"Trick" for a specific potential function defined with an integral

[fatpotato]

Homework Statement

Show that if $\frac{\partial g}{\partial u} = \frac{\partial f}{\partial v}$ then $F(x,y) = \nabla \phi(x,y)$ with $\phi(x,y) = \int_0^1 xf(tx,ty) + yg(tx,ty) dt$

Relevant Equations

Basic theory of fields and potentials.

Hello,

To first clarify what I want to know : I read the answer proposed from the solution manual and I understand it. What I want to understand is how they came up with the solution, and if there is a way to get better at this.

I have to show that, given a vector field $F$ such that $F : \mathbb{R}^2 \rightarrow \mathbb{R}^2$, $F(u,v) = (f(u,v), g(u,v))$, computing the derivatives $\frac{\partial g}{\partial u} = \frac{\partial f}{\partial v}$ must imply that $F(x,y) = \nabla \phi(x,y)$, with $\phi$ defined as :
$$\phi(x,y) = \int_0^1 xf(tx,ty) + yg(tx,ty) dt$$
The solution manual suggests "observing" that :
$$ \frac{\partial}{\partial t} t f(tx,ty) = f(tx,ty) + t [ x \frac{\partial f}{\partial u} + y \frac{\partial f}{\partial v} ]$$
And then "finding" that :
$$ \frac{\partial}{\partial x} x f(tx,ty) + y g(tx,ty) = f(tx,ty) + tx \frac{\partial f}{\partial u} + ty \frac{\partial g}{\partial u} $$
Are the same when we suppose equality between the cross partial derivatives, so in the end, differentiating the integrand of $\phi$ can be expressed with respect to $t$ to find :
$$ \frac{\partial \phi} {\partial x} = \int_0^1 \frac{\partial} {\partial t} t f(tx,ty) dt = f(x,y)$$
Is it supposed to be obvious if you previously never encountered this kind of problem? I feel that I would not be able to find this clever observation in an exam by myself...In books like Schaum, problems are usually way easier and (alas) made to train pure computation and I don't know where to look for similar problems.

So here are my questions :

1. Is this a typical more elaborate exercise?
2. Where can I find books to exercise myself at solving this type of problems?

Thank you for your time.

 

[Office_Shredder]

There are basically three steps here (and the solution manual doesn't present it this way, which probably makes it look absurdly mysterious)
1.) Take the partial derivative using the chain rule
2.) Sub in the partial derivative of g for the partial derivative of g
3.) Notice the whole thing can be written as a derivative of t now.

I think (1) and (2) should be pretty obvious to do. Step 3 seems like a bit of a gotcha. I don't know how you would catch that that's the thing to do, other than I guess realizing you need some sort of derivative wrt t so you can resolve the integral. Probably the best thing to do is just remember that step 3 is a clever thing to try when you have an integral you want to get rid of.

 

[fatpotato]

Thank you very much for your answer. I will indeed keep in mind that step (3) can get rid of the integral in case I encounter a similar problem.

Would you agree that this is a more general problem than the ones usually encountered in introductory books? Problems generally just ask to compute the gradient and to differentiate simple functions (like polynomial or trig. functions) to verify if the field is conservative, but this one is more abstract since no tangible function is involved.

I love solving these, where can I find more?

 

[WWGD]

Thank you very much for your answer. I will indeed keep in mind that step (3) can get rid of the integral in case I encounter a similar problem.

Would you agree that this is a more general problem than the ones usually encountered in introductory books? Problems generally just ask to compute the gradient and to differentiate simple functions (like polynomial or trig. functions) to verify if the field is conservative, but this one is more abstract since no tangible function is involved.

I love solving these, where can I find more?

I suggest considering a "Fake it till you make it" approach mixed with a bottom up. Basically, you do these problems, initially without fully understanding things and then ultimately it sinks in, subconsciously I guess and the tricks start making sense. I don't mean not to try to understand, but don't let this stop you. It has worked for me.

 

[Office_Shredder]

I think your question is, where can you find more exercises where the question asks you to prove something about all functions, not just specific functions.

I'm not sure what a good book is, you might want to ask that question in the textbook section.

https://en.m.wikipedia.org/wiki/Vector_calculus_identities

I wouldn't do all of these, but maybe picking a couple identities here and trying to prove them is a fun start.

 

[FactChecker]

IMO, the solution was not expected to be obvious. Theorems like this about potential functions are not like typical exercises. Potential functions and related integrals are very fundamental and their theory was developed by geniuses over decades. Do not expect to be able to come up with these proofs from scratch at the drop of the hat. It is enough to study what they did. Then you might be able to mimic what they did in similar problems.

 

[WWGD]

Is it a coincidence that Complex-Analytic functions are all potential functions, per Cauchy-Riemann?

 

[FactChecker]

Is it a coincidence that Complex-Analytic functions are all potential functions, per Cauchy-Riemann?

No, it's not a coincidence. The potential functions, analytic functions, Cauchy-Riemann, Taylor series expansion, Cauchy integral formula, etc., all go together. The consequences of any of those conditions are so profound and restrictive that, IMHO, it is no surprise that they are really all the same thing.

 

[fatpotato]

I suggest considering a "Fake it till you make it" approach mixed with a bottom up. Basically, you do these problems, initially without fully understanding things and then ultimately it sinks in, subconsciously I guess and the tricks start making sense. I don't mean not to try to understand, but don't let this stop you. It has worked for me.

I remember the very first time I encountered a problem where the "trick" was to multiply a number by its complex conjugate, and I thought "Yeah I am never able to do this on my own". Perhaps it will stick as well!

I'm not sure what a good book is, you might want to ask that question in the textbook section.

https://en.m.wikipedia.org/wiki/Vector_calculus_identities

I wouldn't do all of these, but maybe picking a couple identities here and trying to prove them is a fun start.

Thank you, I will check the textbook section. I did go through most of them though (fun indeed).

IMO, the solution was not expected to be obvious. Theorems like this about potential functions are not like typical exercises. Potential functions and related integrals are very fundamental and their theory was developed by geniuses over decades. Do not expect to be able to come up with these proofs from scratch at the drop of the hat. It is enough to study what they did. Then you might be able to mimic what they did in similar problems.

The problem is, my instructor won't care at all that these theories were ingeniously crafted over the years...I fear that I will encounter similar problems in an exam, so I am trying to give myself an edge by going through similar proofs.

 

[yucheng]

I have to show that, given a vector field $F$ such that $F : \mathbb{R}^2 \rightarrow \mathbb{R}^2$, $F(u,v) = (f(u,v), g(u,v))$, computing the derivatives $\frac{\partial g}{\partial u} = \frac{\partial f}{\partial v}$ must imply that $F(x,y) = \nabla \phi(x,y)$, with $\phi$ defined as :
$$\phi(x,y) = \int_0^1 xf(tx,ty) + yg(tx,ty) dt$$
The solution manual suggests "observing" that :
$$ \frac{\partial}{\partial t} t f(tx,ty) = f(tx,ty) + t [ x \frac{\partial f}{\partial u} + y \frac{\partial f}{\partial v} ]$$

Just curious, you mentioned "solution manual " which book's solution manual are you referring to? Such a book is rather interesting.

 

[fatpotato]

Just curious, you mentioned "solution manual " which book's solution manual are you referring to? Such a book is rather interesting.

Sure! However I am afraid there are slim chances you will find it unless you live in Switzerland...The problem I described in my first post and its solution are both from :

"Advanced calculus for engineers" (Analyse avancée pour ingénieurs)
Bernard Dacorogna , Chiara Tanteri
PPUR (Presses polytechniques et universitaires romandes)