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
$$\frac{\partial F}{\partial y} =-f(x) \implies \ F\left(x,y,y')= \ -f(x).y \ + \ g(x,y')$$
for some unknown function g

then you know
$$\dfrac{\partial F}{\partial y'} = \dfrac{\partial }{\partial y'} (-f(x).y \ + \ g(x,y')) = \frac{\partial g(x,y')}{\partial y'} = a(x)u'(x)$$
$$\implies g(x,y') = a(x)u'(x)y' \ + \ h(x)$$

 

[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

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

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!

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

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!