Heating transient of a composite rod? A straight rod is made of two parts, $[0,x_1]$ (green in the figure) with thermal diffusivity $\kappa_1$ and $[x_1,x_2]$ (blue) with thermal diffusivity $\kappa_2$. The rod is perfectly insulated. Zero $y$ and $z$ temperature gradients are assumed.
At $x=0$ temperature is maintained at constant $T_0$. At $x=x_2$ the rod is embedded into a perfect insulator ($\kappa=0$). At $t=0$ the rod has a uniform temperature $T(x,0)=T_i$.

Question: what is the temperature evolution of the rod?
1. The simple case where $\kappa_1=\kappa_2=\kappa$:
Let $u(x,t)=T(x,t)-T_0$.
Then Fourier's equation tells us:
$$u_t=\kappa u_{xx}$$
Boundary conditions:
$$u(0,t)=0$$
$$u_x(x_2,t)=0$$
Initial condition:
$$u(x,0)=u_i=T_i-T_0$$
Using the Ansatz $u(x,t)=X(x)\Gamma(t)$, separation constant $-k^2$ and the boundary conditions above, this solves easily to:
$$\Large{u(x,t)=\displaystyle \sum_{n=1}^{+\infty}B_n\sin\Bigg(\frac{n\pi x}{2x_2}\Bigg)e^{-\kappa \Big(\frac{n\pi }{2x_2}\Big)^2t}}$$
(for $n=1,3,5,7,...$)
The $B_n$ coefficients can easily be obtained from the initial condition with the Fourier sine series:
$$B_n=\frac{4u_i}{n\pi}$$
Back-substituting we get:
$$T(x,t)=T_0+\frac{4(T_i-T_0)}{\pi}\displaystyle \sum_{n=1}^{+\infty}\frac{1}{n} \sin\Bigg(\frac{n\pi x}{2x_2}\Bigg)e^{-\kappa \Big(\frac{n\pi }{2x_2}\Big)^2t}$$
(for $n=1,3,5,7,...$)
A plot for the first three terms at $t=0.1$:

2. The case where $\kappa_1\neq\kappa_2$:
We define two functions $u_1(x,t)$ for $[0,x_1]$ and $u_2(x,t)$ for $[x_1,x_2]$. We use the same Ansatz as under $1.$ We'll assume both functions have their own eigenvalues.
Boundary conditions:
$$u_1(0,t)=0\implies X_1(0)=0\tag{1}$$
$$\frac{\partial u_2(x_2)}{\partial x}=0\implies X_2'(x_2)=0\tag{2}$$
In addition (continuity):
$$u_1(x_1,t)=u_2(x_1,t)\tag{3}$$
With Fourier, the heat flux is the same at $x=x_1$:
$$\alpha_1\frac{\partial u_1(x_1)}{\partial x}=\alpha_2\frac{\partial u_2(x_1)}{\partial x}\tag{4}$$
Where $\alpha_i$ are the thermal conductivities.
a. for $u_1(x,t)$:
$$X_1(x)=c_1\cos k_1x+c_2\sin k_1x$$
$$X_1(0)=0\implies c_1=0\implies X_1(x)=c_2\sin k_1x\tag{5}$$
b. for $u_2(x,t)$:
$$X_2(x)=c_3\cos k_2x+c_4\sin k_2x$$
$$X_2'(x_2)=0\tag{2}$$
$$\implies -c_3k_2\sin k_2x_2+c_4k_2\cos k_2x_2=0\tag{6}$$
Using the additional conditions $(3)$ and $(4)$:
$$c_2\sin k_1x_1=c_3\cos k_2x_1+c_4\sin k_2x_1\tag{7}$$
$$c_2\alpha_1k_1\cos k_1x_1=-c_3\alpha_2k_2\sin k_2x_1+c_4\alpha_2k_2\cos k_2x_1\tag{8}$$
Problem:
$(6)$, $(7)$ and $(8)$ form a system of three simultaneous equations but with five unknowns: $c_2$, $c_3$, $c_4$, $k_1$ and $k_2$.
I'm tempted to set $c_3=0$ as it would yield $k_2$ from $(6)$. I think this would yield also the remaining unknowns. But can I a priori assume $c_3=0$? Or is there another approach possible?
I'm also left wondering whether perhaps $k_1=k_2$. The eigenvalues do not depend on $\kappa$, so perhaps the eigenvalues $k$ are common to both functions. Due to $(4)$, $u_1$ and $u_2$ would then still be distinct.
 A: I can see how to do this now.  You have 3 BC equations, and 3 unknown c's plus the unknown time eigenvalue $\lambda^2$ (You know that $k_1=\lambda /\sqrt{\kappa_1}$ and $k_2=\lambda /\sqrt{\kappa_2}$).  You can eliminate 2 of the c's.  Your third equation will give you a relationship with the 3rd c out front and a zero on the other side of the equation.  This will allow you to determine $\lambda$. The last c is then determined by making good on the initial condition, using an infinite series.
ADDENDUM
$$\frac{1}{T}\frac{dT}{dt}=\frac{\kappa}{X}\frac{d^2X}{dx^2}=-\lambda^2$$
So, $$\frac{d^2X}{dx^2}+\frac{\lambda^2}{\kappa}X=0$$
If the time scale for both regions is the same (as expected), then:  $$k_1=\frac{\lambda}{\sqrt{\kappa_1}}$$and$$k_2=\frac{\lambda}{\sqrt{\kappa_2}}$$
A: This uses parameter symbols that I'm more accustomed to:
k is thermal conductivity, $\alpha$ is thermal diffusivity, $\rho$ is density, and $c_p$ is heat capacity.  
I followed the procedure which I outlined in my previous answer and obtained the following: 
The characteristic equation for the n'th eigenvalue is:  $$\tan\left[\frac{\lambda_n(x_2-x_1)}{\sqrt{\alpha_2}}\right]\tan\left[\frac{\lambda_nx_1}{\sqrt{\alpha_1}}\right]=\sqrt{\frac{k_1\rho_1c_{p1}}{k_2\rho_2c_{p2}}}$$
The n'th eigenfunction is:
$$X_n=\cos\left[\frac{\lambda_n(x_2-x_1)}{\sqrt{\alpha_2}}\right]\sin\left[\frac{\lambda_nx}{\sqrt{\alpha_1}}\right]\tag{0 < x <x1}$$
$$X_n=\cos\left[\frac{\lambda_n(x_2-x)}{\sqrt{\alpha_2}}\right]\sin\left[\frac{\lambda_nx_1}{\sqrt{\alpha_1}}\right]\tag{x1<x<x2}$$
I leave it up to you to do the integation to get the n'th coefficient in the infinite series.
