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I am working on a a 4th order linear PDE coming from the modified wave equation of a stiff material. I have radial symmetry which has lead me to a 4th order ODE in $r$:

$r^3 R''''(r) + 2r^2 R'''(r) - rR''(r)+R'(r) = m^4 r^3 R$

where $m^4$ is a constant. This ODE is subject to boundary conditions:

$|R(0)|<\infty$, $|R'(0)|<\infty$, $R''(R)=0$, $R'''(R)=0$,

where $R$ is the radius of my circle. I know that the solution should be in terms of Bessel functions as is typical of problems of this type. Mathematica gives solutions in terms of Bessel functions and Meijer G functions. Looking at the basic formulation (from wikipedia) of the Meijer G functions I do not see how this is obviously a solution (not implying it is obvious, I am implying that I do not see how it fits).

My first question is if someone can help me get to the Meijer G formulation or present to me useful information on how this DE relates to Meijer G functions. Secondly, is there anyway to rewrite this to get it in terms of Bessel functions (I do not think this is the case because they fulfill second order ODEs). Is the analytic solution of this easy (easy in terms of doable at a beginning graduate level) or is it very convoluted?

EDIT: I went back to the original formulation of the problem and found:

$\left( \frac{1}{r}\frac{d}{dr} \left\{ r \frac{d}{dr} \right\} \right) R(r) = m^4 R(r)$.

Now, instead of expanding, I found that I can consider this as two PDEs:

$\left( \frac{1}{r}\frac{d}{dr} \left\{ r \frac{d}{dr} \right\} \right) R(r) = k^2 R(r)$


$\left( \frac{1}{r}\frac{d}{dr} \left\{ r \frac{d}{dr} \right\} \right) R(r) = -k^2 R(r)$.

From this standpoint the solution becomes much easier to understand in terms of Bessel Functions. I will be posting a solution below.

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2 Answers 2

up vote 2 down vote accepted

Perhaps the most direct approach should use the Frobenius method:

Let $R(r)=\sum\limits_{n=0}^\infty a_nr^{n+k}$ ,

Then $R'(r)=\sum\limits_{n=0}^\infty(n+k)a_nr^{n+k-1}$




$\therefore r^3\sum\limits_{n=0}^\infty(n+k)(n+k-1)(n+k-2)(n+k-3)a_nr^{n+k-4}+2r^2\sum\limits_{n=0}^\infty(n+k)(n+k-1)(n+k-2)a_nr^{n+k-3}-r\sum\limits_{n=0}^\infty(n+k)(n+k-1)a_nr^{n+k-2}+\sum\limits_{n=0}^\infty(n+k)a_nr^{n+k-1}=m^4r^3\sum\limits_{n=0}^\infty a_nr^{n+k}$

$\sum\limits_{n=0}^\infty(n+k)(n+k-1)(n+k-2)(n+k-3)a_nr^{n+k-1}+\sum\limits_{n=0}^\infty2(n+k)(n+k-1)(n+k-2)a_nr^{n+k-1}-\sum\limits_{n=0}^\infty(n+k)(n+k-1)a_nr^{n+k-1}+\sum\limits_{n=0}^\infty(n+k)a_nr^{n+k-1}-\sum\limits_{n=0}^\infty m^4a_nr^{n+k+3}=0$

$\sum\limits_{n=0}^\infty(n+k)(n+k-1)^2(n+k-2)a_nr^{n+k-1}-\sum\limits_{n=0}^\infty(n+k)(n+k-2)a_nr^{n+k-1}-\sum\limits_{n=0}^\infty m^4a_nr^{n+k+3}=0$

$\sum\limits_{n=0}^\infty(n+k)((n+k-1)^2-1)(n+k-2)a_nr^{n+k-1}-\sum\limits_{n=0}^\infty m^4a_nr^{n+k+3}=0$

$\sum\limits_{n=0}^\infty(n+k)^2(n+k-2)^2a_nr^{n+k-1}-\sum\limits_{n=4}^\infty m^4a_{n-4}r^{n+k-1}=0$


$\therefore k^2(k-2)^2=0$


Can you take it from here?

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Thank you so much @doraemonpaul, I had never been introduced to this powerful method before this! My only question is here I get two recurrence relations. I haven't gone through it yet, but I assume they are consistent in the series representation. –  Jeremy Upsal Nov 11 '13 at 0:34

Doraemonpaul's answer above is perfectly valid given the ODE originally suggested. However, a simpler method presents itself given the reformulation I have provided in the answers. For anyone interested, this is what I find.

We know that

$\left( \frac{1}{r}\frac{d}{dr} \left\{ r \frac{d}{dr} \right\} \right) R(r) = -k^2 R(r)$

has Bessel function solutions of the first and second kind, i.e. a linear combo of $J_0(kr)$ and $Y_0(kr)$ satisfies the above ODE. Because $n$ is an integer in the Bessel ODE we require both $J_0$ and $Y_0$ for a linearly independent solution. Next, note that

$\left( \frac{1}{r}\frac{d}{dr} \left\{ r \frac{d}{dr} \right\} \right) R(r) = k^2 R(r)$

is solved by a linear combo of $I_0(kr)$ and $K_0(kr)$, modified Bessel functions of the first and second kind. So, to solve

$\left( \frac{1}{r}\frac{d}{dr} \left\{ r \frac{d}{dr} \right\} \right) R(r) = m^4 R(r)$

we simply take a linear combination of all four. The general solution is given by

$R(r) = a_1 J_0(kr) + a_2 Y_0(kr) + a_3 I_0(kr) + a_4K_0(kr)$.

Then, we can apply boundary conditions. Noting that $Y_0(0)\rightarrow -\infty$ at the same rate as $K_0(0)\rightarrow \infty$ and that $J_0(0)=I_0(0)$, we require $a_2 = a_4$. The derivative yields

$R'(r) = -ka_1 J_1(kr) - ka_2 Y_1(kr) + ka_3 I_1(kr) - ka_2 K_1(kr)$.

Noting that $Y_1(0)$ and $K_1(0) \rightarrow \infty$ at the same rate, we find that $a_2 = a_4 \equiv 0$.

I will stop here, the other boundary conditions are easy to apply. You then get a second order system

$\left[\begin{array}{c c} J_0(kR) - J_2(kR) & I_0(kR) + I_2(kR)\\ 3J_1(kr) - J_3(kR) & 3I_1(kR) + I_3(kR) \\ \end{array} \right]$ $\left[\begin{array}{c} a_1\\ a_3\end{array}\right]$ $=\left[\begin{array}{c} 0\\ 0\end{array}\right]$

which can be solved numerically (unless anyone knows how to solve a coupled system of Bessel functions analytically? Or more specifically just the determinant. I would love to hear about it) to find allowable $kR$ values.

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