#### Insulated Boundary Conditions and Differential Equations

Insulated Boundary Conditions:

Insulation:

In lots of the previous sections we have considered fixed boundary conditions, that is u(0) = a,u(L) = b. We executed these simply by assigning uj0 = a and ujn= b for all j.

We as well considered variable boundary conditions such as u(0, t) = g1(t). For instance we mighthave u(0, t) = sin(t) which could stand for periodic heating and cooling of the end at x = 0.

A third significant type of boundary condition is called the insulated boundary condition. It is thereforenamed for the reason that it mimics an insulator at the boundary. Physically the consequence of insulation is thatno heat flows across the boundary. This signifies that the temperature gradient is zero which impliesthat must require the mathematical boundary condition u′(L) = 0.

To utilize it in a program we should replace u′(L) = 0 by a discrete version. Evoke that in our discreteequations we typically have L = xn. Evoke from the section on numerical derivatives that there arethree different manners to replace a derivative by a difference equation left, right as well as central differences the three of them at xn would be:

u′(xn) ≈ (un− un−1)/ h≈ (un+1− un) h ≈ (un+1− un−1) 2h

If xn is the last node of our grid, then it is clear that we cannot use the right or central difference, but are stuck with the first of these. Setting that expression to zero entails:

un= un−1.

This restriction is able to be easily execute in a program simply by putting a statement u(n+1)=u(n) inside the loop that updates values of the profile. Nevertheless since this method replaces u′(L) = 0 by an expression that is only accurate to first order it is not extremely accurate and is usually avoided.

Instead we want to utilize the most accurate version the central difference. For that we must have:

u′(L) = u′(xn) = (un+1− un−1)/ 2h = 0.

or simply

un+1 = un−1.

Nevertheless un+1 would represent u(xn+1) and xn+1 would be L + h, which is outside the domain. This nevertheless isn’t an obstacle in a program. We can merely extend the grid to one more node xn+1, and let un+1 always equal un−1 by copying un−1 into un+1 whenever un−1 changes. The point xn+1 is fictional however a computer doesn’t know the difference between fiction and reality!

This idea is carried out in the calculation.

Example- of an insulated boundary condition using a fictional point xn+1 with un+1 = un−1.

A manner to think of an insulated boundary that makes sense of the point L+h is to think of two bars joined end to end where you let the second bar be mirror image of the first bar. If you do this after that no heat will flow across the joint which is exactly the same effect as insulating.

Another practical manner to implement an insulated boundary is to let the grid points straddle the boundary. For illustration suppose we want to impose insulated boundary at the left end of a bar that is u′(0) = 0 then you could let the first two grid points be at x0 = −h/2 and x1 = h/2. Then you can let,
u0 = u1.

This will once more force the central difference at x = 0 to be 0.

Execution in a linear equation by elimination:

Consider the BVP:

uxx= −1, u(0) = 5, u′(1) = 0. (38.1)

This symbolizes the steady state temperature of a bar with a uniformly applied heat source, with one end held at a fixed temperature and the other end insulated.

If we utilize 4 equally spaced intervals then

m = 4, L = 1, ⇒ h = L/m=1/4

and

x0 = 0, x1 = .25, x2 = .5, x3 = .75, x4 = 1, and x5 = 1.25.

The point x5 = 1.25 is outside the region and therefore fictional and the boundary condition at x0 = 0 is implemented as
u0 = 5.

For the insulated condition we will necessitate:

u5 = u3.

This creates the central difference for u′(x4) be zero. We are able to write the differential equation as a difference equation:

(ui−1− 2ui+ ui+1)/ h2= −1

or

ui−1− 2ui+ ui+1 = −0.0625, i = 1, 2, 3, 4.

For i = 1, recalling that u0 = 5 we have:

5 − 2u1 + u2 = −.0625 or − 2u1 + u2 = −5.0625.

For i = 2 and i = 3 we have:

u1− 2u2 + u3 = −.0625 and u2 − 2u3 + u4 = −.0625.

For i = 4 we have:

u3− 2u4 + u5 = −.5.

Note that we now have 5 unknowns in our problem: u1, . . . , u5 Nevertheless from the boundary condition u5 = u3 and thus we can eliminate u5 from our equations and write

2u3− 2u4 = −.0625.

Summarizing we are able to put the unknown quantities in a vector u = (u1, u2, u3, u4)′and write the equations as a matrix equation Au = b where:

and b = (−5.0625,−.0625,−.0625. − .0625)′. Resolve this system and plot the results

> u = A\b
> u = [5 ; u]
> x = 0:.25:1
>plot(x,u,’d’)

the interpolate with a spline.

The precise solution of this BVP is:

U(x) = 5 + x − .5x2.

Use holds on and plots this function on the same graph to compare. >xx = 0:.01:1; > uu = 5 + xx - .5*xx.^2; must see that our approximate solution almost perfect!

Insulated boundary conditions in time-dependent problems:

To execute the insulated boundary condition in an explicit difference equation with time we need to copy values from inside the region to fictional points just outside the region.
An example
The stable state temperature u(r) (given in polar coordinates) of a disk subjected to a radially symmetric heat load g(r) and cooled by conduction to the rim of the disk as well as radiation to its environment is determined by the boundary value problem

2u/∂r2+(1/r)(∂u/∂r)= d(u4 − u4b) − g(r) with u(R) = uRand u′(0) = 0.

Here ub is the (fixed) background temperature as well as uRis the (fixed) temperature at the rim of the disk.

It is an example of information flow for the explicit method with an insulated boundary condition. An open circles are the fictional points as well as the curved arrows represent copying the value.

The class website has a program myheatdisk.m that executes these equations for parameter values R = 5, d = .1, uR= ub= 10 and g(r) = (r − 5)2. Notice that the equations comprise a singularity (discontinuity) at r = 0. How does the program evade this problem? How does the program execute uR= 10 and u′(0) = 0? Run the program.

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