Liouville's formula

Consider the n-dimensional first-order homogeneous linear differential equation

${\displaystyle y'=A(x)y}$

on an interval I of the real line, where A(x) for x ∈ I denotes a square matrix of dimension n with real or complex entries. Let Φ denote a matrix-valued solution on I, meaning that each Φ(x) is a square matrix of dimension n with real or complex entries and the derivative satisfies

${\displaystyle \Phi '(x)=A(x)\Phi (x),\qquad x\in I.}$

Let

${\displaystyle \mathrm {tr} \,A(\xi )=\sum _{i=1}^{n}a_{i,i}(\xi ),\qquad \xi \in I,}$

denote the trace of A(ξ) = (a_i, j(ξ))_{i, j ∈ {1,...,n}}, the sum of its diagonal entries. If the trace of A is a continuous function, then the determinant of Φ satisfies

${\displaystyle \det \Phi (x)=\det \Phi (x_{0})\,\exp {\biggl (}\int _{x_{0}}^{x}\mathrm {tr} \,A(\xi )\,{\textrm {d}}\xi {\biggr )}}$

for all x and x₀ in I.

This example illustrates how Liouville's formula can help to find the general solution of a first-order system of homogeneous linear differential equations. Consider

${\displaystyle y'=\underbrace {\begin{pmatrix}1&-1/x\\1+x&-1\end{pmatrix}} _{=\,A(x)}y}$

on the open interval I = (0, ∞). Assume that the easy solution

${\displaystyle y(x)={\begin{pmatrix}1\\x\end{pmatrix}},\qquad x\in I,}$

is already found. Let

${\displaystyle y(x)={\begin{pmatrix}y_{1}(x)\\y_{2}(x)\end{pmatrix}}}$

denote another solution, then

${\displaystyle \Phi (x)={\begin{pmatrix}y_{1}(x)&1\\y_{2}(x)&x\end{pmatrix}},\qquad x\in I,}$

is a square-matrix-valued solution of the above differential equation. Since the trace of A(x) is zero for all x ∈ I, Liouville's formula implies that the determinant

${\displaystyle c_{1}:=\det \Phi (x)=x\,y_{1}(x)-y_{2}(x),\qquad x\in I,}$

(1)

is actually a constant independent of x. Writing down the first component of the differential equation for y, we obtain using (1) that

${\displaystyle y'_{1}(x)=y_{1}(x)-{\frac {y_{2}(x)}{x}}={\frac {x\,y_{1}(x)-y_{2}(x)}{x}}={\frac {c_{1}}{x}},\qquad x\in I.}$

Therefore, by integration, we see that

${\displaystyle y_{1}(x)=c_{1}\ln x+c_{2},\qquad x\in I,}$

involving the natural logarithm and the constant of integration c₂. Solving equation (1) for y₂(x) and substituting for y₁(x) gives

${\displaystyle y_{2}(x)=x\,y_{1}(x)-c_{1}=\,c_{1}x\ln x+c_{2}x-c_{1},\qquad x\in I,}$

which is the general solution for y. With the special choice c₁ = 0 and c₂ = 1 we recover the easy solution we started with, the choice c₁ = 1 and c₂ = 0 yields a linearly independent solution. Therefore,

${\displaystyle \Phi (x)={\begin{pmatrix}\ln x&1\\x\ln x-1&x\end{pmatrix}},\qquad x\in I,}$

is a so-called fundamental solution of the system.

We omit the argument x for brevity. By the Leibniz formula for determinants, the derivative of the determinant of Φ = (Φ_i, j)_{i, j ∈ {0,...,n}} can be calculated by differentiating one row at a time and taking the sum, i.e.

${\displaystyle (\det \Phi )'=\sum _{i=1}^{n}\det {\begin{pmatrix}\Phi _{1,1}&\Phi _{1,2}&\cdots &\Phi _{1,n}\\\vdots &\vdots &&\vdots \\\Phi '_{i,1}&\Phi '_{i,2}&\cdots &\Phi '_{i,n}\\\vdots &\vdots &&\vdots \\\Phi _{n,1}&\Phi _{n,2}&\cdots &\Phi _{n,n}\end{pmatrix}}.}$

(2)

Since the matrix-valued solution Φ satisfies the equation Φ' = AΦ, we have for every entry of the matrix Φ'

${\displaystyle \Phi '_{i,k}=\sum _{j=1}^{n}a_{i,j}\Phi _{j,k}\,,\qquad i,k\in \{1,\ldots ,n\},}$

or for the entire row

${\displaystyle (\Phi '_{i,1},\dots ,\Phi '_{i,n})=\sum _{j=1}^{n}a_{i,j}(\Phi _{j,1},\ldots ,\Phi _{j,n}),\qquad i\in \{1,\ldots ,n\}.}$

When we subtract from the i^th row the linear combination

${\displaystyle \sum _{\scriptstyle j=1 \atop \scriptstyle j\not =i}^{n}a_{i,j}(\Phi _{j,1},\ldots ,\Phi _{j,n}),}$

of all the other rows, then the value of the determinant remains unchanged, hence

${\displaystyle \det {\begin{pmatrix}\Phi _{1,1}&\Phi _{1,2}&\cdots &\Phi _{1,n}\\\vdots &\vdots &&\vdots \\\Phi '_{i,1}&\Phi '_{i,2}&\cdots &\Phi '_{i,n}\\\vdots &\vdots &&\vdots \\\Phi _{n,1}&\Phi _{n,2}&\cdots &\Phi _{n,n}\end{pmatrix}}=\det {\begin{pmatrix}\Phi _{1,1}&\Phi _{1,2}&\cdots &\Phi _{1,n}\\\vdots &\vdots &&\vdots \\a_{i,i}\Phi _{i,1}&a_{i,i}\Phi _{i,2}&\cdots &a_{i,i}\Phi _{i,n}\\\vdots &\vdots &&\vdots \\\Phi _{n,1}&\Phi _{n,2}&\cdots &\Phi _{n,n}\end{pmatrix}}=a_{i,i}\det \Phi }$

for every i ∈ {1, . . . , n} by the linearity of the determinant with respect to every row. Hence

${\displaystyle (\det \Phi )'=\sum _{i=1}^{n}a_{i,i}\det \Phi =\mathrm {tr} \,A\,\det \Phi }$

(3)

by (2) and the definition of the trace. It remains to show that this representation of the derivative implies Liouville's formula.

Fix x₀ ∈ I. Since the trace of A is assumed to be continuous function on I, it is bounded on every closed and bounded subinterval of I and therefore integrable, hence

${\displaystyle g(x):=\det \Phi (x)\exp \left(-\int _{x_{0}}^{x}\mathrm {tr} \,A(\xi )\,{\textrm {d}}\xi \right),\qquad x\in I,}$

is a well defined function. Differentiating both sides, using the product rule, the chain rule, the derivative of the exponential function and the fundamental theorem of calculus, we obtain

${\displaystyle g'(x)={\bigl (}(\det \Phi (x))'-\det \Phi (x)\,\mathrm {tr} \,A(x){\bigr )}\exp {\biggl (}-\int _{x_{0}}^{x}\mathrm {tr} \,A(\xi )\,{\textrm {d}}\xi {\biggr )}=0,\qquad x\in I,}$

due to the derivative in (3). Therefore, g has to be constant on I, because otherwise we would obtain a contradiction to the mean value theorem (applied separately to the real and imaginary part in the complex-valued case). Since g(x₀) = det Φ(x₀), Liouville's formula follows by solving the definition of g for det Φ(x).

• Chicone, Carmen (2006), Ordinary Differential Equations with Applications (2 ed.), New York: Springer-Verlag, pp. 152–153, ISBN 978-0-387-30769-5, MR 2224508, Zbl 1120.34001
• Teschl, Gerald (2012), Ordinary Differential Equations and Dynamical Systems, Providence: American Mathematical Society, MR 2961944, Zbl 1263.34002