Decoupling coupled linear ODEs by constant linear combinations

Coupled systems of linear second-order ordinary differential equations appear everywhere in mathematical physics: coupled oscillators, multi-component wave equations, small perturbations in gravity, and so on. A very natural question is:

When can a coupled system be transformed into independent scalar equations by taking constant linear combinations of the original unknown functions?

Mathematically, the key idea is that decoupling is equivalent to simultaneous diagonalization of a matrix-valued coefficient function $Q(x)$ by a constant change of basis.

---

1. Matrix formulation

Consider $n$ scalar functions $y_i(x)$ and a coupled linear system of second-order ODEs of the form

$$y_i''(x) + \sum_{j=1}^n Q_{ij}(x)\,y_j(x) = 0, \qquad i = 1,\dots,n.$$

Introduce the vector

$$\mathbf y(x) = \begin{pmatrix} y_1(x)\\ \vdots\\ y_n(x) \end{pmatrix},$$

and the $n\times n$ matrix

$$Q(x) = \bigl(Q_{ij}(x)\bigr)_{1\le i,j \le n}.$$

Then the system can be written compactly as

$$\mathbf y''(x) + Q(x)\,\mathbf y(x) = 0.$$

We will look for a constant invertible matrix $S\in GL(n)$ and new unknowns

$$\mathbf z(x) = S^{-1}\mathbf y(x),$$

such that in the $\mathbf z$–variables the system is diagonal:

$$z_i''(x) + \lambda_i(x)\,z_i(x) = 0, \qquad i = 1,\dots,n.$$

Since $S$ is constant, $\mathbf y'' = S\mathbf z''$, and the equation becomes

$$S\mathbf z''(x) + Q(x)S\mathbf z(x) = 0,$$

or equivalently

$$\mathbf z''(x) + S^{-1}Q(x)S\,\mathbf z(x) = 0.$$

Thus:

The system is decoupled in the $\mathbf z$–variables if and only if the matrix $S^{-1}Q(x)S$ is diagonal for all $x$.

Equivalently, there exists a constant basis in which $Q(x)$ is always diagonal. This is what we mean by simultaneous diagonalization of the family of matrices $\{Q(x)\}$ by a single constant matrix $S$.

---

2. The two-equation case in detail

Let us now specialize to $n=2$ and work everything out explicitly.

2.1 The system and its matrix

Consider

$$\begin{aligned} y_1''(x) + q_1(x)y_1(x) + r_1(x)y_2(x) &= 0,\\ y_2''(x) + q_2(x)y_1(x) + r_2(x)y_2(x) &= 0, \end{aligned}$$

where $q_1,q_2,r_1,r_2$ are given functions of $x$.

In matrix form,

$$\mathbf y''(x) + Q(x)\,\mathbf y(x) = 0,$$

with

$$\mathbf y(x) = \begin{pmatrix} y_1(x)\\ y_2(x) \end{pmatrix}, \qquad Q(x) = \begin{pmatrix} q_1(x) & r_1(x)\\ q_2(x) & r_2(x) \end{pmatrix}.$$

We want to find constant linear combinations of $y_1,y_2$ that decouple this system.

2.2 Decoupling via a constant change of basis

Suppose there is a constant invertible matrix $S$ such that

$$\mathbf z = S^{-1}\mathbf y$$

satisfies

$$z_1''(x) + \lambda_1(x)\,z_1(x) = 0,\qquad z_2''(x) + \lambda_2(x)\,z_2(x) = 0.$$

As explained above, this is equivalent to the existence of a constant $S$ such that

$$S^{-1}Q(x)S = \begin{pmatrix} \lambda_1(x) & 0\\ 0 & \lambda_2(x) \end{pmatrix} \quad\text{for all }x.$$

Thus $\{Q(x)\}$ must have an $x$-independent eigenbasis: there must exist two constant vectors $v_1,v_2$ and scalar functions $\lambda_1(x),\lambda_2(x)$ such that

$$Q(x)v_i = \lambda_i(x)\,v_i,\qquad i=1,2,\ \text{for all }x.$$

We now derive explicit conditions on the scalar functions $q_1,q_2,r_1,r_2$ that guarantee this.

2.3 Method of constant linear combinations

Instead of working directly with eigenvectors, let us take a more “ODE-centric” view. We search for combinations of the form

$$u(x) = y_1(x) + \kappa\,y_2(x),$$

with $\kappa$ constant, such that $u(x)$ satisfies a decoupled equation

$$u''(x) + \lambda(x)\,u(x) = 0,$$

for some coefficient function $\lambda(x)$.

From the original system we have

$$\begin{aligned} y_1'' &= - q_1 y_1 - r_1 y_2,\\ y_2'' &= - q_2 y_1 - r_2 y_2. \end{aligned}$$

Differentiate $u$ twice:

$$u'' = y_1'' + \kappa y_2'' = -(q_1+\kappa q_2)y_1 - (r_1+\kappa r_2)y_2.$$

We want $u'' + \lambda u = 0$, i.e.

$$-(q_1+\kappa q_2)y_1 - (r_1+\kappa r_2)y_2 + \lambda(y_1+\kappa y_2) = 0$$

for all $y_1,y_2$. Equating coefficients of $y_1$ and $y_2$ gives

$$\begin{aligned} \lambda &= q_1 + \kappa q_2,\\ \lambda\kappa &= r_1 + \kappa r_2. \end{aligned}$$

Eliminating $\lambda$ yields

$$(q_1 + \kappa q_2)\,\kappa = r_1 + \kappa r_2,$$

or

$$q_2(x)\,\kappa^2 + \bigl(q_1(x)-r_2(x)\bigr)\,\kappa - r_1(x) = 0.$$

This is a quadratic equation in $\kappa$ whose coefficients are functions of $x$. For a given choice of $\kappa$, we need this identity to hold for all $x$. For decoupling, we need two distinct constant solutions $\kappa_1,\kappa_2$ (one for each independent scalar mode).

That is only possible if the coefficients are proportional as functions of $x$. Assume $q_2(x)\not\equiv 0$ on the interval of interest. Then the condition is:

There exist constants $\alpha,\beta$ such that

$$\frac{r_1(x)}{q_2(x)} = \alpha,\qquad \frac{q_1(x)-r_2(x)}{q_2(x)} = \beta,$$

for all $x$ where $q_2(x)\neq 0$.

Equivalently,

$$r_1(x) = \alpha\,q_2(x),\qquad q_1(x) - r_2(x) = \beta\,q_2(x).$$

Under these conditions, the quadratic simplifies to

$$q_2(x)\left(\kappa^2 + \beta\kappa - \alpha\right) = 0.$$

The roots of $\kappa^2 + \beta\kappa - \alpha = 0$,

$$\kappa_{1,2} = \frac{-\beta\pm\sqrt{\beta^2+4\alpha}}{2},$$

are constants, independent of $x$. Thus we obtain two constant linear combinations

$$u_1 = y_1 + \kappa_1 y_2,\qquad u_2 = y_1 + \kappa_2 y_2,$$

each of which satisfies a decoupled second-order ODE

$$u_i''(x) + \lambda_i(x)\,u_i(x) = 0,$$

with $\lambda_i(x) = q_1(x) + \kappa_i q_2(x)$.

Special cases like $q_2\equiv 0$ can be treated similarly by swapping the roles of $y_1$ and $y_2$ (or equivalently interchanging $q_2\leftrightarrow r_1$).

2.4 Matrix interpretation

The conditions

$$r_1(x) = \alpha\,q_2(x),\qquad q_1(x) - r_2(x) = \beta\,q_2(x)$$

have a nice matrix interpretation. Using

$$Q(x) = \begin{pmatrix} q_1(x) & r_1(x)\\ q_2(x) & r_2(x) \end{pmatrix},$$

we can write

$$Q(x) = \begin{pmatrix} r_2(x) + \beta q_2(x) & \alpha q_2(x)\\ q_2(x) & r_2(x) \end{pmatrix} = r_2(x)\,I + q_2(x)\,M,$$

where

$$M = \begin{pmatrix} \beta & \alpha\\ 1 & 0 \end{pmatrix}$$

is a constant $2\times 2$ matrix.

Thus every $Q(x)$ in this family lies in the two-dimensional commutative algebra

$$\mathcal A = \mathrm{span}\{I,M\}.$$

The eigenvectors of $M$ are constant and provide the desired decoupling. Indeed, if $M v_i = \mu_i v_i$ with $v_1,v_2$ linearly independent and constant, then

$$Q(x) v_i = \bigl(r_2(x) + \mu_i q_2(x)\bigr)\,v_i,$$

so $v_1,v_2$ form an eigenbasis of $Q(x)$ for all $x$.

In summary:

Two-equation system (generic case).
On an interval where $q_2$ does not vanish identically and $Q(x)$ has distinct eigenvalues, the system can be decoupled by constant linear combinations if and only if there exist constants $\alpha,\beta$ such that

$$\frac{r_1(x)}{q_2(x)} = \alpha,\qquad \frac{q_1(x)-r_2(x)}{q_2(x)} = \beta$$

for all $x$ in the interval.

The two conditions above are precisely two independent functional constraints among the four coefficient functions $q_1,q_2,r_1,r_2$.

---

3. General $n$-equation systems

Let us return to the general case

$$\mathbf y''(x) + Q(x)\,\mathbf y(x) = 0, \qquad \mathbf y(x)\in\mathbb C^n,\quad Q(x)\in\mathrm{Mat}_{n\times n}(\mathbb C).$$

We ask: when can we find a constant invertible $S$ such that $\mathbf z = S^{-1}\mathbf y$ satisfies the decoupled system

$$z_i''(x) + \lambda_i(x)\,z_i(x) = 0,\qquad i=1,\dots,n?$$

As before, the system in $\mathbf z$ reads

$$\mathbf z''(x) + S^{-1}Q(x)S\,\mathbf z(x) = 0.$$

Thus we require that

$$S^{-1}Q(x)S = \mathrm{diag}\!\bigl(\lambda_1(x),\dots,\lambda_n(x)\bigr)$$

for all $x$. Equivalently,

There exist $n$ linearly independent constant vectors $v_1,\dots,v_n$ and scalar functions $\lambda_1(x),\dots,\lambda_n(x)$ such that

$$Q(x)\,v_i = \lambda_i(x)\,v_i,\qquad i=1,\dots,n,\ \text{for all }x.$$

The vectors $v_i$ form a common eigenbasis for the entire family $\{Q(x)\}$, and stacking them as columns of $S$ gives the desired transformation.

3.1 Projector / Cartan subalgebra viewpoint

A convenient way to package this condition is in terms of projectors.

Let $v_i$ be a common eigenbasis for all $Q(x)$, and let $w_i^{\mathsf T}$ be the corresponding dual basis so that $w_i^{\mathsf T}v_j = \delta_{ij}$. Then the rank-one projectors

$$P_i = v_i w_i^{\mathsf T}$$

satisfy

$$P_i^2 = P_i,\qquad P_iP_j = 0\ (i\ne j),\qquad \sum_{i=1}^n P_i = I.$$

The matrix $Q(x)$ can then be written as

$$Q(x) = \sum_{i=1}^n \lambda_i(x)\,P_i.$$

If the system is decouplable by a constant change of basis, there exist such constant projectors $P_i$ and coefficient functions $\lambda_i(x)$.

In Lie-algebra language, the set

$$\mathfrak h = \mathrm{span}\{P_1,\dots,P_n\}$$

is a Cartan subalgebra of $\mathfrak{gl}_n$, conjugate to the subalgebra of diagonal matrices. The condition above says precisely that

$Q(x)$ lies in a fixed Cartan subalgebra for all $x$.

Choosing a basis that diagonalizes $\mathfrak h$ is exactly the constant change of basis $S$ that decouples the system.

3.2 Commuting families (a useful sufficient condition)

A standard sufficient condition, under mild diagonalizability assumptions, is:

1. The matrices $Q(x)$ are diagonalizable (e.g. Hermitian/symmetric), and
2. They commute pairwise: $$[Q(x_1),Q(x_2)] = 0 \quad\text{for all }x_1,x_2.$$

Then there exists a constant basis of joint eigenvectors, and the system can be decoupled by a constant change of variables.

This is often how decoupling arises in physics: the coupling matrices are all functions of $x$ built from a fixed set of commuting operators.

---

4. How many independent conditions for $n$ equations?

In the two-equation case we found two independent relations among $q_1,q_2,r_1,r_2$. For general $n$, a simple dimension count tells us how many independent functional constraints are needed.

• A completely general $Q(x)$ is an $n\times n$ matrix of functions: there are $n^2$ independent scalar coefficient functions $Q_{ij}(x)$.

• If the system is decouplable by a constant $S$, then

  $$Q(x) = S\,\Lambda(x)\,S^{-1},\qquad \Lambda(x) = \mathrm{diag}(\lambda_1(x),\dots,\lambda_n(x)).$$

  The only $x$-dependent data are the $n$ scalar functions $\lambda_i(x)$; the matrix $S$ itself is constant.

Thus the space of decouplable families $Q(x)$ has $n$ functional degrees of freedom, while a general $Q(x)$ has $n^2$ functional degrees of freedom. Therefore, we need

$$n^2 - n$$

independent functional constraints on the coefficients $Q_{ij}(x)$ in order for the system to be decouplable into $n$ scalar second-order equations by constant linear combinations.

For example:

• $n=2$: $n^2-n = 4-2 = 2$ constraints, matching the two ratio conditions in the two-equation case.
• $n=3$: $n^2-n = 9-3 = 6$ constraints on the nine functions $(Q_{ij})$, expressing the fact that $Q(x)$ lies in a fixed three-dimensional Cartan subalgebra.

This count is generic: it assumes a full decoupling into $n$ one-dimensional modes (no persistent degeneracy of eigenvalues). If there are exact degeneracies, the effective conditions may be weaker because one can decouple only into blocks corresponding to the degenerate eigenspaces.

---

5. Summary

• A system of $n$ coupled linear second-order ODEs

  $$\mathbf y''(x) + Q(x)\,\mathbf y(x) = 0$$

  can be decoupled by constant linear combinations of the unknowns if and only if there exists a constant invertible matrix $S$ such that

  $$S^{-1}Q(x)S \quad\text{is diagonal for all }x.$$

• Equivalently, there exists an $x$-independent eigenbasis $v_1,\dots,v_n$ and functions $\lambda_i(x)$ such that $Q(x)v_i = \lambda_i(x)v_i$ for all $x$.

• In the two-equation case, this condition can be written explicitly as the two ratio conditions

  $$\frac{r_1(x)}{q_2(x)} = \text{const},\qquad \frac{q_1(x)-r_2(x)}{q_2(x)} = \text{const}$$

  (up to symmetric special cases). These ensure the existence of two constant linear combinations $u_1,u_2$ that obey decoupled scalar equations.

• In the general $n$-equation case, the decoupling condition can be expressed invariantly as

  $$Q(x) = \sum_{i=1}^n \lambda_i(x)\,P_i,$$

  where the projectors $P_i$ are constant and sum to the identity. In Lie-algebra terms, $Q(x)$ must lie in a fixed Cartan subalgebra (conjugate to the diagonal matrices) for all $x$.

• Generically, there are

  $$n^2 - n$$

  independent functional conditions on the entries $Q_{ij}(x)$ for full decoupling into $n$ scalar second-order equations.

From the point of view of applications, this is exactly the familiar problem of finding normal modes: decoupling is possible precisely when the “shape” of the coupling between components does not change with $x$—only the eigenvalues of the coupling matrix are allowed to vary, while its eigenvectors remain fixed.