# Variation Problem with Euler-Lagrange Differential Equation

I'm just trying to understand that type of equations, and I can't solve this, kind of a simple minimization problem. Maybe someone can help me ? Here is my equation

\eqalign{ & A(y(x)) = \int_{ - 1}^1 {{{\sqrt {1 + {{\left[ {y'(x)} \right]}^2}} } \over {y(x)}}dx} \cr & y( - 1) = 1 \cr & y(1) = 1 \cr}

Thanks!

• welcome to MSE. your problem is not clearly stated. :) – H. R. Nov 7 '15 at 15:03
• Try to edit your question. :) It is not clear that what is your problem. :) – H. R. Nov 7 '15 at 15:06
• @H. R. I presume he's trying to find a function y with the specified Dirichlet boundary, which minimises the functional A (presumably A stands for Action). – p0llard Nov 7 '15 at 15:07
• @Edgar What are you going to do with $A(y)$? Minimize it maybe? – A.Γ. Nov 7 '15 at 15:07
• Use the Euler-Lagrange equation. With $L(y,y') = \sqrt{1+y'^2}/y$ the EL equation is $\frac{dL}{dy} - \frac{d}{dx}\frac{dL}{dy'} = 0$. This will give you a ODE that you need to solve. – Winther Nov 7 '15 at 15:17

Here the Lagrangian $L(t,y,v)=\frac{\sqrt{1+v^2}}{y}$ does not depend directly on the variable $t$, in this case is useful to apply the second formulation of Euler-Lagrange equation:

$$L(y(t),y'(t))-y'L_v(y(t),y'(t))=cost.$$

In your case it produces the following differential equation $$y(t)\sqrt{1+y'(t)^2}=cost.=C$$

We can rewrite the previous equation in the form:

$$\pm \frac{y'(t)y(t)}{\sqrt{C^2-y(t)^2}}=1$$ and this is equivalent to $$\pm\big(\sqrt{C^2-y(t)^2}\big)'=1.$$ Hence $$\pm \sqrt{C^2-y(t)^2}=t+D$$ and finally squaring we get $$C^2-y(t)^2=(t+D)^2.$$ Now just plug in the conditions $y(1)=y(-1)=1$ to get $D=0$ and $C^2=2$.

• It applies if $\frac{dL}{dx}=0$ and that's the case here. – mrprottolo Nov 7 '15 at 15:52
• I meant the partial derivative respect to $x$ :$\frac{\partial L}{\partial x}=0$. – mrprottolo Nov 7 '15 at 15:58
• Sorry, I confused the variables. You are of course correct. – Winther Nov 7 '15 at 16:00
• answer : "x^2+y^2=2" but i still have problems, probably in the first steps – Edgar Nov 7 '15 at 16:00
• $x^2+y(x)^2=2$ is indeed a solution of the differential equation I wrote. – mrprottolo Nov 7 '15 at 16:06

Just to make a complementing remark to @mrprottolo's solution.

It happens in some variational problems that the Beltrami identity gives more complicated equations to solve than the canonical E-L equation. This problem is one of the examples. If we take the classical E-L we get after elementary simplifications that $$y'(1+y'^2+yy'')=0,\quad y(-1)=y(1)=1.$$ It gives one possibility $y'=0$ and another to be $$1+y'^2+yy''=1+(yy')'=1+\frac{1}{2}(y^2)''=0\qquad\Leftrightarrow\qquad (y^2)''=-2.$$ Both DE's are very elementary (without square roots etc).