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I am interested in solving the linear PDE for $f(r,t)$ $$ (\partial_{tt}+a\partial_t-b\nabla^2)f(r,t)=0 $$ $$ \nabla^2\equiv \frac{1}{r}\partial_r(r\partial_r)-\frac{1}{r^2}=\partial_{rr}+\frac{1}{r}\partial_r-\frac{1}{r^2} $$ with conditions $$ \frac{\partial f(0,t)}{\partial t}=0,\quad \frac{\partial f(R,t)}{\partial t}=d\cos (\omega t) $$ where $a,b,d,R>0$. You can see the laplacian like term is written in a cylindrical basis, so I assume Bessel function will arise. I would like a complete solution which includes the steps to find $f(r,t)$ and details in evaluating the expansion coefficients of the fourier-bessel series that may arise in $f(r,t)$.

Thank you! enjoy the bounty!

Jeff Faraci
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  • Bessel J is a real function which means that $\bar J_1(z)=J_1(z)$. – Nikolay Gromov Sep 16 '15 at 23:31
  • Bessel J is not a real function if $ z\in \mathbb{C}$. Is that not true? I do not know what you mean Bessel J is a real function. Thanks! – Jeff Faraci Sep 16 '15 at 23:33
  • real function you can define in a number of ways: for example if you represent your function by its Teylor expansion - real function will have real coefficients in the expansion

    Another definition - it takes real values for real arguments.

     – Nikolay Gromov Sep 16 '15 at 23:35
  • That is not correct if $z\in \mathbb{C}$. Thanks though. – Jeff Faraci Sep 16 '15 at 23:36
  • What exactly is not correct? – Nikolay Gromov Sep 16 '15 at 23:36
  • If z is a complex variable, thus the argument of the bessel function is not real and its taylor series is an expansion in terms of a complex variable. Thank you. – Jeff Faraci Sep 16 '15 at 23:38
  • Let me make the statement more precise (you are also welcome to read https://en.wikipedia.org/wiki/Analytic_function): we can write $J_1(z)=\sum_{n=0}^\infty c_n z^n$ with $c_n\in{\mathbb R}$. $z$ can be comlex of course and the values of $J_1$ at a generic point of the complex plane are complex, but this is a real function (sorry may be the name is a bit misleading, but that's not my invention). – Nikolay Gromov Sep 16 '15 at 23:41
  • @Integrals I believe that Nikolay is suggesting that $\bar J_1(z)=J_1(\bar z)$. But that does not get you the desired result. – Mark Viola Sep 17 '15 at 00:03
  • Oh Okay, Thanks @Dr.MV. I stated in the post that $\bar{J}_1(z)=J_1(\bar z)$, but wasn't sure what to do after. thanks for the clarification. Do you have any suggestion on how to approach this integral? Thanks again for your help on multiple problems I post. – Jeff Faraci Sep 17 '15 at 00:13
  • @NikolayGromov I didn't understand what you meant, sorry about that. However I am still not sure how that helps. Thanks – Jeff Faraci Sep 17 '15 at 00:13
  • No I was suggesting $\overline{J_1(z)}=J_1(\bar z)$. At the same time $\bar J_1(z)\neq J_1(\bar z)$ for complex $z$. Could it be that you are confusing notations $\overline{J_1(z)}$ with $\overline{J_1}(z)$? (in the first case you first evaluate $J_1$ on $z$ and then conjugate the result, in the second case you conjugate only the coefficients in the Taylor expansion but not the argument. Some other identities you may find useful $\overline{f(z)} = \bar f(\bar z)$ and $\overline{f(\bar z)} = \bar f(z)$ and for real functions (such as $J_1$) one also have $\bar f(z)=f(z)$ – Nikolay Gromov Sep 17 '15 at 00:19
  • Are you looking for an indefinite integral (i.e. an anti-derviative)? In that case, it can't be done, since $z^2 \overline{J_1(z)}$ is not holomorphic. – mrf Sep 17 '15 at 14:42
  • @NikolayGromov Thank you for the clarification, that's very helpful! I understand what you're talking about now. – Jeff Faraci Sep 17 '15 at 15:03
  • @mrf Yes, that's what I was looking for. Thank you for telling me that! Just a question, how do you know that $z^2\bar{J_1}$ is not holomorphic? Do you suggest then doing this integral numerically? I was trying to do the integral as a definite integral but since no indefinite integral exists, this demands numerics, correct? Thank you. – Jeff Faraci Sep 17 '15 at 15:05

1 Answers1

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In a previous topic, $(\partial_{tt}+\partial_t-\nabla^2)f(r,t)=0$ , the solution was explicitely expressed on the form : $$f(r,t)=g(r)\cos(\omega t)+h(r)\sin(\omega t)$$ where the functions $g(r)$ and $h(r)$ are complicated terms involving a Bessel function. But there is an hitch because the coefficients and the Bessel functions are in the complex range.

Instead of trying to express the solution with the Bessel functions, another approach consists in looking for the Taylor series of $g(r)$ and $h(r)$. An attempt is done below :

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The recurence relationships allow to compute $A_{2n+1}$ and $B_{2n+1}$ as functions of $A_3$ and $B_3$ :

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Then, the last step is the computation of $A_3$ and $B_3$ according to the boundary condition :

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JJacquelin
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  • Wow this is amazing! I just will check everything and see if I can agree with all of the details. I posted in our comments here :http://math.stackexchange.com/questions/1440553/partial-tt-partial-t-nabla2fr-t-0, a possible simplification I found for your first answer involving the Bessel functions. Thanks so much for showing another powerful method, as well as a complete solution involving the two different methods. Very useful for learning! +500 :) – Jeff Faraci Sep 22 '15 at 14:40