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Find all the solutions of the form $y(x)= x^m \sum_{n=0}^{\infty} a_nx^n, \ x>0 (m \in \mathbb{R})$ of the differential equation $3x^2y''+5xy'+3xy=0$.

That's what I have tried:

Since $x>0$ the differential equation can be written as follows.

$$y''+ \frac{5}{3x}y'+ \frac{1}{x}y=0$$

$$p(x)=\frac{5}{3x}, q(x)= \frac{1}{x}$$

The point $0$ is regular singular, i.e. the fuctions $xp(x), x^2q(x)$ can be written as power series at a region of $0$.

We are looking for solutions of the form $y(x)=x^m \sum_{n=0}^{\infty} a_n x^n$ for a suitable $m \in \mathbb{R}$ and for suitable $a_n \in \mathbb{R}$ and for $x \in (0,R)$ where $R$ is a suitable positive number.

Then we have:

$$y'(x)= \sum_{n=0}^{\infty} a_n (n+m) x^{n+m-1} \\ \Rightarrow xy'=\sum_{n=0}^{\infty} a_n (n+m) x^{n+m} \\ \Rightarrow 5xy'=\sum_{n=0}^{\infty} 5a_n (n+m) x^{n+m}$$

and

$$y''(x)= \sum_{n=0}^{\infty} a_n (n+m)(n+m-1) x^{n+m-2} \\ \Rightarrow x^2y''=\sum_{n=0}^{\infty} a_n (n+m)(n+m-1) x^{n+m} \\ \Rightarrow 3x^2y''=\sum_{n=0}^{\infty} 3a_n (n+m)(n+m-1) x^{n+m}$$

$$3xy= \sum_{n=0}^{\infty} 3a_n x^{n+m+1}$$

So it has to hold the following:

$$\sum_{n=0}^{\infty} \left[ 3a_n (n+m)(n+m-1) x^{n+m}+5a_n (n+m) x^{n+m}+ 3a_n x^{n+m+1}\right]=0 \Rightarrow \sum_{n=0}^{\infty} \left[ 3a_n (n+m)(n+m-1) +5a_n (n+m) + 3a_n x\right]x^{n+m}=0 $$

So it has to hold that:

$$a_n=\frac{-3a_{n-1}}{3m+3n+2}, \forall n=1,2,3, \dots$$

EDIT: I am looking again at the exercise. For $m=0$ I got the following:

$$a_1=-\frac{3a_0}{5} \\ a_2=\frac{3^2 a_0}{5 \cdot 8} \\ a_3=-\frac{3^3 a_0}{5 \cdot 8 \cdot 11} \\ a-4= \frac{3^4 a_0}{5 \cdot 8 \cdot 11 \cdot 14}$$

So isn't for $m=0$ the general formula for $a_n$ the following? $$$$

$$a_n=(-1)^n \frac{a_0}{\prod_{i=0}^{n-1} (3i+5)}$$

And for $m=-\frac{3}{2}$ isn't the formula for $a_n$ the following?

$$a_n=(-1)^n \frac{3^n a_0}{ \prod_{i=0}^{n-1} \frac{(6i+1)}{2}}$$

If so, then could we say the following?

$$y_1(x)= x^0 \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{\prod_{i=0}^{n-1}(3i+5)}$$

and

$$y_2(x)=x^{-\frac{3}{2}} \sum_{n=0}^{\infty} \frac{(-1)^n 3^n}{ \prod_{i=0}^{n-1} \frac{(6i+1)}{2}}x^n$$

are solutions of the differential equation for $a_0=1$.

$$\left| \frac{\frac{(-1)^{n+1} x^{n+1}}{\prod_{i=0}^n (3i+5)}}{\frac{(-1)^{n} x^n}{\prod_{i=0}^{n-1} (3i+5)}}\right|=\left| \frac{x}{3n+5}\right| \to 0<1$$

Do we deduce from the latter that the radius of convergence is $+\infty$. If so, do we continue as follows?

Similarly we show that the radius of convergence of $y_2(x)$ is $+\infty$.

$$$$

$y_1, y_2$ are linearly independent in $(0,+\infty)$. Because if $c_1, c_2 \in \mathbb{R}$ with $c_1y_1(x)+c_2y_2(x)=0 \forall x \in (0,+\infty)$ then since $c_1 y_1(x)+ c_2y_2(x)$ is a power series with radius of convergence $+\infty$ we have $0= c_1 y_1(x)+c_2y_2(x)= \sum_{n=0}^{\infty} d_n x^n$ for some $d_n \in \mathbb{R}$ and thus $d_n=0 \forall n=0,1,2, \dots$

However $d_0=c_1=0$ and $d_1=-\frac{3}{5} c_2=0 \Rightarrow c_2=0$.

Thus, the general solution of the differntial equation is:

$$y(x)=c_1 \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{\prod_{i=0}^{n-1}(3i+5)}+ c_2 x^{-\frac{3}{2}} \sum_{n=0}^{\infty} \frac{(-1)^n 3^n}{ \prod_{i=0}^{n-1} \frac{(6i+1)}{2}}x^n, c_1, c_2 \in \mathbb{R}$$

EDIT: I remade the calculations for $m=0$ and now I got the following: $$$$ For $n=1$: $a_1=-\frac{3a_0}{1 \cdot 5}$

$$$$ For $n=2$: $a_2=\frac{3^2 a_0}{2 \cdot 5 \cdot 8}$ $$$$ For $n=3$: $a_3=-\frac{3^3 a_0}{2 \cdot 3 \cdot 5 \cdot 8 \cdot 11}$ $$$$ For $n=4$: $a_4=\frac{3^4 a_0}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 8 \cdot 11 \cdot 14}$ $$$$

Are my current calculations right or were the previous one correct?

$$$$ If they are right how could we write the formula of $a_n$ without the use of the Gamma function?

EDIT:I retried it again. Couldn't we write the general formula for $a_n$ when $m=0$ as follows? $$$$ $$a_n=\frac{(-1)^n a_0}{n! \prod_{i=1}^n (3i+2)}$$ Or am I wrong?

Also it should be $m_2=-\frac{2}{3}$. Or am I wrong?

If it is like that isn't the general formula for $a_n$ in this case the following?

$$a_n=(-1)^n \frac{a_0}{n! \prod_{i=0}^{n-2} (2 \cdot 2+3 \cdot i)}$$

Or am I wrong?

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I think your elaboration provided in your question together with some of your comments is quite near to a correct solution.

Let's calculate $y_1(x)$.

We start with $a_n$ which was correctly specified in one of your comments:

\begin{align*} a_n=\frac{-3a_{n-1}}{(m+n)(3m+3n+2)}\qquad\forall n=1,2,3,\ldots \end{align*} Setting $m=0$ we obtain \begin{align*} a_n&=\frac{-3a_{n-1}}{n(3n+2)}\\ &=\frac{-a_{n-1}}{n\left(n+\frac{2}{3}\right)}\\ &=\frac{a_{n-2}}{n(n-1)\left(n+\frac{2}{3}\right)\left(n-\frac{1}{3}\right)}\\ &=\ldots\\ &=\frac{(-1)^na_0}{n!\left(n+\frac{2}{3}\right)\left(n+\frac{2}{3}-1\right)\cdots\left(n+\frac{2}{3}-(n-1)\right)}\\ &=\frac{(-1)^na_0}{n!{\left(n+\frac{2}{3}\right)}_n}\tag{1}\\ &=(-1)^n\frac{a_0}{n!}\frac{\Gamma\left(\frac{5}{3}\right)}{\Gamma\left(n+\frac{5}{3}\right)}\tag{2} \end{align*}

In (1) we apply the Pochhammer symbol $(x)_n=x(x-1)\cdots (x-n+1)$ and in (2) we use the identity $$(x)_n=\frac{\Gamma(x+1)}{\Gamma(x-n+1)}$$

So, we find the solution

$$y_1(x)=a_0\Gamma\left(\frac{5}{3}\right)\sum_{n=0}^{\infty}\frac{(-1)^n}{\Gamma\left(n+\frac{5}{3}\right)}\frac{x^n}{n!}$$

According to a comment of @JackDAurizio let's have a look at the Bessel function of the first kind

\begin{align*} J_\alpha(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\Gamma(n+\alpha+1)}\left(\frac{x}{2}\right)^{2n+\alpha} \end{align*}

We observe

\begin{align*} x^{-\frac{1}{3}}J_{\frac{2}{3}}(2\sqrt{x})=x^{-\frac{1}{3}}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\Gamma\left(n+\frac{5}{3}\right)}x^{n+\frac{1}{3}} =\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\Gamma\left(n+\frac{5}{3}\right)}x^{n} \end{align*}

We conclude

\begin{align*} y_1(x)=a_0\Gamma\left(\frac{5}{3}\right)x^{-\frac{1}{3}}J_{\frac{2}{3}}(2\sqrt{x}) \end{align*}

Similarly we can find $$y_2(x)=a_0\Gamma\left(\frac{1}{3}\right)x^{-\frac{1}{3}}J_{-\frac{2}{3}}(2\sqrt{x})$$

and the general solution is therefore $$y(x)=c_1\Gamma\left(\frac{5}{3}\right)x^{-\frac{1}{3}}J_{\frac{2}{3}}(2\sqrt{x})+c_2\Gamma\left(\frac{1}{3}\right)x^{-\frac{1}{3}}J_{-\frac{2}{3}}(2\sqrt{x})$$


[2015-06-04]: Note according to OPs comment

Avoiding the Gamma function $\Gamma(x)$ a representation via the Pochhammer symbol (1) is convenient:

$$a_n=\frac{(-1)^na_0}{n!{\left(n+\frac{2}{3}\right)}_n}$$

We observe for $n=3$:

\begin{align*} a_3&=-\frac{3^3a_0}{2\cdot3\cdot5\cdot8\cdot11}\\ &=-\frac{3^3a_0}{3!\cdot5\cdot8\cdot11}\\ &=-\frac{a_0}{3!\cdot\frac{5}{3}\cdot\frac{8}{3}\cdot\frac{11}{3}}\\ &=-\frac{a_0}{3!\cdot\left(3-\frac{4}{3}\right)\left(3-\frac{1}{3}\right)\left(3+\frac{2}{3}\right)}\\ &=-\frac{a_0}{3!\left(3+\frac{2}{3}\right)_3} \end{align*} Similarly the case $n=4$: \begin{align*} a_4&=\frac{3^4a_0}{2\cdot3\cdot4\cdot5\cdot8\cdot11\cdot14}\\ &=\frac{3^4a_0}{4!\cdot5\cdot8\cdot11\cdot14}\\ &=\frac{a_0}{4!\cdot\frac{5}{3}\cdot\frac{8}{3}\cdot\frac{11}{3}\cdot\frac{14}{3}}\\ &=\frac{a_0}{4!\left(4-\frac{7}{3}\right)\left(4-\frac{4}{3}\right)\left(4-\frac{1}{3}\right)\left(4+\frac{2}{3}\right)}\\ &=\frac{a_0}{4!\left(4+\frac{2}{3}\right)_4} \end{align*}

We obtain the following representation for $y_1(x)$: \begin{align*} y_1(x)&=x^m\sum_{n=0}^{\infty}a_nx^n\\ &=x^0\sum_{n=0}^{\infty}\frac{(-1)^na_0}{n!{\left(n+\frac{2}{3}\right)}_n}x^n\\ &=a_0\sum_{n=0}^{\infty}\frac{(-1)^n}{{\left(n+\frac{2}{3}\right)}_n}\frac{x^n}{n!} \end{align*}


[2015-06-04]: Additional note according to OPs comments

Since

\begin{align*} \prod_{j=1}^{n}(3j+2)&=5\cdot8\cdot11\cdot\ldots\cdot(3n+2)\\ &=(3\cdot1+2)(3\cdot2+2)(3\cdot3+2)\cdot\ldots\cdot(3n+2)\\ &=3^n\left(1+\frac{2}{3}\right)\left(2+\frac{2}{3}\right)\left(3+\frac{2}{3}\right)\cdot\ldots\cdot\left(n+\frac{2}{3}\right)\\ &=3^n\left(n+\frac{2}{3}\right)_n \end{align*} it's perfectly valid, to write \begin{align*} a_n=\frac{(-1)^n3^na_0}{n!\prod_{j=1}^{n}(3j+2)}=\frac{(-1)^na_0}{n!\left(n+\frac{2}{3}\right)_n} \end{align*}

Note: Usually, it's more convenient to use the Pochhammer symbol, since consecutive factors have the nice difference $1$ and we do not need to write the product symbol together with factor $3^n$. It's similar to the situation that we prefer to write $n!$ instead of $\prod_{j=1}^nj$.

When looking at the coefficient \begin{align*} a_n=\frac{-3a_{n-1}}{(n+m)(3n+3m+2)}\qquad n\geq 1,m\in \mathbb{R} \end{align*} calculated by OP in one of his comments, we observe that $m=-\frac{2}{3}$ is a proper choice besides $m=0$ in order to get a second linearly independent solution, since in this case we have the simple recurrence relation \begin{align*} a_n&=\frac{-3a_{n-1}}{(n-\frac{2}{3})3n}=\frac{-a_{n-1}}{(n-\frac{2}{3})n}\\ &=\frac{a_{n-2}}{\left(n-\frac{2}{3}\right)\left(n-\frac{5}{3}\right)n(n-1)}\\ &=\ldots\\ &=\frac{(-1)^na_0}{\left(n-\frac{2}{3}\right)\left((n-1)-\frac{2}{3}\right)\cdot\ldots\cdot\left((n-(n-1))-\frac{2}{3}\right)n!}\\ &=\frac{(-1)^na_0}{\left(n-\frac{2}{3}\right)\left(n-\frac{5}{3}\right)\cdot\ldots\cdot\left(\frac{1}{3}\right)n!}\\ &=\frac{(-1)^na_0}{n!\left(n-\frac{2}{3}\right)_n}\tag{3} \end{align*} Of course, we can also use the product symbol in (3) instead of the Pochhammer symbol \begin{align*} \left(n-\frac{2}{3}\right)_n&=\prod_{j=0}^{n-1}\left(n-j-\frac{2}{3}\right)=\prod_{j=0}^{n-1}\left(n-(n-1-j)-\frac{2}{3}\right)\\ &=\prod_{j=0}^{n-1}\left(j+\frac{1}{3}\right)=\frac{1}{3^n}\prod_{j=0}^{n-1}\left(3j+1\right) =\frac{1}{3^n}\prod_{j=1}^{n}\left(3j-2\right)\\ \end{align*}

So, we get $y_2(x)$ \begin{align*} y_2(x)&=a_0x^{-\frac{2}{3}}\sum_{n=0}^{\infty}\frac{(-1)^n}{\left(n-\frac{2}{3}\right)_n}\frac{x^n}{n!}\\ &=a_0x^{-\frac{2}{3}}\sum_{n=0}^{\infty}\frac{(-1)^n3^n}{\prod_{j=1}^{n}\left(3j-2\right)}\frac{x^n}{n!} \end{align*}

Conclusion: Alternately to the representation of $y(x)$ via Bessel functions of the first kind we obtain as general solution for $y(x)$:

\begin{align*} y(x)&= c_0\sum_{n=0}^{\infty}\frac{(-1)^n}{{\left(n+\frac{2}{3}\right)}_n}\frac{x^n}{n!}+c_1x^{-\frac{2}{3}}\sum_{n=0}^{\infty}\frac{(-1)^n}{\left(n-\frac{2}{3}\right)_n}\frac{x^n}{n!}\\ &=c_0\sum_{n=0}^{\infty}\frac{(-1)^n3^n}{\prod_{j=1}^{n}(3j+2)}\frac{x^n}{n!}+c_1x^{-\frac{2}{3}}\sum_{n=0}^{\infty}\frac{(-1)^n3^n}{\prod_{j=1}^{n}\left(3j-2\right)}\frac{x^n}{n!}\qquad c_0,c_1\in\mathbb{R}\\ \end{align*}

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  • $\begingroup$ I should not use the Gamma function. $$$$ I remade the calculations for $m=0$ and now I got the following: $$$$ For $n=1$: $a_1=-\frac{3a_0}{1 \cdot 5}$ $$$$ For $n=2$: $a_2=\frac{3^2 a_0}{2 \cdot 5 \cdot 8}$ $$$$ For $n=3$: $a_3=-\frac{3^3 a_0}{2 \cdot 3 \cdot 5 \cdot 8 \cdot 11}$ $$$$ For $n=4$: $a_4=\frac{3^4 a_0}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 8 \cdot 11 \cdot 14}$ $$$$ Are my current calculations right or where the previous one correct? $$$$ If they are right how could we write the formula of $a_n$ without the use of the Gamma function? $\endgroup$ – evinda Jun 4 '15 at 0:09
  • $\begingroup$ @Evinda: Your current calculations are correct. I've added some more details which should be helpful. Do you think you could calculate $y_2(x)$ by your own efforts? $\endgroup$ – Markus Scheuer Jun 4 '15 at 6:15
  • $\begingroup$ I retried it again. Couldn't we write the general formula for $a_n$ when $m=0$ as follows? $$$$ $$a_n=\frac{(-1)^n a_0}{n! \prod_{i=1}^n (3i+2)}$$ Or am I wrong? $\endgroup$ – evinda Jun 4 '15 at 14:04
  • $\begingroup$ Also it should be $m_2=-\frac{2}{3}$. Or am I wrong? If it is like that isn't the general formula for $a_n$ in this case the following? $$a_n=(-1)^n \frac{a_0}{n! \prod_{i=0}^{n-2} (2 \cdot 2+3 \cdot i)}$$ Or am I wrong? $\endgroup$ – evinda Jun 4 '15 at 14:11
  • $\begingroup$ @evinda: In your first comment a factor $3^n$ is missing in the numerator. Compare it with your examples for $n=1,2,3,4$. The setting $m=-\frac{2}{3}$ is quite ok. But, we also have to use the factor $3^n$ in the numerator and observe that the product contains only $n-1$ factors instead of $n$. I've added some more detailed info, which should clarify your questions. $\endgroup$ – Markus Scheuer Jun 4 '15 at 17:05
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Hint: if you replace $y(x)$ with $x^{-1/3}g(2\sqrt{x})$, then $g$ satisfies a Bessel differential equation.

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  • $\begingroup$ I retried it again. Couldn't we write the general formula for $a_n$ when $m=0$ as follows? $$$$ $$a_n=\frac{(-1)^n a_0}{n! \prod_{i=1}^n (3i+2)}$$ Or am I wrong? $\endgroup$ – evinda Jun 4 '15 at 14:06
  • $\begingroup$ @evinda: yes, Markus Scheuer has already computed the coefficients of the Bessel function that gives the solution. $\endgroup$ – Jack D'Aurizio Jun 4 '15 at 14:08
  • $\begingroup$ Nice :) Also it should be $m_2=-\frac{2}{3}$. Or am I wrong? If it is like that isn't the general formula for $a_n$ in this case the following? $$a_n=(-1)^n \frac{a_0}{n! \prod_{i=0}^{n-2} (2 \cdot 2+3 \cdot i)}$$ Or am I wrong? $\endgroup$ – evinda Jun 4 '15 at 14:10
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You need to perform an index shift so that all powers of $x$ have the same exponent in each term. $$ 3xy= \sum_{n=0}^{\infty} 3a_n x^{n+m+1}= \sum_{n=1}^{\infty} 3a_{n-1} x^{n+m}= \sum_{n=0}^{\infty} 3a_{n-1} x^{n+m} $$ The last has to define an auxillary coefficient $a_{-1}=0$. If you do not like coefficients with negative index, you have to treat the term for $n=0$ separately.

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  • $\begingroup$ I tried to treat the term for $n=0$ separately. That's what I got: $$$$ It holds: $$\left[ 3a_0m(m-1)+5a_0m\right] x^m + \sum_{n=1}^{\infty} \left[ 3a_n(n+m)(n+m-1)+5a_n (n+m)+3a_{n-1}\right] x^{n+m}=0$$ So it has to hold: $$3a_0m(m-1)+5a_0m=0 \Rightarrow m=0 \text{ or } m=\frac{-3}{2}$$ Also it has to hold: $$3a_n(n+m)(n+m-1)+5a_n (n+m)+3a_{n-1}=0 \Rightarrow a_n=- \frac{3a_{n-1}}{(n+m)(3n+3m+2)}$$ Is it right so far? If so how could we continue? Do we have to distinguish the cases $m=0$ and $m=-\frac{3}{2}$ and find for each of these $m$ the general formula of $a_n$? $\endgroup$ – evinda Mar 23 '15 at 20:07
  • $\begingroup$ Yes. I think there is a notation $(a)_{\bigl[n\bigr]}=a(a-1)(a-2)...(a-n+1)$ that would simplify the formula of the solution of the formula. Cancel the factor $3$. $\endgroup$ – LutzL Mar 23 '15 at 20:31
  • $\begingroup$ I found the following for $m=0$: $$a_1= -\frac{3a_0}{5} \\ a_2=- \frac{3a_1}{16} \\ a_3= \frac{3a_1}{171} \\ a_4= \frac{-3a_1}{1064}$$ How could I use the above formula to find the general formula for $a_n$ ? $\endgroup$ – evinda Mar 23 '15 at 20:45
  • $\begingroup$ $$a_n=\frac{(-1)^n}{(n+m)_{[\,n\,]}·(n+m+\frac23)_{[\,n\,]}}.$$ Note also that $$(a)_{\small{[n]}}=\frac{a!}{(a-n)!}$$ for positive integer $a$ and $n\le a$. Write the general case in terms of Gamma functions at your leisure. (mathjax does not like brackets in subscripts.) $\endgroup$ – LutzL Mar 23 '15 at 20:50
  • $\begingroup$ I applied this formula for $n=2$ and I got : $$a_2= \frac{1}{\frac{(n+m)!}{m!} \frac{\left( n+m+\frac{2}{3} \right)!}{\left( \frac{2}{3}\right)!}}$$ but it isn't the same result as above... Or am I wrong? $\endgroup$ – evinda Mar 23 '15 at 20:56

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