# Diffusion equation

1. Apr 6, 2012

### zezima1

1. The problem statement, all variables and given/known data
Solve the diffusion equation with the boundary conditions v(0,t)=0 for t > 0 and v(x,0) = c for t=0. The method should be separation of variables.

2. Relevant equations
The separation of variables method.

3. The attempt at a solution
Attempting a solution of the form XT leads you to an exponential for T and a sinusoidal for X:
X = Asin(kx) + Bcos(kx)
where -k^2 was the constant used for solving the two separated differential equations.

However. My solution manuals writes the constants A and B as a continious function of the parameter k, and I don't understand why. Why do the constants, which are chosen from the boundary conditions have anything to do with k?

And going further the full solution is then written as a fourier integral from 0 to ∞ of XTB(k)dk

Where on earth does this come from? Note that A(k)=0 from the boundary conditions.

Can someone try to explain why you most impose a continuous superposition like the above to get the general solution?

2. Apr 6, 2012

### HallsofIvy

Staff Emeritus
In general there is NOT a function of the form X= A(t)sin(kx)+ B(t)cos(kx) satisfying the boundary conditions. What you then need to do is to look for a sum of such things (if the boundary conditions require that k be discrete) or an integral of such thing (if the boundary conditions allow k to be continuous- any number on some interval).

3. Apr 7, 2012

### zezima1

hmm I see. So the discrete superposition comes from if you get multiple k's satisfying X(0)=0 for instance (as a sinusoidal term).

But I still don't see how my boundary conditions give a continuous set of parameters k. We wonna solve the diffusion equation for v(0,t)=0 for t>0 and v(x,0)=u. Can you show me exactly how this integral arises?

I'll do the work I can:

We get:

X(x) = Acos(kx) + Bsin(kx)
T(t) = Cexp(k2t) + Dexp(-k2t)

v(0,t) implies A=0
We also want C=0 (I think) since that term wouldn't really make sense, when we are considering thermal conduction (which we are). That leaves us with:

v(x,t) = B'sin(kx)exp(-k2t) , where B'=BD

Now my book writes this as:

v(x,t) = ∫0B(k)sin(kx)exp(-k2t) dk

I don't see how on earth this arises!!

4. Apr 7, 2012

### Ray Vickson

In general, the solution is a sum (or integral) of terms of the form $B(k) \sin(kx) \exp(-k^2 t),$ where you vary k and the coefficients B(k) are chosen to make v(x,0) have the desired form f(x); that is, you need to be able to say that $f(x) = \sum_{k} B(k) \sin(kx) \text{ or } f(x) = \int B(k) \sin(kx) \, dk$ for all x. If you have a condition on an infinite interval, as in your case of f(x) = c for all x > 0 (or is it for all x ≠ 0 in (-∞,∞)?) then a finite sum (= a Fourier series) will not work; you need an integral (= Fourier integral).

RGV

5. Apr 7, 2012

### zezima1

hmm.. I still don't really see it. First of all a minor worry: The integral doesn't have the same units as the sum since you are multiplying by dk. So isn't that a problem.
Secondly, let's try to do it pieceweise:
The condition v(x,0)=u implies:
B(k)sin(kx) = u
How does solving this equation lead you to the integral above?

I can see how a sum arises, for instance say:

B(k)sin(kb) = 0
=>
kb=n∏
=>
k=n∏/b

So a solution would be:

f = ƩB(k)sin(n∏x/b)

But how does an integral arise...

6. Apr 7, 2012

### Ray Vickson

I already told you. It is the difference between Fourier series and Fourier integrals. You need to represent a function f(x) in terms of a sum or integral of sin(k*x) and/or cos(k*x), and whether you should use a sum or whether you should use an integral depends on f(x). It is all readily available in books and articles; just look it up, using Google for example.

RGV

7. Apr 7, 2012

### Telemachus

You use the fourier integral when the eigenvalues are continuous, this kind of eigenvalues arises generally when you work with problems extended on an infinite interval. When you have a finite interval, the border conditions gives discrete eigenvalues, so you make a sum, because of the superposition principle, this way a fourier series arises. When you have infinite eigenvalues (for example any positive real is an eigenvalue), you have to make an integral, which is like a continuos sum, if you know the Riemann definition of integral it will be easy to you to get this (the integral can be interpreted as a limit of sums), if not, you can look for it at wikipedia: http://en.wikipedia.org/wiki/Riemann_integral in this case the fourier integral arises.

Last edited: Apr 7, 2012