I am working on the following problem:
Suppose
$\displaystyle \frac{\partial T}{\partial t} = \kappa \frac{\partial ^{2}T}{\partial x^{2}}, \, 0<x<L,$
where $\kappa > 0$ is constant and $T(x=0,t)=A$, $\displaystyle \frac{\partial T}{\partial x}(x=L, t) = B$ with $A$ and $B$ constant. Find the steady state, also called the equilibrium, temperature.
So far, I have rewritten the equation in the form $\displaystyle T_{t}=\kappa T_{x x}$, $0 < x < L$, simply because I find it easier to work with that notation.
From there, I surmised that since a steady state temperature does not change in time, this means that $T_{t}=0$.
Therefore, our equation becomes instead $\displaystyle 0 = \kappa T_{x x } \to 0 = T_{x x}$, which is an ODE we can solve by integrating twice:
- $\displaystyle \int T^{\prime \prime}(x) dx = \int 0 dx \to T^{\prime}(x) + a = b$, where $a$ and $b$ are constants of integration.
- Integrating again, we obtain $\displaystyle \int \left( T^{\prime}(x) + a \right)dx = \int b dx \to T(x) + ax + c = bx + d$, where $c$ and $d$ are constants of integration. Isolating $T(x)$ on the LHS, we obtain $T(x) = (b-a)x + (d-c)$, and letting $b-a = m$, $d-c=n$, we have $\mathbf{T(x) = mx+n}$.
Now, applying the initial conditions, we obtain:
- $\mathbf{T(x=0,t) = A}: \, T(0) = m(0) + n = A$, which implies that $T(0)=n=A \, \to \, n = A$.
- $\displaystyle \mathbf{\frac{\partial T}{\partial x}\left( x = L, t \right) = B}: \, \frac{\partial T}{\partial x} = m = B$.
So, we have that $\mathbf{T(x) = Bx+A}$ is the steady state.
Is that all there is to this problem? Did I even go about it correctly? If not, how should I have done this problem?
I'm pretty new to figuring out PDEs this way, so if, in answering this question, you pretend like you're explaining this to someone who is completely and utterly clueless, you won't be far off. I warn you, though, that hints and Socratic-type questions will probably not be productive - I doubt I would be able to answer such questions and instead will probably become increasingly frustrated and lost.