Cauchy-Euler equation

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.

The equation

Let $y^\{(n)\}(x)$ be the "n"th 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.

Second order

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

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

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, "m"₁ and "m"₂
* Case #2: One real repeated root, "m"
* Case #3: Complex roots, α ± "i"β

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:

:

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.

Example

Given : $x^2u"-3xu\text{'}+3u=0,$we substitute the simple solution "x"^α:: $x^2(alpha(alpha-1)x^\{alpha-2\})-3x(alpha(x^\{alpha-1\}))+3(x^alpha)=alpha(alpha-1)x^alpha-3alpha\; x^alpha+3x^alpha,$: $(alpha(alpha-1)-3alpha+3)x^alpha.,$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: $u=Ax+Bx^3.,$

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 $ln$,

: $varphi(n)\; =\; sum\_\{k=1\}^n\; frac\{1\}\{k\; -\; m\_1\}.$

(Compare with: $ln\; (x\; -\; m\_1)\; =\; int\_\{1+m\_1\}^x\; frac\{1\}\{t\; -\; m\_1\}\; ,\; dt.$)

In cases where fractions become involved, one may use

: $f\_m(n)\; :=\; frac\{Gamma(n+m)\}\{Gamma(n)\}$

instead (or simply use it in all cases), which coincides with the definition before for integer "m".

ee also

* Hypergeometric differential equation