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In [[mathematics]], '''Liouville's formula''', also known as the Abel-Jacobi-Liouville Identity, is an equation that expresses the [[determinant]] of a [[matrix (mathematics)#Square matrices|square-matrix]] solution of a first-order system of homogeneous [[linear differential equation]]s in terms of the sum of the diagonal coefficients of the system. The formula is named after the [[France|French]] [[mathematician]] [[Joseph Liouville]]. [[Jacobi's formula]] provides another representation of the same mathematical relationship.

Liouville's formula is a generalization of [[Abel's identity]] and can be used to prove it. Since Liouville's formula relates the different [[linear independence|linearly independent]] solutions of the system of differential equations, it can help to find one solution from the other(s), see the example application below.

==Statement of Liouville's formula==
Consider the {{math|''n''}}-dimensional first-order homogeneous linear differential equation

:<math>y'=A(x)y\,</math>

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

:<math>\Phi'(x)=A(x)\Phi(x),\qquad x\in I.</math>

Let

:<math>\mathrm{tr}\,A(\xi)=\sum_{i=1}^n a_{i,i}(\xi),\qquad \xi\in I,</math>

denote the [[trace (linear algebra)|trace]] of {{math|''A''(''ξ'') {{=}} (''a''''i'', ''j'' (''ξ''))''i'', ''j'' ∈ {1,...,''n''}}}, the sum of its diagonal entries. If the trace of {{math|''A''}} is a [[continuous function]], then the determinant of {{math|Φ}} satisfies

:<math>\det\Phi(x)=\det\Phi(x_0)\,\exp\biggl(\int_{x_0}^x \mathrm{tr}\,A(\xi) \,\textrm{d}\xi\biggr)</math>

for all {{math|''x''}} and {{math|''x''0}} in {{math|''I''}}.

==Example application==
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

:<math>y'=\underbrace{\begin{pmatrix}1&-1/x\\1+x&-1\end{pmatrix}}_{=\,A(x)}y</math>

on the open interval {{math|''I'' {{=}} }}{{open-open|0, ∞}}. Assume that the easy solution

:<math>y(x)=\begin{pmatrix}1\\x\end{pmatrix},\qquad x\in I,</math>

is already found. Let

:<math>y(x)=\begin{pmatrix}y_1(x)\\y_2(x)\end{pmatrix}</math>

denote another solution, then

:<math>\Phi(x)=\begin{pmatrix}y_1(x)&1\\y_2(x)&x\end{pmatrix},\qquad x\in I,</math>

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

{{NumBlk|:|<math>c_1:=\det\Phi(x)=x\,y_1(x)-y_2(x),\qquad x\in I,</math>|{{EquationRef|1}}}}

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

:<math>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.</math>

Therefore, by integration, we see that

:<math>y_1(x)=c_1\ln x+c_2,\qquad x\in I,</math>

involving the [[natural logarithm]] and the [[constant of integration]] {{math|''c''2}}. Solving equation ({{EquationNote|1}}) for {{math|''y''2(''x'')}} and substituting for {{math|''y''1(''x'')}} gives

:<math>y_2(x)=x\,y_1(x)-c_1=\,c_1x\ln x+c_2x-c_1,\qquad x\in I,</math>

which is the general solution for {{math|''y''}}. With the special choice {{math|''c''1 {{=}} 0}} and {{math|''c''2 {{=}} 1}} we recover the easy solution we started with, the choice {{math|''c''1 {{=}} 1}} and {{math|''c''2 {{=}} 0}} yields a linearly independent solution. Therefore,

:<math>\Phi(x)=\begin{pmatrix}\ln x&1\\x\ln x-1&x\end{pmatrix},\qquad x\in I,</math>

is a so-called fundamental solution of the system.

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

{{NumBlk|:|<math>(\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}.</math>|{{EquationRef|2}}}}

Since the matrix-valued solution {{math|Φ}} satisfies the equation {{math|Φ' = ''A''Φ}}, we have for every entry of the matrix {{math|Φ'}}

:<math>\Phi'_{i,k}=\sum_{j=1}^n a_{i,j}\Phi_{j,k}\,,\qquad i,k\in\{1,\ldots,n\},</math>

or for the entire row

:<math>(\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\}.</math>

When we subtract from the {{math|''i''}} th row the linear combination

:<math>\sum_{\scriptstyle j=1\atop\scriptstyle j\not=i}^n a_{i,j}(\Phi_{j,1},\ldots,\Phi_{j,n}),</math>

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

:<math>\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</math>

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

{{NumBlk|:|<math>(\det\Phi)'=\sum_{i=1}^n a_{i,i}\det\Phi=\mathrm{tr}\,A\,\det\Phi</math>|{{EquationRef|3}}}}

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

Fix {{math|''x''0 ∈ ''I''}}. Since the trace of {{math|''A''}} is assumed to be continuous function on {{math|''I''}}, it is bounded on every closed and bounded subinterval of {{math|''I''}} and therefore integrable, hence

:<math>g(x):=\det\Phi(x) \exp\left(-\int_{x_0}^x \mathrm{tr}\,A(\xi) \,\textrm{d}\xi\right), \qquad x\in I,</math>

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

:<math>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,</math>

due to the derivative in ({{EquationNote|3}}). Therefore, {{math|''g''}} has to be constant on {{math|''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 {{math|''g''(''x''0) {{=}} det Φ(''x''0)}}, Liouville's formula follows by solving the definition of {{math|''g''}} for {{math|det Φ(''x'')}}.

==References==
* {{Citation
| first = Carmen
| last = Chicone
| title = Ordinary Differential Equations with Applications
| place = New York
| publisher = Springer-Verlag
| year = 2006
| edition = 2
| pages = 152–153
| isbn = 978-0-387-30769-5
| mr = 2224508
| zbl = 1120.34001}}
* {{Citation
| last = Teschl
| first = Gerald
| authorlink=Gerald Teschl
| title = Ordinary Differential Equations and Dynamical Systems
| publisher=[[American Mathematical Society]]
| place = [[Providence, Rhode Island|Providence]]
| year = 2012
| url = http://www.mat.univie.ac.at/~gerald/ftp/book-ode/
| mr = 2961944
| zbl = 06054089}}

[[Category:Mathematical identities]]
[[Category:Ordinary differential equations]]
[[Category:Articles containing proofs]]

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