How do I minimize a function with a constraint using Lagrange-Euler method?

[fery]

I am working on a functional and I need to find its minimum, the conventional procedure is to use Lagrange-Euler method and find the minimum state of the function, but if I need to impose a constraint to the function, I don't know what I need to do

J=int(F(t, f(t), a, b)) minimize(f) and int(G(t, f(t), a, b))=M,
It should be very elementary, but I am confused about what I need to do.

Your help will be very appreciated.
Farshad

 

[Stephen Tashi]

I don't understand your notation. See if I've guess the problem correctly:

$$J(f)$$ is a linear functional defined by $$J(f) = \int_a^b F(x,f(x)) dx$$ where $$F(x,y)$$ is a given function of two variables. We wish to find the minimum value of $$J$$ over all functions $$f$$ that satisfy $$\int_a^b G(t,f(t)) = M$$ where $$M$$ is a given constant and $$G(x,y)$$ is a given function of two variables.

 

[fery]

All true but F(x,f(x)) is a functional not a function, which is mapping of a function to R. For minimization of the functional Euler-Lagrange is the conventional method, but when there is constraint (int(G(t,f(t),a,b)=M) I am not sure what should I do.

Farshad

 

[Stephen Tashi]

All true but F(x,f(x)) is a functional not a function,

Then I don't understand the notation F(x,f(x)). If F is a functional and f is a function then
F(f) is a real number correct? We don't need the argument 'x'.

For example, in the calculus of variations an arc length problem is to minimize the functional $$J$$ given by
$$J[f] = \int_a^b \sqrt{1 + (f'(x))^2} dx$$
The expression $$\sqrt{1 + (f'(x))^2)}$$ is a function not a functional.

 

[fery]

Yes F(f) is a real number, as I said it is a mapping in R, the reason I put x in F(f,x) is that sometimes x is explicitly in the functional. for example it is possible to write
int((f^2+x^2),a,b) as a functional
http://mathworld.wolfram.com/Euler-LagrangeDifferentialEquation.html
The inclusion of constraint is my concern
Farshad

 

[Stephen Tashi]

I think the problem you describe is treated in section 2.4 of this PDF:
http://www.google.com/url?sa=t&sour...sg=AFQjCNFhHnoE14HmK8g5HfBRMZxUBXaqbw&cad=rja

The link you gave to the Euler-Lagrange equations does not say that the integrand is a "functional". The integrand is a function.

 

[homology]

I agree, the integrand does not return a number given a function. It returns an expression. You integrate and then you have a number.

You might look up the 'isoperimetric problem' which is an example, or 'variational problems with subsidiary conditions' more generally. Gelfand and Fomin's little book on the calculus of variations has a section on it.

that is, Given the functional:

$$J[y]=\int_a^b F(x,y,y')dx$$
let the admissable curves satisfy the conditions:

$$y(a)=A,y(b)=B, K[y]=\int_a^b G(x,y,y')dx=M$$

Where K[y] is another functional and let J[y] have an extremum for y=y(x). Then, if y=y(x) is not an extremal of K[y], there exists a constant $$\lambda$$ such that y=y(x) is an extremal of the functional:

$$\int_a^b (F+\lambda G)dx$$

That's from the text and probably is enough to get you started.