calculus of variations question

Question

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?

Answer 1

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, $$I=8y+\frac{32}3+constant$$

Answer 2

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

Answer 3

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.