Hey all

I do Calculas of Variations, and have an exam in around nine hours, so would appreciate help before then. After is good too thhough.

I got a solution, but it has a bit I don't understand, was hoping you guys could help.

My problem is:

Find extremal of J(y)=$\displaystyle \int_0^1(y(x))dx$,subject to $\displaystyle \int_0^1\sqrt(1+y'(x)^2)dx = L$.

$\displaystyle y(0) = y(1) =0$

Solution that I have gotten is:

$\displaystyle F'= F + \lambda*G$

$\displaystyle F= y(x), G =\sqrt(1+(y')^2).$

$\displaystyle \frac{dF*}{dy}-\frac{d}{dx}(\frac{dF*}{dy'})=0$

$\displaystyle 1-\frac{d}{dx}(\lambda*y'/\sqrt(1+(y')^2)$

Integrate:

$\displaystyle \rightarrow \lambda*y'/\sqrt(1+(y')^2)=x+C $ C arbitary constant

so

$\displaystyle (y')^2/(1+(y')^2)=(x+C)^2/\lambda^2$

$\displaystyle y'^2 = ((x+c)^2/\lambda^2)/(1-(x+C)^2/\lambda^2);$

$\displaystyle y'=(x+C/\lambda)/(\sqrt(1+(x+C)^2)/\lambda^2).$

$\displaystyle u=1-(x+C)^2/\lambda^2$

Put u in, integrate, change the result around a small bit and you get

$\displaystyle (x+C)^2/\lambda^2+(y+D)^2/\lambda^2=1 $ D arbitary constant

inset bound conditions, you get $\displaystyle C = -1/2$

$\displaystyle D=\sqrt(\lambda^2-1/4) =\sqrt(\lambda^2-C^2)$

So, $\displaystyle (x-1/2)^2+(y+\sqrt(\lambda^2-1/4)^2=\lambda^2$

Diffretiate with respect to x;

$\displaystyle (x+1/2)+(y+\sqrt(\lambda^2-1/4)*y'=0

=> y'=-(x-1/2)/(y+\sqrt(\lambda^2-1/4).$

Then (And herin lies the problem), I get

$\displaystyle 1+y'^2=1+(x-1/2)^2/((y+\sqrt(\lambda^2-1/4))^2) = \lambda^2/(\lambda^2-(x-1/2)^2.$

How does he change $\displaystyle 1+(x-1/2)^2/((y+\sqrt(\lambda^2-1/4))^2) $ into $\displaystyle \lambda^2/(\lambda^2-(x-1/2)^2)?$

Thanks in advance.