# Finding the Euler Lagrange equation - differentiation

I'm teachin myself the basics of Calculus of variations. So far I know how to calculate the Euler Lagrange equation for simple functionals. I'm now stuck on how to compute the total differentiation of the following problem: $$I[y]=\int_0^1 (y\frac{dy}{dx})^2 -\lambda y^2 \ dx$$

To calculate the Euler Lagrange equation I have the following: $$\frac{d}{dx}(2y^2y')-2y^2\frac{dy}{dx}+2\lambda y=0$$.

Is this correct? If so I'm unsure of how to evaluate the total differentiation part. That is taking the total derivative in this case.

For a functional $$I[y]=\int_a^bf(x,y,y')$$ the Euler-Lagrange equation is defined as $$\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)=0$$

For the $I$ you gave above, $f(x,y,y')=(yy')^2-\lambda y^2$, hence the E-L equation for $I$ is $$2y\left(\frac{dy}{dx}\right)^2-2\lambda y-\frac{d}{dx}\left(2y^2\frac{dy}{dx}\right)=0$$ Your only mistake is that $\frac{\partial f}{\partial y}=2\color{red}{y(y')^2}-2\lambda y$, not $2\color{red}{y^2y'}-2\lambda y$ as you have written. Does this help you go from here?

Edit. Since $f$ does not depend upon $x$, we can simplify the problem.

Lemma. For a functional $$I[y]=\int_a^b f(y,y')$$ which is independent of $x$, we have that $$\frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}-f\right)=0$$

Proof. Just calculate the derivative $$\frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}-f\right)=y''\frac{\partial f}{\partial y'}+y'\frac{d}{x}\frac{\partial f}{\partial y'}-\frac{\partial f}{\partial x}-y'\frac{\partial f}{\partial y}-y''\frac{\partial y}{\partial y'}$$ Since $f$ is independent of $x$, $\frac{\partial f}{\partial x}=0$, and we can take out a factor of $y'$ $$\frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}-f\right)=y'\left(\frac{d}{dx}\frac{\partial f}{\partial y'}-\frac{\partial f}{\partial y}\right)$$ which is clearly equal to $0$ by E-L. QED

This means that $$y'\frac{\partial f}{\partial y'}-f=\text{constant}$$ Applying this to your function $I$ gives \begin{align*} y'(2y^2y')-(yy')^2+\lambda y^2&=C \\ y^2\left(\left(\frac{dy}{dx}\right)^2+\lambda\right)&=C \\ \frac{dy}{dx}&=\sqrt{\frac{C}{y^2}-\lambda} \end{align*}

• Yes, thank you. My main problem now is to evaluate the total derivative part of the Euler Lagrange. Do I treat y and y' as separate variables again? – user45503 May 17 '13 at 15:13
• @user45503 Do you mean how to calculate $\frac{dy}{dx}$? I've edited my answer. – Warren Moore May 17 '13 at 15:44