Lagrangian for two coupled second order linear differential equations

Question

Consider a system of two coupled linear differential equations $$ \left( \begin{bmatrix} \Omega \end{bmatrix}^{-1} + \frac{d^2}{dt^2} \right) \vec{V}(t) = \begin{bmatrix} C \end{bmatrix}^{-1} \vec{J}(t) + \begin{bmatrix} \Omega \end{bmatrix}^{-1} \vec{K}(t) $$ where $\vec{V}(t)$ is a two-element vector describing the degree of freedom of the system, $\vec{J}(t)$ and $\vec{K}(t)$ are drive sources, and $[\Omega]^{-1}$ and $[C]^{-1}$ are constant 2x2 matrices. This system represents two coupled harmonic resonators with time-dependent (but position independent) drive forces. For whatever it's worth, suppose we can decompose $[\Omega]^{-1}$ as $$ [\Omega]^{-1} = [C]^{-1}[L]^{-1}$$ where $[L]^{-1}$ is another 2x2 matrix$^{[1]}$. Both $[L]$ and $[C]$ are symmetric.

Is there a systematic way to find the Lagrangian for this system of equations?

[1]: Both $[C]$ and $[L]$ have the property that their off-diagonal elements are smaller than their diagonal elements, which is probably useful for approximations.

Answer 1

$\boldsymbol{\S}$ A. A special case : symmetric $\Omega^{\boldsymbol{-}1}$

Let the $2\times2$ real symmetric matrices \begin{equation} C^{\boldsymbol{-}1}\boldsymbol{=} \begin{bmatrix} \xi_1 & \xi \vphantom{\dfrac{a}{b}}\\ \xi &\xi_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \quad \text{and} \quad L^{\boldsymbol{-}1}\boldsymbol{=} \begin{bmatrix} \eta_1 & \eta \vphantom{\dfrac{a}{b}}\\ \eta &\eta_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-01}\label{A-01} \end{equation} Then \begin{equation} \Omega^{\boldsymbol{-}1}\boldsymbol{=}C^{\boldsymbol{-}1}L^{\boldsymbol{-}1}\boldsymbol{=} \begin{bmatrix} \xi_1\eta_1 \boldsymbol{+} \xi\eta & \xi_1\eta \boldsymbol{+} \xi\eta_2 \vphantom{\dfrac{a}{b}}\\ \hphantom{_1}\hphantom{_2}\xi\eta_1 \boldsymbol{+} \xi_2\eta & \hphantom{_1}\hphantom{_2}\xi\eta \boldsymbol{+}\xi_2\eta_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-02}\label{A-02} \end{equation} With respect to the coordinates \begin{equation} \mathbf{V} \boldsymbol{=} \begin{bmatrix} V_1\vphantom{\dfrac{a}{b}}\\ V_2\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-03}\label{A-03} \end{equation}
the two coupled equations are \begin{equation} \dfrac{\mathrm d}{\mathrm dt}\left(\mathbf{\dot{V}}\right)\boldsymbol{-}\left(C^{\boldsymbol{-}1}\mathbf{J}\boldsymbol{+}\Omega^{\boldsymbol{-}1}\mathbf{K}\boldsymbol{-}\Omega^{\boldsymbol{-}1}\mathbf{V}\right)\boldsymbol{=}\boldsymbol{0} \tag{A-04}\label{A-04} \end{equation} Now, if there exists a Lagrangian $\mathrm L\left(\mathbf{V},\mathbf{\dot{V}},t\right)$ for the problem then the Euler-Lagrange equations are \begin{equation} \dfrac{\mathrm d}{\mathrm dt}\left(\dfrac{\partial \mathrm L}{\partial \mathbf{\dot{V}}}\right)\boldsymbol{-}\dfrac{\partial \mathrm L}{\partial \mathbf{V}}\boldsymbol{=}\boldsymbol{0} \tag{A-05}\label{A-05} \end{equation} where \begin{equation} \dfrac{\partial \mathrm L}{\partial \mathbf{V}}\boldsymbol{=} \begin{bmatrix} \dfrac{\partial \mathrm L}{\partial V_1} \vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \dfrac{\partial \mathrm L}{\partial V_2} \vphantom{\dfrac{a}{b}} \end{bmatrix} \quad \text{and} \quad \dfrac{\partial \mathrm L}{\partial \mathbf{\dot{V}}}\boldsymbol{=} \begin{bmatrix} \dfrac{\partial \mathrm L}{\partial \dot{V}_1} \vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \dfrac{\partial \mathrm L}{\partial \dot{V}_2} \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-06}\label{A-06} \end{equation} Comparing equations \eqref{A-04} and \eqref{A-05} we note that the Lagrangian $\mathrm L\left(\mathbf{V},\mathbf{\dot{V}},t\right)$ must satisfy, except constants, the following two equations \begin{align} \dfrac{\partial \mathrm L}{\partial \mathbf{\dot{V}}} & \boldsymbol{=}\mathbf{\dot{V}}\vphantom{\dfrac{a}{\dfrac{a}{b}}} \tag{A-07a}\label{A-07a}\\ \dfrac{\partial \mathrm L}{\partial \mathbf{V}} & \boldsymbol{=}C^{\boldsymbol{-}1}\mathbf{J}\boldsymbol{+}\Omega^{\boldsymbol{-}1}\mathbf{K}\boldsymbol{-}\Omega^{\boldsymbol{-}1}\mathbf{V} \tag{A-07b}\label{A-07b} \end{align} From equation \eqref{A-07a} and partly because of the first two terms in the rhs of equation \eqref{A-07b} we note that one part $\mathrm L_1\left(\mathbf{V},\mathbf{\dot{V}},t\right)$ of the Lagrangian would be \begin{equation} \mathrm L_1\left(\mathbf{V},\mathbf{\dot{V}},t\right)\boldsymbol{=}\frac12\left(\mathbf{\dot{V}}\boldsymbol{\cdot}\mathbf{\dot{V}}\right)\boldsymbol{+}\left[\left(C^{\boldsymbol{-}1}\mathbf{J}\right)\boldsymbol{\cdot}\mathbf{V}\right]\boldsymbol{+}\left[\left(\Omega^{\boldsymbol{-}1}\mathbf{K}\right)\boldsymbol{\cdot}\mathbf{V}\right] \tag{A-08}\label{A-08} \end{equation} while a second part $\mathrm L_2\left(\mathbf{V},\mathbf{\dot{V}},t\right)$ of the Lagrangian must satisfy the equation \begin{equation} \dfrac{\partial \mathrm L_2}{\partial \mathbf{V}} \boldsymbol{=}\boldsymbol{-}\Omega^{\boldsymbol{-}1}\mathbf{V} \tag{A-09}\label{A-09} \end{equation} If the matrix $\Omega^{\boldsymbol{-}1}$ of equation \eqref{A-02} is symmetric, that is if the elements of the matrices $C^{\boldsymbol{-}1}$ and $L^{\boldsymbol{-}1}$ satisfy the condition \begin{equation} \left(\xi_1\boldsymbol{-}\xi_2\right)\eta\boldsymbol{=}\left(\eta_1\boldsymbol{-}\eta_2\right)\xi \tag{A-10}\label{A-10} \end{equation} then \begin{equation} \mathrm L_2\left(\mathbf{V},\mathbf{\dot{V}},t\right) \boldsymbol{=}\boldsymbol{-}\frac12\left[\left(\Omega^{\boldsymbol{-}1}\mathbf{V}\right)\boldsymbol{\cdot}\mathbf{V}\right] \tag{A-11}\label{A-11} \end{equation} and so \begin{align} &\mathrm L\left(\mathbf{V},\mathbf{\dot{V}},t\right) \boldsymbol{=}\mathrm L_1\left(\mathbf{V},\mathbf{\dot{V}},t\right)\boldsymbol{+}\mathrm L_2\left(\mathbf{V},\mathbf{\dot{V}},t\right) \qquad \textbf{for symmetric } \Omega^{\boldsymbol{-}1} \nonumber\\ & \boldsymbol{=}\frac12\left(\mathbf{\dot{V}}\boldsymbol{\cdot}\mathbf{\dot{V}}\right)\boldsymbol{-}\frac12\left[\left(\Omega^{\boldsymbol{-}1}\mathbf{V}\right)\boldsymbol{\cdot}\mathbf{V}\right]\boldsymbol{+}\left[\left(C^{\boldsymbol{-}1}\mathbf{J}\right)\boldsymbol{\cdot}\mathbf{V}\right]\boldsymbol{+}\left[\left(\Omega^{\boldsymbol{-}1}\mathbf{K}\right)\boldsymbol{\cdot}\mathbf{V}\right] \tag{A-12}\label{A-12} \end{align}

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

$\boldsymbol{\S}$ B. The general case : A systematic way to find the Lagrangian for two coupled second order linear differential equations

The efforts to find a Lagrangian for two coupled second order linear differential equations (as in the question) would be unsuccessful because of the so called $^{\prime\prime}$cross terms$^{\prime\prime}$ that appear at an intermediate step , for example terms like $V_1 V_2, \dot{V}_1 \dot{V}_2, \dot{V}_1 V_2$ etc. These terms "couple" the two equations. So we must find a method to eliminate terms of this kind. This will give us at first two uncoupled second order linear differential equations and next a well-defined Lagrangian.

Because of linearity we make a change of the variables from old $V_1, V_2$ to new $q_1, q_2$ via a linear transformation \begin{align} V_1 & \boldsymbol{=}a_{11}q_1\boldsymbol{+}a_{12}q_2 \tag{B-01a}\label{B-01a}\\ V_2 & \boldsymbol{=}a_{21}q_1\boldsymbol{+}a_{22}q_2 \tag{B-01b}\label{B-01b} \end{align} or \begin{equation} \mathbf{V}\boldsymbol{=} \begin{bmatrix} V_1\vphantom{\dfrac{a}{b}}\\ V_2\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} a_{11} & a_{12}\vphantom{\dfrac{a}{b}}\\ a_{21} & a_{22}\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} p_1\vphantom{\dfrac{a}{b}}\\ p_2\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=}A\mathbf{q} \tag{B-02}\label{B-02} \end{equation}
that is \begin{equation} \mathbf{V}\boldsymbol{=}A\mathbf{q} \,,\qquad A\boldsymbol{=} \begin{bmatrix} a_{11} & a_{12}\vphantom{\dfrac{a}{b}}\\ a_{21} & a_{22}\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{B-03}\label{B-03} \end{equation} and we'll try to find, if there exists, an invertible transformation $\:A\:$ that eliminates the cross terms so uncoupling the two equations.

If on our initial equation
\begin{equation} \mathbf{\ddot{V}}\boldsymbol{+}\Omega^{\boldsymbol{-}1}\mathbf{V}\boldsymbol{=}C^{\boldsymbol{-}1}\mathbf{J}\boldsymbol{+}\Omega^{\boldsymbol{-}1}\mathbf{K} \tag{B-04}\label{B-04} \end{equation} we apply from the left the transformation $\:A^{\boldsymbol{-}1}\:$ we have \begin{equation} A^{\boldsymbol{-}1}\mathbf{\ddot{V}}\boldsymbol{+}A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1}\mathbf{V}\boldsymbol{=}A^{\boldsymbol{-}1}C^{\boldsymbol{-}1}\mathbf{J}\boldsymbol{+}A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1}\mathbf{K} \tag{B-05}\label{B-05} \end{equation} Making use of \eqref{B-03} we replace $\:\mathbf{V}\:$ by $\:A\mathbf{q}\:$ so \begin{equation} A^{\boldsymbol{-}1}\left(A\mathbf{\ddot{q}}\right)\boldsymbol{+}A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1}\left(A\mathbf{q}\right)\boldsymbol{=}A^{\boldsymbol{-}1}C^{\boldsymbol{-}1}\mathbf{J}\boldsymbol{+}A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1}\mathbf{K} \nonumber \end{equation} that is \begin{equation} \mathbf{\ddot{q}}\boldsymbol{+}\left(A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1} A\right)\mathbf{q}\boldsymbol{=}\left(A^{\boldsymbol{-}1}C^{\boldsymbol{-}1 }A\right)\mathbf{j}\boldsymbol{+}\left(A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1 }A\right)\mathbf{k} \tag{B-06}\label{B-06} \end{equation} or \begin{align} &\mathbf{\ddot{q}}\boldsymbol{+}W\,\mathbf{q} \boldsymbol{=}U\,\mathbf{j}\boldsymbol{+}W\,\mathbf{k} \tag{B-07a}\label{B-07a}\\ &\text{where} \nonumber\\ &W\boldsymbol{=}A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1}A\,, \quad U\boldsymbol{=}A^{\boldsymbol{-}1}C^{\boldsymbol{-}1}A\,, \quad \mathbf{j}\boldsymbol{=}A^{\boldsymbol{-}1}\mathbf{J}\,,\quad \mathbf{k}\boldsymbol{=}A^{\boldsymbol{-}1}\mathbf{K} \tag{B-07b}\label{B-07b} \end{align} Now, the two second order linear differential equations \eqref{B-07a} would be uncoupled if the matrix $\:W\:$ could be diagonal \begin{equation} W\boldsymbol{=}A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1 }A\boldsymbol{=} \begin{bmatrix} \mathrm w_1 & 0 \vphantom{\dfrac{a}{b}}\\ 0 & \mathrm w_2\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{B-08}\label{B-08} \end{equation} This uncoupling is shown explicitly below \begin{align} \ddot{q}_1\boldsymbol{+}\mathrm w_1 p_1 &\boldsymbol{=}\left(U\,\mathbf{j}\right)_1 \boldsymbol{+}\left(W\,\mathbf{k}\right)_1 \tag{B-09a}\label{B-09a}\\ \ddot{q}_2\boldsymbol{+}\mathrm w_2 p_2 &\boldsymbol{=}\left(U\,\mathbf{j}\right)_2 \boldsymbol{+}\left(W\,\mathbf{k}\right)_2 \tag{B-09b}\label{B-09b} \end{align} These two independent $^{\prime\prime}$motions$^{\prime\prime}$ are called normal modes and the variables $q_1,q_2$ normal coordinates.

Now, from \eqref{B-08} the constants $\:\mathrm w_1,\mathrm w_2\:$ are the eigenvalues of the matrix $\:\Omega^{\boldsymbol{-}1}\:$ while the columns of the matrix $\:A\:$ are the eigenvectors respectively \begin{align} \mathbf{a}_1 & \boldsymbol{=} \begin{bmatrix} a_{11} \vphantom{\dfrac{a}{b}}\\ a_{21} \vphantom{\dfrac{a}{b}} \end{bmatrix}\boldsymbol{=}\text{eigenvector of eigenvalue } \mathrm w_1 \tag{B-10a}\label{B-10a}\\ \mathbf{a}_2 & \boldsymbol{=} \begin{bmatrix} a_{12} \vphantom{\dfrac{a}{b}}\\ a_{22} \vphantom{\dfrac{a}{b}} \end{bmatrix}\boldsymbol{=}\text{eigenvector of eigenvalue } \mathrm w_2 \tag{B-10b}\label{B-10b} \end{align} Note that depending on the matrix $\:\Omega^{\boldsymbol{-}1}\:$ the eigenvalues $\:\mathrm w_1,\mathrm w_2\:$ could be either both real or both complex conjugates.

Now, since the diagonal matrix $\:W\:$ is symmetric we make use of the results of $\boldsymbol{\S}$ A and we build the Lagrangian for the Euler-Lagrange equations \eqref{B-09a},\eqref{B-09b} according to equation \eqref{A-12}
\begin{equation} \mathrm L\left(\mathbf{q},\mathbf{\dot{q}},t\right) \boldsymbol{=} \tfrac12\left(\mathbf{\dot{q}}\boldsymbol{\cdot}\mathbf{\dot{q}}\right)\boldsymbol{-}\tfrac12\left[\left(W\mathbf{q}\right)\boldsymbol{\cdot}\mathbf{q}\vphantom{\dfrac{a}{b}}\right]\boldsymbol{+}\left[\left(U\mathbf{j}\right)\boldsymbol{\cdot}\mathbf{q}\vphantom{\dfrac{a}{b}}\right]\boldsymbol{+}\left[\left(W\mathbf{k}\right)\boldsymbol{\cdot}\mathbf{q}\vphantom{\dfrac{a}{b}}\right] \tag{B-11}\label{B-11} \end{equation} Explicitly \begin{align} \mathrm L\left(\mathbf{q},\mathbf{\dot{q}},t\right) & \boldsymbol{=} \tfrac12\left(\dot{q}^2_1\boldsymbol{+}\dot{q}^2_2\right)\boldsymbol{-}\tfrac12\left(\mathrm w_1 q^2_1\boldsymbol{+}\mathrm w_2 q^2_2\right) \tag{B-12}\label{B-12}\\ &\boldsymbol{+} \left[\left(U\mathbf{j}\right)_1\boldsymbol{+}\left(W\mathbf{k}\right)_1\vphantom{\dfrac{a}{b}}\right]q_1\boldsymbol{+} \left[\left(U\mathbf{j}\right)_2\boldsymbol{+}\left(W\mathbf{k}\right)_2\vphantom{\dfrac{a}{b}}\right]q_2 \nonumber \end{align} Note that the above Lagrangian doesn't contain $^{\prime\prime}$cross terms$^{\prime\prime}$ like $q_1 q_2, \dot{q}_1 \dot{q}_2, \dot{q}_1 q_2$ etc. Use of this Lagrangian in the equations below \begin{align} \dfrac{\mathrm d}{\mathrm dt}\left(\dfrac{\partial \mathrm L}{\partial \dot{q}_1}\right)\boldsymbol{-}\dfrac{\partial \mathrm L}{\partial q_1}\boldsymbol{=}0 \tag{B-13a}\label{B-13a}\\ \dfrac{\mathrm d}{\mathrm dt}\left(\dfrac{\partial \mathrm L}{\partial \dot{q}_2}\right)\boldsymbol{-}\dfrac{\partial \mathrm L}{\partial q_2}\boldsymbol{=}0 \tag{B-13b}\label{B-13b} \end{align} yields equations \eqref{B-09a} and \eqref{B-09b} as expected.

Now, based on \eqref{B-11} we can build the Lagrangian $\:\mathrm L\left(\mathbf{V},\mathbf{\dot{V}},t\right)\:$ for the initial coordinates $\:V_1,V_2\:$ from $\:\mathrm L\left(\mathbf{q},\mathbf{\dot{q}},t\right)$. We simply replace $\:\mathbf{q}\:$ by $\:A^{\boldsymbol{-}1}\mathbf{V}\:$ in \eqref{B-11} and we have \begin{align} &\mathrm L\left(\mathbf{V},\mathbf{\dot{V}},t\right)\boldsymbol{=} \tag{B-14}\label{B-14}\\ &\tfrac12\left[\left(A^{\boldsymbol{-}1}\mathbf{\dot{V}}\right)\boldsymbol{\cdot}\left(A^{\boldsymbol{-}1}\mathbf{\dot{V}}\right)\vphantom{\dfrac{a}{b}}\right]\boldsymbol{-}\tfrac12\left[\left(A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1}\mathbf{V}\right)\boldsymbol{\cdot}\left(A^{\boldsymbol{-}1}\mathbf{V}\right)\vphantom{\dfrac{a}{b}}\right] \nonumber\\ &\boldsymbol{+}\left[\left(A^{\boldsymbol{-}1}C^{\boldsymbol{-}1}\mathbf{J}\right)\boldsymbol{\cdot}\left(A^{\boldsymbol{-}1}\mathbf{V}\right)\vphantom{\dfrac{a}{b}}\right]\boldsymbol{+}\left[\left(A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1}\mathbf{K}\right)\boldsymbol{\cdot}\left(A^{\boldsymbol{-}1}\mathbf{V}\right)\vphantom{\dfrac{a}{b}}\right] \nonumber \end{align} If $\:\Omega^{\boldsymbol{-}1}\:$ is (real) symmetric then the Lagrangian of \eqref{B-14} must yield that of \eqref{A-12}. But these two expressions are very different and it seems that we have a contradiction here. But there is no contradiction : in case of symmetric matrix $\:\Omega^{\boldsymbol{-}1}\:$ the eigenvalues $\:\mathrm w_1,\mathrm w_2\:$ are both real, the eigenvectors $\:\mathbf{a}_1,\mathbf{a}_2 $ of equations \eqref{B-10a},\eqref{B-10b} are orthogonal and the matrix $\:A\:$ of equations \eqref{B-02},\eqref{B-03} is orthogonal . For this matrix we have $\:A^{\boldsymbol{-}1}\boldsymbol{=}A^{\boldsymbol{\top}}\boldsymbol{=}\text{transpose of }A$. Replacing $\:A^{\boldsymbol{-}1}\:$ by $\:A^{\boldsymbol{\top}}\:$ the expression \eqref{B-14} becomes identical to \eqref{A-12}.In other words, since $\:A^{\boldsymbol{-}1}\:$ is also orthogonal it leaves the inner product of two vectors invariant, so in \eqref{B-14} we could replace any inner product $\:\left(A^{\boldsymbol{-}1}\mathbf{x}\right)\boldsymbol{\cdot}\left(A^{\boldsymbol{-}1}\mathbf{y}\right)\vphantom{\dfrac{a}{b}}\:$ by $\:\left(\mathbf{x}\boldsymbol{\cdot}\mathbf{y}\right)\vphantom{\dfrac{a}{b}}$.

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

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