5.4: Euler Equation

Definition: Euler Equation

An Euler equation is a second order differential equation of the form \(x^2 y^{\prime \prime}+\alpha x y^{\prime}+\beta y=0.\)

 

Let \(L(y):=x^2 y^{\prime \prime}+\alpha x y^{\prime}+\beta y\)
Recall that \(L\) is a linear function and if \(f\) is a solution to the euler equation, then \(L(f)=0\).

Note

If \(x \neq 0\), then \(x\) is an ordinary point and if \(x=0\), then \(x\) is a singular point.

Theorem \(\PageIndex{1}\)

Suppose \(x>0\), then \(L\left(x^r\right)=0\) for some value of \(r.\)

Proof

Observe that 

\[ \notag
\begin{array}{l}
y=x^r, y^{\prime}=r x^{r-1}, y^{\prime \prime}=r(r-1) x^{r-2} \\
x^2 y^{\prime \prime}+\alpha x y^{\prime}+\beta y=0 \\
x^2 r(r-1) x^{r-2}+\alpha x r x^{r-1}+\beta x^r=0 \\
\left(r^2-r\right) x^r+\alpha r x^r+\beta x^r=0 \\
x^r\left[r^2-r+\alpha r+\beta\right]=0 \\
x^r\left[r^2+(\alpha-1) r+\beta\right]=0
\end{array}
\]

Thus \(x^r\) is a solution if and only if \(r^2+(\alpha-1) r+\beta=0.\) Thus \(r=\frac{-(\alpha-1) \pm \sqrt{(\alpha-1)^2-4 \beta}}{2}.\)

 

Theorem \(\PageIndex{2}\)

If \(x<0\), then \(L\left((-x)^r\right)=0\) for some value of \(r\).

Note

Observe that
\[ \notag
\begin{array}{l}
y=(-x)^r, y^{\prime}=-r(-x)^{r-1}, y^{\prime \prime}=r(r-1)(-x)^{r-2} \\
x^2 y^{\prime \prime}+\alpha x y^{\prime}+\beta y=0 \\
x^2 r(r-1)(-x)^{r-2}-\alpha x r(-x)^{r-1}+\beta(-x)^r=0 \\
\left(r^2-r\right)(-x)^r+\alpha r(-x)^r+\beta(-x)^r=0\\
(-x)^r\left[r^2-r+\alpha r+\beta\right]=0 \\
(-x)^r\left[r^2+(\alpha-1) r+\beta\right]=0
\end{array}
\]

Thus \((-x)^r\) is a solution if and only if \(r^2+(\alpha-1) r+\beta=0\)
Thus \(r=\frac{-(\alpha-1) \pm \sqrt{(\alpha-1)^2-4 \beta}}{2}\)
Recall \(|x|=\left\{\begin{array}{ll}x & \text { if } x>0 \\ -x & \text { if } x<0\end{array}\right.\)
Thus \(|x|^r=\left\{\begin{array}{ll}x^r & \text { if } x>0 \\ (-x)^r & \text { if } x<0\end{array}\right.\)
Thus if \(r=\frac{-(\alpha-1) \pm \sqrt{(\alpha-1)^2-4 \beta}}{2}\), then \(y=|x|^r\) is a solution to Euler's equation for \(x \neq 0\).

Case 1: 2 real distinct roots, \(r_1, r_2\) : General solution is \(y=c_1|x|^{r_1}+c_2|x|^{r_2}\).

Case 2: 2 complex solutions \(r_i=\lambda \pm i \mu\) : Convert solution to form without complex numbers.
\[ \notag
\begin{array}{l}
\text { Note }|x|^{\lambda \pm i \mu}=e^{l n\left(|x|^{\lambda \pm i \mu}\right)}=e^{(\lambda \pm i \mu) l n|x|}=e^{\lambda l n|x|} e^{i( \pm \mu l n|x|)} \\
=|x|^\lambda[\cos ( \pm \mu l n|x|)+i \sin ( \pm \mu l n|x|)] \\
=|x|^\lambda[\cos (\mu l n|x|) \pm i \sin (\mu l n|x|)]
\end{array}
\]

Case 3: repeated root: Find 2nd solution. 

 

Example \(\PageIndex{1}\)

Solve \(x^2 y^{\prime \prime}-2 x y^{\prime}=0\left(^*\right)\).

Solution

We could solve by letting \(v=y^{\prime}\), but we will instead use 5.4 methods. Note \(x\) is an ordinary point if and only if \(x \neq 0 \quad\left(y^{\prime \prime}-\frac{2}{x} y^{\prime}=0.\right)\) \(x=0\) is a singular point.

Note \(x^2 x^{r-2} r(r-1)-2 x x^{r-1} r=0\) implies \(r^2-r-2 r=0\) and recall \(y=(-x)^r\) gives same equation for \(r\) as \(y=x^r\).

Thus \(y=|x|^r\) implies \(r^2+(\alpha-1) r+\beta=r^2-3 r+0=r(r-3)=0.\) Thus \(r=0,3\). Thus \(y=|x|^0=1\) and \(y=|x|^3\) are solutions to \(\left({ }^*\right).\) Since \(\left(^*\right)\) is a linear equation, the general solution is \(y=c_1+c_2|x|^3\). Note an equivalent general solution is \(y=k_1+k_2 x^3\). Both forms are valid for all \(x\).

 

From the above example we want to try and answer the question:

When is a unique solution to the following initial value problem guaranteed?

\[\notag
\begin{array}{c}
x^2 y^{\prime \prime}-2 x y^{\prime}=0, \quad y\left(t_0\right)=y_0, \quad y^{\prime}\left(t_0\right)=y_1(* *) \\
y^{\prime \prime}-\frac{2}{x} y^{\prime}=0, \quad y\left(t_0\right)=y_0, \quad y^{\prime}\left(t_0\right)=y_1
\end{array}
\]

Since \(\frac{2}{x}\) and the zero constant function are continuous on\((-\infty, 0) \cup(0, \infty)\)

\((* *)\) has a unique solution for \(t_0<0\) and this solution exists on\((-\infty, 0)\)

\((* *)\) has a unique solution for \(t_0>0\) and this solution exists on \( (0, \infty)\)

There are an infinite number of solutions for \(y(0)=a, y^{\prime}(0)=0\).

How is \(x^r\) defined:

If \(n\) is a positive integer: \(x^n=x \cdot x \cdot \ldots \cdot x\)
If \(m\) is a positive integer: If \(f(x)=x^m\), then \(f^{-1}(x)=x^{\frac{1}{m}}\) and \(x^{\frac{n}{m}}=\left(x^n\right)^{\frac{1}{m}}\)

Let \(r \geq 0\). Let \(r_n\) be any sequence consisting of positive rational numbers such that \(\lim _{n \rightarrow \infty} r_n=r\). Then
\[ \notag
x^r=\lim _{n \rightarrow \infty} x^{r_n} .
\]

See more advanced class for why the above is well-defined.
If \(r<0\), then \(x^r=x^{-r}\).

If \(x\) is a real number, when is \(x^r\) a real number?

\(x^n=x \cdot x \cdot \ldots \cdot x\) is a real number when \(n\) is a positive integer. If \(f(x)=x^n\), then the image of

\[\notag
f=\left\{\begin{array}{ll}\text { real numbers } & n \text { odd } \\ {[0, \infty)} & n \text { even }\end{array}\right.
\]

Thus if \(f^{-1}(x)=x^{\frac{1}{n}}\) is real-valued, then the domain of \(f^{-1}\) is

\[ \notag
D(f^{-1})=\left\{\begin{array}{ll}\text { real numbers } & n \text { odd } \\ {[0, \infty)} & n \text { even }\end{array}\right.
\]

Complex Analysis

In complex analysis, \(\left(\frac{1+i \sqrt{3}}{2}\right)^3=-1,(-1)^3=-1,\left(\frac{1-i \sqrt{3}}{2}\right)^3=-1.\) Recall \(\left(e^{\frac{i \pi}{3}}\right)^3=\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^3=-1.\)

Definition: Root of Unity

A complex number is a root of unity if there exists some positive integer \(n\) such that \( (x+ iy)^n=1.\)

Some examples of roots of unity:
\[ \notag
\left(e^{\frac{2 i \pi}{3}}\right)^3=1 \quad\left(e^{\frac{-2 i \pi}{3}}\right)^3=1, \quad(1)^3=1
\]

Example \(\PageIndex{2}\)

Solve \(x^2 y^{\prime \prime}+\alpha x y^{\prime}+\beta y=0\)

Solution

Let \(y=x^r\), \(y^{\prime}=r x^{r-1}, y^{\prime \prime}=r(r-1) x^{r-2}\) (case when \(y=(-x)^r\) is similar). \(x^2 x^{r-2} r(r-1)+\alpha x x^{r-1} r+\beta x^r=0\)
\(x^r\left[r^2-r+\alpha r+\beta\right]=0\) for all \(x\) implies \(r^2+(\alpha-1) r+\beta=0\) Thus \(x^r\) is a solution iff \(r=\frac{-(\alpha-1) \pm \sqrt{(\alpha-1)^2-4 \beta}}{2}\)

Case 1: Two real roots, \(r_1, r_2\).
General solution is \(y=c_1|x|^{r_1}+c_2|x|^{r_2}\)

Case 2: Two complex roots, \(r_i=\lambda \pm i \mu\) :
Convert solution to form without complex numbers.
Note \(|x|^{ \pm i \mu}=e^{l n\left(|x|^{ \pm i \mu}\right)}=e^{( \pm i \mu) l n|x|}=e^{i( \pm \mu l n|x|)} =\cos ( \pm \mu l n|x|)+i \sin ( \pm \mu l n|x|) =\cos (\mu l n|x|) \pm i \sin (\mu l n|x|)\)

General solution is \(y=c_1|x|^{r_1}+c_2|x|^{r_2}=c_1|x|^{\lambda+i \mu}+c_2|x|^{\lambda-i \mu}\)

\(\begin{array}{l}=|x|^\lambda\left(c_1|x|^{i \mu}+c_2|x|^{-i \mu}\right) \\
=|x|^\lambda\left(c_1[\cos (\mu \ln |x|)+i \sin (\mu \ln |x|)]+c_2[\cos (\mu \ln |x|)-i \sin (\mu \ln |x|)]\right)\\
=|x|^\lambda\left(\left[c_1+c_2\right] \cos (\mu l n|x|)+i\left[c_1-c_2\right] \sin (\mu l n|x|)\right)\\
=|x|^\lambda\left(k_1 \cos (\mu l n|x|)+k_2 \sin (\mu l n|x|)\right)\\
=k_1|x|^\lambda \cos (\mu \ln |x|)+k_2|x|^\lambda \sin (\mu \ln |x|)
\end{array}\)

Case 3: one repeated root, \(r_1=\frac{-(\alpha-1)}{2}\). (i.e., \(\sqrt{(\alpha-1)^2-4 \beta}=0\) ): Thus \(|x|^{r_1}\) is a solution. Find 2nd solution.

Method 1. Reduction of order: Suppose \(y=u(x)|x|^{r_1}\) is a solution to \(x^2 y^{\prime \prime}+\alpha x y^{\prime}+\beta y=0\). Plug in and determine \(u(x)\)

Method 2: Let \(L(y)=x^2 y^{\prime \prime}+\alpha x y^{\prime}+\beta y\) where \(y^{\prime}=\frac{d y}{d x}\).
\[ \notag
\begin{array}{l}
L\left(|x|^r\right)=|x|^r\left(r-r_1\right)^2 \\
\frac{\partial}{\partial r}\left[L\left(|x|^r\right)\right]=\frac{\partial}{\partial r}\left[|x|^r\left(r-r_1\right)^2\right]=\left(|x|^r\right)^{\prime}\left(r-r_1\right)^2+2|x|^r\left(r-r_1\right)=0 \\
\text { if } r=r_1 .
\end{array}
\]

Suppose \(x\) is constant with respect to \(r\) and all the partial derivatives are continuous. Then
\[ \notag
\begin{array}{l} 
\frac{\partial}{\partial r}[L(y)]=\frac{\partial}{\partial r}\left[x^2 y^{\prime \prime}+\alpha x y^{\prime}+\beta y\right]= x^2 \frac{\partial y^{\prime \prime}}{\partial r}+\alpha x \frac{\partial y^{\prime}}{\partial r}+\beta \frac{\partial y}{\partial r} \\
= x^2 \frac{\partial}{\partial r}\left[\frac{\partial^2 y}{\partial x^2}\right]+\alpha x \frac{\partial}{\partial r}\left[\frac{\partial y}{\partial x}\right]+\beta \frac{\partial y}{\partial r} \\
= x^2 \frac{\partial^2}{\partial x^2}\left[\frac{\partial y}{\partial r}\right]+\alpha x \frac{\partial}{\partial x}\left[\frac{\partial y}{\partial r}\right]+\beta \frac{\partial y}{\partial r} \\
=L\left(\frac{\partial y}{\partial r}\right) \text { for all } r
\end{array}
\]

\begin{array}{l}
L\left(\frac{\partial|x|^r}{\partial r}\right)=\frac{\partial}{\partial r}\left[L\left(|x|^r\right)\right]=0 \text { for } r=r_1 . \\
\frac{\partial|x|^r}{\partial r}=\frac{\partial e^{l n|x|^r}}{\partial r} \frac{\partial e^{r l n|x|}}{\partial r}=\left(e^{r l n|x|}\right) \ln |x|=|x|^r \ln |x|
\end{array}

Thus \(|x|^{r_1} \ln |x|\) is a solution.
Thus general solution is \(y=c_1|x|^{r_1}+c_2|x|^{r_1} \ln |x|\)

since by the Wronskian, \(|x|^{r_1}\) and \(|x|^{r_1} \ln |x|\) are linearly independent. Suppose \(x>0\) and \(r_1 \neq 0\).
\[ \notag
\left|\begin{array}{cc}
x^{r_1} & x^{r_1} \ln |x| \\
r_1 x^{r_1-1} & r_1 x^{r_1-1} \ln |x|+x^{r_1-1}
\end{array}\right|\\
\]

\[ \notag
\begin{array}{l}
=x^{r_1}\left(r_1 x^{r_1-1} \ln |x|+x^{r_1-1}\right)-x^{r_1} \ln |x| r_1 x^{r_1-1}\\
=x^{2 r_1-1}\left[r_1 \ln |x|+1-\ln |x| r_1\right]=x^{2 r_1-1} \neq 0 \text { for } x \neq 0
\end{array}
\]

Other cases for Wronskian are similar.