Transient heat equation in a tube (polar coordinates) with uniform initial temperature and uniform boundary temperature

I tried to solve this problem:

I have a tube of inner and outer radius $$R_1$$ and $$R_2$$ (like coppers tube for fridges). The temperature at the exterior of the tube ($$\rho = R_2$$) is maintained to $$T_e$$. We denote $$T_1$$ the temperature inside the tube and $$T_2$$ the temperature between $$R_1$$ and $$R_2$$. The diffusivity coefficients of the material are $$k_1$$ (like the fluid inside the tube) and $$k_2$$ (copper coefficient). $T_e$ outside the tube" />

So the heat equations are:

$$\left[ k_i \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial}{\partial \rho} \right) - \frac{\partial}{\partial t} \right] T_i = 0$$

then boundary conditions are:

$$T_1(R_1,t) = T_2(R_1,t)$$ $$l_1 \frac{\partial}{\partial \rho} T_1(\rho = R_1,t) = l_2 \frac{\partial}{\partial \rho} T_2(\rho = R_1,t)$$ $$T_2(R_2,t) = T_e$$

with $$l_i$$ the thermal conductivity, and initial condition is:

$$T_1(\rho,0) = T_2(\rho,0) = T_0$$

By separation of variables we find general solutions:

$$T_1 = A + \sum_{n = 1} a_n e^{- \lambda_n t} J_0 \left( \sqrt{\lambda_n/k_1} \rho \right)$$ $$T_2 = B + \sum_{n = 1} b_n e^{- \mu_n t} J_0 \left( \sqrt{\mu_n/k_2} \rho \right) + c_n e^{- \nu_n t} Y_0 \left( \sqrt{\nu_n/k_2} \rho \right)$$

as $$t \rightarrow + \infty$$ the steady state is reached so logically we have: $$A = B = T_e$$ but I have difficulties to find the eigenvalues $$\lambda_n$$, $$\mu_n$$ and $$\nu_n$$ and the coefficients of the series. Does anyone know the solution of this problem?

What I know

For a simpler case, a disk of radius $$R$$, with initial temperature inside the disk $$T_0$$ and boundary temperature $$T(R,t) = T_e$$ we find the solution:

$$T(\rho,t) = T_e + \sum_{n=1}^{+ \infty} a_n J_0 \left( \sqrt{\lambda_n/k_1} \rho \right) e^{- \lambda_n t}$$

with the $$\lambda_n$$ solving (linked to the zeros of $$J_0$$ Bessel function):

$$J_0 \left( \sqrt{\lambda_n/k_1} R \right) = 0$$

and coefficients $$a_n$$ are (using weighted scalar product):

$$a_n = (T_0 - T_e) \frac{ \int_{0}^{R_1} \rho J_0 \left( \sqrt{\lambda_n/k_1} \rho \right) d \rho}{ \int_{0}^{R_1} \rho J_0^2 \left( \sqrt{\lambda_n/k_1} \rho \right) d \rho }$$

• I have on idea, using the 3 B.C and assuming $\nu_n = \lambda_n = \mu_n$ to get rid of exponential time terms, I found a determinental equation on $\lambda_n$ which gives the $\lambda_n$ (involving of course numerical calculations). To have the temperature inside, $T_1$ (so the coefficients $a_n$), I can integrate on radius $R_n$, which correspond to $J_0(R_n \sqrt(\lambda_n/k_1)) = 0$ and I get rid of the infinite sum because of orthogonality. What do you think folks? Dec 31, 2021 at 15:17

As I said in comment we need to apply BC and we find a matrix equality to zero with coefficients, which gives a zero determinant equation, able to find the $$\lambda_n$$ eigenvalues.