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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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    I see, you have some endpoints. โ€“ 2013-02-11

3 Answers 3

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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 $dx=\frac{dy}{y'}=2\frac{C^2\sin\phi}{(\cos\phi)^3}\frac{\cos\phi}{\sin\phi}d\phi=2ะก^2\frac{d\phi}{(\cos\phi)^2}$ $x=2C^2\tan \phi +A$

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    it's $\phi$ indeed, corrected โ€“ 2012-09-23
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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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    But what about the Euler-Lagrange equation? โ€“ 2012-09-22
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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.