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I want to solve a PDE given in cylindrical coordinates as: $$\frac{\partial u}{\partial t}=\frac{1}{r}\frac{\partial u}{\partial r} + \frac{\partial^2 u}{\partial r^2} + \frac{\partial^2 u}{\partial z^2}$$ where $u(r,z,t)$ and boundary conditions are given as:

  1. $u(r,z,t\rightarrow 0)=K_{constant} \delta(r-r_0)\delta(z)$
  2. $\\lim_{r\to\infty}~~~~u(r,z,t)=0$
  3. $ \\lim_{z\to\pm\infty} ~~u(r,z,t)=0$
  4. $D\frac{\partial u(r,z,t)}{\partial r}= wu(r,z,t)$ when $r=r_r $and$z'\leq z\leq z'+d$
  5. $D\frac{\partial u(r,z,t)}{\partial r}= wu(r,z,t)$ when $r=r_r $and$-z'\geq z\geq -z'-d$

I have found the solution of following equation with different boundary conditions (given as: Steady Temperatures in a Circular Cylinder page 480) in DIFFERENTIAL EQUATIONS with Boundary-Value Problems $7^{th}$ edition (Zill, Cullen): $$\frac{1}{r}\frac{\partial u}{\partial r} + \frac{\partial^2 u}{\partial r^2} + \frac{\partial^2 u}{\partial z^2}=0$$ But it is steady state as $u(r,z)$ and not a function of $t$. Further it is mentioned there: "The method of separation of variables is a powerful but not universally applicable method for solving boundary-value problems.If the partial differential equation is nonhomogeneous, if the boundary conditions are timedependent, or if the domain of the spatial variable is an infinite interval ($\infty$,$\infty$) or a semi-infinite interval (a,$\infty$) we may be able to use an integral transform to solve the problem." Since this equation looks to me a more like a heat equation I am assuming it solution must be solved somewhere with different boundary conditions, if you can tell me how should I proceed to solve this PDE equation or help me provide some reference.

Edit:- In response to comment I am posting my scenario. I am trying to model Diffusion equation basically Ficks' second law in the presence of single molecule point source and two cylindrical receivers. As can be seen hereenter image description here the transmitter is a point source, two receivers are assumed to be cylindrical. Ficks' second law equation as in cylindrical coordinates: $$\frac{\partial p(r,z,t)}{\partial t}=\frac{D}{r} \frac{\partial}{\partial r} \frac{r \partial p(r,z,t) }{\partial r} +D \frac{\partial^2 p(r,z,t)}{\partial z^2} $$ where $p(r,z,t)$ is the concentration function, $D$ is Diffusion coefficient. Now about boundary conditions:

  • First initial condition tell us about the origin of concentration at radial distance $r_0$ from origin and $z=0$
  • second and third boundary condition tells us the concentration is zero at radial distance and vertical distance $\infty$.
  • We assume receiver 1 and receiver 2 both of which are basically cylinders covered with receptors and the flux at there surface is defined by these Neumann boundary condition. The receivers of length $d$ are symmetrically placed at $z'$ distance from origin, further the axis of cylindrical coordinate system coincide with cylindrical receiver axis; $w$ is the reaction rate which when $\infty$ tells that the concentration is perfectly absorbed at the surface of receiver(we need to find for $w=\infty$ scenario). This is given by $4^{th}$ and $5^{th}$ condition where flux is given at the cylindrical receivers. if any further clarification is needed please tell me.
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Edit Given the updated/clarified system & boundary conditions, the below transform is not really appropriate and most likely needs to be modified to the 'finite' Hankel transform..although i doubt that will help given the structure of the system

Consider the Hankel Transform defined here to be $$U\left( k,z,t \right)=\int\limits_{0}^{\infty }{r{{J}_{0}}\left( rk \right)u\left( r,z,t \right)dr}$$ where ${{J}_{0}}\left( z \right)$ is the zero order Bessel function of the first kind. Apply the transform to the PDE, and we have $$\frac{\partial U}{\partial t}=\int\limits_{0}^{\infty }{\frac{\partial }{\partial r}\left( r\frac{\partial u}{\partial r} \right){{J}_{0}}\left( rk \right)dr}+\frac{{{\partial }^{2}}U}{\partial {{z}^{2}}}$$ For the integral, integrate by parts and we have $$\int\limits_{0}^{\infty }{\frac{\partial }{\partial r}\left( r\frac{\partial u}{\partial r} \right){{J}_{0}}\left( rk \right)dr}=\left[ r\frac{\partial u}{\partial r}{{J}_{0}}\left( rk \right) \right]_{0}^{\infty }+\int\limits_{0}^{\infty }{\frac{\partial u}{\partial r}kr{{J}_{1}}\left( rk \right)dr}$$ The first terms are zero, and therefore integrating by parts once more yields $$\int\limits_{0}^{\infty }{\frac{\partial }{\partial r}\left( r\frac{\partial u}{\partial r} \right){{J}_{0}}\left( rk \right)dr}=-\int\limits_{0}^{\infty }{\left( k{{J}_{1}}\left( rk \right)+\frac{1}{2}{{k}^{2}}r\left( {{J}_{0}}\left( rk \right)-{{J}_{2}}\left( rk \right) \right) \right)udr}$$ Well known properties of Bessel functions (namely their recurrence properties etc) allows use to condense the integrand and so we find $$\int\limits_{0}^{\infty }{\frac{\partial }{\partial r}\left( r\frac{\partial u}{\partial r} \right){{J}_{0}}\left( rk \right)dr}=-{{k}^{2}}\int\limits_{0}^{\infty }{r{{J}_{0}}\left( rk \right)u\left( r,z,t \right)dr}=-{{k}^{2}}U\left( k,z,t \right)$$ The differential equation becomes therefore $${{U}_{t}}={{U}_{zz}}-{{k}^{2}}U$$

For a very general method of solving this very equation (althought different boundary conditions) see here. Is that enough to continue – if not I’ll edit the post and provide more.

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  • $\begingroup$ I am able to follow your answer except for (namely their recurrence properties etc) this comment for which I think I need to read Bessel properties. But my issue is while following the answer in that linked problem I have difficulties do I need to read "Fokas’ Universal Transform" to understand the method because right now what i got is just that you perform some sort of Fourier transform over$ u$ and rest is not clear to me. Can you please give me any quick reference which can help me to understand your solution. $\endgroup$ – Userhanu Jul 27 '17 at 10:59
  • $\begingroup$ Also I forget to mention one constant involved in equation so my equation become $$U_t=DU_{zz}-K'^2U$$ where $K'=k\sqrt(D)$. I hope this will not change the way for solving PDE. $\endgroup$ – Userhanu Jul 27 '17 at 10:59
  • $\begingroup$ One more question if $U(k,t)$ is just a function of $k,t$ then where is $z$. How can we compute$U_{zz}(k,t)$? $\endgroup$ – Userhanu Jul 27 '17 at 12:26
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    $\begingroup$ Yes my apologies, very sloppy of me. I have updated the post and added the z. The transform and the transformed equation remain the same however. How you proceed from here depends on a few things: a) how much you know about complex analysis (or are prepared to learn), and b) your boundary conditions. I don't fully understand what it is you are trying to model here. what is z' and what is d? $\endgroup$ – mathstackuser12 Jul 27 '17 at 21:43
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    $\begingroup$ Looking at the system i suspect you are going to find it quite difficult to obtain an exact solution (in terms of an integral/series or otherwise). Also you havent specified boundary conditions on the ends of the cylinders, so as it stands it is still not well-posed. $\endgroup$ – mathstackuser12 Jul 29 '17 at 0:29

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