5
$\begingroup$

Consider a functional $J[y]$ defined by: $$J[y] = \int_a^b F(x, y, y') dx \tag{1}$$

Here, $F$ is a function that depends on the independent variable $x$, the function $y(x)$, and its derivative $$y' = \frac{dy}{dx} \tag{2}$$.

In the calculus of variations, the operation of differentiating $F$ with respect to $y'$ is involved:
$$\frac{\partial F}{\partial y'} \tag{3}$$

This represents the rate of change of the function $F$ with respect to the derivative of $y$, $y'$. Operationally, since $y'$ is $\frac{dy}{dx}$, this differentiation is investigating how sensitive $F$ is to changes in the rate at which $y$ changes with respect to $x$.

I find the notation $\frac{\partial F}{\partial y'}$ a bit confusing in the sense that we are differentiating a function with respect to the derivative. If we just think $y'$ is just another variable symbol and proceed normally as most books do it does not cause much problems, but my question is :
What is the mathematical meaning of $\frac{\partial F}{\partial y'}$ in terms of limits?

In ordinary differential calculus, we don't encounter differentiation with respect to a derivative itself. Thanks.

$\endgroup$
2
  • $\begingroup$ In calculus of variations $y'$ itself is treated as a variable, instead of being the derivative of $y$. $\endgroup$
    – blomp
    Apr 13 at 16:10
  • 1
    $\begingroup$ Related but not quite a duplicate: math.stackexchange.com/questions/1205829/… $\endgroup$
    – whpowell96
    Apr 13 at 21:51

3 Answers 3

12
$\begingroup$

Consider a functional $J[y]$ defined by: $$J[y] = \int_a^b F(x, y, y') dx \tag{1}$$

I think it will be helpful to remind you right away that your notation implies that you have already agreed to consider the function $F$ to be a function of three independent variables. $$ F=F(a,b,c)\;. $$

It may also be helpful to consider an explicit example $F_{ex}$. You could, for example, have a function like: $$ F_{ex}(a,b,c) = a^2 + b^5 + c^9\;, $$ which means you have: $$ F_{ex}(x, y(x), y'(x)) = x^2 + (y(x))^5 + (y'(x))^9\;. $$

The notation $F(x,y,y')$ simply means that you are evaluating $F(a,b,c)$ at $a=x$, $b=y$, and $c=y'$.

Although, as you correctly note, $y' = \frac{dy}{dx}$ and therefore $y'$ is not generally independent of $y$, we do not really care about that. We only care that $$ \frac{\partial F}{\partial c} $$ is supposed to remind us we are differentiating $F$ with respect to its third argument.

This is also what $$ \frac{\partial F}{\partial y'} $$ is supposed to mean.

...I find the notation $\frac{\partial F}{\partial y'}$ a bit confusing in the sense that we are differentiating a function with respect to the derivative.

Yes, it is a bit confusing. But remember, what $$ \frac{\partial F}{\partial y'}\tag{A} $$ really means is just differentiation with respect to only the third argument. It might be easier to understand if written like this: $$ \left.{\left(\frac{\partial F}{\partial c}\right)}\right|_{a=x, b=y(x), c=y'(x)}\;.\tag{B} $$ But, of course, it takes a lot more pencil marks to write B than to write A.

If we just think $y'$ is just another variable symbol and proceed normally as most books do it does not cause much problems, but my Doubt/Question is :
What is the mathematical meaning of $\frac{\partial F}{\partial y'}$ in terms of limits?

It is the usual definition of the partial deriviative: $$ \frac{\partial F}{\partial y'}\equiv \left.{\left(\lim_{h\to 0}\frac{F(a,b,c + h) - F(a,b,c)}{h}\right)}\right|_{a=x,b=y(x), c=y'(x)} $$

For example, using our example $F_{ex}$ specified above: $$ \frac{\partial F}{\partial y'} = 9(y'(x))^8\;. $$

$\endgroup$
0
3
$\begingroup$

The "functional" is a way to convert functions to real numbers : It maps a function (of 1 variable or 2 variables or more variables) to a real number.

Here the functional is $J[y]$ to convert the given function $y$ to a real number.
Current functional is to take the Integration of some $F$ which is a "common" or "ordinary" function of 3 independent variables.

When we consider $F$ in isolation , it is a "common" or "ordinary" function of 3 variables , like $F(V_1,V_2,V_3)=V_1+V_2^2+ 3V_3$ : nothing confusing there.

We then "Substitute" those variables with terms involving the "Input" to the functional : given the function $y$ , we extract the independent variable $x$ & then evaluate the derivative $y'$.
With that "Substitution" , we will have only $x$ terms in the Integral & we can get the real number for the functional.

Coming to the confusing $\partial F / \partial y'$ , it is a short-hand for $\partial F / \partial V_3$ , which is straight-forward & not confusing.

Like-wise , $\partial F / \partial y$ is a short-hand for $\partial F / \partial V_2$

Of course , $\partial F / \partial x$ is a short-hand for $\partial F / \partial V_1$

In each case [ $\partial F / \partial V_1$ , $\partial F / \partial V_2$ , $\partial F / \partial V_3$ ] the other 2 variables are held "Constant" to get the Partial Derivatives.

With that , there is no over-all confusion.

OP : What is the mathematical meaning of $\frac{\partial F}{\partial y'}$ in terms of limits?

There is no meaning in terms of limits involving $x$ & $y$ & $y'$ simultaneously.
It is simply a short-hand notation for $\partial F / \partial V_3$ where $V_3$ is the third "Input" variable to the function $F$ , keeping the other variables unchanged.

Later , we will be setting it to $V_3=y'$ & we will have a Differential Equation involving $y',y,x$. Solving that , we will get $y$ in terms of $x$.

$\endgroup$
4
  • $\begingroup$ I am a huge fan of introducing temporary auxiliary variables in situations like these. +1. $\endgroup$
    – JonathanZ
    Apr 14 at 18:43
  • $\begingroup$ Thanks for the "Supportive Comment" , @JonathanZ , though I wonder whether you forgot the +1 at the end ! $\endgroup$
    – Prem
    Apr 14 at 18:59
  • $\begingroup$ Hmmm, I'm seeing an up arrow registered at my end. Maybe it's taking a while to update on some server? (When I was as close to 10k rep as you are, I counted every up vote. 🙂) $\endgroup$
    – JonathanZ
    Apr 14 at 19:22
  • $\begingroup$ I think that vote disappeared into the void , since there is no increment on my side. No worries though , @JonathanZ , It is the thought that counts ! More-over the other answer is very similar to mine & came up 80 minutes later , yet it has got lots of upvotes : unpredictable ! $\endgroup$
    – Prem
    Apr 15 at 6:45
1
$\begingroup$

$F$ is a function from $\mathbb R^3$ to $\mathbb R$. If you label points in $\mathbb R^3$ as $(x,y,z)$, then $\frac{\partial F}{\partial x}$, $\frac{\partial F}{\partial y}$ and $\frac{\partial F}{\partial z}$ make sense as functions from $\mathbb R^3$ to $\mathbb R$.

Now, among all points in $\mathbb R^3$, you only look at the points that are of the form $(x,y(x),y^\prime(x))$ for some $x$. The notation $\frac{\partial F}{\partial y^\prime}$ simply means $\frac{\partial F}{\partial z}$ evaluated at $(x,y(x),y^\prime(x))$ (remember $\frac{\partial F}{\partial z}$ is a function from $\mathbb R^3$ to $\mathbb R$).

Hope this helps. :)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .