Determine minimum value of integral

[skrat]

Homework Statement

Determine minimum value of integral $B(y)=\int_{0}^{2}({y}')^2dx$ for function $y\in C^1(\mathbb{R})$ and $y(0)=y(2)=0$ and $\int_{0}^{2}y^2dx=4$

Homework Equations

The Attempt at a Solution

IF I am not mistaken, the idea is to first find the function $y$ for given conditions and than calculate $B(y)$.

My idea:

Since $y\in C^1(\mathbb{R})$ I simply started with polynomial $y(x)=ax^2+bx+c$. Of course $y(0)=0$ so immediately $c=0$ than $y(2)=0=4a+2b=0$ so $b=-2a$.

Third condition says $\int_{0}^{2}y^2dx=\int_{0}^{2}(ax^2+bx)^2dx=4$

$\frac{32a^2}{5}+\frac{ab16}{2}+8b^2=4$ using $b=-2a$ gives me $a=\sqrt{\frac{5}{28}}$ and $b=-\sqrt{\frac{5}{7}}$.

So using $y(x)=\sqrt{\frac{5}{28}}x^2-\sqrt{\frac{5}{7}}x$ the $B(y)=\int_{0}^{2}({y}')^2dx=\frac{5}{3}$.

Or is it not?

 

[LCKurtz]

Homework Statement

Determine minimum value of integral $B(y)=\int_{0}^{2}({y}')^2dx$ for function $y\in C^1(\mathbb{R})$ and $y(0)=y(2)=0$ and $\int_{0}^{2}y^2dx=4$

Homework Equations

The Attempt at a Solution

IF I am not mistaken, the idea is to first find the function $y$ for given conditions and than calculate $B(y)$.

My idea:

Since $y\in C^1(\mathbb{R})$ I simply started with polynomial $y(x)=ax^2+bx+c$. Of course $y(0)=0$ so immediately $c=0$ than $y(2)=0=4a+2b=0$ so $b=-2a$.

Third condition says $\int_{0}^{2}y^2dx=\int_{0}^{2}(ax^2+bx)^2dx=4$

$\frac{32a^2}{5}+\frac{ab16}{2}+8b^2=4$ using $b=-2a$ gives me $a=\sqrt{\frac{5}{28}}$ and $b=-\sqrt{\frac{5}{7}}$.

So using $y(x)=\sqrt{\frac{5}{28}}x^2-\sqrt{\frac{5}{7}}x$ the $B(y)=\int_{0}^{2}({y}')^2dx=\frac{5}{3}$.

Or is it not?

I'm not sure how to work that problem, but I'm pretty sure that isn't it. There are lots of functions that satisfy $y(0)=y(2)=0$ such that $\int_0^2 y^2~dx = 4$. You are to find the one that minimizes $\int_0^2 y'^2~dx$ over all such functions. You must be pretty lucky if the one function you found is it. This looks like a calculus of variations problem. I'm not very familiar with such problems, but if that's what you are studying, surely your text must give a method for solving them.

 

[Ray Vickson]

Homework Statement

Determine minimum value of integral $B(y)=\int_{0}^{2}({y}')^2dx$ for function $y\in C^1(\mathbb{R})$ and $y(0)=y(2)=0$ and $\int_{0}^{2}y^2dx=4$

Homework Equations

The Attempt at a Solution

IF I am not mistaken, the idea is to first find the function $y$ for given conditions and than calculate $B(y)$.

My idea:

Since $y\in C^1(\mathbb{R})$ I simply started with polynomial $y(x)=ax^2+bx+c$. Of course $y(0)=0$ so immediately $c=0$ than $y(2)=0=4a+2b=0$ so $b=-2a$.

Third condition says $\int_{0}^{2}y^2dx=\int_{0}^{2}(ax^2+bx)^2dx=4$

$\frac{32a^2}{5}+\frac{ab16}{2}+8b^2=4$ using $b=-2a$ gives me $a=\sqrt{\frac{5}{28}}$ and $b=-\sqrt{\frac{5}{7}}$.

So using $y(x)=\sqrt{\frac{5}{28}}x^2-\sqrt{\frac{5}{7}}x$ the $B(y)=\int_{0}^{2}({y}')^2dx=\frac{5}{3}$.

Or is it not?

You made some errors: after solving $y(2) = 0$ to get $b = -2a$, your other condition should be
$$\int_0^2 (ax^2 - 2ax)^2 \, dx = 4$$

More seriously: why would you assume a quadratic polynomial? How do you know that a polynomial of degree 1000 won't produce better results? How do you know that you could not do better using a non-polynomial?
You need to use a Lagrange multiplier method:
$$\min \left( \int_0^2 y'^2 \, dx + \lambda \int_0^2 y^2 \, dx \right)$$
where $\lambda$ is a constant. This gives a Lagrangian of $L = y'^2 + \lambda y^2$. Now apply the Euler-Lagrange equations.

 

[skrat]

There are lots of functions that satisfy $y(0)=y(2)=0$ such that $\int_0^2 y^2~dx = 4$. You are to find the one that minimizes $\int_0^2 y'^2~dx$ over all such functions. You must be pretty lucky if the one function you found is it. This looks like a calculus of variations problem. I'm not very familiar with such problems, but if that's what you are studying, surely your text must give a method for solving them.

Hmmm, I completely agree with you. I guess I should go the opposite way and firstly find a general function that minimizes the integral and than find one that satisfies given conditions.

And yes I am dealing with variations problems.

 

[HallsofIvy]

It's not clear to me what course this is for or what knowledge you have. If this we in a "Calculus of Variations" class then I would think that a good "Relevant Equation" would be the Euler-Lagrange equation: if f minimizes
$$\int_{x_1}^{x_2} L(x, y, y')dx$$
then
$$\frac{\partial L}{\partial y}- \frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right)= 0$$

But you seem to be saying this is just for a Calculus class. My question then is do you have any reason to assume the solution can be written as a polynomial? And even so, why quadratic? There appear to be four conditions to be met:
(1)y(0)= 0.
(2)y(2)= 0.
(3)$$\int_0^2 y^2 dx= 4$$ and
(4)$$\int_0^2 (y')^2dy$$ is a minimum.

If I had reason to believe that y was a polynomial function of x, seeing that there were four conditions to be met, I would expect a polynomial with four constants to be determined, a cubic polynomial.

 

[skrat]

You made some errors: after solving $y(2) = 0$ to get $b = -2a$, your other condition should be
$$\int_0^2 (ax^2 - 2ax)^2 \, dx = 4$$

I integrated here $$\int_0^2 y(x)^2 \, dx = 4$$ where I used $y(x)=ax^2+bx+c$, however one of the conditions already gave me $c=0$ so $\int_0^2 (ax^2+bx)^2 dx = 4$. After integrating I used the fact that $b=-2a$.

Since $b$ and $a$ are not functions of $x$, I don't think this actually plays a huge role here. Unless I am mistaken.

More seriously: why would you assume a quadratic polynomial? How do you know that a polynomial of degree 1000 won't produce better results? How do you know that you could not do better using a non-polynomial?
You need to use a Lagrange multiplier method:
$$\min \left( \int_0^2 y'^2 \, dx + \lambda \int_0^2 y^2 \, dx \right)$$
where $\lambda$ is a constant. This gives a Lagrangian of $L = y'^2 + \lambda y^2$. Now apply the Euler-Lagrange equations.

Haha, I know that simply by having only two conditions. :D Thanks for the hint. Will do that and publish it here.

EDIT: This is in a way also answer to Hallsoflvy. Why two and not four as you stated?... Yup, that's what I am asking myself at this very moment when you are making it obvious. :D

 

[skrat]

Version 2.0:

$F(y,{y}')=\int_{0}^{2}({y}')^2dx+\lambda \int_{0}^{2}y^2dx=\int_{0}^{2}(({y}')^2+\lambda y^2)dx$

Let's say that $L=({y}')^2+\lambda y^2$, than Euler-Lagrange equation tells me that

$2\lambda y-2{y}''=0$ which is a differential equation with solutions $y(x)=Ae^{\sqrt{\lambda }x}+Be^{-\sqrt{\lambda }x}$.

Condition $y(0)=0$ gives me $A+B=0$ and from $y(2)=0$ I get

$Ae^{\sqrt{\lambda }2}+Be^{-\sqrt{\lambda }2}=0$

$Ae^{4\sqrt{\lambda }}=-B$

$-Be^{4\sqrt{\lambda }}=-B$

$4\sqrt{\lambda }=ln(1)=0$ therefore $\lambda =0$.This has to be wrong but I can't find a mistake...

 

[pasmith]

Version 2.0:

$F(y,{y}')=\int_{0}^{2}({y}')^2dx+\lambda \int_{0}^{2}y^2dx=\int_{0}^{2}(({y}')^2+\lambda y^2)dx$

Let's say that $L=({y}')^2+\lambda y^2$, than Euler-Lagrange equation tells me that

$2\lambda y-2{y}''=0$ which is a differential equation with solutions $y(x)=Ae^{\sqrt{\lambda }x}+Be^{-\sqrt{\lambda }x}$.

Condition $y(0)=0$ gives me $A+B=0$ and from $y(2)=0$ I get

$Ae^{\sqrt{\lambda }2}+Be^{-\sqrt{\lambda }2}=0$

$Ae^{4\sqrt{\lambda }}=-B$

$-Be^{4\sqrt{\lambda }}=-B$

$4\sqrt{\lambda }=ln(1)=0$ therefore $\lambda =0$.This has to be wrong but I can't find a mistake...

You've made the error of assuming that $\lambda > 0$. If $\lambda = -k^2 < 0$ you get a solution in terms of sines and cosines, and you can choose $k > 0$ such that $\sin(2k) = 0$.

(There are a countable infinity of such $k$, so you'll have to decide which of them minimizes $B(y)$.)

Finding $\lambda$ such that $y'' = \lambda y$ has a non-zero solution with $y(0) = y(L) = 0$ is a standard eigenvalue problem which you will probably see again in other contexts.

 

[skrat]

Ok, so for $\lambda >0$ nothing interesting happens.

Now let's assume that $\lambda <0$, than $y(x)=Ae^{i\sqrt{\lambda}x}+Be^{-i\sqrt{\lambda}x}=Asin(\sqrt{\lambda}x)+Bcos(\sqrt{\lambda}x)$.

Due to $y(0)=0$ we find out that $B=0$.

Also $y(2)=0=Asin(2\sqrt{\lambda})$. Since we are looking for non trivial solutions, let's assume that $A\neq 0$, than $2\sqrt{\lambda}=n\pi$.

$\sqrt{\lambda}=\frac{n\pi }{2}$ so $y(x)=Asin(\frac{n\pi }{2}x)$.

Using condition $\int_{0}^{2}y^2dx=4$ we get value for $A$ which is $A=2\sqrt{2}$.

Now finally $y(x)=\left\{\begin{matrix} 0 ;& \lambda \geq 0\\ 2\sqrt{2}sin(\frac{n\pi }{2}x) ; &\lambda <0 \end{matrix}\right.$

Now $B(y)=\int_{0}^{2}({y}')^2dx$ to determine $n$ and finally the right $y(x)$.

$B(y)=2n^2\pi ^2$ and obviously $n=1$ if we want $B(y)$ to be minimzed but not zero.

Therefore $y(x)=2\sqrt{2}sin(\frac{\pi }{2}x)$.


Right or wrong... You guys are being very helpful! Thanks to all of you!

 

[skrat]

Since nobody had any complaints, I assume everything is more or less ok and the next step is to say thanks to you.

So, thanks!