as a result $V = f(A)$.
I would be really grateful to you if you could provide any of you thoughts regarding my reasoning.
I am an engineer - not a mathematician. So, I apologize if the way I work with equations seems barbarian to you.
Thank you in advance.
You certainly cannot!
First of all, it's not really clear what the operation $\partial/\partial A$ even means. (What quantity is supposed to be held constant as $A$ varies?)
Secondly, the right-hand side of the new equation, $-\partial x/\partial t$, is identically zero (if $\partial/\partial t$ has the same meaning as in the term $\partial A/\partial t$ in the original equation, namely that $x$ is held constant as $t$ varies), and whatever the left-hand side $\partial (A \cdot V)/\partial A$ means, it's not reasonable that it should be unconditionally equal to zero.