Condition for f(x,y,z) = f(x,y,z(x,y)) being extremized

[PhDeezNutz]

Homework Statement

Given $f(x,y,z) = f(x,y,z(x,y))$ then what are the conditions involving the partial derivatives of $f$

Relevant Equations

Apparently they are

$\left(\frac{\partial f}{\partial x} \right)_y =\left(\frac{\partial f}{\partial y} \right)_x = \left(\frac{\partial f}{\partial z} \right)_z = 0$

where the subscript denotes the independent variable being held constant

As far as I know when a function is extremized its partial derivatives are all equal to 0 (provided we aren't dealing with a constraint)

$\left(\frac{\partial f}{\partial x} \right)_{yz} = \left(\frac{\partial f}{\partial y}\right)_{xz} = \left(\frac{\partial f}{\partial z}\right)_{xy} =0$
Let's start off
$\left(\frac{\partial f}{\partial x} \right)_{y} = \left(\frac{\partial f}{\partial x}\right)_{yz} + \left( \frac{\partial f}{\partial z} \right)_{xy} \left( \frac{\partial z}{\partial x} \right)_{y}$

I really don't know where to go from here. I'm not even sure my understanding of implicit functions is up to par to even begin addressing this question.

 

[fresh_42]

... then what are the conditions involving the partial derivatives of $f$ ...

What do you mean by conditions? The implicit function theorem goes as:
https://en.wikipedia.org/wiki/Implicit_function_theorem
The notation of the subscript isn't necessary, because this is automatically implied by the use of partial $\partial$ derivatives.

 

[Orodruin]

What do you mean by conditions? The implicit function theorem goes as:
https://en.wikipedia.org/wiki/Implicit_function_theorem
The notation of the subscript isn't necessary, because this is automatically implied by the use of partial $\partial$ derivatives.

It is not implied. It is a bastard notation often used in physics (mainly thermodynamics) when you have the choice to use any two out of three variables as your coordinates.

 

[PhDeezNutz]

What do you mean by conditions? The implicit function theorem goes as:
https://en.wikipedia.org/wiki/Implicit_function_theorem
The notation of the subscript isn't necessary, because this is automatically implied by the use of partial $\partial$ derivatives.

It is not implied. It is a bastard notation often used in physics (mainly thermodynamics) when you have the choice to use any two out of three variables as your coordinates.

I understand everything up until the line right after (22.63)

Why are Arken and Weber setting

$\left( \frac{\partial f}{\partial x}\right)_y = 0$ instead of $\left(\frac{\partial f}{\partial x} \right)_{yz} = \left(\frac{\partial f}{\partial y}\right)_{xz} = \left(\frac{\partial f}{\partial z}\right)_{xy} =0$?

And presumably the other 2 equations in (22.66) come from setting

$\left( \frac{\partial f}{\partial y}\right)_z = 0$

$\left( \frac{\partial f}{\partial z}\right)_ x= 0$

after cyclically permuting. (It says the second equation comes from swapping x and y but I don't see it, also that doesn't explain the third equation)

Why these specific derivatives?

Given $f(x,y,z(x,y))$

Hopefully I was able to clarify my question. Any help is appreciated because I am lost. So thanks in advance.

 

[fresh_42]

I find the usage of $f$ a bit confusing.

We have a function $f$ to be minimized and a constraint $g$. What we actually do is to define a new function $\mathcal{L}=f-\lambda g$ such that we can minimize $\mathcal{L}$ now which has no constraint. Thus we can as usual solve $\nabla \mathcal{L} = \operatorname{grad}\mathcal{L}=0$.

It is better explained here (including examples):
https://en.wikipedia.org/wiki/Lagrange_multiplier(22.64) is the first coordinate of $\nabla \mathcal{L}=0$.

 

[PhDeezNutz]

I find the usage of $f$ a bit confusing.

We have a function $f$ to be minimized and a constraint $g$. What we actually do is to define a new function $\mathcal{L}=f-\lambda g$ such that we can minimize $\mathcal{L}$ now which has no constraint. Thus we can as usual solve $\nabla \mathcal{L} = \operatorname{grad}\mathcal{L}=0$.

It is better explained here (including examples):
https://en.wikipedia.org/wiki/Lagrange_multiplier(22.64) is the first coordinate of $\nabla \mathcal{L}=0$.

I know how to use the Lagrangian multiplier method. I'm trying to prove it.

Why is the first coordinate $\left(\frac{\partial f}{\partial x}\right)_y$? Why is the second coordinate $\left(\frac{\partial f}{\partial y}\right)_z$? Why is the third coordinate $\left(\frac{\partial f}{\partial z}\right)_x$? I understand the chain rule helps us confirm these expressions.

I want to know why we set them = 0? I know that once we do we get the system of equations $\nabla f = \lambda \nabla g$.

 

[Orodruin]

It is much easier to think geometrically about the Lagrange multiplier method. If a constraint surface is given by $g = c_0$. For any function $f$ it holds that the infinitesimal change $df$ of $f$ when making a displacement $d\vec x$ is given by
$$
df = \vec d\vec x \cdot \nabla f.
$$
Restricting $d\vec x$ to the constraint surface means that $d\vec x \cdot \vec n = 0$, where $\vec n \propto \nabla g$ is the surface normal. Thus, in order for $df = 0$ when restricting displacements to lie in the constraint surface results in
$$
df = d\vec x \cdot \nabla f = 0
$$
for all $d\vec x$ in the constraint surface, i.e., the projection of $\nabla f$ to the local tangent space of the surface is zero. This is true if $\nabla f = \lambda \nabla g$ for some $\lambda$, since $\nabla g$ is normal to the surface. Therefore
$$
df = d\vec x \cdot (\nabla f - \lambda \nabla g) = 0
$$
and we can adjust $\lambda$ such that this is true not only for displacements in the constraint surface, but for all displacements.

 

[PhDeezNutz]

It is much easier to think geometrically about the Lagrange multiplier method. If a constraint surface is given by $g = c_0$. For any function $f$ it holds that the infinitesimal change $df$ of $f$ when making a displacement $d\vec x$ is given by
$$
df = \vec d\vec x \cdot \nabla f.
$$
Restricting $d\vec x$ to the constraint surface means that $d\vec x \cdot \vec n = 0$, where $\vec n \propto \nabla g$ is the surface normal. Thus, in order for $df = 0$ when restricting displacements to lie in the constraint surface results in
$$
df = d\vec x \cdot \nabla f = 0
$$
for all $d\vec x$ in the constraint surface, i.e., the projection of $\nabla f$ to the local tangent space of the surface is zero. This is true if $\nabla f = \lambda \nabla g$ for some $\lambda$, since $\nabla g$ is normal to the surface. Therefore
$$
df = d\vec x \cdot (\nabla f - \lambda \nabla g) = 0
$$
and we can adjust $\lambda$ such that this is true not only for displacements in the constraint surface, but for all displacements.

The reason I want to prove it in the manner shown above is because I eventually want to use the same approach for calculus of variations when a functional that depends on coordinates has constraints on coordinates.

Where $\lambda_j$ is a function of t instead of a mere scalar factor. And in the calculus of variation case there is no clear geometric meaning. Furthermore the derivative definitions are more complicated.

Edit the constraints are of the form $g_j(q_i(t)) = C$

 

[Orodruin]

And in the calculus of variation case there is no clear geometric meaning.

What makes you think that? The derivation is completely analogous.

 

[PhDeezNutz]

What makes you think that? The derivation is completely analogous.

Maybe I misspoke but I would think to set the functional derivatives equal to each other except for a scalar multiple (I.e. the Lagrange Euler equations) which would work except for the fact lambda is a function of t. I want to do the analysis that reveals lambda is an actual function of t.

Edit: I don’t understand how derivatives can be parallel to each other but differ by a multiplicative function of t.

 

[vela]

Why are Arken and Weber setting

$\left( \frac{\partial f}{\partial x}\right)_y = 0$ instead of $\left(\frac{\partial f}{\partial x} \right)_{yz} = \left(\frac{\partial f}{\partial y}\right)_{xz} = \left(\frac{\partial f}{\partial z}\right)_{xy} =0$?

I believe it's because the constraint removes one degree of freedom. You have the freedom to hold $y$ constant, but once you impose that condition, a change in $x$ generally requires a change in $z$ to remain on the surface $g(x,y,z)=C$.

 

[PhDeezNutz]

I believe it's because the constraint removes one degree of freedom. You have the freedom to hold $y$ constant, but once you impose that condition, a change in $x$ generally requires a change in $z$ to remain on the surface $g(x,y,z)=C$.

Let me see if I understand

$\left( \frac{\partial f}{\partial x} \right)_y = 0$ z is a function of x and y so only x and y are allowed to vary, x is varying so y is constant.

$\left(\frac{\partial f}{\partial y} \right)_x = 0$ z is a function of x and y so only x and y are allowed to vary, y is varying so x is constant.

I don't see how $\left(\frac{\partial f}{\partial z}\right)_x = 0$ is consistent with these choices. And therefore I don't see how we get the third equation in 22.64.

Either way, your post definitely helped. Even If I don't totally understand it.

 

[vela]

You seem to be confusing yourself because you accorded $z$ special status. The implicit function theorem doesn't single out $z$ as being a function of the other variables. You could just as easily say $x=x(y,z)$.

 

[PhDeezNutz]

You seem to be confusing yourself because you accorded $z$ special status. The implicit function theorem doesn't single out $z$ as being a function of the other variables. You could just as easily say $x=x(y,z)$.

I think you're right.

With your help I think I've got a grasp on it. Would you mind giving my work a quick look through?

Lemma 1 (Implicit Function Theorem)

Let $g : R^n \rightarrow R$ and let $L_c(g)$ be a level set of $g$, i.e.

$L_c(g) = \left\{ \vec{x} \epsilon R^n | g(\vec{x}) = c \right\}$

it follows that

$dg = \sum \limits_{i = 1}^{n} \left(\frac{\partial g}{\partial x_i} \right) \left(x_i - a_i \right) = 0$

implies

$\left( \frac{\partial g}{\partial x_n}\right) \left( x_n - a_n \right) = - \sum \limits_{i=1}^{n-1} \left( \frac{\partial g}{\partial x_i}\right) \left( x_i - a_i \right)$

which then gives us

$x_n = a_n + \sum \limits_{i=1}^{n-1} - \frac{\left( \frac{\partial g }{\partial x_i}\right)}{\left( \frac{\partial g}{\partial x_n}\right)} \left( x_i - a_i \right)$

Linearizing $x_n$

$x_n = a_n + \sum \limits_{i=1}^{n-1} \left(\frac{\partial x_n}{\partial x_i}\right) \left( x_i - a_i \right)$

Therefore

$\left(\frac{\partial x_n}{\partial x_i}\right) = - \frac{\left( \frac{\partial g }{\partial x_i}\right)}{\left( \frac{\partial g}{\partial x_n}\right)}$

Lemma 2 (We are going to need this later to show that the the Lagrange Multiplier is indeed a constant)

If $p(x_i) - \sum \limits_{j=1}^{m} \lambda_j \left( x_n \right) s_j \left(x_i \right) = 0 \Rightarrow \lambda_j \left( x_n \right) = C_j$ where $C_j$ is a constant.Proof:

$p \left( x_i \right) = \sum \limits_{j=1}^{m} \lambda_j \left( x_n \right) s_j \left(x_i \right) = \sum \limits_{j=1}^{m} \Lambda_j \left( x_i \right)$

$\Lambda_j \left( x_i \right) = \lambda_j \left( x_n \right) s_j \left(x_i \right)$ implies $\lambda_j \left( x_n \right)$ is either a function of $x_i$ or a constant. It can't be a function $x_i$ so it must be a constant.

So we have

$\lambda_j \left( x_n \right) = constant$

Proof of Lagrange Multiplier Method for Real-Valued Functions:(Multiple Constraints)
Let $f: R^n \rightarrow R$ and $g_j: R^n \rightarrow R$. Want to constrain $f \left( \vec{x} \right)$ by $g_j \left( \vec{x} \right) = C_j$. Suppose there are $m$ such constraints.

$dg_j = 0$ so by the Implicit Function Theorem $\left( \frac{\partial x_n}{\partial x_i}\right) = - \sum\limits_{j=1}^{m} \frac{\left( \frac{\partial g_j}{\partial x_i }\right)}{\left( \frac{\partial g_j}{\partial x_n}\right)}$ (This part I'm unsure about but it seems necessary to get the result we want)

If $f$ is extremized then $\left( \frac{\partial f}{\partial x_i} \right)_{total} = 0$

$\left( \frac{\partial f}{\partial x_i} \right)_{total} = \left( \frac{\partial f}{\partial x_i} \right) + \left( \frac{\partial f}{\partial x_n}\right) \left( \frac{\partial x_n}{\partial x_i}\right) = 0$

$\left( \frac{\partial f}{\partial x_i} \right)- \sum \limits_{j=1}^{m} \frac{\left( \frac{\partial f}{\partial x_i}\right)}{\left( \frac{\partial g}{\partial x_n}\right)} \left(\frac{\partial g_j}{\partial x_i} \right) = \left( \frac{\partial f}{\partial x_i} \right)- \sum \limits_{j=1}^{m} \lambda_j \left( x_n \right) \left(\frac{\partial g_j}{\partial x_i} \right) = 0$

implies $\lambda_j(x_n) = constant$ by Lemma 2

Therefore we have a system of equations

$\nabla f - \sum\limits_{j=1}^{m} \lambda_j \nabla g_j = 0$ Q.E.D.

Hopefully I didn't do anything egregiously wrong.

 

[PhDeezNutz]

Can anyone confirm, deny, or motivate a proof for the following statement for the intersection of a level sets of different functions?

$g: R^n \rightarrow R$
$h: R^n \rightarrow R$
$j: R^n \rightarrow R$

and $j(\vec{x}) = g(\vec{x}) + h(\vec{x})$

Define the following level sets

$L_c(g) = \left\{ \vec{x} \epsilon R^n | g(\vec{x}) = c \right\}$

$L_d(h) = \left\{ \vec{x} \epsilon R^n | h(\vec{x}) = d \right\}$

Is the following a true statement? How would I prove it. Keep in mind I'm a physics student not a math student so forgive me if I'm a little slow in understanding.

$L_c(g) \cap L_d(h) = \left\{ \vec{x} \epsilon R^n | g(\vec{x}) = c \wedge h(\vec{x}) = d \right\} = \left\{ \vec{x} \epsilon R^n | j(\vec{x}) = c + d \right\} = L_{(c+d)} (j)$

 

[George Jones]

Can anyone confirm, deny, or motivate a proof for the following statement for the intersection of a level sets of different functions?

$g: R^n \rightarrow R$
$h: R^n \rightarrow R$
$j: R^n \rightarrow R$

and $j(\vec{x}) = g(\vec{x}) + h(\vec{x})$

Define the following level sets

$L_c(g) = \left\{ \vec{x} \epsilon R^n | g(\vec{x}) = c \right\}$

$L_d(h) = \left\{ \vec{x} \epsilon R^n | h(\vec{x}) = d \right\}$

Is the following a true statement? How would I prove it. Keep in mind I'm a physics student not a math student so forgive me if I'm a little slow in understanding.

$L_c(g) \cap L_d(h) = \left\{ \vec{x} \epsilon R^n | g(\vec{x}) = c \wedge h(\vec{x}) = d \right\} = \left\{ \vec{x} \epsilon R^n | j(\vec{x}) = c + d \right\} = L_{(c+d)} (j)$

Consider $n=2$. What happens when $g\left(x,y\right)=x$, $h\left(x,y\right)=y$, $c=2$, and $d=3$?