# Errors in computing the Crank-Nicolson Method

1. Sep 2, 2011

### danielakkerma

Hi all!
Having decided to return to coding some of the physical problems that I require solving(in lieu say, of Mathematica), I endeavored to adapt the Crank-Nicolson to my needs, but, in order to test my algorithm, I had first attempted to resolve the common and trivial heat equation, in one dimension, e.g $$\alpha\frac{\partial^2u}{\partial x^2} = \frac{\partial u}{\partial t}$$, with some plain boundary conditions, that is u(x, 0) = some trivial function(sine times cosine) and u(0, t) = 10(or some value), and at the end-point(10, t in this case), the same(essentially a rod kept at both ends at an equal temperature).
The results however, are unsatisfactory; For sake of my equation, I used:
$$\beta u^{i+1}_{j+1}+(1+2\beta)u^{i+1}_{j}+\beta u^{i+1}_{j-1}=u^{i}_{j}$$
Where beta is a constant derived from my step sizes(i.e, dx, dt) and i & j denote the step in time and displacement respectively.
This, as is well known, creates a tridiagonal system, which is readily evaluated for all U(i+1)ns(in my case, I've got a grid with a size of 100 units).
I've attached two graphs corresponding to the solutions; Image2 was created by Mathematica, and Image1 is done using my program. You can see the very disturbing groove that's created, that seems to wedge and ebb towards the margins of the computational region. This is, first-and-foremost, physically inaccurate, and should not occur, and in fact, using Mathematica, does not.
In short, I am reliant on your aid to solve this discrepancy,
I thank you for your attention,
Beholden,
Daniel

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2. Sep 10, 2011

### AlephZero

One obvious difference between the plots is that the slope of the function at the left and right edges is different.

In the right hand plot the slopes apparear to be constrained to be 0, in the left hand plot they are not.

It's not very clear from your description of the problem or from the the plots (no labels on the axes) exactly what you were trying to do, but I would guess the boundary conditions are different in your code and in Mathematica.

3. Sep 17, 2011

### danielakkerma

You're absolutely right on the matter, and indeed, correct, or rather, setting the proper Dirichlet boundary conditions aligned the two!