Boundary Conditions
When solving the heat equation from the last section, we require two boundary conditions, each end of the rod.
Boundary conditions
There are many types of boundary conditions
You can mix and match boundary conditions however you please.
Equilibruim temperature distribution
An equilibruim tempature distribution would or steady-state solution is a solution independent of time.
A very typical problem in heat flow is one where the thermal coefficients are constant and there are no sources of thermal energy. In this case, the tempature in a one-dimensional rod 0 \leq x \leq L satisfies
\frac{\partial u}{\partial t} = k\frac{\partial^2 u}{\partial x^2}
With the initial conditions
\begin{align*} u(0,t) &= T_1\\ u(L,t) &= T_2\\ u(x,0) &= f(x)\\ \end{align*}
We know that if our desired function u_{eq}(x) is not dependent on time, then
\frac{\partial u}{\partial t} = 0 = k\frac{\partial^2 u}{\partial x^2}
Integrating twice gives us
u_{eq}(x) = C_1x + C_2
Using our boundaries, we have that u_{eq}(0) = T_1 and u_{eq}(L) = T_2. Thus T_1 = C_2 and T_2 = C_1L + C_2. Solving for the constants we have
u_{eq}(x) = \frac{T_2 - T_1}{L}x + T_1.
Lets try the same problem with an insulated rod. That is
\begin{align*} \frac{\partial u}{\partial t} &= k\frac{\partial^2 u}{\partial x^2}\\ \phi(0,t) &= 0\\ \phi(L,t) &= 0\\ u(x,0) &= f(x) \end{align*}.
\frac{d}{dt} \int_0^L e(x,t)= c \int_0^L u(x,t \;dx) = 0 .
If f(x) \geq 0 and Q(x,t) \geq 0, then u(x,t) \geq 0 for all t.
If f(x) \leq M and Q(x,t) \leq 0, then u(x,t) \leq M.