Lagrange Multipliers: Understand Why \nabla f = \lambda \nabla g

[Gregg]

Homework Statement

Why is
\nabla f = \lambda \nabla g

where f is the function you want to find the extrema of and g is the contraint?

Also how would you identify the above in the following






Determine the least real number M such that the inequality
<br /> |ab(a^2-b^2) + bc(b^2-c^2) + ca(c^2-a^2)|\le M(a^2 + b^2 + c^2)^2




holds for all real numbers a, b and c.

The Attempt at a Solution

It is the first part of the problem which I cannot do so there is no working to show

You have to minimise M but subject to what? Also, there is no explicit definition for M, only the inequality.

 

[LCKurtz]

Why is
\nabla f = \lambda \nabla g

where f is the function you want to find the extrema of and g is the contraint?

I will answer your first question.

This example might help you see that. Suppose you have a circular wire in the shape of the unit circle x²+y²=1. Suppose the temperature at a point (x,y) in the plane is given by T = f(x,y) = y + 2x, and you want to know what is the warmest point on the wire. Now make a sketch showing the circle and several of the level curves where

T = y + 2x = c

for various values of c. Plot them for at least c = 0, 1/2, 1, 3/2, 2. They will be parallel lines cutting through or near your wire, each of on which the temperature is constant. With this picture you will be able to estimate the max temperature on the wire although the level curve giving it may not have been drawn. But you should be able to see that the level curve that does the trick will be tangent to the circle. It has to at least touch the circle, and if it crosses the circle you can back away some to get a higher temperature. This means that the normal to the circle and the level curve must be parallel at the point giving max temperature, so their normals are proportional. That gives the gradient condition. The same idea works in 3D. Your text likely gives a mathematical argument to buttress this hueristic argument.

 

[Gregg]

That's really clear, thanks. Without getting involved in the second part of my question is there anything you can say about using Lagrange multipliers in inequalities?

 

[snipez90]

Well you aren't actually minimizing M, since M is a constant. Instead, notice that M is really the supremum of

\frac{|ab(a^2-b^2) + bc(b^2-c^2) + ca(c^2-a^2)|}{(a^2 + b^2 + c^2)^2}

over reals a,b,c.

Note maximizing/minimizing the above quantity is the same as maximizing/minimizing the expression inside the absolute values of the numerator. The other thing you need is a constraint, and one way to get a constraint in any inequality problem is to check if your function is homogeneous.

 

[Gregg]

f(ka,kb,kc)=f(a,b,c) so it is homogenous of degree 1? If it is just the numerator we want to maximise or minimise then it is of degree 4. I don't understand why the denominator is not needed, though.

 

[Gregg]

Also I found this ab(a^2-b^2)+bc(b^2-c^2)+ac(c^2-a^2) = \left|<br /> \begin{array}{ccc}<br /> 1 &amp; 1 &amp; -1 \\<br /> -a^2 &amp; -b^2 &amp; c^2 \\<br /> -\text{bc} &amp; -\text{ac} &amp; \text{ab}<br /> \end{array}<br /> \right| = (a - b) (a - c) (b - c) (a + b + c)

 

[snipez90]

f(ka,kb,kc)=f(a,b,c) so it is homogenous of degree 1? If it is just the numerator we want to maximise or minimise then it is of degree 4. I don't understand why the denominator is not needed, though.

Hmm I see the logical connection I didn't make now. The ratio is homogeneous of degree 1, so we may assume that a^2 + b^2 + c^2 = 1. This reduces the problem to finding the maximum of just the numerator over the unit sphere, which is compact.

 

[Gregg]

\nabla f(a,b,c) - \lambda \nabla g(a,b,c) = 0

\frac{\partial f(a,b,c)}{\partial a} - \lamba\frac{\partial g(a,b,c)}{\partial a}=0

\frac{\partial f(a,b,c)}{\partial b} - \lamba\frac{\partial g(a,b,c)}{\partial b}=0

\frac{\partial f(a,b,c)}{\partial c} - \lamba\frac{\partial g(a,b,c)}{\partial c}=0

f(a,b,c) = (a-b)(a-c)(b-c)(a+b+c)

g(a,b,c)= a^2 + b^2 + c^2 = 1

I find it difficult solving for a,b,c without getting a=b=c which would make M = 0. Is this the correct approach?

 

[snipez90]

It should be obvious that M is not zero right? It's not hard to think of a, b, c for which the ratio of the absolute value expression to the squared expression in the original inequality is positive, implying that M is > 0. The very condition \nabla f = \lambda \nabla should give you the following equality between vectors:

(3a^{2}b-b^{3}+c^{3}-3ca^{2},3b^{2}c-c^{3}+a^{3}-3ab^{2},b^{2}-3bc^{2}+3ca^{2}-a^{3})=\lambda(2a,2b,2c) .

This is not a particularly nice set of equations to solve, but doing the algebra is typically the bulk of any lagrange multipliers problem.

 

[Gregg]

It should be obvious that M is not zero right? It's not hard to think of a, b, c for which the ratio of the absolute value expression to the squared expression in the original inequality is positive, implying that M is > 0. The very condition \nabla f = \lambda \nabla should give you the following equality between vectors:

(3a^{2}b-b^{3}+c^{3}-3ca^{2},3b^{2}c-c^{3}+a^{3}-3ab^{2},b^{2}-3bc^{2}+3ca^{2}-a^{3})=\lambda(2a,2b,2c) .

This is not a particularly nice set of equations to solve, but doing the algebra is typically the bulk of any lagrange multipliers problem.

Yeah that's what I did to start the problem and the algebra is as you said

 

[Gregg]

The set of equations are inpenetrable for me. I find a quadratic in a,b and c and I have the constraint also. So I have 4 equations and 4 unknowns. But I just can't seem to reduce it.

I get

a = \frac{\lambda\pm\sqrt{\lambda^2-3(b-c)(c^3-b^3)}}{3(b-c)}

b = \frac{\lambda\pm\sqrt{\lambda^2-3(a+c)(a^3-c^3)}}{3(a+c)}

c = \frac{\lambda\pm\sqrt{\lambda^2-3(a-b)(b^3-a^3)}}{3(a-b)}

From

\nabla f = \lambda \nabla g