Imagine a very long cylinder, made of an inner core ($r\leq R_1$) and an outer shell ($R_1<r\leq R_0$), both with different thermal diffusivities $\kappa$. The outer surface ($r=R_0$) is kept at a constant temperature $T_0$.

Composite cylinder

Initially, the whole object is at a uniform temperature $T_i$.

What I'm interested in is the temperature evolution at the material interface $r=R_1$, i.e. $T(R_1,t)$:

In the interval $[R_1,R_0]$ I'll call the temperature $u$, where:


Assuming no $z$ temperature gradient ($u_z=0$, $u_t=0$), then Fourier's heat equation becomes: $$\frac{1}{\kappa}u_t=\frac1r u_r+u_{rr}$$ With Ansatz $u(r,t)=R(r)\Gamma(t)$ and separation constant $-k^2$ this separates into the ODEs:

$$\frac{1}{\kappa}\frac{\Gamma'(t)}{\Gamma(t)}=-k^2 \tag{1}$$ $$rR''(r)+R'(r)+k^2rR(r)=0 \tag{2}$$ Spatial ODE:

The boundary condition translates to: $$u(R_0,t)=0\implies R(R_0)=0\tag{3}$$

$(2)$ solves to: $$R(r)=c_1J_0(kr)+c_2Y_0(kr)$$ Because $Y_0(kr)\to\infty$ for $r\to 0$, $\implies c_2=0$

With $(3)$ we get:

$$c_1J_0(k_nR_0)=0\implies J_0(k_nR_0)=0\tag{4}$$ So the eigenvalues $k_n$ are the roots of $(4)$.

The particular solution for $R_n(r)$ is thus: $$R_n(r)=C_nJ_0(k_nr)$$ Time evolution:

$(1)$ solves to: $$\Gamma_n(t)=A_ne^{-\kappa k_n^2t}$$ The particular solution becomes:

$$u_n(r,t)=B_nJ_0(k_nr)e^{-\kappa k_n^2t}$$ And with superposition: $$u(r,t)=\displaystyle \sum_{n=1}^{+\infty}B_nJ_0(k_nr)e^{-\kappa k_n^2t}$$ Using the initial condition $u(r,0)=T_i-T_0=u_i$: $$u(r,0)=u_i=\displaystyle \sum_{n=1}^{+\infty}B_nJ_0(k_nr)$$ Now I believe that to obtain $B_n$ we use the Fourier sine series: $$B_n=\frac{2}{R_0-R_1}\int_{R_1}^{R_0} u_iJ_0(k_nr)dr=\frac{2u_i}{R_0-R_1}\int_{R_1}^{R_0} J_0(k_nr)dr\tag{5}$$ Now I have two questions:

  1. Does $(5)$ need integrating between $R_1$ and $R_0$ or between $0$ and $R_0$?
  2. I had fully expected to have to define two $u(r,t)$ functions, one for the shell and one for the core, to match them up, determine eigenvalues for the core etc. I seem to have done nothing of the sort. Is my approach above satisfactory or wholly incorrect? It seems hard to believe the shell temperature's evolution is wholly independent of the core's temperature.

1 Answer 1


You have only one boundary condition applicable to each region. You can't, say, use the boundary condition for r = 0, and expect it to apply to the annular region. So each of the two regions will have a separate solution and one constant that still needs to be determined. These constants can be obtained by setting both the temperature and the heat flux values continuous at the interface between the two regions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.