# Can we solve two heat equations simultaneously?

Consider a basic 1D heat equation

$$\frac{\partial^{2} T(x, t)}{\partial x^{2}}=\kappa \frac{\partial T(x, t)}{\partial t}$$

for two isolated 1D rods with different thermal diffusivities $$\kappa$$ and lengths ($$AB$$ and $$CD$$), one at 0°C and the other at 100°C.

Then connecting them at points $$B$$ and $$C$$. We don't have fixed boundary conditions, as the temperature changes by time in both $$B$$ and $$C$$.

Initial Conditions: $$\begin{array}{ll} \mathrm{IC\ (rod\ 1)}: & T(x, t=0)=0\\ \mathrm{IC\ (rod\ 2)}: & T(x, t=0)=100 \end{array}$$

How can we set the boundary conditions? We have to define boundaries conditions based on solving the other equation:

$$\begin{array}{ll} \mathrm{BC\ (rod\ 1)}: & T(x=B)=f(\tau) \\ \mathrm{BC\ (rod\ 2)}: & T(x=C)=g(\tau) \\ \end{array}$$

How can we solve the heat equations for these boundary conditions, as $$f(\tau)$$ is given by solving the heat equation for rod 2, and $$g(\tau)$$ is given by solving the heat equation for rod 1.

The focus is not the difference between $$f$$ and $$g$$. How can we solve the heat equation if assuming $$f$$=$$g$$?

• How $f$ and $g$ are defined? Aug 30, 2021 at 12:25
• @Andrew they are not defined. I just used them as arbitrary notations. When connecting points $B$ and $C$, how does the temperature at the connecting point change? Aug 30, 2021 at 13:01
• What does it mean to "connect" $B$ and $C$ if they don't even have a common temperature?
– Ali
Aug 30, 2021 at 13:29
• You probably want to assume the heat flux is given by Newton’s law, or a similar jump condition. Aug 30, 2021 at 13:33
• Usually in such a case it is assumed two conditions in the point (surface) of contact: jump of temperature is zero and jump of heat flow (i.e. $T_x$) is zero. In other words ($B=C$) the conditions are the following: $T(B-0,t)=T(B+0,t)$ (and therefore $f=g$) and $\kappa_1 T_x(B-0,t)=\kappa_2 T(B+0,t)$ (the coefficients of the 2nd condition to be checked though). Such problems are called contact problems. Aug 30, 2021 at 13:36

For ease of notation let's say the spatial domain is $$[a,b]$$ and the contact point is $$c \in (a,b)$$.
There is some boundary condition at $$a$$ and another at $$b$$ which are not really relevant here. The boundary condition at $$c$$ is where the two communicate. There is one automatic equation here based on conservation of energy, and that is $$\kappa_1 T_x(x=c^-)=\kappa_2 T_x(x=c^+)$$. The LHS is the flux into the left rod on its right side; the RHS is the flux out of the right rod on its left side. This equation says they're the same.
But this alone does not close the system, since you need two equations at the contact point since it is a boundary of two essentially unrelated PDEs. You need a further equation to do that. The conventional way is a Robin condition, which says the flux into the left rod is proportional to how much hotter the right rod is than the left rod, i.e. $$\kappa_1 T_x(x=c^-)=\lambda (T(x=c^+)-T(x=c^-))$$. This $$\lambda$$ is kind of like the conductivity of the interface between the rods, though that analogy is a bit loose because it has different units.