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Hi I am trying to find the kernel of the linear differential operator $D$ $$ D\equiv\partial_{tt}+b\partial_t-a\nabla^2,\quad a,b>0. $$ We have $$ \nabla^2\equiv \frac{1}{r}\partial_r(r\partial_r)-\frac{1}{r^2}=\partial_{rr}+\frac{1}{r} \partial_r -\frac{1}{r^2}. $$ So I am trying to solve $$ Df(r,t)=\left(\partial_{tt}+b\partial_t-a\nabla^2\right)f(r,t)=0 $$

$$ \partial_t f(0,t)=0, \quad \partial_t(R,t)=v_0\cos \omega t $$ What is the general way to do this for this linear PDE? Thanks.

I assume we can just solve the pde by an ansatz for $f(r,t)$, however I am unsure of what to use. Since this problem is in cylindrical geometry (hence the Laplacian like operator), maybe there will be Bessel functions as the Kernel? I am not sure but trying to prove something. Thanks!

Jeff Faraci
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Solutions can be found, thanks to the separation of variables. Not really a smart method, but effective anyways.

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In addition, the boundary condition $\left(\frac{\partial(f(r,t)}{\partial t} \right)_{r=R}=v_0 \cos(\omega t)$ leads to look for a solution made with real sinusoidal functions of $\omega t$ . This implies a particular form of the exponential terms :

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JJacquelin
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  • Note, my solution using the initial condition $\partial_t x(0,t)=0$ and involving the unknown coefficient $\alpha_\lambda$ is $$ x(r,t)=\sum_\lambda \alpha_\lambda J_1 (\sqrt \lambda r) e^{-\xi t} $$ where $\xi$ is defined above. But, I do not now how to apply the other boundary condition $\partial_t x(R,t)=v_0\cos \omega t$ to this, in order to find $\alpha_\lambda$. Thanks. – Jeff Faraci Sep 18 '15 at 17:12
  • Thank you! I am stuck finding $\alpha_\lambda$. How can I use the two boundary conditions $\partial_t x(0,t)=0, \partial_t x(R,t)=v_0\cos \omega t$? When I do this, I obtain $\beta_\lambda=0$ since $Y_1$ is not finite at the origin. However i'm stuck on the second boundary condition in order to find $\alpha_\lambda$, specifically using $\partial_t x(R,t)=v_0\cos \omega t$, I obtain$$ \partial_t x(R,t)=-\xi \sum_\lambda \alpha_\lambda J_1(\sqrt \lambda R)e^{-\xi t}=v_0\cos \omega t$$ where $\xi \equiv \beta/2 \pm \sqrt{\beta^2-4a \lambda}/2$. How can I solve for $\alpha_\lambda$? – Jeff Faraci Sep 18 '15 at 17:16
  • Let $(\beta^2-4a\lambda)/2=-\omega^2$. This permits to have too complex conjugate terms which a convenient combination reduces to $\sin(\omega t)$. The derivative is a term with $\cos(\omega t)$. Then compute $\alpha$ so that the coefficient of $\cos(\omega t)$ be equal to $v_0$. – JJacquelin Sep 19 '15 at 10:39
  • If you had given the boundary conditions in the wording of the question, it had avoid to express the general solutions with $\Sigma$ and/or $\int$.The boundary conditions are the most important information to chose a method of PDE solving. – JJacquelin Sep 19 '15 at 10:46
  • With these boundary conditions, you don't need all this stuff. Just set $x(r,t)=g(r)\sin(\omega t)$, put it into the PDE and solve for $g(r)$ which will give the Bessel function. – JJacquelin Sep 19 '15 at 11:09
  • Thanks for your reply. I still am working on finishing this problem. I will post a reply later as to the progress I make based on what you jus told me. However, can't we still use the series expansion and just apply boundary conditions to it? Also, why does $$x(r,t)=g(r)\sin(\omega t)$$work, should we use $$x(r,t)=g(r)\sin \omega t+d(r)\cos \omega t$$ instead? I want the solution in terms of a series expansion, and I need to compute the coefficients of the expansion, $\alpha_\lambda$. This is what I obtained for $\alpha_\lambda$ using the boundary condition: – Jeff Faraci Sep 19 '15 at 13:53
  • $${\boxed{\alpha_\lambda=-\frac{2v_0e^{\xi t}}{\xi J^2_2(\sqrt \lambda)}\cos(\omega t)\int_0^1 J_1(\sqrt \lambda R)RdR}}$$ Does this look correct? Thanks. I may make repost this question and make it into a bounty and award you. – Jeff Faraci Sep 19 '15 at 13:55
  • The exact question given to me was as written: Solve $$ \left(\partial_{tt}+b\partial_t-a\nabla^2\right)x(r,t)=0 $$ with $\partial_t x(0,t)=0, \partial_t x(R,t)=v_0\cos \omega t$, and calculate the coefficients of the expansion in the solution. Plot the coefficients as functions of $\omega$. (I will do this part myself obviously). However, this was the exact problem. – Jeff Faraci Sep 19 '15 at 14:05
  • Yes, your atempt to use a series expansion is well suggested by the complete wording of the problem. – JJacquelin Sep 19 '15 at 15:05
  • No need for reposting the question. Don't waste time. I don't run behind rewards. Don't mention it. – JJacquelin Sep 19 '15 at 15:18
  • Do you think the expansion coefficients are correct that I posted, $\alpha_\lambda$? Thanks! I just used http://mathworld.wolfram.com/Fourier-BesselSeries.html. – Jeff Faraci Sep 19 '15 at 17:48
  • Posted it as a bounty, enjoy! Thanks for your assistance. – Jeff Faraci Sep 20 '15 at 16:37
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    Sorry for the late answer, I was aways for a few days. Even more sorry for a mistake in my comments (not in the main answer). I completely forgot the exponential term $e^{-bt/2}$. This term doesn't exist in the bourdary conditions. This makes the boundary conditions inconsistent with the formula of general solution if the parameters $\lambda$ and $\alpha_\lambda$ are real. This case is considered in the new addendum to my first answer. Of course, it is very complicated because the Bessel functions are now in the complex range. The use of Bessel functions might be not a good idea. – JJacquelin Sep 21 '15 at 06:53
  • This is very clear, thanks so much for your work on solving this and all the little details involved. I am going through everything now, I only have one question. I can follow it all, I am just curious as to why you know to write $$ f(r,t)=\alpha_1 J_1(\sqrt{\lambda_1} r)e^{i\omega t}+\alpha_2 J_1(\sqrt{\lambda_2} r)e^{-i\omega t} $$ or in other words, I see why we have two roots for $\lambda$, however, how did you know to just write the solution as a sum of bessel functions involving both those roots? . Is it just because of that boundary condition at $r=R$? Thanks again ! – Jeff Faraci Sep 21 '15 at 16:28
  • By the way, seeing all of this helped clear some up of my other general misunderstandings on solving these linear PDE's. You also mentioned that the boundary conditions are the most important information in order to choose a PDE solving method. Is there any books you have on linear PDE's or references? Thanks. – Jeff Faraci Sep 21 '15 at 16:36
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    This is more by experience than from books. Often, the most difficult part of the task is to make fit the gereral solution to the boundary conditions. A same general solution can be expressed on various forms, depending on the chosen system (polar, Cartesian, etc.), depending on the use special functions or series, etc. These various possibilities can make either arduous or easy the fitting to the boundary conditions. It is important to know the boundary conditions in order to look for a favourable form of the general solution. Or to imagine shortcuts which avoid full solving of the PDE. – JJacquelin Sep 21 '15 at 18:03
  • Your solution can be very nicely simplified (I think). Since $\alpha_1=\bar{\alpha}_2$ and $\lambda_1=\bar{\lambda}_2$ where the bar is complex conjugation, we can write $$ f(r,t)=\alpha_1 J_1 (\sqrt \lambda_1 r)e^{i\omega t}+\alpha_2 J_1(\sqrt \lambda_2 r) e^{-i\omega t}=2 \Re(\alpha_1(t)J_1(\sqrt \lambda_1 r)) $$ where $\alpha_1(t)\equiv \alpha_1 e^{i\omega t}$. – Jeff Faraci Sep 22 '15 at 14:31
  • Thank you very much for your help and thorough explanations. I think this simplification works since This works since $$ \alpha_1 J_1 (\sqrt \lambda_1 r)e^{i\omega t}+\alpha_2 J_1(\sqrt \lambda_2 r) e^{-i\omega t}=\alpha_1 J_1 (\sqrt \lambda_1 r)e^{i\omega t}+{\bar{\alpha}}_1 J_1(\sqrt {\bar{\lambda}}_1 r) e^{-i\omega t}=(\alpha_1(t) J_1)+{(\bar \alpha_1(t) \bar J_1)} $$ and $z+\bar z=2\Re (z)$. – Jeff Faraci Sep 22 '15 at 14:35