Consider a solid cylinder of length L and radius a in which the thermal conductivity has a jump discontinuity at a point along its axis. The two bases of the cylinder are maintained at zero temperature while the lateral surface is at a constant temperature. We wish to find the steady state temperature in the cylinder.
The problem has axial symmetry so the temperature u is a function only of the cylindrical coordinates(r,z). Selecting the origin of a cylindrical coordinate system at the point of discontinuity, the governing equations and boundary condition are: $$\nabla^2u_1=0, -l_1<z<0$$ $$\nabla^2u_2=0, 0<z<l_2$$ $u_1(r,-l_1)=0,u_2(r,l_2)=0,u_1(a,z)=u_0,u_2(a,z)=u_0$
In the section of the cylinder $-l_1<z<0,$ the thermal conductivity is $k_1$ while in the other section $0<z<l_2$, the thermal conductivity is $k_2$. In addition to the above boundary conditions, we need to impose the two continuity conditions
$u_1=u_2$ and $k_1 \frac{\partial u_1}{\partial z}=k_2 \frac{\partial u_2}{\partial z}$ at z=0.
Following the usual method of separation of variables and enforcing the homogeneous boundary conditions, we find $$u_1(r,z)=\sum _{n=1}^{\infty } A_n I_0\left(\lambda _n r\right) \left(\cos \left(\lambda _n z\right)+\cot \left(\lambda _n l_1\right) \sin \left(\lambda _n z\right)\right)$$ $$u_2(r,z)=\sum _{n=1}^{\infty } A_n I_0\left(\lambda _n r\right) \left(\cos \left(\lambda _n z\right)-\cot \left(\lambda _n l_2\right) \sin \left(\lambda _n z\right)\right)$$ where $I_0(x)$ is the modified Bessel function of order zero of the first kind and $\lambda _n$ is the nth positive roots of the equation $$k_1 \cot \left(\lambda l_1\right)+k_2 \cot \left(\lambda l_2\right)=0$$ Setting $u_1(a,z)=u_0$ we obtain an expansion that looks like a Fourier series but cannot directly determine the coefficients $A_n$ since we do not have an orthogonality condition. Fuththermore, setting $u_2(a,z)=u_0$ we obtain a similar situation. This posses two problems
- How do we compute the coefficients $A_n$
- Do the two alterenatives give the same solution
We can solve this problem by the superposition of two harmonic functions $\tilde{u}(r,z)$ and $\hat{u}(r,z)$ where $\tilde{u}(r,z)$ vanishes on the two bases of the cylinder and assumes the value $u_0$ on the lateral surface.The function $\tilde{u}(r,z)$ (as computed below) does not satisfy the continuity requirements at the interface, so we find a harmonic function $\hat{u}(r,z)$ such that $\tilde{u}(r,z)$+$\hat{u}(r,z)$ vanishes at $z=-l_1$ and $z=l_2$, assumes the value $u_0$ on the lateral surface and satisfies the continuity conditions at $z=0$. Using separtion of variables in cylindrical coordinates, it is readily found that $$\tilde{u_1}(r,z)=\frac{4u_0}{\pi }\sum _{n=1}^{\infty } \frac{I_0\left(\frac{(2 n-1) \pi r}{l_1}\right) \sin \left((2 n-1) \pi \left(\frac{z}{l_1}+1\right )\right)}{I_0\left(\frac{(2 n-1) \pi a}{l_1}\right) (2 n-1)}\hspace{4 mm}-l_1<z<0$$ $$\tilde{u_2}(r,z)=\frac{4u_0}{\pi }\sum _{n=1}^{\infty } \frac{I_0\left(\frac{(2 n-1) \pi r}{l_2}\right) \sin \left(\frac{(2 n-1) \pi z}{l_2}\right)}{I_0\left(\frac{(2 n-1) \pi a}{l_2}\right) (2 n-1)}\hspace{4 mm}0<z<l_2$$ It follows that $$\frac{\partial \tilde{u}_1}{\partial z}\Bigr\rvert_{z = 0}=-\frac{4 u_0}{l_1}\sum _{n=1}^{\infty } \frac{I_0\left(\frac{(2 n-1) \pi r}{l_1}\right)}{I_0\left(\frac{(2 n-1) \pi a}{l_1}\right)}\doteq \sigma _1(r),\hspace{4 mm}\frac{\partial \tilde{u}_2}{\partial z}\Bigr\rvert_{z = 0}=\frac{4 u_0}{l_2}\sum _{n=1}^{\infty } \frac{I_0\left(\frac{(2 n-1) \pi r}{l_2}\right)}{I_0\left(\frac{(2 n-1) \pi a}{l_2}\right)}\doteq \sigma _2(r)$$ Now write $$u_1(r,z)=\tilde{u}_1(r,z)+\hat{u}_1(r,z)\hspace{4 mm}-l_1<z<0$$ $$u_2(r,z)=\tilde{u}_2(r,z)+\hat{u}_2(r,z)\hspace{4 mm}0<z<l_2$$
In order for $u_1$ and $u_2$ to satisfy the given boundary conditions,we require $$\hat{u}_1\left(r,-l_1\right)=\hat{u}_2\left(r,l_2\right)= \hat{u}_1(a,z)=\hat{u}_2(a, z)=0,\hspace{4 mm}\hat{u}_1(r,0)=\hat{u}_2(r,0)\hspace{4 mm}$$ and $$k_1 \frac{\partial \hat{u}_1}{\partial z}\Bigr\rvert_{z = 0}-k_2 \frac{\partial \hat{u}_2}{\partial z}\Bigr\rvert_{z = 0}=k_2 \sigma _2(r)-k_1 \sigma _1(r)$$ Assume the following forms for $\hat{u}_1(r,z)$ and $\hat{u}_2(r,z)$ $$\hat{u}_1(r,z)=A_{10}+B_{10} z+\sum _{n=1}^{\infty } J_0\left(\alpha _n r\right) \left(A_{1 n} \cosh \left(\alpha _n z\right)+B_{1 n} \sinh \left(\alpha _n z\right)\right)\hspace{4 mm}-l_1<z<0$$ $$\hat{u}_2(r,z)=A_{20}+B_{20} z+\sum _{n=1}^{\infty } J_0\left(\alpha _n r\right) \left(A_{2 n} \cosh \left(\alpha _n z\right)+B_{2 n} \sinh \left(\alpha _n z\right)\right)\hspace{4 mm}0<z<l_2$$ Since $\hat{u}_1(a,z)=\hat{u}_2(a,z)=0,A_{10}=B_{10}=A_{20}=B_{20}=0$ and $\alpha _n=\frac{j_{0,n}}{a}$ where $j_{0,n}$ is the nth positive zero of $J_0(x).$ It follows that $$\hat{u}_1\left(r,-l_1\right)=\sum _{n=1}^{\infty } J_0\left(\alpha _n r\right) \left(A_{1 n} \cosh \left(\alpha _n l_1\right)-B_{1 n} \sinh \left(\alpha _n l_1\right)\right)=0$$ Hence $A_{1 n}=B_{1 n} \tanh \left(\alpha _n l_1\right)$. Similarly, since $\hat{u}_2\left(r,l_2\right)=0,A_{2 n}=-B_{2 n} \tanh \left(\alpha _n l_2\right).$ Thus we have the series expansions $$\hat{u}_1(r,z)=\sum _{n=1}^{\infty } B_{1 n} J_0\left(\alpha _n r\right) \left(\tanh \left(\alpha _n l_1\right) \cosh \left(\alpha _n z\right)+\sinh \left(\alpha _n z\right)\right)$$ $$\hat{u}_2(r,z)=\sum _{n=1}^{\infty } B_{2 n} J_0\left(\alpha _n r\right) \left(\sinh \left(\alpha _n z\right)-\tanh \left(\alpha _n l_2\right) \cosh \left(\alpha _n z\right)\right)$$ It now follows from the continuity condition $\hat{u}_1(r,0)=\hat{u}_2(r,0)$ that $B_{1 n}=-B_{2 n} \tanh \left(\alpha _n l_2\right) \coth \left(\alpha _n l_1\right)$ Finally the heat flux continuity condition lead to the equation $$\sum _{n=1}^{\infty } B_{2 n} \alpha _n J_0\left(\alpha _n r\right) \left(k_1 \coth \left(\alpha _n l_1\right) \tanh \left(\alpha _n l_2\right)+k_2\right)=f(r)$$ where $f(r)=k_1 \sigma _1(r)-k_2 \sigma _2(r).$ To determine the coefficients $B_{2 n},$ we recall the expansion formula for the Fourier-Bessel series $$f(r)=\sum _{n=1}^{\infty } c_n J_0\left(\alpha _n r\right)$$ $$c_n=\frac{2 \int_0^a r f(r) J_0\left(\alpha _n r\right) \, dr}{a^2 J_1\left(j_{0,n}\right){}^2 }$$ From which it follows that $B_{2 n}=\frac{c_n}{\alpha _n \left(k_1 \coth \left(\alpha _n l_1\right) \tanh \left(\alpha _n l_2\right)+k_2\right)}$