Liouville's formula

Template:Short description In mathematics, Liouville's formula, also known as the Abel–Jacobi–Liouville identity, is an equation that expresses the determinant of a square-matrix solution of a first-order system of homogeneous linear differential equations in terms of the sum of the diagonal coefficients of the system. The formula is named after the 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 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 Template:Math-dimensional first-order homogeneous linear differential equation

$y^{\prime }=A(t)y$

on an interval Template:Math of the real line, where Template:Math for Template:Math denotes a square matrix of dimension Template:Math with real or complex entries. Let Template:Math denote a matrix-valued solution on Template:Math, meaning that Template:Math is the so-called fundamental matrix, a square matrix of dimension Template:Math with real or complex entries and the derivative satisfies

${\mathrm{\Phi }}^{\prime }(t)=A(t)\mathrm{\Phi }(t),t\in I.$

Let

$\mathrm{tr}A(s)=\sum _{i=1}^n{a}_{i,i}(s),s\in I,$

denote the trace of Template:Math, the sum of its diagonal entries. If the trace of Template:Math is a continuous function, then the determinant of Template:Math satisfies

$\det \mathrm{\Phi }(t)=\det \mathrm{\Phi }(t_0)\exp (\underset{t_0}{\overset{t}{\int }}\mathrm{tr}A(s)\text{d}s)$

for all Template:Math and Template:Math in Template:Math.

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

$y^{\prime }=\underset{=A(x)}{\underbrace{\left(\begin{array}{cc}1& -1/x\\ 1+x& -1\end{array}\right)}}y$

on the open interval Template:Math. Assume that the easy solution

$y(x)=\left(\begin{array}{c}1\\ x\end{array}\right),x\in I,$

is already found. Let

$y(x)=\left(\begin{array}{c}{y}_1(x)\\ y_2(x)\end{array}\right)$

denote another solution, then

$\mathrm{\Phi }(x)=\left(\begin{array}{cc}{y}_1(x)& 1\\ y_2(x)& x\end{array}\right),x\in I,$

is a square-matrix-valued solution of the above differential equation. Since the trace of Template:Math is zero for all Template:Math, Liouville's formula implies that the determinant

is actually a constant independent of Template:Math. Writing down the first component of the differential equation for Template:Math, we obtain using (Template:EquationNote) that

$y{\text{'}}_1(x)=y_1(x)-\frac{y_2(x)}{x}=\frac{x{y}_1(x)-y_2(x)}{x}=\frac{c_1}{x},x\in I.$

Therefore, by integration, we see that

$y_1(x)=c_1\ln x+c_2,x\in I,$

involving the natural logarithm and the constant of integration Template:Math. Solving equation (Template:EquationNote) for Template:Math and substituting for Template:Math gives

$y_2(x)=x{y}_1(x)-c_1=c_{1}x\ln x+c_{2}x-c_1,x\in I,$

which is the general solution for Template:Math. With the special choice Template:Math and Template:Math we recover the easy solution we started with, the choice Template:Math and Template:Math yields a linearly independent solution. Therefore,

$\mathrm{\Phi }(x)=\left(\begin{array}{cc}\ln x& 1\\ x\ln x-1& x\end{array}\right),x\in I,$

is a so-called fundamental solution of the system.

Proof of Liouville's formula

We omit the argument Template:Math for brevity. By the Leibniz formula for determinants, the derivative of the determinant of Template:Math can be calculated by differentiating one row at a time and taking the sum, i.e.

Since the matrix-valued solution Template:Math satisfies the equation Template:Math, we have for every entry of the matrix Template:Math

$\mathrm{\Phi }{\text{'}}_{i,k}=\sum _{j=1}^n{a}_{i,j}{\mathrm{\Phi }}_{j,k},i,k\in \{1,\dots ,n\},$

or for the entire row

$(\mathrm{\Phi }{\text{'}}_{i,1},\dots ,\mathrm{\Phi }{\text{'}}_{i,n})=\sum _{j=1}^n{a}_{i,j}({\mathrm{\Phi }}_{j,1},\dots ,{\mathrm{\Phi }}_{j,n}),i\in \{1,\dots ,n\}.$

When we subtract from the Template:Math-th row the linear combination

$\sum _{\genfrac{}{}{0ex}{}{j=1}{j\ne i}}^n{a}_{i,j}({\mathrm{\Phi }}_{j,1},\dots ,{\mathrm{\Phi }}_{j,n}),$

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

$\det \left(\begin{array}{cccc}{\mathrm{\Phi }}_{1,1}& {\mathrm{\Phi }}_{1,2}& \cdots & {\mathrm{\Phi }}_{1,n}\\ ⋮& ⋮& & ⋮\\ \mathrm{\Phi }{\text{'}}_{i,1}& \mathrm{\Phi }{\text{'}}_{i,2}& \cdots & \mathrm{\Phi }{\text{'}}_{i,n}\\ ⋮& ⋮& & ⋮\\ {\mathrm{\Phi }}_{n,1}& {\mathrm{\Phi }}_{n,2}& \cdots & {\mathrm{\Phi }}_{n,n}\end{array}\right)=\det \left(\begin{array}{cccc}{\mathrm{\Phi }}_{1,1}& {\mathrm{\Phi }}_{1,2}& \cdots & {\mathrm{\Phi }}_{1,n}\\ ⋮& ⋮& & ⋮\\ a_{i,i}{\mathrm{\Phi }}_{i,1}& a_{i,i}{\mathrm{\Phi }}_{i,2}& \cdots & a_{i,i}{\mathrm{\Phi }}_{i,n}\\ ⋮& ⋮& & ⋮\\ {\mathrm{\Phi }}_{n,1}& {\mathrm{\Phi }}_{n,2}& \cdots & {\mathrm{\Phi }}_{n,n}\end{array}\right)=a_{i,i}\det \mathrm{\Phi }$

for every Template:Math} by the linearity of the determinant with respect to every row. Hence

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

Fix Template:Math. Since the trace of Template:Math is assumed to be continuous function on Template:Math, it is bounded on every closed and bounded subinterval of Template:Math and therefore integrable, hence

$g(x):=\det \mathrm{\Phi }(x)\exp (-\underset{x_0}{\overset{x}{\int }}\mathrm{t}\mathrm{r}A(\xi )\text{d}\xi ),x\in I,$

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

$g^{\prime }(x)=((\det \mathrm{\Phi }(x){)}^{\prime }-\det \mathrm{\Phi }(x)\mathrm{t}\mathrm{r}A(x))\exp (-\underset{x_0}{\overset{x}{\int }}\mathrm{tr}A(\xi )\text{d}\xi )=0,x\in I,$

due to the derivative in (Template:EquationNote). Therefore, Template:Math has to be constant on Template:Math, 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 Template:Math, Liouville's formula follows by solving the definition of Template:Math for Template:Math.

References

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