Cauchy-Euler equation

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In mathematics, a Cauchy-Euler equation (also Euler-Cauchy equation) is a linear homogeneous ordinary differential equation with variable coefficients. They are sometimes known as equi-dimensional equations. Because of its simple structure the equation can be replaced with an equivalent equation with constant coefficients which can then be solved explicitly.

1 The equation
- 1.1 Second order
- 1.2 Example
2 Difference equation analogue
3 See also

[edit] The equation

Let $y (n) (x)$ be the nth derivative of the unknown function $y (x)$ . Then a Cauchy-Euler equation of order n has the form

$x^n y^{(n)}(x) + a_{n-1} x^{n-1} y^{(n-1)}(x) + \cdots + a_0 y(x) = 0.$

The substitution $\scriptstyle x = e^u$ reduces this equation to a linear differential equation with constant coefficients.

[edit] Second order

The Euler-Cauchy equation crops up in a number of engineering applications. It is given by the equation:

$x^2\frac{d^2y}{dx^2} + ax\frac{dy}{dx} + by = 0 \,$

We assume a trial solution given by

$y = x^m. \,$

Differentiating, we have:

$\frac{dy}{dx} = mx^{m-1} \,$

and

$\frac{d^2y}{dx^2} = m(m-1)x^{m-2}. \,$

Substituting into the original equation, we have:

$x^2( m(m-1)x^{m-2} ) + ax( mx^{m-1} ) + b( x^m ) = 0 \,$

Or rearranging gives:

$m^2 + (a-1)m + b = 0. \,$

We then can solve for m. There are three particular cases of interest:

Case #1: Two distinct roots, m₁ and m₂
Case #2: One real repeated root, m
Case #3: Complex roots, α ± iβ

In case #1, the solution is given by:

$y = c_{1}x^{m_{1}} + c_{2}x^{m_{2}} \,$

In case #2, the solution is given by:

$y = c_{1}x^{m}\ln(x) + c_{2}x^{m} \,$

To get to this solution, the method of reduction of order must be applied after having found one solution y = x^m.

In case #3, the solution is given by:

$y = c_{1}x^{\alpha}\cos(\beta \ln(x)) + c_{2}x^{\alpha}\sin(\beta \ln(x)) \,$

This equation also can be solved with $x = e t$ transformation.

This particular case is of no great practical importance and hence this has been left as a challenge for the reader.

[edit] Example

Given

$x^2u''-3xu'+3u=0\,$

we substitute the simple solution x^α:

$x^2(\alpha(\alpha-1)x^{\alpha-2})-3x(\alpha(x^{\alpha-1}))+3(x^\alpha)=\alpha(\alpha-1)x^\alpha-3\alpha x^\alpha+3x^\alpha\,$

$(\alpha(\alpha-1)-3\alpha+3)x^\alpha.\,$

For this to indeed be a solution, either x=0 giving the trivial solution, or the coefficient of x^α is zero, so solving that quadratic, we get α=1,3. So, the general solution is

$u=Ax+Bx^3.\,$

[edit] Difference equation analogue

There is a difference equation analogue to the Cauchy–Euler equation. For a fixed $m > 0$ , define the sequence $f m (n)$ as

$f_m(n) := n(n+1)\cdots (n+m-1).$

Applying the difference operator to $f m$ , we find that

$Df_m(n) = f_{m+1}(n) - f_m(n) = m(n+1)(n+2) \cdots (n+m-1) = \frac{m}{n} f_m(n).$

If we do this k times, we will find that

$f_m^{(k)}(n) = \frac{m(m-1)\cdots(m-k+1)}{n(n+1)\cdots(n+k-1)} f_m(n) = m(m-1)\cdots(m-k+1) \frac{f_m(n)}{f_k(n)},$

where the superscript ^(k) denotes applying the difference operator k times. Comparing this to the fact that the k-th derivative of $x m$ equals $m(m-1)\cdots(m-k+1)\frac{x^m}{x^k}$ suggests that we can solve the N-th order difference equation

$f_N(n) y^{(N)}(n) + a_{N-1} f_{N-1}(n) y^{(N-1)}(n) + \cdots + a_0 y(n) = 0,$

in a similar manner to the differential equation case. Indeed, substituting the trial solution

y (n) = f m (n)

brings us to the same situation as the differential equation case,

$m(m-1)\cdots(m-N+1) + a_{N-1} m(m-1) \cdots (m-N+2) + \cdots + a_1 m + a_0 = 0.$

One may now proceed as in the differential equation case, since the general solution of an N-th order linear difference equation is also the linear combination of N linearly independent solutions. Applying reduction of order in case of a multiple root $m 1$ will yield expressions involving a discrete version of $ln$ ,

$\varphi(n) = \sum_{k=1}^n \frac{1}{k - m_1}.$