Liouville's formula

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

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

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

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

is already found. Let

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

denote another solution, then

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

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

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

${\displaystyle y_1^{\prime }(x)=y_1(x)-\frac{y_2(x)}{x}=\frac{x{y}_1(x)-y_2(x)}{x}=\frac{c_1}{x},x\in I.}$

Therefore, by integration, we see that

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

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

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

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,

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

is a so-called fundamental solution of the system.

Proof of Liouville's formula

We omit the argument 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.

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

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

or for the entire row

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

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

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

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

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

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

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

${\displaystyle g(x):=det\mathrm{\Phi }(x)\exp \left(-{\int }_{x_0}^x\mathrm{t}\mathrm{r}A(\xi )\text{d}\xi \right),x\in I,}$

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

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

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).