FDM Heat Equation FTCS Scheme with NBC

[Adyssa]

Homework Statement

I have a quiz question that I'm struggling with. We've been working on using a FTCS scheme with two Essential Boundary Conditions, and now I have a problem with one EBC (ie static) and a Natural Boundary Condition (ie a derivative). The condensed problem statement:

$u_{t} = 2u_{xx}, 0 < x < 1, t > 0$
$u_{x}(0,t) = 0.5$ <- NBC
$u(1,t) = 3$ <- EBC

and grid spacing $Δx = 0.2, Δt = 0.01$

Construct a Finite Difference Approximation to the PDE using FTCS scheme. Approximate the derivative boundary condition using a central difference approximation at x = 0.

Homework Equations

As above. [itex][/itex]

The Attempt at a Solution

The basic scheme I have been using is this:

$U^{n+1}_{i} = sU^{n}_{i-1} + (1-2s)U^{n}_{i}+sU^{n}_{i+1}$, where $s = \kappa \frac{Δt}{(Δx)^{2}} = 2 \frac{0.01}{0.04} = 0.5$

With regards to the derivative boundary condition, I believe that because $u_{x}(0,t) = 0.5$ is a positive gradient, that there is heating (not cooling) at the boundary. In my notes I have the following scheme for a Neumann Condition (heat flux at boundary):

Second order central dierence approximation to the boundary condition. Need grid point
outside the boundary to satisfy the boundary condition.

$k \frac{U^{n}_{j+1} - U^{n}_{j-1}}{2Δx} = Fl$ where $k$ is the conductivity and $Fl$ is the heat flux.

If this is even the correct expression, I'm not sure how to integrate it into my FTCS scheme in order to take into account the derivative boundary condition. In class we did some algebraic rearrangement of the equation to get $U^{n}_{j-1}$ by itself (LHS) and then substituted the RHS into the scheme, which seems appropriate:

$U^{n+1}_{i} = \frac{2sΔx.Fl}{k} + (1-2s)U^{n}_{i}+2sU^{n}_{i+1}$

but we never worked the solution. I'm not sure which values to use for $k$ and $Fl$. I think one of them will be 0.5? Also, it appears to me that I have now defined a constant heat flux at each point $Δx$ as I compute along the spatial dimension for $U^{n}_{i}, (i=0,1,2,3,4,5) (n = time step)$, but this is just speculation, I'm not a physics/eng major and this is all very new to me.

 

[voko]

You basically need another grid line at x = -h. Then you have the usual equation relating u at -h, 0, and +h, plus you approximate the derivative with [u(+h, t) - u(-h, t)]/2h = a, where a is whatever constant the x-derivative must be at x = 0. So you get 2xN additional equations.