How should I find the value of $a$ such that the functional can be written as the sum of two functionals?

Question

Let the two coupled Euler-Lagrange equations giving the stationary path of the functional $S[y_{1}, y_{2}]=\int [y_{1}'^2+2y_{2}'^2+(2y_{1}+y_{2})^2]dx$ be $y_{1}''-2(2y_{1}+y_{2})=0$ and $2y_{2}''-(2y_{1}+y_{2})=0$. By making the linear transformation $z_{1}=y_{1}+ay_{2}, z_{2}=2y_{1}+y_{2}$ to the new dependent variables $(z_{1}, z_{2})$, where $a$ is a constant, find the value of $a$ such that the functional can be written as the sum of two functionals, one of which depends only on $z_{1}$ and the other only on $z_{2}$.

Here's my work:

Consider the functional $S[y_{1}, y_{2}]=\int [y_{1}'^2+2y_{2}'^2+(2y_{1}+y_{2})^2]dx$ with the linear transformation $z_{1}=y_{1}+ay_{2}, z_{2}=2y_{1}+y_{2}$ to the new dependent variables $(z_{1}, z_{2})$, where $a$ is a constant.

Then $y_{1}=z_{1}-ay_{2}$ and $y_{2}=z_{2}-2y_{1}$.

This means $y_{1}'=z_{1}'-ay_{2}'$ and $y_{2}'=z_{2}'-2y_{1}'$.

Observe that $S[y_{1}, y_{2}]=\int [y_{1}'^2+2y_{2}'^2+(2y_{1}+y_{2})^2]dx\implies S[y_{1}, y_{2}]=\int [(z_{1}'-ay_{2}')^2+2(z_{2}'-2y_{1}')^2+(2y_{1}+y_{2})^2]dx\implies S[y_{1}, y_{2}]=\int (z_{1}'^2-2az_{1}'y_{2}'+a^2y_{2}'^2+2z_{2}'^2-8z_{2}'y_{1}'+8y_{1}'^2+4y_{1}^2+4y_{1}y_{2}+y_{2}^2)dx.$

From here I'm stuck. How should I find the value of $a$?

Answer 1

We have the functional $S$ given by

$$S[y_1,y_2]=\int_{x_0}^{x_1}(y_1'(x)^2+2y_2'(x)^2+(2y_1(x)+y_2(x))^2)\,\mathrm{d}x.$$

We make the change of variables

$$\begin{pmatrix} z_1\\ z_2 \end{pmatrix}=\begin{pmatrix} 1&a\\ 2&1 \end{pmatrix}\begin{pmatrix} y_1\\ y_2 \end{pmatrix},$$

where $a\in\mathbb{R}\setminus\left\{\frac{1}{2}\right\}$. The square matrix above can easily be inverted to give us that

$$\begin{pmatrix} y_1\\ y_2 \end{pmatrix}=\frac{1}{2a-1}\begin{pmatrix} -1&a\\ 2&-1 \end{pmatrix}\begin{pmatrix} z_1\\ z_2 \end{pmatrix}.$$

Observing quickly that

$$S[\alpha y_1,\alpha y_2]=\alpha^2S[y_1,y_2]$$

for all $\alpha\in\mathbb{R}$ we compute

$$S[y_1,y_2]=\frac{1}{(2a-1)^2}S[-z_1+az_2,2z_1-z_2].$$

Furthermore then

\begin{align*} S[-z_1+az_2,2z_1-z_2] &=\int_{x_0}^{x_1}((-z_1'(x)+az_2'(x))^2+2(2z_1'(x)-z_2'(x))^2+(2a-1)^2z_2(x)^2)\,\mathrm{d}x \\ &=\int_{x_0}^{x_1}(9z_1'(x)^2+(a^2+2)z_2'(x)^2+(2a-1)^2z_2^2(x)-2(a+4)z_1'(x)z_2'(x))\,\mathrm{d}x. \end{align*}

Now for this to be able to be written as the sum of two functionals where only depends only on $z_1$ and the other only on $z_2$, it must be the case that the term $2(a+4)z_1'(x)z_2'(x)$ vanishes for all $z_1',z_2'$, and so it must be the case that $a=-4$. With $a=-4$ we then also get that

$$S[y_1,y_2]=\frac{1}{9}\left(\int_{x_0}^{x_1}z_1'(x)^2\,\mathrm{d}x+\int_{x_0}^{x_1}\left(2z_2'(x)^2+9z_2(x)^2\right)\,\mathrm{d}x\right),$$

giving you the two desired functionals as well.

Answer 2

(Partial answer)

You got stuck because you stopped halfway when substituting the old variables with the new ones. Indeed, let's take $y_1 = z_1 - ay_2$. You have replaced $y_1$ by $z_1$, but $y_2$ is still present in the expression. You can go on as follows : $y_1 = z_1 - ay_2 = z_1 - a(z_2 - 2y_1)$, hence $(1+2a)y_1 = z_1 - az_2$ by isolating $y_1$ and finally $y_1 = \frac{z_1-az_2}{1+2a}$. And $y_2$ is determined in a similar way.

I must admit I haven't carried the computations until the end, but the transformation you consider might be not general enough to permit decouple the two variables inside the functional. Actually, they will be separated by the same transformation which decouples the equations of motion, i.e. $$ \frac{\mathrm{d}^2}{\mathrm{d}t^2} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}, $$ by diagonalizing the matrix in front of dependent variables.