# Thread: Calculus of Variation - Euler Lagrange Question

1. ## Calculus of Variation - Euler Lagrange Question

Find the minimum value taken by the following two integral expressions where $y$ and $y'$ are both functions of $x$

(a) $\int_{0}^{1}((y')^2-6y^2)e^{-5x}dx$ $, y(0)=1, y'(0)=2$

(b) $\int_{0}^{1}\frac{(y')^2}{(1+y)^2}dx, y(0)=0, y(1)=e^3-1$

I've had a few stabs at this but my answers are looking dodgy. Could someone please show me the way? ^^

Here is what I've done so far:
(a) Using Euler Lagrange Equation $\frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{\partial F}{\partial y}$, I get
$\frac{d}{dx}(2e^{-5x}y')=-12e^{-5x}y$
$(2e^{-5x}y')=\frac{-12e^{-5x}y}{5}+A$
$y'=\frac{6y}{5}+\frac{A}{2e^{-5x}}$
$y(0)=1, y'(0)=2$ so $A=\frac{8}{5}$
$y'=\frac{dy}{dx}$ so by integrating the last expression for $y'$ I get $y=\frac{6xy}{5}+\frac{4e^{5x}}{25}+B$
Applying boundary conditions again, I get $B=\frac{21}{25}$
Substituting and rearranging, I end up with $y(1-\frac{6x}{5})=\frac{4e^5x+21}{25}\rightarrow y(x)=\frac{4e^5x+21}{25-30x}$ as the extremal funcation.

For part (b) I did something similar using $F-y'\frac{\partial F}{\partial y'}=constant$ as the Euler Langrage equation and ended up with $y=\frac{x}{x-1}$
Here is my working out for part (b):
$I=\int_{0}^{1}\frac{(y')^2}{(1+y)^2}dx, y(0)=0, y(1)=e^3-1$
so using $F-y'\frac{\partial F}{\partial y'}=constant$, I get
$\frac{(y')^2-2(y')^2}{(1+y)^2}=k$
Taking square roots of both sides and multiplying the whole equation by -1, I get
$\frac{y'}{1+y}=A\rightarrow y'=A+Ay \rightarrow y=Ax+Axy+B$
Using boundary condition $y(0)=0$, I get $B=0$
Substituting and rearranging, I get
$y=\frac{Ax}{1+Ax}$
Using the second boundary condition, I get
$e^3-1=\frac{A}{1+A}$
$(e^3-1)(1+A)=A$
$A=\frac{e^3-2}{2-e^3}=-1$
Hence, $y=\frac{x}{x-1}$

2. ## Re: Calculus of Variation - Euler Lagrange Question

Is y funcion of x?
Otherwise, the two integral are just numbers.

3. ## Re: Calculus of Variation - Euler Lagrange Question

Sorry, yes. y is a function of x

4. ## Re: Calculus of Variation - Euler Lagrange Question

Originally Posted by yaohui88
Find the minimum value taken by the following two integral expressions where $y$ and $y'$ are both functions of $x$

(a) $\int_{0}^{1}((y')^2-6y^2)e^{-5x}dx$, y(0)=1, y'(0)=2

(b) $\int_{0}^{1}\frac{(y')^2}{(1+y)^2}dx$, y(0)=0, y(1)= $e^3-1$

I've had a few stabs at this but my answers are looking dodgy. Could someone please show me the way? ^^

Here is what I've done so far:
(a) Using Euler Lagrange Equation $\frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{\partial F}{\partial y}$, I get
$\frac{d}{dx}(2e^{-5x}y')=-12e^{-5x}y$
Good so far, but the next step is, I think, incorrect. Try taking the LHS derivative again. You can't just integrate the way you did, because you don't know what y is.

$(2e^{-5x}y')=\frac{-12e^{-5x}y}{5}+A$
$y'=\frac{6y}{5}+\frac{A}{2e^{-5x}}$
y(0)=1, y'(0)=2 so A= $\frac{8}{5}$
$y'=\frac{dy}{dx}$ so by integrating the last expression for $y'$ I get $y=\frac{6xy}{5}+\frac{4e^{5x}}{25}+B$
Applying boundary conditions again, I get $B=\frac{21}{25}$
Substituting and rearranging, I end up with $y(1-\frac{6x}{5})=\frac{4e^5x+21}{25}\rightarrow y(x)=\frac{4e^5x+21}{25-30x}$ as the extremal funcation.

For part (b) I did something similar using $F-y'\frac{\partial F}{\partial y'}=constant$ as the Euler Langrage equation and ended up with $y=\frac{x}{x-1}$
Hmm. That's not what I get. What is your differential equation?

5. ## Re: Calculus of Variation - Euler Lagrange Question

Hmmm...okay. I've updated the original post to show you what I've done with part (b), and I'm pretty certain I have gone wrong somewhere...Is it the same issue with not being able to integrate like I did because I don't know what $y$ is?

As with part (a), I gave it another shot considering your advice. So, I differentiate again and end up with:
$2e^{-5x}(y''-5y'+6y)=0$
Which gives me a second order differential equation:
$y''-5y'+6y=0$
The solution I get from this is:
$y=e^{2x-2}+e^{3x}$ which I assume is the extremal function? Then I differentiate to get $y'=2e^{2x-2}+3e^{3x}$?

6. ## Re: Calculus of Variation - Euler Lagrange Question

Concerning part a):

I don't know if I agree with your solution of the DE. You've got the right basic functions, but your constants are off. The solution looks like

$y=Ae^{2x}+Be^{3x},$ and hence

$y'=2Ae^{2x}+3Be^{3x}.$ Plugging in the initial conditions yields the system of equations

$1=A+B$
$2=2A+3B.$

Can you proceed?

As for part b), I can see that you're using the equivalent of Troutman's Equation (4) on page 150 of Variational Calculus and Optimal Control, valid when x does not show up explicitly in the integrand of the functional.

Again, your problem is in solving the DE. You cannot simply integrate the way you have done, when y = y(x) is unknown. Your DE is

$y'-Ay=A.$

Try using the integrating factor method.