Euler–Cauchy equation

The general second-order Euler–Cauchy equation is:

$ax^{2}y''+bxy'+cy=0$

where a, b, and c can be any (real) constants.

Solving by the Method of Frobenius

The leading coefficient function $ax^{2}$ is 0 at $x=0$; so there is a singular point at $x=0$. Normal form is:

$y''+p(x)y'+q(x)y=0$

where $p(x)={b \over ax}$ and $q(x)={c \over ax^{2}}$.

Then $xp(x)={b \over a}$ and $x^{2}q(x)={c \over a}$. These two functions, $xp(x)$ and $x^{2}q(x)$, have no discontinuities; so the singular point is regular. Thus the method of Frobenius is applicable to this E.C.e. at $x=0$.

Let

$y=\sum _{n=0}^{\infty }k_{n}x^{n+r}$

be a tentative Frobenius-series solution. Then

$y'=\sum _{n=0}^{\infty }k_{n}(n+r)x^{n+r-1}$

$y''=\sum _{n=0}^{\infty }k_{n}(n+r)(n+r-1)x^{n+r-2}$

Substituting these into the E.C.e. yields

$\sum _{k=0}^{\infty }k_{n}[a(n+r)(n+r-1)+b(n+r)+c]x^{n+r}=0$

Since the L.H.S. is identically equal to 0, then each coefficient of the Frobenius series must equal 0:

$k_{n}[a(n+r)(n+r-1)+b(n+r)+c]=0$

$k_{n}[a(n+r)^{2}+(b-a)(n+r)+c]=0$

When $n=0$,

$k_{0}(ar^{2}+(b-a)r+c)=0$

Letting $k_{0}\neq 0$, then

$ar^{2}+(b-a)r+c=0$

is the indicial equation. Its roots are

$r_{1}={-(b-a)+{\sqrt {(b-a)^{2}-4ac}} \over 2a}$

$r_{2}={-(b-a)-{\sqrt {(b-a)^{2}-4ac}} \over 2a}$

If there exists some integer $n_{1}$ such that

$a(n_{1}+r)^{2}+(b-a)(n_{1}+r)+c=0$

then either $r_{2}=n_{1}+r_{1}$ or $r_{1}=n_{1}+r_{2}$ (since $r_{1}$ and $r_{2}$ are the only two solutions of the indicial equation); and there can be at most one such $n_{1}$. For all other values of $n$, the equation

$a(n+r)^{2}+(b-a)(n+r)+c=0$

is not satisfied, which means that $k_{n}=0$ for all $n$ such that $n\neq 0$ and $n\neq n_{1}$. (N.B.: If $r_{2}-r_{1}$ is not an integer then such an $n_{1}$ does not exist.)

$y_{1}=k_{0}x^{r_{1}}$ or $y_{1}=k_{0}x^{r_{1}}+k_{n_{1}}x^{r_{1}+n_{1}}=k_{0}x^{r_{1}}+k_{n_{1}}x^{r_{2}}$

$y_{2}=h_{0}x^{r_{2}}$ or $y_{2}=h_{0}x^{r_{2}}+h_{n_{1}}x^{r_{2}+n_{1}}=h_{0}x^{r_{2}}+h_{n_{1}}x^{r_{1}}$

The two right sides of the two logical disjunctions cannot be simultaneously true. Since $k_{n_{1}}$ is independent of $k_{0}$, then a solution such as $y_{1}=k_{0}x^{r_{1}}+k_{n_{1}}x^{r_{2}}$ is a linear combination of linearly independent solutions.

${\Bigg |}{\begin{matrix}x^{r_{1}}&x^{r_{2}}\\r_{1}x^{r_{1}-1}&r_{2}x^{r_{2}-1}\end{matrix}}{\Bigg |}=r_{2}x^{r_{1}}x^{r_{2}-1}-r_{1}x^{r_{1}-1}x^{r_{2}}=(r_{2}-r_{1})x^{r_{1}+r_{2}-1}$

Assuming that $r_{2}\neq r_{1}$ then this Wronskian is never zero, so the two terms of the solution $y_{1}$ are actually linearly independent solutions. (The case when $r_{2}=r_{1}$ will be dealt with further down from here.) The second term of $y_{1}$ is just “echoing” the other solution $y_{2}=h_{0}x^{r_{2}}$, so to speak. So the general solution is

$y=k_{0}x^{r_{1}}+h_{0}x^{r_{2}}$

The case of complex conjugate indicial roots

If $r_{1,2}$ are a complex conjugate pair,

$r_{1,2}=\alpha \pm i\beta$

then

$y=k_{0}x^{\alpha }x^{i\beta }+h_{0}x^{\alpha }x^{-i\beta }$

$y=k_{0}x^{\alpha }e^{i\beta \ln x}+h_{0}x^{\alpha }e^{-i\beta \ln x}$

If $k_{0}$ is set to $A/2$ and $h_{0}$ is set to $A/2$ as well then

$y=Ax^{\alpha }\cos(\beta \ln x)$

as may be derived through Euler's formula $e^{i\theta }=\cos \theta +i\sin \theta$. If $k_{0}={B \over 2i}$ and $h_{0}={-B \over 2i}$ then

$y=Bx^{\alpha }\sin(\beta \ln x)$

These two solutions are linearly independent as can be verified by calculating their Wronskian:

${\Bigg |}{\begin{matrix}x^{\alpha }\cos(\beta \ln x)&x^{\alpha }\sin(\beta \ln x)\\x^{\alpha -1}(\alpha \cos(\beta \ln x)-\beta \sin(\beta \ln x))&x^{\alpha -1}(\alpha \sin(\beta \ln x)+\beta \cos(\beta \ln x))\end{matrix}}{\Bigg |}$

$=x^{2\alpha -1}\beta$

which is non-zero almost everywhere. Thus

$y=Ax^{\alpha }\cos(\beta \ln x)+Bx^{\alpha }\sin(\beta \ln x)$

is a general solution.

The case of equal indicial roots

If $r_{1}=r_{2}$, let $r=r_{1}=r_{2}$;

$y_{1}=x^{r}$

is one solution. The other one may be determined through the method of undetermined coefficients. Let

$y_{2}=ux^{r}$

where $u$ is a function of $x$. Now derive:

$y_{2}'=u'x^{r}+urx^{r-1}$

$y_{2}''=u''x^{r}+2u'rx^{r-1}+ur(r-1)x^{r-2}$

The original E.C.e. is

$ax^{2}y''+bxy'+cy=0$

Substitute $y_{2}$ and its derivatives into the E.C.e.:

$ax^{2}(u''x^{r}+2u'rx^{r-1}+ur(r-1)x^{r-2})+bx(u'x^{r}+urx^{r-1})+cux^{r}=0$

Distribute:

$au''x^{r+2}+2au'rx^{r+1}+aur(r-1)x^{r}+bu'x^{r+1}+burx^{r}+cux^{r}=0$

Group by derivatives of $u$:

$u''ax^{r+2}+u'(2arx^{r+1}+bx^{r+1})+u[ar(r-1)x^{r}+brx^{r}+cx^{r}]=0$

Factor out powers of $x$:

$u''ax^{r+2}+u'(2ar+b)x^{r+1}+u[ar(r-1)+br+c]x^{r}=0$

The expression within the square brackets is the indicial equation, for which $r$ is the root, so the third term in the sum of the L.H.S. vanishes.

$u''ax^{r+2}+u'(2ar+b)x^{r+1}=0$

For $x>0$, $x^{r+1}\neq 0$, so divide by $x^{r+1}$:

$u''ax+u'(2ar+b)=0$

And here note well that $2ar+b=a$, thus

$u''ax+u'a=0$

$u''x+u'=0$

Let $u'=v$, so that $v'=u''$:

$v'x+v=0$

${v' \over v}=-{1 \over x}$

$\int {dv \over v}=-\int {dx \over x}$

$\ln |v|=-(\ln |x|+k_{0})$

$|v|=e^{-\ln |x|-k_{0}}={k_{1} \over |x|}$

$v={k_{1} \over x}=u'$

assuming that $x>0$.

$u=\int {k_{1} \over x}dx=k_{1}\ln x$

So $y_{2}=ux^{r}=k_{1}x^{r}\ln x$ and the general solution is

$y=k_{0}x^{r}+k_{1}x^{r}\ln x$

Solving another way

Recalling the second order Euler–Cauchy equation:

$ax^{2}y''+bxy'+cy=0$

Let $z=\ln x$. This is suggested by the solution $x^{\alpha }\cos(\beta \ln x)$ corresponding to the indicial root $\alpha +i\beta$, with $\cos(\beta \ln x)$ instead of $\cos(\beta x)$ and $\sin(\beta \ln x)$ instead of $\sin(\beta x)$. So, letting $z=\ln x$, what happens when $y$ is derived with respect to $z$ instead of w.r.t. $x$?

${dy \over dx}={dy \over dz}{dz \over dx}={1 \over x}{dy \over dz}$

$Dy={1 \over x}{\mathfrak {D}}y$

where ${\mathfrak {D}}=d/dz$.

$xDy={\mathfrak {D}}y$

${\mathfrak {D}}{\Big (}{1 \over x}{\Big )}=D{\Big (}{1 \over x}{\Big )}{\mathfrak {D}}x$

$x=e^{z}$ so ${\mathfrak {D}}x=x$

${\mathfrak {D}}{\Big (}{1 \over x}{\Big )}=-{1 \over x^{2}}x=-{1 \over x}$

$D^{2}y={1 \over x}{\mathfrak {D}}{\Big (}{1 \over x}{\mathfrak {D}}y{\Big )}$

$D^{2}y={1 \over x}{\Bigg [}{\mathfrak {D}}{\Big (}{1 \over x}{\Big )}{\mathfrak {D}}y+{1 \over x}{\mathfrak {D}}^{2}y{\Bigg ]}={1 \over x}{\Bigg [}-{1 \over x}{\mathfrak {D}}y+{1 \over x}{\mathfrak {D}}^{2}y{\Bigg ]}$

$D^{2}y={1 \over x^{2}}{\mathfrak {D}}({\mathfrak {D}}-1)y$

$x^{2}D^{2}y={\mathfrak {D}}({\mathfrak {D}}-1)y$

So

$(ax^{2}D^{2}+bxD+c)y=0$

becomes

$(a{\mathfrak {D}}({\mathfrak {D}}-1)+b{\mathfrak {D}}+c)y=0$

$(a{\mathfrak {D}}^{2}+(b-a){\mathfrak {D}}+c)y=0$

which is a “H.O.L.D.E.” (Homogeneous Ordinary Linear Differential Equation) with constant coefficients and characteristic equation

$ar^{2}+(b-a)r+c=0$

If its two roots $r_{1,2}$ are real and distinct then the H.O.L.D.E.w.c.c. has solution

$y=k_{1}e^{r_{1}z}+k_{2}e^{r_{2}z}$

where $z=\ln x$ so the E.C.e. has solution

$y=k_{1}x^{r_{1}}+k_{2}x^{r_{2}}$

If the characteristic equation has two equal (and thus real) roots then the H.O.L.D.E.w.c.c. has solution

$y=k_{1}e^{rz}+k_{2}ze^{rz}$

so the E.C.e. has solution

$y=k_{1}x^{r}+k_{2}x^{r}\ln x$

If the roots are a complex conjugate pair then the H.O.L.D.E.w.c.c. has solution

$y=k_{1}e^{\alpha z}\cos(\beta z)+k_{2}e^{\alpha z}\sin(\beta z)$

where, again, $z=\ln x$ so the E.C.e. has solution

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

Higher-order cases

${d \over dz}{\Big (}{1 \over x^{2}}{\Big )}={d \over dz}(e^{-2z})=-2e^{-2z}={-2 \over x^{2}}$

$D^{3}y=DD^{2}y=D{\Big (}{1 \over x^{2}}{\mathfrak {D}}({\mathfrak {D}}-1)y{\Big )}$

$\qquad ={1 \over x}{\mathfrak {D}}{\Bigg (}{1 \over x^{2}}{\Big (}{\mathfrak {D}}^{2}-{\mathfrak {D}})y{\Bigg )}$

$\qquad ={1 \over x}{\Bigg [}{\mathfrak {D}}{\Big (}{1 \over x^{2}}{\Big )}({\mathfrak {D}}^{2}-{\mathfrak {D}})y+{1 \over x^{2}}{\mathfrak {D}}({\mathfrak {D}}^{2}-{\mathfrak {D}})y{\Bigg ]}$

$\qquad ={1 \over x}{\Bigg [}{-2 \over x^{2}}({\mathfrak {D}}^{2}-{\mathfrak {D}})y+{1 \over x^{2}}({\mathfrak {D}}^{3}-{\mathfrak {D}}^{2})y{\Bigg ]}$

$\qquad ={1 \over x^{3}}({\mathfrak {D}}^{3}-3{\mathfrak {D}}^{2}+2{\mathfrak {D}})y$

$\qquad D^{3}y={1 \over x^{3}}{\mathfrak {D}}({\mathfrak {D}}-1)({\mathfrak {D}}-2)y$

$x^{3}D^{3}y={\mathfrak {D}}({\mathfrak {D}}-1)({\mathfrak {D}}-2)y$

This can be generalized:

$x^{n}D^{n}y={\mathfrak {D}}({\mathfrak {D}}-1)({\mathfrak {D}}-2)...({\mathfrak {D}}-(n-1))y$

To prove it, use mathematical induction. The base cases have already been proven. As inductive hypothesis, take this very formula that is to be proven. Divide it by $x^{n}$ and apply $D$ to (both sides of) it:

$D(D^{n})y=D{\Bigg [}{1 \over x^{n}}{\mathfrak {D}}({\mathfrak {D}}-1)...({\mathfrak {D}}-(n-1))y{\Bigg ]}$

$\qquad ={1 \over x}{\mathfrak {D}}{\Bigg [}{1 \over x^{n}}{\mathfrak {D}}({\mathfrak {D}}-1)...({\mathfrak {D}}-(n-1))y{\Bigg ]}$

$\qquad ={1 \over x}{\Bigg [}{-n \over x^{n}}+{1 \over x^{n}}{\mathfrak {D}}{\Bigg ]}{\mathfrak {D}}({\mathfrak {D}}-1)...({\mathfrak {D}}-(n-1))y$

$D^{n+1}y={1 \over x^{n+1}}({\mathfrak {D}}-n){\mathfrak {D}}({\mathfrak {D}}-1)...({\mathfrak {D}}-(n-1))y$

$D^{n+1}y={1 \over x^{n+1}}{\mathfrak {D}}({\mathfrak {D}}-1)({\mathfrak {D}}-2)...({\mathfrak {D}}-n)y$

$x^{n+1}D^{n+1}y={\mathfrak {D}}({\mathfrak {D}}-1)({\mathfrak {D}}-2)...({\mathfrak {D}}-n)y$

Thus the inductive step has been proven; the generalization is true:

$x^{n}D^{n}y={\mathfrak {D}}({\mathfrak {D}}-1)({\mathfrak {D}}-2)...({\mathfrak {D}}-(n-1))y$

This rule can be used to convert a higher-order E.C.e. into a H.O.L.D.E. with constant coefficients whose independent variable is z. Once the H.O.L.D.E.w.c.c. is solved, replace z with $\ln x$.

External links

• How do I use Frobenious' power series method to solve a Cauchy-Euler's equation ? at Socratic Q&A