Consider a homogeneous linear second-order ordinary differential equation on an intervalI of the real line with real- or complex-valued continuous functionsp and q. Abel's identity states that the Wronskian of two real- or complex-valued solutions and of this differential equation, that is the function defined by the determinant satisfies the relation for every point x0 in I, where C is an arbitrary constant.
Remarks
In particular, the Wronskian is either always the zero function or always different from zero with the same sign at every point in. In the latter case, the two solutions and are linearly independent.
It is not necessary to assume that the second derivatives of the solutions and are continuous.
Abel's theorem is particularly useful if, because it implies that W=const.
Proof
the Wronskian using the product rule gives Solving for in the original differential equation yields Substituting this result into the derivative of the Wronskian function to replace the second derivatives of and gives 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 at. Since the function is continuous on, it is bounded on every closed and bounded subinterval of and therefore integrable, hence 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 due to the differential equation for. Therefore, has to be constant on, because otherwise we would obtain a contradiction to the mean value theorem. Since, Abel's identity follows by solving the definition of for.
Generalization
Consider a homogeneous linear th-order ordinary differential equation on an interval of the real line with a real- or complex-valued continuous function. The generalisation of Abel's identity states that the Wronskian of real- or complex-valued solutions of this th-order differential equation, that is the function defined by the determinant satisfies the relation for every point in.
Direct proof
For brevity, we write for and omitthe argument. It suffices to show that the Wronskian solves the first-order linear differential equation because the remaining part of the proof then coincides with the one for the case. In the case we have and the differential equation for coincides with the one for. Therefore, assume in the following. The derivative of the Wronskian 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 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: Since every solves the ordinary differential equation, we have for every. Hence, adding to the last row of the above determinant times its first row, times its second row, and so on until times its next to last row, the value of the determinant for the derivative of is unchanged and we get
Proof using Liouville's formula
The solutions form the square-matrix valued solution of the -dimensional first-order system of homogeneous linear differential equations The trace of this matrix is, hence Abel's identity follows directly from Liouville's formula.
