How to solve this partial differential (transient temperature in circular cylinder like ) equation?

Question

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 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.

Answer 1

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.