7.4.1: Regular Singular Points Euler Equations (Exercises)

Last updated
Save as PDF

Page ID: 30754

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

Q7.4.1

In Exercises 7.4.1-7.4.18 find the general solution of the given Euler equation on $(0,\infty)$.

1. $x^2y''+7xy'+8y=0$

2. $x^2y''-7xy'+7y=0$

3. $x^2y''-xy'+y=0$

4. $x^2y''+5xy'+4y=0$

5. $x^2y''+xy'+y=0$

6. $x^2y''-3xy'+13y=0$

7. $x^2y''+3xy'-3y=0$

8. $12x^2y''-5xy''+6y=0$

9. $4x^2y''+8xy'+y=0$

10. $3x^2y''-xy'+y=0$

11. $2x^2y''-3xy'+2y=0$

12. $x^2y''+3xy'+5y=0$

13. $9x^2y''+15xy'+y=0$

14. $x^2y''-xy'+10y=0$

15. $x^2y''-6y=0$

16. $2x^2y''+3xy'-y=0$

17. $x^2y''-3xy'+4y=0$

18. $2x^2y''+10xy'+9y=0$

Q7.4.2

19.

Adapt the proof of Theorem 7.4.3 to show that $y=y(x)$ satisfies the Euler equation \[ax^2y''+bxy'+cy=0\tag{A} \] on $(-\infty,0)$ if and only if $Y(t)=y(-e^t)$ \[a {d^2Y\over dt^2}+(b-a){dY\over dt}+cY=0.\nonumber \] on $(-\infty,\infty)$.
Use (a) to show that the general solution of Equation A on $(-\infty,0)$ is \[\begin{aligned} y&=c_1|x|^{r_1}+c_2|x|^{r_2}\mbox{ if $r_1$ and $r_2$ are distinct real numbers; } \\[4pt] y&=|x|^{r_1}(c_1+c_2\ln|x|)\mbox{ if $r_1=r_2$; } \\[4pt] y&=|x|^{\lambda}\left[c_1\cos\left(\omega\ln|x|\right)+ c_2\sin\left(\omega\ln|x| \right)\right]\mbox{ if $r_1,r_2=\lambda\pm i\omega$ with $\omega>0$}.\end{aligned}\nonumber \]

20. Use reduction of order to show that if

\[ar(r-1)+br+c=0\nonumber \]

has a repeated root $r_1$ then $y=x^{r_1}(c_1+c_2\ln x)$ is the general solution of

\[ax^2y''+bxy'+cy=0\nonumber \]

on $(0,\infty)$.

21. A nontrivial solution of

\[P_0(x)y''+P_1(x)y'+P_2(x)y=0\nonumber \]

is said to be oscillatory on an interval $(a,b)$ if it has infinitely many zeros on $(a,b)$. Otherwise $y$ is said to be nonoscillatory on $(a,b)$. Show that the equation

\[x^2y''+ky=0 \quad (k=\; \mbox{constant})\nonumber \]

has oscillatory solutions on $(0,\infty)$ if and only if $k>1/4$.

22. In Example 7.4.2 we saw that $x_0=1$ and $x_0=-1$ are regular singular points of Legendre’s equation

\[(1-x^2)y''-2xy'+\alpha(\alpha+1)y=0. \tag{A} \]

Introduce the new variables $t=x-1$ and $Y(t)=y(t+1)$, and show that $y$ is a solution of (A) if and only if $Y$ is a solution of \[t(2+t){d^2Y\over dt^2}+2(1+t){dY\over dt}-\alpha(\alpha+1)Y=0,\nonumber \] which has a regular singular point at $t_0=0$.
Introduce the new variables $t=x+1$ and $Y(t)=y(t-1)$, and show that $y$ is a solution of (A) if and only if $Y$ is a solution of \[t(2-t){d^2Y\over dt^2}+2(1-t){dY\over dt}+\alpha(\alpha+1)Y=0,\nonumber \] which has a regular singular point at $t_0=0$.

23. Let $P_0,P_1$, and $P_2$ be polynomials with no common factor, and suppose $x_0\ne0$ is a singular point of

\[P_0(x)y''+P_1(x)y'+P_2(x)y=0. \tag{A} \]

Let $t=x-x_0$ and $Y(t)=y(t+x_0)$.

Show that $y$ is a solution of (A) if and only if $Y$ is a solution of \[R_0(t){d^2Y\over dt^2}+R_1(t){dY\over dt}+R_2(t)Y=0. \tag{B} \] where \[R_i(t)=P_i(t+x_0),\quad i=0,1,2.\nonumber \]
Show that $R_0$, $R_1$, and $R_2$ are polynomials in $t$ with no common factors, and $R_0(0)=0$; thus, $t_0=0$ is a singular point of (B).

Search

Text Color

Text Size

Margin Size

Font Type

Q7.4.1

Q7.4.2