Euler's differential equation

The Euler equation (by Leonhard Euler ) is a linear differential equation of higher order with non-constant coefficients of the particular form

${\displaystyle \sum _{k=0}^{N}a_{k}\,(cx+d)^{k}\;y^{(k)}(x)=b(x)\ ,\ cx+d>0}$

to given and inhomogeneity . If one knows a fundamental system of the homogeneous solution, one can determine the general solution of the inhomogeneous equation with the procedure of the variation of the constants . Therefore only needs to be considered. ${\displaystyle N\in \mathbb {N} ,\ a_{0},\ldots ,a_{N},c,d\in \mathbb {R} ,\ c\neq 0}$ ${\displaystyle b}$ ${\displaystyle b\equiv 0}$

Euler's differential equation is converted into a linear differential equation with constant coefficients by means of the transformation . ${\displaystyle z(t):=y\left({\tfrac {e^{t}-d}{c}}\right)}$

Motivation of the transformation

Be a sufficiently smooth function and ${\displaystyle y}$

${\displaystyle z(x):=y\left({\frac {e^{x}-d}{c}}\right)}$, so . ${\displaystyle \ y(x)=z(\ln(cx+d))}$

Then applies

${\displaystyle {\begin{array}{lll}y'(x)&=&{\frac {c}{cx+d}}z'(\ln(cx+d))\ ,\\y''(x)&=&{\frac {c^{2}}{(cx+d)^{2}}}z''(\ln(cx+d))-{\frac {c^{2}}{(cx+d)^{2}}}z'(\ln(cx+d))\ ,\\\end{array}}}$

so

${\displaystyle {\begin{array}{lll}(cx+d)y'(x)&=&c\cdot z'(\ln(cx+d))\ ,\\(cx+d)^{2}y''(x)&=&c^{2}\cdot [z''-z'](\ln(cx+d))\ .\\\end{array}}}$

In this respect, Euler's second order differential equation would transform into a linear differential equation with constant coefficients. The following questions now arise:

• Does this transformation also convert the higher order terms into those with constant coefficients? ${\displaystyle (cx+d)^{k}y^{(k)}(x)}$
• How can one calculate the coefficients on the right-hand side more easily without deriving the transformation enough times each time?

These questions are clarified by the following transformation theorem:

The transformation set

Let the linear differential equation be solved with constant coefficients ${\displaystyle z}$

${\displaystyle \sum _{k=0}^{n}a_{k}c^{k}\left(\left[\prod _{j=0}^{k-1}\left({\frac {\rm {d}}{{\rm {d}}x}}-j\right)\right]z\right)(x)=0\ .}$

Then

${\displaystyle \ y(x):=z(\ln(cx+d))}$

a solution of Euler's (homogeneous) differential equation

${\displaystyle \sum _{k=0}^{N}a_{k}(cx+d)^{k}y^{(k)}(x)=0\ ,\ cx+d>0\ .}$

Explanation of the notation

Here, the differential operators are first linked with each other (comparable to multiplying) before they are applied to a function, for example:

${\displaystyle \left[\prod _{j=0}^{-1}\left({\frac {\rm {d}}{{\rm {d}}x}}-j\right)\right]z=z\ ,}$

${\displaystyle \left[\prod _{j=0}^{0}\left({\frac {\rm {d}}{{\rm {d}}x}}-j\right)\right]z=\left({\frac {\rm {d}}{{\rm {d}}x}}-0\right)z=z'\ ,}$

${\displaystyle \left[\prod _{j=0}^{1}\left({\frac {\rm {d}}{{\rm {d}}x}}-j\right)\right]z=\left({\frac {\rm {d}}{{\rm {d}}x}}-0\right)\left({\frac {\rm {d}}{{\rm {d}}x}}-1\right)z=\left({\frac {\rm {d^{2}}}{{\rm {d}}x^{2}}}-{\frac {\rm {d}}{{\rm {d}}x}}\right)z=z''-z'\ ,}$

${\displaystyle \left[\prod _{j=0}^{2}\left({\frac {\rm {d}}{{\rm {d}}x}}-j\right)\right]z=\left({\frac {\rm {d}}{{\rm {d}}x}}-0\right)\left({\frac {\rm {d}}{{\rm {d}}x}}-1\right)\left({\frac {\rm {d}}{{\rm {d}}x}}-2\right)z=\left({\frac {\rm {d^{3}}}{{\rm {d}}x^{3}}}-3{\frac {\rm {d^{2}}}{{\rm {d}}x^{2}}}+2{\frac {\rm {d}}{{\rm {d}}x}}\right)z=z'''-3z''+2z'\ .}$

proof

To show is only for everyone . This is done by means of complete induction . The beginning of induction is trivial. Provided that the identity is valid for , this identity can be differentiated. It turns out ${\displaystyle c^{k}\left(\left[\prod _{j=0}^{k-1}\left({\frac {\rm {d}}{{\rm {d}}x}}-j\right)\right]z\right)(\ln(cx+d))=(cx+d)^{k}y^{(k)}(x)}$ ${\displaystyle k\in \mathbb {N} _{0}}$ ${\displaystyle k=0}$ ${\displaystyle k_{0}\in \mathbb {N} _{0}}$

${\displaystyle (cx+d)^{k_{0}}y^{(k_{0}+1)}(x)+ck_{0}(cx+d)^{k_{0}-1}y^{(k_{0})}(x)={\frac {c^{k_{0}+1}}{cx+d}}\left({\frac {\rm {d}}{{\rm {d}}x}}\left[\prod _{j=0}^{k_{0}-1}\left({\frac {\rm {d}}{{\rm {d}}x}}-j\right)\right]z\right)(\ln(cx+d))\ .}$

Applying the induction hypothesis implies

${\displaystyle {\begin{array}{lll}(cx+d)^{k_{0}+1}y^{(k_{0}+1)}(x)&=&c^{k_{0}+1}\left({\frac {\rm {d}}{{\rm {d}}x}}\left[\prod _{j=0}^{k_{0}-1}\left({\frac {\rm {d}}{{\rm {d}}x}}-j\right)\right]z\right)(\ln(cx+d))-ck_{0}(cx+d)^{k_{0}}y^{(k_{0})}(x)\\&=&c^{k_{0}+1}\left({\frac {\rm {d}}{{\rm {d}}x}}\left[\prod _{j=0}^{k_{0}-1}\left({\frac {\rm {d}}{{\rm {d}}x}}-j\right)\right]z\right)(\ln(cx+d))\\&&\quad -c^{k_{0}+1}k_{0}\left(\left[\prod _{j=0}^{k_{0}-1}\left({\frac {\rm {d}}{{\rm {d}}x}}-j\right)\right]z\right)(\ln(cx+d))\\&=&c^{k_{0}+1}\left(\left[\prod _{j=0}^{k_{0}}\left({\frac {\rm {d}}{{\rm {d}}x}}-j\right)\right]z\right)(\ln(cx+d))\ .\\\end{array}}}$

${\displaystyle \Box }$

Conclusion: construction of a fundamental system

The characteristic equation for the differential equation of is ${\displaystyle z}$

${\displaystyle \chi (\lambda )=\sum _{k=0}^{n}a_{k}c^{k}\prod _{j=0}^{k-1}(\lambda -j)=0\ .}$

Now denote the zeros of the characteristic polynomial and the multiplicity of , so forms ${\displaystyle \lambda _{1},\ldots ,\lambda _{M}}$ ${\displaystyle \chi (\lambda )}$ ${\displaystyle R_{j}}$ ${\displaystyle \lambda _{j}}$

${\displaystyle \{z_{j,k}(x)=e^{\lambda _{j}z}z^{k}\ |\ j=1,\ldots ,M\ ,\ k=0,\ldots ,R_{j}-1\}}$

a fundamental system of the equation for . So is ${\displaystyle z}$

${\displaystyle \{y_{j,k}(x)=(cx+d)^{\lambda _{j}}[\ln(cx+d)]^{k}\ |\ j=1,\ldots ,M\ ,\ k=0,\ldots ,R_{j}-1\}}$

a fundamental system of the (homogeneous) Euler differential equation.

example

Euler's differential equation is given

${\displaystyle a_{2}x^{2}y''(x)+a_{1}xy'(x)+a_{0}y(x)=0\ ,\ a_{2}\neq 0\ ,\ x>0\ .}$

According to the above sentence, the following linear differential equation with constant coefficients has to be solved

${\displaystyle a_{2}(z''(x)-z'(x))+a_{1}z'(x)+a_{0}z(x)=0\ ,}$

so

${\displaystyle a_{2}z''(x)+(a_{1}-a_{2})z'(x)+a_{0}z(x)=0\ .}$

The characteristic polynomial belonging to this differential equation is

${\displaystyle \chi (\lambda )=\ a_{2}\lambda ^{2}+(a_{1}-a_{2})\lambda +a_{0}}$

and has the zeros

${\displaystyle \lambda _{1,2}={\frac {a_{2}-a_{1}}{2a_{2}}}\pm {\sqrt {{\frac {(a_{2}-a_{1})^{2}}{4a_{2}^{2}}}-{\frac {a_{0}}{a_{2}}}}}\ .}$

Case 1 :, both real. ${\displaystyle \lambda _{1}\neq \lambda _{2}}$

Then there is a fundamental system for the transformed linear differential equation. The inverse transformation yields that there is a fundamental system for the original Euler's differential equation. ${\displaystyle \{e^{\lambda _{1}z},e^{\lambda _{2}z}\}}$ ${\displaystyle \{x^{\lambda _{1}},x^{\lambda _{2}}\}}$

Case 2: . ${\displaystyle \ \lambda _{1}=\lambda _{2}}$

Then there is a double zero of the characteristic polynomial. Hence is a fundamental system for the transformed linear differential equation. The inverse transformation yields that there is a fundamental system for the original Euler's differential equation. ${\displaystyle \lambda :={\frac {a_{2}-a_{1}}{2a_{2}}}}$ ${\displaystyle \ \{e^{\lambda z},ze^{\lambda z}\}}$ ${\displaystyle \ \{x^{\lambda },x^{\lambda }\ln x\}}$

Case 3: both not real. ${\displaystyle \ \lambda _{1},\lambda _{2}}$

Then are complex conjugate to each other. So is a (complex) fundamental system. Be , . Then is a real fundamental system of the transformed linear differential equation. Inverse transformation provides the fundamental system for the original Euler's differential equation. ${\displaystyle \ \lambda _{1},\lambda _{2}}$ ${\displaystyle \ \{e^{\lambda _{1}z},e^{\lambda _{2}z}\}}$ ${\displaystyle \ \lambda _{1}=\mu +i\nu }$ ${\displaystyle \mu ,\nu \in \mathbb {R} }$ ${\displaystyle \ \{e^{\mu z}\sin(\nu z),e^{\mu z}\cos(\nu z)\}}$ ${\displaystyle \ \{x^{\mu }\sin(\nu \ln x),x^{\mu }\cos(\nu \ln x)\}}$

${\displaystyle \Box }$

literature

• Earl A. Coddington, Norman Levinson : Theory of Ordinary Differential Equations . McGraw-Hill, New York 1955, ISBN 978-0-07-011542-2 .
• Harro Heuser : Ordinary differential equations, page 240, Vieweg + Teubner, Stuttgart 2009, ISBN 978-3-8348-0705-2