Abel's identity

• In particular, the Wronskian ${\displaystyle W(y_{1},y_{2})}$ is either always the zero function or always different from zero with the same sign at every point ${\displaystyle x}$ in ${\displaystyle I}$. In the latter case, the two solutions ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ are linearly independent (see that article about the Wronskian for a proof).
• It is not necessary to assume that the second derivatives of the solutions ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ are continuous.
• Abel's theorem is particularly useful if ${\displaystyle p(x)=0}$, because it implies that ${\displaystyle W}$ is constant.

Differentiating the Wronskian using the product rule gives (writing ${\displaystyle W}$ for ${\displaystyle W(y_{1},y_{2})}$ and omitting the argument ${\displaystyle x}$ for brevity)

${\displaystyle {\begin{aligned}W'&=y_{1}'y_{2}'+y_{1}y_{2}''-y_{1}''y_{2}-y_{1}'y_{2}'\\&=y_{1}y_{2}''-y_{1}''y_{2}.\end{aligned}}}$

Solving for ${\displaystyle y''}$ in the original differential equation yields

${\displaystyle y''=-(py'+qy).}$

Substituting this result into the derivative of the Wronskian function to replace the second derivatives of ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ gives

${\displaystyle {\begin{aligned}W'&=-y_{1}(py_{2}'+qy_{2})+(py_{1}'+qy_{1})y_{2}\\&=-p(y_{1}y_{2}'-y_{1}'y_{2})\\&=-pW.\end{aligned}}}$

This is a first-order linear differential equation, and it remains to show that Abel's identity gives the unique solution, which attains the value ${\displaystyle W(x_{0})}$ at ${\displaystyle x_{0}}$. Since the function ${\displaystyle p}$ is continuous on ${\displaystyle I}$, it is bounded on every closed and bounded subinterval of ${\displaystyle I}$ and therefore integrable, hence

${\displaystyle V(x)=W(x)\exp \left(\int _{x_{0}}^{x}p(\xi )\,{\textrm {d}}\xi \right),\qquad x\in I,}$

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, one obtains

${\displaystyle V'(x)={\bigl (}W'(x)+W(x)p(x){\bigr )}\exp {\biggl (}\int _{x_{0}}^{x}p(\xi )\,{\textrm {d}}\xi {\biggr )}=0,\qquad x\in I,}$

due to the differential equation for ${\displaystyle W}$. Therefore, ${\displaystyle V}$ has to be constant on ${\displaystyle I}$, 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 ${\displaystyle V(x_{0})=W(x_{0})}$, Abel's identity follows by solving the definition of ${\displaystyle V}$ for ${\displaystyle W(x)}$.

For brevity, we write ${\displaystyle W}$ for ${\displaystyle W(y_{1},\ldots ,y_{n})}$ and omit the argument ${\displaystyle x}$. It suffices to show that the Wronskian solves the first-order linear differential equation

${\displaystyle W'=-p_{n-1}\,W,}$

because the remaining part of the proof then coincides with the one for the case ${\displaystyle n=2}$.

In the case ${\displaystyle n=1}$ we have ${\displaystyle W=y_{1}}$ and the differential equation for ${\displaystyle W}$ coincides with the one for ${\displaystyle y_{1}}$. Therefore, assume ${\displaystyle n\geq 2}$ in the following.

The derivative of the Wronskian ${\displaystyle W}$ is the derivative of the defining determinant. It follows from the Leibniz formula for determinants that this derivative can be calculated by differentiating every row separately, hence

${\displaystyle {\begin{aligned}W'&={\begin{vmatrix}y'_{1}&y'_{2}&\cdots &y'_{n}\\y'_{1}&y'_{2}&\cdots &y'_{n}\\y''_{1}&y''_{2}&\cdots &y''_{n}\\y'''_{1}&y'''_{2}&\cdots &y'''_{n}\\\vdots &\vdots &\ddots &\vdots \\y_{1}^{(n-1)}&y_{2}^{(n-1)}&\cdots &y_{n}^{(n-1)}\end{vmatrix}}+{\begin{vmatrix}y_{1}&y_{2}&\cdots &y_{n}\\y''_{1}&y''_{2}&\cdots &y''_{n}\\y''_{1}&y''_{2}&\cdots &y''_{n}\\y'''_{1}&y'''_{2}&\cdots &y'''_{n}\\\vdots &\vdots &\ddots &\vdots \\y_{1}^{(n-1)}&y_{2}^{(n-1)}&\cdots &y_{n}^{(n-1)}\end{vmatrix}}\\&\qquad +\ \cdots \ +{\begin{vmatrix}y_{1}&y_{2}&\cdots &y_{n}\\y'_{1}&y'_{2}&\cdots &y'_{n}\\\vdots &\vdots &\ddots &\vdots \\y_{1}^{(n-3)}&y_{2}^{(n-3)}&\cdots &y_{n}^{(n-3)}\\y_{1}^{(n-2)}&y_{2}^{(n-2)}&\cdots &y_{n}^{(n-2)}\\y_{1}^{(n)}&y_{2}^{(n)}&\cdots &y_{n}^{(n)}\end{vmatrix}}.\end{aligned}}}$

However, note that every determinant from the expansion contains a pair of identical rows, except the last one. Since determinants with linearly dependent rows are equal to 0, one is only left with the last one:

${\displaystyle W'={\begin{vmatrix}y_{1}&y_{2}&\cdots &y_{n}\\y'_{1}&y'_{2}&\cdots &y'_{n}\\\vdots &\vdots &\ddots &\vdots \\y_{1}^{(n-2)}&y_{2}^{(n-2)}&\cdots &y_{n}^{(n-2)}\\y_{1}^{(n)}&y_{2}^{(n)}&\cdots &y_{n}^{(n)}\end{vmatrix}}.}$

Since every ${\displaystyle y_{i}}$ solves the ordinary differential equation, we have

${\displaystyle y_{i}^{(n)}+p_{n-2}\,y_{i}^{(n-2)}+\cdots +p_{1}\,y'_{i}+p_{0}\,y_{i}=-p_{n-1}\,y_{i}^{(n-1)}}$

for every ${\displaystyle i\in \lbrace 1,\ldots ,n\rbrace }$. Hence, adding to the last row of the above determinant ${\displaystyle p_{0}}$ times its first row, ${\displaystyle p_{1}}$ times its second row, and so on until ${\displaystyle p_{n-2}}$ times its next to last row, the value of the determinant for the derivative of ${\displaystyle W}$ is unchanged and we get

${\displaystyle W'={\begin{vmatrix}y_{1}&y_{2}&\cdots &y_{n}\\y'_{1}&y'_{2}&\cdots &y'_{n}\\\vdots &\vdots &\ddots &\vdots \\y_{1}^{(n-2)}&y_{2}^{(n-2)}&\cdots &y_{n}^{(n-2)}\\-p_{n-1}\,y_{1}^{(n-1)}&-p_{n-1}\,y_{2}^{(n-1)}&\cdots &-p_{n-1}\,y_{n}^{(n-1)}\end{vmatrix}}=-p_{n-1}W.}$

The solutions ${\displaystyle y_{1},\ldots ,y_{n}}$ form the square-matrix valued solution

${\displaystyle \Phi (x)={\begin{pmatrix}y_{1}(x)&y_{2}(x)&\cdots &y_{n}(x)\\y'_{1}(x)&y'_{2}(x)&\cdots &y'_{n}(x)\\\vdots &\vdots &\ddots &\vdots \\y_{1}^{(n-2)}(x)&y_{2}^{(n-2)}(x)&\cdots &y_{n}^{(n-2)}(x)\\y_{1}^{(n-1)}(x)&y_{2}^{(n-1)}(x)&\cdots &y_{n}^{(n-1)}(x)\end{pmatrix}},\qquad x\in I,}$

of the ${\displaystyle n}$-dimensional first-order system of homogeneous linear differential equations

${\displaystyle {\begin{pmatrix}y'\\y''\\\vdots \\y^{(n-1)}\\y^{(n)}\end{pmatrix}}={\begin{pmatrix}0&1&0&\cdots &0\\0&0&1&\cdots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&\cdots &1\\-p_{0}(x)&-p_{1}(x)&-p_{2}(x)&\cdots &-p_{n-1}(x)\end{pmatrix}}{\begin{pmatrix}y\\y'\\\vdots \\y^{(n-2)}\\y^{(n-1)}\end{pmatrix}}.}$

The trace of this matrix is ${\displaystyle -p_{n-1}(x)}$, hence Abel's identity follows directly from Liouville's formula.