Template:Short description 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 equationEdit
Let Template:Math be the nth derivative of the unknown function Template:Math. Then a Cauchy–Euler equation of order n has the form <math display="block">a_{n} x^n y^{(n)}(x) + a_{n-1} x^{n-1} y^{(n-1)}(x) + \dots + a_0 y(x) = 0.</math>
The substitution <math>x = e^u</math> (that is, <math>u = \ln(x)</math>; for <math>x < 0</math>, in which one might replace all instances of <math>x</math> by <math>|x|</math>, extending the solution's domain to <math>\reals \setminus \{0\}</math>) can be used to reduce this equation to a linear differential equation with constant coefficients. Alternatively, the trial solution <math>y = x^m</math> can be used to solve the equation directly, yielding the basic solutions.<ref name="kreyszig">Template:Cite book</ref>
Second order – solving through trial solutionEdit
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<ref name="kreyszig" /><ref>Template:Cite book</ref>
<math display="block">x^2\frac{d^2y}{dx^2} + ax\frac{dy}{dx} + by = 0.</math>
We assume a trial solution<ref name="kreyszig" /> <math display="block">y = x^m.</math>
Differentiating gives <math display="block">\frac{dy}{dx} = mx^{m-1} </math> and <math display="block">\frac{d^2y}{dx^2} = m\left(m-1\right)x^{m-2}. </math>
Substituting into the original equation leads to requiring that <math display="block">x^2\left( m\left(m-1 \right)x^{m-2} \right) + ax\left( mx^{m-1} \right) + b\left( x^m \right) = 0</math>
Rearranging and factoring gives the indicial equation <math display="block">m^2 + \left(a-1\right)m + b = 0.</math>
We then solve for m. There are three cases of interest:
- Case 1 of two distinct roots, Template:Math and Template:Math;
- Case 2 of one real repeated root, Template:Mvar;
- Case 3 of complex roots, Template:Math.
In case 1, the solution is <math display="block">y = c_1 x^{m_1} + c_2 x^{m_2}</math>
In case 2, the solution is <math display="block">y = c_1 x^m \ln(x) + c_2 x^m </math>
To get to this solution, the method of reduction of order must be applied, after having found one solution Template:Math.
In case 3, the solution is <math display="block">y = c_1 x^\alpha \cos(\beta \ln(x)) + c_2 x^\alpha \sin(\beta \ln(x)) </math> <math display="block">\alpha = \operatorname{Re}(m)</math> <math display="block">\beta = \operatorname{Im}(m)</math>
For <math>c_1, c_2 \isin \R</math>.
This form of the solution is derived by setting Template:Math and using Euler's formula.
Second order – solution through change of variablesEdit
<math display="block">x^2\frac{d^2y}{dx^2} +ax\frac{dy}{dx} + by = 0 </math>
We operate the variable substitution defined by
<math display="block">t = \ln(x). </math> <math display="block">y(x) = \varphi(\ln(x)) = \varphi(t). </math>
Differentiating gives <math display="block">\frac{dy}{dx}=\frac{1}{x}\frac{d\varphi}{dt}</math> <math display="block">\frac{d^2y}{dx^2}=\frac{1}{x^2}\left(\frac{d^2\varphi}{dt^2}-\frac{d\varphi}{dt}\right).</math>
Substituting <math>\varphi(t)</math> the differential equation becomes <math display="block">\frac{d^2\varphi}{dt^2} + (a-1)\frac{d\varphi}{dt} + b\varphi = 0.</math>
This equation in <math>\varphi(t)</math> is solved via its characteristic polynomial <math display="block">\lambda^2 + (a-1)\lambda + b = 0.</math>
Now let <math>\lambda_1</math> and <math>\lambda_2</math> 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 <math display="block">\varphi(t)=c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t},</math> where the exponentials may be complex.
If the roots are equal, the general solution is <math display="block">\varphi(t)=c_1 e^{\lambda_1 t} + c_2 t e^{\lambda_1 t}.</math>
In both cases, the solution <math>y(x)</math> can be found by setting <math>t = \ln(x)</math>.
Hence, in the first case, <math display="block">y(x) = c_1 x^{\lambda_1} + c_2 x^{\lambda_2},</math> and in the second case, <math display="block">y(x) = c_1 x^{\lambda_1} + c_2 \ln(x) x^{\lambda_1}.</math>
Second order - solution using differential operatorsEdit
Observe that we can write the second-order Cauchy-Euler equation in terms of a linear differential operator <math> L </math> as <math display="block">Ly = (x^2 D^2 + axD + bI)y = 0,</math> where <math> D = \frac{d}{dx} </math> and <math> I </math> is the identity operator.
We express the above operator as a polynomial in <math> xD </math>, rather than <math> D </math>. By the product rule, <math display="block"> (x D)^2 = x D(x D) = x(D + x D^2) = x^2D^2 + x D.</math> So, <math display="block"> L = (xD)^2 + (a-1)(xD) + bI.</math>
We can then use the quadratic formula to factor this operator into linear terms. More specifically, let <math> \lambda_1, \lambda_2 </math> denote the (possibly equal) values of <math display="block">-\frac{a-1}{2} \pm \frac{1}{2}\sqrt{(a-1)^2 - 4b}. </math> Then, <math display="block">L = (xD - \lambda_1 I)(xD - \lambda_2 I).</math>
It can be seen that these factors commute, that is <math>(xD - \lambda_1 I)(xD - \lambda_2 I) = (xD - \lambda_2 I)(xD - \lambda_1 I)</math>. Hence, if <math> \lambda_1 \neq \lambda_2 </math>, the solution to <math> Ly = 0 </math> is a linear combination of the solutions to each of <math> (xD - \lambda_1 I)y = 0 </math> and <math> (xD - \lambda_2 I)y = 0 </math>, which can be solved by separation of variables.
Indeed, with <math> i \in \{1,2\} </math>, we have <math> (xD - \lambda_i I)y = x\frac{dy}{dx} - \lambda_i y = 0 </math>. So, <math display="block">\begin{align} x\frac{dy}{dx} &= \lambda_i y\\ \int \frac{1}{y}\, dy &= \lambda_i \int \frac{1}{x}\, dx\\ \ln y &= \lambda_i \ln x + C\\ y &= c_i e^{\lambda_i \ln x} = c_i x^{\lambda_i}.\end{align}</math> Thus, the general solution is <math> y = c_1 x^{\lambda_1} + c_2 x^{\lambda_2} </math>.
If <math> \lambda = \lambda_1 = \lambda_2 </math>, then we instead need to consider the solution of <math>(xD - \lambda I)^2y = 0 </math>. Let <math> z = (xD-\lambda I)y </math>, so that we can write <math display="block"> (xD - \lambda I)^2y = (xD - \lambda I)z = 0.</math> As before, the solution of <math> (xD- \lambda I)z = 0 </math> is of the form <math> z = c_1x^\lambda </math>. So, we are left to solve <math display="block"> (xD - \lambda I)y = x\frac{dy}{dx} - \lambda y = c_1x^\lambda.</math> We then rewrite the equation as <math display="block"> \frac{dy}{dx} - \frac{\lambda}{x} y = c_1x^{\lambda-1},</math> which one can recognize as being amenable to solution via an integrating factor.
Choose <math> M(x) = x^{-\lambda} </math> as our integrating factor. Multiplying our equation through by <math> M(x) </math> and recognizing the left-hand side as the derivative of a product, we then obtain <math display="block">\begin{align} \frac{d}{dx}(x^{-\lambda} y) &= c_1x^{-1}\\ x^{-\lambda} y &= \int c_1x^{-1}\, dx\\ y &= x^\lambda (c_1\ln(x) + c_2)\\ &= c_1\ln(x)x^\lambda +c_2 x^\lambda.\end{align}</math>
ExampleEdit
Given <math display="block">x^2 u - 3xu' + 3u = 0\,,</math> we substitute the simple solution Template:Math: <math display="block">x^2\left(m\left(m-1\right)x^{m-2}\right)-3x\left(m x^{m-1}\right) + 3x^m = m\left(m-1\right)x^m - 3m x^m+3x^m = \left(m^2 - 4m + 3\right)x^m = 0\,.</math>
For Template:Math to be a solution, either Template:Math, which gives the trivial solution, or the coefficient of Template:Math is zero. Solving the quadratic equation, we get Template:Math. The general solution is therefore
- <math>u=c_1 x+c_2 x^3\,.</math>
Difference equation analogueEdit
There is a difference equation analogue to the Cauchy–Euler equation. For a fixed Template:Math, define the sequence Template:Math as <math display="block">f_m(n) := n (n+1) \cdots (n+m-1).</math>
Applying the difference operator to <math>f_m</math>, we find that <math display="block">\begin{align} Df_m(n) & = f_{m}(n+1) - f_m(n) \\ & = m(n+1)(n+2) \cdots (n+m-1) = \frac{m}{n} f_m(n). \end{align}</math>
If we do this Template:Mvar times, we find that <math display="block">\begin{align} 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)}, \end{align}</math>
where the superscript Template:Math denotes applying the difference operator Template:Mvar times. Comparing this to the fact that the Template:Mvar-th derivative of Template:Math equals <math display="block">m(m-1) \cdots (m-k+1)\frac{x^m}{x^k}</math> suggests that we can solve the N-th order difference equation <math display="block">f_N(n) y^{(N)}(n) + a_{N-1} f_{N-1}(n) y^{(N-1)}(n) + \cdots + a_0 y(n) = 0,</math> in a similar manner to the differential equation case. Indeed, substituting the trial solution <math display="block">y(n) = f_m(n) </math> brings us to the same situation as the differential equation case, <math display="block">m(m-1)\cdots(m-N+1) + a_{N-1} m(m-1) \cdots (m-N+2) + \dots + a_1 m + a_0 = 0.</math>
One may now proceed as in the differential equation case, since the general solution of an Template:Mvar-th order linear difference equation is also the linear combination of Template:Mvar linearly independent solutions. Applying reduction of order in case of a multiple root Template:Math will yield expressions involving a discrete version of Template:Math, <math display="block">\varphi(n) = \sum_{k=1}^n \frac{1}{k - m_1}.</math>
(Compare with: <math display="inline">\ln (x - m_1) = \int_{1+m_1}^x \frac{dt}{t - m_1} .</math>)
In cases where fractions become involved, one may use <math display="block">f_m(n) := \frac{\Gamma(n+m)}{\Gamma(n)}</math> instead (or simply use it in all cases), which coincides with the definition before for integer Template:Mvar.
See alsoEdit
ReferencesEdit
<references />