Finding the Function F in the Euler-Lagrange Equation

[ephedyn]

Homework Statement

Find the function F in
J\left[y\right]={\displaystyle \int}_{a}^{b}F\left(x,y,y'\right)\ dx
such that the resulting Euler's equation is
f-\left(-\dfrac{d}{dx}\left(a\left(x\right)u'\right)\right)=0
for x\in\left(a,b\right) where a\left(x\right) and f\left(x\right) are given. Solve the equation in the special case a=0 , b=1 , a\left(x\right)\equiv 1, f\left(x\right)\equiv 1, u\left(a\right)=A, u\left(b\right)=B

Homework Equations

From the Euler-Lagrange equation,

F_{y}-\dfrac{d}{dx}F_{y'}=0

The Attempt at a Solution

we observe that F_{y}=f and F_{y'}=-a\left(x\right)u' or F_{y}=-f and F_{y'}=a\left(x\right)u' .

\dfrac{\partial F}{\partial y'}=a\left(x\right)u'\left(x\right)
\dfrac{\partial F}{\partial y}=-f\left(x\right)\implies F\left(x,y,y'\right)=-f\left(x\right)+c_{2}\qquad c_{2}\in\mathbb{R}

Suppose f\left(x\right)\equiv1 and a\left(x\right)\equiv1,

-\dfrac{d}{dx}u'=1\implies u'=-x
u\left(x\right)=-\dfrac{x^{2}}{2}
u\left(a=0\right)=-\dfrac{a^{2}}{2}=0=A
u\left(b=1\right)=-\dfrac{b^{2}}{2}=-\dfrac{1}{2}=B

Hence, we have A=0 and B=-\dfrac{1}{2}.

2 questions at this point... Sorry if this sounds silly, but what now? Also, did I get those 2 lines with the partial derivatives correct? How should I go around finding F after that?

Thanks in advance!

 

[lanedance]

are you using a as both and integration limit and a different function of x?

\dfrac{\partial F}{\partial y'}=a\left(x\right)u'\left(x\right)
\dfrac{\partial F}{\partial y}=-f\left(x\right)\implies F\left(x,y,y'\right)=-f\left(x\right)+c_{2}\qquad c_{2}\in\mathbb{R}

also that implies statement doesn't make sense to me, something better would be
\frac{\partial F}{\partial y}<br /> =-f(x)<br /> \implies \ F\left(x,y,y&#039;)= \ -f(x).y \ + \ g(x,y&#039;)
for some unknown function g

then you know
\dfrac{\partial F}{\partial y&#039;} <br /> = \dfrac{\partial }{\partial y&#039;} (-f(x).y \ + \ g(x,y&#039;)) <br /> = \frac{\partial g(x,y&#039;)}{\partial y&#039;}<br /> = a(x)u&#039;(x)
\implies <br /> g(x,y&#039;)<br /> = a(x)u&#039;(x)y&#039; \ + \ h(x) <br />

 

[ephedyn]

My professor has a very way of writing it, I suppose a(x) and the integration limit x=a refer to different 'a's. Let me read through the rest of your post for a while and think...

 

[lanedance]

now as for your problem

-\dfrac{d}{dx}u&#039;=1\implies u&#039;=-x
u\left(x\right)=-\dfrac{x^{2}}{2}
u\left(a=0\right)=-\dfrac{a^{2}}{2}=0=A
u\left(b=1\right)=-\dfrac{b^{2}}{2}=-\dfrac{1}{2}=B

Hence, we have A=0 and B=-\dfrac{1}{2}.

2 questions at this point... Sorry if this sounds silly, but what now? Also, did I get those 2 lines with the partial derivatives correct? How should I go around finding F after that?

Thanks in advance!

first you should get
-\dfrac{d}{dx}u&#039;=1\implies u&#039;=-x + c \ , c \in \mathhbb{R}
u\left(x\right)=-\dfrac{x^{2}}{2} + cx + d \ , c,d \in \mathhbb{R}

this will allow you to satisfy the given endpoints A & B (you are not meant to be solving for these you are meant to solve for u)

 

[ephedyn]

^Oops I did some very sloppy integration there, all right I think I have enough hints to solve this! Thanks a lot!