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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?

Rod Carvalho
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mimi
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  • 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
  • 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? http://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation – Rod Carvalho Sep 22 '12 at 19:24
  • Of course I did. I just got stuck at one point. Also, if the function is f=f(y,y'), the Euler-Lagrange equation changes into f - y'* df/dy = c (c=const.) right? – mimi Sep 22 '12 at 19:43
  • What about $y=0$? Note that in order to take a square root, $y(\sqrt{(y')^2+1})$ must be non-negative (otherwise you have a complex valued function, which upon integration may or may not give a real number). – Baby Dragon Sep 22 '12 at 20:01
  • @BabyDragon That's not the problem. – mimi Sep 22 '12 at 20:28
  • are u sure, there is y' instead of y? – Rutvij Kotecha Sep 22 '12 at 19:01
  • 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
  • @Rut welcome to math.se! Posts of this nature should usually be comments on the main question rather than answers. The latter are generally reserved for complete solutions to the problem at hand. – Andrew Christianson Sep 22 '12 at 19:47
  • I see, you have some endpoints. – Baby Dragon Feb 11 '13 at 18:42

3 Answers3

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

Community
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Valentin
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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
  • How can the x in your result depend on x? – mimi Sep 23 '12 at 16:53
  • it's $\phi$ indeed, corrected – Valentin Sep 23 '12 at 17:53
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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, $$I=8y+\frac{32}3+constant$$

Ayush Khemka
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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
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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.

user12477
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