Cauchy–Euler equation explained

In mathematics, an Euler–Cauchy equation, or Cauchy–Euler equation, or simply Euler's equation, is a linear homogeneous ordinary differential equation with variable coefficients. It is sometimes referred to as an equidimensional equation. Because of its particularly simple equidimensional structure, the differential equation can be solved explicitly.

The equation

Let be the nth derivative of the unknown function . Then a Cauchy–Euler equation of order n has the form$$a\_\; x^n\; y^(x)\; +\; a\_\; x^\; y^(x)\; +\; \backslash dots\; +\; a\_0\; y(x)\; =\; 0.$$

The substitution

x = e ^{u} (that is, u = ln (x) ; for x < 0 , in which one might replace all instances of x by |x| , extending the solution's domain to \reals \setminus \{ 0 \} ) can be used to reduce this equation to a linear differential equation with constant coefficients. Alternatively, the trial solution y = x ^{m} can be used to solve the equation directly, yielding the basic solutions.^[1]

Second order – solving through trial solution

The most common Cauchy–Euler equation is the second-order equation, which appears in a number of physics and engineering applications, such as when solving Laplace's equation in polar coordinates. The second order Cauchy–Euler equation is^[2]

$$x^2\backslash frac\; +\; ax\backslash frac\; +\; by\; =\; 0.$$

We assume a trial solution $$y\; =\; x^m.$$

Differentiating gives $$\backslash frac\; =\; mx^$$ and $$\backslash frac\; =\; m\backslash left(m-1\backslash right)x^.$$

Substituting into the original equation leads to requiring that$$x^2\backslash left(m\backslash left(m-1\; \backslash right)x^\; \backslash right)\; +\; ax\backslash left(mx^\; \backslash right)\; +\; b\backslash left(x^m\; \backslash right)\; =\; 0$$

Rearranging and factoring gives the indicial equation$$m^2\; +\; \backslash left(a-1\backslash right)m\; +\; b\; =\; 0.$$

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

• Case 1 of two distinct roots, and ;
• Case 2 of one real repeated root, ;
• Case 3 of complex roots, .

In case 1, the solution is $$y\; =\; c\_1\; x^\; +\; c\_2\; x^$$

In case 2, the solution is $$y\; =\; c\_1\; x^m\; \backslash ln(x)\; +\; c\_2\; x^m$$

To get to this solution, the method of reduction of order must be applied, after having found one solution .

In case 3, the solution is $$y\; =\; c\_1\; x^\backslash alpha\; \backslash cos(\backslash beta\; \backslash ln(x))\; +\; c\_2\; x^\backslash alpha\; \backslash sin(\backslash beta\; \backslash ln(x))$$$$\backslash alpha\; =\; \backslash operatorname(m)$$$$\backslash beta\; =\; \backslash operatorname(m)$$

For

c _{1,} c _{2} \isin \R .

This form of the solution is derived by setting and using Euler's formula

Second order – solution through change of variables

$$x^2\backslash frac\; +ax\backslash frac\; +\; by\; =\; 0$$

We operate the variable substitution defined by

$$t\; =\; \backslash ln(x).$$$$y(x)\; =\; \backslash varphi(\backslash ln(x))\; =\; \backslash varphi(t).$$

Differentiating gives$$\backslash frac=\backslash frac\backslash frac$$$$\backslash frac=\backslash frac\backslash left(\backslash frac-\backslash frac\backslash right).$$

Substituting

\varphi(t) the differential equation becomes$$\backslash frac\; +\; (a-1)\backslash frac\; +\; b\backslash varphi\; =\; 0.$$

This equation in

\varphi(t) is solved via its characteristic polynomial$$\backslash lambda^2\; +\; (a-1)\backslash lambda\; +\; b\; =\; 0.$$

Now let

λ _{1} and λ _{2} denote the two roots of this polynomial. We analyze the case in which there are distinct roots and the case in which there is a repeated root:

If the roots are distinct, the general solution is $$\backslash varphi(t)=c\_1\; e^\; +\; c\_2\; e^,$$ where the exponentials may be complex.

If the roots are equal, the general solution is $$\backslash varphi(t)=c\_1\; e^\; +\; c\_2\; t\; e^.$$

In both cases, the solution

y(x) can be found by setting t = ln (x) .

Hence, in the first case, $$y(x)\; =\; c\_1\; x^\; +\; c\_2\; x^,$$ and in the second case, $$y(x)\; =\; c\_1\; x^\; +\; c\_2\; \backslash ln(x)\; x^.$$

Second order - solution using differential operators

L as $$Ly\; =\; (x^2\; D^2\; +\; axD\; +\; bI)y\; =\; 0,$$ where D = (d)/(dx) and I is the identity operator.

We express the above operator as a polynomial in

xD , rather than D . By the product rule, $$(x\; D)^2\; =\; x\; D(x\; D)\; =\; x(D\; +\; x\; D^2)\; =\; x^2D^2\; +\; x\; D.$$ So, $$L\; =\; (xD)^2\; +\; (a-1)(xD)\; +\; bI.$$

We can then use the quadratic formula to factor this operator into linear terms. More specifically, let

λ _{1,} λ _{2} denote the (possibly equal) values of $$-\backslash frac\; \backslash pm\; \backslash frac\backslash sqrt.$$ Then, $$L\; =\; (xD\; -\; \backslash lambda\_1\; I)(xD\; -\; \backslash lambda\_2\; I).$$

It can be seen that these factors commute, that is

(xD - λ _{1} I)(xD - λ _{2} I) = (xD - λ _{2} I)(xD - λ _{1} I) . Hence, if λ _{1} ≠ λ _{2} , the solution to Ly = 0 is a linear combination of the solutions to each of (xD - λ _{1} I)y = 0 and (xD - λ _{2} I)y = 0 , which can be solved by separation of variables.

Indeed, with

i \in \{ 1 , 2 \} , we have (xD - λ _{i} I)y = x (dy)/(dx) - λ _{i} y = 0 . So, Thus, the general solution is y = c _{1} x ^{λ _{1}} + c _{2} x ^{λ _{2}} .

If

λ = λ _{1} = λ _{2} , then we instead need to consider the solution of (xD - λ I) ^{2y} = 0 . Let z = (xD -λ I)y , so that we can write $$(xD\; -\; \backslash lambda\; I)^2y\; =\; (xD\; -\; \backslash lambda\; I)z\; =\; 0.$$ As before, the solution of (xD - λ I)z = 0 is of the form z = c ^{λ} _{1 x} . So, we are left to solve $$(xD\; -\; \backslash lambda\; I)y\; =\; x\backslash frac\; -\; \backslash lambda\; y\; =\; c\_1x^\backslash lambda.$$ We then rewrite the equation as $$\backslash frac\; -\; \backslash frac\; y\; =\; c\_1x^,$$ which one can recognize as being amenable to solution via an integrating factor.

Choose

M(x) = x ^{-λ} as our integrating factor. Multiplying our equation through by M(x) and recognizing the left-hand side as the derivative of a product, we then obtain

Example

Given$$x^2\; u$$ - 3xu' + 3u = 0\,,we substitute the simple solution :$$x^2\backslash left(m\backslash left(m-1\backslash right)x^\backslash right)-3x\backslash left(m\; x^\backslash right)\; +\; 3x^m\; =\; m\backslash left(m-1\backslash right)x^m\; -\; 3m\; x^m+3x^m\; =\; \backslash left(m^2\; -\; 4m\; +\; 3\backslash right)x^m\; =\; 0\backslash ,.$$

For to be a solution, either, which gives the trivial solution, or the coefficient of is zero. Solving the quadratic equation, we get . The general solution is therefore

u = c _{1} x + c _{2} x ^{3.}

Difference equation analogue

There is a difference equation analogue to the Cauchy–Euler equation. For a fixed, define the sequence as$$f\_m(n)\; :=\; n\; (n+1)\; \backslash cdots\; (n+m-1).$$

Applying the difference operator to

f _{m} , we find that

If we do this times, we find that

where the superscript denotes applying the difference operator times. Comparing this to the fact that the -th derivative of equals$$m(m-1)\; \backslash cdots\; (m-k+1)\backslash frac$$suggests that we can solve the N-th order difference equation$$f\_N(n)\; y^(n)\; +\; a\_\; f\_(n)\; y^(n)\; +\; \backslash cdots\; +\; a\_0\; y(n)\; =\; 0,$$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,$$m(m-1)\backslash cdots(m-N+1)\; +\; a\_\; m(m-1)\; \backslash cdots\; (m-N+2)\; +\; \backslash dots\; +\; a\_1\; m\; +\; a\_0\; =\; 0.$$

One may now proceed as in the differential equation case, since the general solution of an -th order linear difference equation is also the linear combination of linearly independent solutions. Applying reduction of order in case of a multiple root will yield expressions involving a discrete version of ,$$\backslash varphi(n)\; =\; \backslash sum\_^n\; \backslash frac.$$

(Compare with: $\backslash ln\; (x\; -\; m\_1)\; =\; \backslash int\_^x\; \backslash frac\; .$)

In cases where fractions become involved, one may use $$f\_m(n)\; :=\; \backslash frac$$ instead (or simply use it in all cases), which coincides with the definition before for integer .

See also

• Hypergeometric differential equation
• Cauchy–Euler operator

References

1. Book: Kreyszig, Erwin. Advanced Engineering Mathematics . Wiley. May 10, 2006. 978-0-470-08484-7.
2. Book: Elementary Differential Equations and Boundary Value Problems. Boyce. William E.. 272–273. DiPrima. Richard C.. Rosatone. Laurie. 10th. 2012. 978-0-470-45831-0.