Let $P_1=(a,y_a),P_2=(b,y_b), y\in C^1 (a,b), y_a>0,y_b>0$

And the area integral: $\int^b_a y(x) \sqrt{1+y'(x)}dx$

From the Euler differential-equation we obtain:

$$y'=1/\alpha \sqrt{y^2-\alpha^2}, \quad \alpha\in \mathbb R_0$$

Now the Author Book (Hans Sagan,Introduction to the calculus of Variations) concludes "Separation of Variables and substitution of $y=\alpha \cosh(t)$ yield

$$\alpha t+\beta =x$$ and hence,

$$y=\alpha \left(\cosh\left(\frac{(x-\beta)}{\alpha}\right)\right)$$

I can't reproduce this results, maybe you can see where i do go wrong:

$$y'=1/\alpha \sqrt{y^2-\alpha^2}, \quad \alpha\in \mathbb R_0$$

I do separate the variables:

$$\int \frac{dy}{\sqrt{y^2-1}}=\int dx$$

$\Longleftrightarrow \operatorname{arccosh}(y/\alpha)+\beta=x$, $\beta\in \mathbb R$ now theres obviously an $\alpha$ missing

but lets just go on and substitute like he does:" $y=\alpha \cosh(t)$"

$$\Longleftrightarrow \mbox{arccosh}(\alpha \cosh(t/\alpha )\alpha)+\beta=\mbox{arccosh}(t)+\beta=x$$

$\Longleftrightarrow t=\cosh(x-\beta)$ Which is wrong.

Also shouldn't he calculate the solution with the help of the limits a,b of the integral? I guess he just uses $\alpha$ and $\beta$ , because thats more convient for him in this case.


Tiny mistakes. First of all, separation of variables should read $$ \alpha \int \frac{dy}{\sqrt{y^2-\alpha^2}} = \int dx $$ Taking $y = \alpha \cosh (t)$, then $dy = \alpha \sinh (t) \, dt$ and the integral becomes $$ \alpha\int \frac{dy}{\sqrt{y^2-\alpha^2}} = \alpha \int dt = \alpha t + \beta = \alpha \,\mbox{arccosh} \big(\frac{y}{\alpha}\big) + \beta, $$ hence $$ \alpha\,\mbox{arccosh}\big(\frac{y}{\alpha}\big) + \beta = x \Longleftrightarrow \mbox{arccosh}\big(\frac{y}{\alpha}\big) = \frac{x-\beta}{\alpha} $$ Finally $$ y = \alpha \cosh\left(\frac{x-\beta}{\alpha}\right). $$

About the limits, I guess you are right. It might be more convenient to do it this way.


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