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I would like to find a continuous function $y : [0,4] \to \mathbb{R}$ that minimizes the following functional

$$I (y) := \displaystyle\int_{0}^4\sqrt{y\left(1+(y^{\prime})^2\right)} dx$$

subject to the boundary conditions $y (0) = 5/4$ and $y (4) = 13/4$. How do I solve this minimization problem? I tried and tried, but I can't get rid of the $y^{\prime}$. Whatever I do, I still have a big ugly equation with $y$ and $y^{\prime}$, and even if I change it to $\frac{dy}{dx}$, it doesn't get any better. Anyone has an idea?

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are u sure, there is y' instead of y? – Rut Sep 22 '12 at 19:01
Equation? What equation? I see none. Moreover, what are the limits of integration? You should integrate from $a$ to $b$ so that you obtain a scalar instead of a function, which is what a functional does. – Rod Carvalho Sep 22 '12 at 19:04
I checked it, that's how it's written in my book. Also, for an integral from 0 to 4, and y(0)=5/4, y(4)=13/4, the answer should be y = ((x-1)/2)^2 + 1, but like I said, I can't get rid of y' or y in order to solve the equation. – mimi Sep 22 '12 at 19:10
Sorry for being so not-precise. So, again, for an integral from 0 to 4, and y(0)=5/4, y(4)=13/4, the answer should be y = ((x-1)/2)^2 + 1, but like I said, I can't get rid of y' or y in order to solve the equation. – mimi Sep 22 '12 at 19:18
@mimi: Have you tried using the Euler-Lagrange equations? – Rod Carvalho Sep 22 '12 at 19:24

Since the integrand $F(y,y')$ does not depend on $x$ the Euler-Lagrange equation has a first integral

$$y'F_{y'}-F=C \tag{1}$$ where $$F=\sqrt{y(1+y'^2)}$$ $$F_{y'}=\frac{y'y^{1/2}}{\sqrt{1+y'^2}}$$ Inserting in (1): $$\frac{y'^2y^{1/2}}{\sqrt{1+y'^2}}-\sqrt{y(1+y'^2)}=C$$ $$\frac{y'^2y^{1/2}-y^{1/2}(1+y'^2)}{\sqrt{1+y'^2}}=C$$ $$\frac{y^{1/2}}{\sqrt{1+y'^2}}=C$$ $$\frac{y}{1+y'^2}=C^2$$ I like to use the following change of variable to solve this type of equations in parametric form: $$y'=\tan \phi$$ We derive $$y=\frac{C^2}{(\cos \phi)^2}$$ Now $$dx=\frac{dy}{y'}=2\frac{C^2\sin\phi}{(\cos\phi)^3}\frac{\cos\phi}{\sin\phi}d\phi=2С^2\frac{d\phi}{(\cos\phi)^2}$$ $$x=2C^2\tan \phi +A$$

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Isn't it $ F_{y'}=\frac{y'(1+y'^2)}{2 \sqrt{y(1+y'^2)}} $ ? – mimi Sep 22 '12 at 20:59
no, since the derivative is taken with respect to $y'$ with $y$ being regarded as constanе – Valentin Sep 22 '12 at 21:59
Sorry, I got lost in my papers. It's my fault, I forgot the prime. But how did you get $\frac{y}{1+y'^2}=C^2$ ? I got $ \frac{y(1 + y'^2)^2 - yy'^2}{1+y'^2}=C^2$ – mimi Sep 22 '12 at 22:50
I have filled in a few lines of calculation – Valentin Sep 22 '12 at 22:58
Ok, now it's clear. Thanks for the explanation. It's very important for me to understand getting to the y=.... solution (for my mathematical physics course). I'll try to find the non-parametric solution on my own (otherwise I'll ask for some more help). – mimi Sep 22 '12 at 23:33

Let $$\int\sqrt{y\left(1+(y^{\prime})^2\right)}dx=I$$ Then, $$\left(\frac{dI}{dx} \right)^2={y\left(1+\left (\frac{dy}{dx} \right)^2\right)}$$So,$$\left(dI \right)^2=y\left(\left(dx \right)^2 + \left(dy \right)^2\right)=y\left(dx \right)^2 + y\left(dy \right)^2$$Therefore, $$\int\left(\int dI\right) dI=\int\left(\int y dx\right) dx+\int\left(\int y dy \right)dy$$This will eventually give you, $${I}=\int_0^4{yx}dx+\int_0^4 \frac{y^2}2dy+constant$$Therefore, $$I=8y+\frac{32}3+constant$$

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I don't get it. I have to find the extremal of the functional. I have to find y. The integral is from 0 to 4 (I just didn't know how to put it right in the first post), and for y(0)=5/4, y(4)=13/4, the answer should be y = ((x-1)/2)^2 + 1 – mimi Sep 22 '12 at 19:32
But what about the Euler-Lagrange equation? – mimi Sep 22 '12 at 19:47

Turning the handle on the Euler-Lagrange equations, I get $$2yy''=1+(y')^2.$$ Let $u=y^{1/2}$ to write this as $$u''=\frac{1}{u^3}.$$ (This arises by trying to write the first and second derivative terms together as a derivative of $y^ky'$ for some $k$.) Multiply by $u'$ and integrate. This yields a first order equation for $u$ of the form $$ u'=\frac{\sqrt{cu^2-1}}{u},$$ which should be solveable.

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