Technically, Fourier's equation is parabolic; see the classification section in https://en.wikipedia.org/wiki/Partial_differential_equation .

One could also say that it is parabolic because of the single time derivative. The solution evolves in time where a single derivative, rather than a second derivative, controls that evolution.

However, if the conduction problem reaches steady state, then the Laplacian which results then is indeed an elliptic equation. So the steady-state version of Fourier's equation is elliptic, but Fourier's equation itself (which involves both time and spatial derivatives) is parabolic.I think that some say that Fourier is parabolic in time simply to emphasize the fact that time is the marching direction, i.e. that the solution evolves with time.

Technically, Fourier's equation is parabolic; see the classification section in https://en.wikipedia.org/wiki/Partial_differential_equation .

One could also say that it is parabolic because of the single time derivative. The solution evolves in time where a single derivative, rather than a second derivative, controls that evolution.

However, if the conduction problem reaches steady state, then the Laplacian which results then is indeed an elliptic equation. So the steady-state version of Fourier's equation is elliptic, but Fourier's equation itself (which involves both time and spatial derivatives) is parabolic.I think that some say that Fourier is parabolic in time simply to emphasize the fact that time is the marching direction, i.e. that the solution evolves with time.

thank u sir

Sir, can you please clarify conditional stability of explicit finite difference method.  That is when r>1/2 [ where r =Δt/(Δx)^2] how it becomes unbounded

Dear Anishchandran,

I hope that the attached file makes the analysis and the reason clear. I have given you a quick rundown of the analysis for the 1D Fourier's equation. Hopefully there will be no problem in extending this to 2D or to 3D.

my gratitude sir.

parabolic, elliptic or hyperbolic character of the second order PDE is a mathematical topic as already addressed. It implies to find the nature of the characteristic curves in the time-space domain. For parabolic equation you discover real coincident solutions for the characteristic curves that produce a privileged direction of propagation

 sir

Could you please clarify that how the value of 2n comes in an explicit scheme step as mentioned in the below quote

"An implicit scheme, such as Crank–Nicolson, requires solving an

n × n nonlinear tridiagonal system at every step.  An explicit scheme, needs to solve 2n single (uncoupled) nonlinear equations in every forward/reverse sweep pair ".

It looks as though you have quoted from Programming Projects in C for Students of Engineering, Science, and Mathematics By Rouben Rostamian - a google search for a phrase in your quotation indicates this.

He seems to be advocating the Seidman sweep as an elegant alternative to Crank-Nicolson. I have to be honest and say that I hadn't even heard of this method before! On page 261 of the book, and referring to figure 20.6, is a diagram showing how the Seldman method works. Unlike the forward difference method, an explicit scheme which generates the values of u_{n+1} (i.e. the values at the next time step) directly from  the values of u_n, it uses a two-stage process which involves intermediate solutions halfway between t_n and t_{n+1}. So if the forward difference scheme solves n nonlinear equations per time step, the Seidman method must solve 2n equations, n from t_n to t_{n+1/2} and another n from t_{n+1/2} to t_{n+1}.

It would be interesting to determine how accurate the forward Euler and the Seidman methods are relative to one another. For a given initial condition on a finite spatial domain, one can always write down an analytical solution (using separation of variables and Fourier series) and hence this will give an essentially exact solution. My suspicion without trying it out, is that the accuracies will be roughly comparable with each other for the same dt and dx, and then the question is whether 2n equations per timestep is worth doing compared with n per timestep.

I have to warn you that his third point on page 262 is confusing and possibly wrong. He states that Crank-Nicholson solves an nxn nonlinear tridiagonal system at each timestep. My understanding is that the basic C.N. method is designed to solve a linear equation (the heat equation) and that the tridiagonal matrix/vector system is linear. Indeed, every equation on that page is linear.

However, C.N. can be used to solve a nonlinear version of Fourier's equation by adopting a Newton-Raphson iteration approach. In that case we still get a linear tridiagonal system to solve, but it is solved as part of an iteration scheme, the number of iterations for which is generally unknown in advance. So the precise comparison between all of these alternative methods is somewhat more complicated than the author seems to indicate in the book.

I prefer to use BDF2 for the time derivative term instead of C.N. It is also second order accurate, but one doesn't get slowly decaying transients when the timestep is too large.

Sir

First of all let me express my gratitude for your answer.                                                      The quote was taken from the same book. which I am using as reference for writing the code. I am trying to develop a computing framework for efficient optimization of PDE Stencils.

I would highligt some points:

1) Crank-Nicolson scheme is nothing else that a trapezoidal rule applied in the discretization of the time integral. It has nothing to do with the type of resulting matrix that, actually, depends on a) number of spatial dimensions b) type of spatial discretization of the operators.

2) The resulting tri-diagonal form is a result of a 1D problem along with a second order central discretization. 5-diagonals (2D) and 7-diagonals (3D) would appear in multidimensional extension. However, the algebric system require the problem to be linear or to be linearized.

3) Higher order discretizations increase the numbers of non-zero matrix entry  even in the 1D case.

4) Once you have a standard tri-diagonal matrix, the Thomas algorithm resolves exactly in a well definite number of steps.

Thanks sir

Dear Anishchandran,

I hope that the attached file helps you.

Thank you dear Javad Rahmannezhad