How to obtain a solution for the following IBVP

Question

I am trying to solve the following advection-diffusion equation for transient flow conditions for radial flow.

The governing equation is as follows. $$\frac{\partial T}{\partial t} = \frac{\partial^2 T}{\partial x^2} + \frac{1-2v(t)}{x} \frac{\partial T}{\partial x}$$ $$\frac{\partial T}{\partial t} = \frac{\partial^2 T}{\partial x^2} + \frac{f(t)}{x} \frac{\partial T}{\partial x}$$

where $$f(t)=1 -2 v(t)$$

Initial condition $$T(x,t=0)=0$$ BCs. $$T(x=0,t)=1$$ $$\lim_{x \to \infty} T(x,t)=0$$

I have tried to solve the problem using the following solution procedure.

Assume the solution takes the following form. $$T(x,t)=\left ( e^{-\frac{x^2}{4t}}\right) F(t)$$

The similarity variable $-\frac{x^2}{4t}$ is appropriate selection for solving diffusion equation for radial flow.

The partial derivatives of $T(x,t)$ are as follows. $$\frac{\partial T}{\partial x} =-\frac{x}{2t}\left ( e^{-\frac{x^2}{4t}}\right) F(t)$$ $$\frac{\partial^2 T}{\partial x^2} =F(t)\left( -\frac{1}{2t}\left ( e^{-\frac{x^2}{4t}}\right) + \left( \frac{x}{2t} \right)^2 \left ( e^{-\frac{x^2}{4t}}\right) \right) $$ $$\frac{\partial T}{\partial t} = \left (\left( \frac{x}{2t} \right)^2 e^{-\frac{x^2}{4t}}\right)F(t) + \left ( e^{-\frac{x^2}{4t}}\right)\frac{\partial F(t)}{\partial t} $$

By substituting into the governing equation, the following ODE in $F(t)$ is obtained. $$\frac{dF(t)}{dt}=-\left ( \frac{1 + f(t)}{2t}\right)F(t)$$

The solution of the ODE is as follow. $$F(t) = \exp\left ( -\int_{0} ^{t} \left ( \frac{1 + f(u)}{2u}\right) \, du\right)$$

Finally, the solution of the governing is as follow. $$T(x,t) =\left ( e^{-\frac{x^2}{4t}}\right) \exp\left ( -\int_{0} ^{t} \left ( \frac{1 + f(u)}{2u}\right) \, du\right)$$

This solution is the same as that given in Handbook of Linear Partial Differential Equations for Engineers and Scientists - Page 367 (when $a = 1$)(https://www.taylorfrancis.com/books/9780429166464). Unfortunately, this solution satisfies the initial condition as well as the outer BC, however it doesn't satisfy the inner BC. When $x$ is put equals to zero, the resulting solution will be as follows. $$IBC \rightarrow T(x = 0,t) = \exp\left ( -\int_{0} ^{t} \left ( \frac{1 + f(u)}{2u}\right) \, du\right) ≠ 1$$

I was thinking of how to use the given solution to obtain a solution that satisfies the governing equations, initial conditions, and all boundary conditions of my problem. The resulting solution seems to be a solution of the same problem, however with time dependent inner BC. Duhamel's integral can be used to obtain a solution for time-dependent BC problem given the corresponding solution for constant BC problem, however the problem here seems to the opposite. Can anyone give me a hint of how to proceed to obtain the solution that satisfies the inner BC?

Answer 1

I have developed the following solution to my problem, however I am not sure if my procedure is mathematically fully right or not. It will be appreciated if anyone give his/her feedback about the following solution procedure. Sorry for that, since my major is not mathematics.

The solution that satisfies the inner BC of the problem (say $y(x,t)$) should satisfy the following Duhamel's integral rule (assuming that Duhamel's integral is valid for the governing equation of this problem).

$$T(x,t) = g(t)*\frac{\partial }{\partial t}y(t) = \int_{0}^{t} g(t - \tau)\frac{\partial }{\partial \tau} y(\tau) d\tau$$ where $g(t)$ is the time-dependent inner BC. $$g(t)= \exp\left ( -\int_{0} ^{t} \left ( \frac{1 + f(u)}{2u}\right) \, du\right)$$

The objective is to get $y(t)$ using the above-mentioned equations. I think this doable using deconvolution. I don't know much about deconvolution, however I think the following Laplace transformation rule can be used. $$\mathcal{L}[g(t)*\frac{\partial }{\partial t}y(t)]=\mathcal{L}[g(t)]\mathcal{L}[\frac{\partial }{\partial t}y(t)]$$
$$\mathcal{L}[g(t)*\frac{\partial }{\partial t}Y(t)]= T(s) = G(s) (sY(s) - y(t=0))=s G(s)Y(s)$$

Consequently, the solution of the problem in Laplace domain will be as follows. $$Y(s) = \frac{T(s)}{sG(s)}$$

and the solution in the real-time domain will be as follows. $$y(x,t) = \mathscr{L}^{-1}\left ( \frac{T(s)}{s G(s)}\right)= \mathscr{L}^{-1}\left ( \frac{\mathcal{L}[\left ( e^{-\frac{x^2}{4t}}\right) \exp\left ( -\int_{0} ^{t} \left ( \frac{1 + F(u)}{2u}\right) \, du\right)]}{s\mathcal{L}[\exp\left ( -\int_{0} ^{t} \left ( \frac{1 + F(u)}{2u}\right) \, du\right)]}\right)$$

Now this solution also satisfies the inner BC because when $x$ is put equals to zero, the solution in Laplace domain will be simplified to $\frac{1}{s}$ which is corresponding to Inner BC $(= 1)$ in the real-time domain. I would like to know if this procedure is mathematically right or not. It will also be appreciated if you give hints for another solution procedure.

Answer 2

I have updated the first order ODE in my solution procedure to the following correct form. $$\frac{dF}{dt}=-\left ( \frac{1 + f(t)}{2t}\right)F(t)$$

where $$f(u)=1 -2 v(u)$$

The solution of ODE is as follow.

$$F(t) = \exp\left ( -\int_{0} ^{t} \left ( \frac{1 + f(u)}{2u}\right) \, du\right)$$

The final solution is as follows. $$T(x,t) =\left ( e^{-\frac{x^2}{4t}}\right) \exp\left ( -\int_{0} ^{t} \left ( \frac{1 + f(u)}{2u}\right) \, du\right)$$ Sorry for the inconvenience.