Solving Diffusion Equation with Separation of Variables Method

In summary, the diffusion equation with the given boundary conditions is solved using the method of separation of variables. This leads to a solution in the form of a Fourier integral, where the coefficients are chosen to satisfy the boundary conditions. The use of an integral instead of a sum arises from the fact that the eigenvalues in this case are continuous, as opposed to discrete in a finite interval.
  • #1
zezima1
123
0

Homework Statement


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.


Homework Equations


The separation of variables method.


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?
 
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  • #2
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
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
zezima1 said:
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!

In general, the solution is a sum (or integral) of terms of the form 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 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
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
zezima1 said:
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...

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
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.
 
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FAQ: Solving Diffusion Equation with Separation of Variables Method

What is the diffusion equation and why is it important in science?

The diffusion equation is a mathematical representation of the process of diffusion, which is the movement of particles from an area of high concentration to an area of low concentration. This equation is important in science because it can be used to model and understand various physical and chemical processes, such as the spread of pollutants, the diffusion of gases, and the movement of heat.

What is the separation of variables method and how does it apply to solving the diffusion equation?

The separation of variables method is a mathematical technique used to solve differential equations, such as the diffusion equation. It involves separating the variables in the equation into individual functions and then solving for each function separately. In the case of the diffusion equation, the variables are typically separated into time and space, and then solved using specific boundary conditions.

What are the advantages of using the separation of variables method to solve the diffusion equation?

One advantage of using the separation of variables method is that it can simplify complex differential equations into a series of simpler equations, making them easier to solve. Additionally, this method can provide more insight into the behavior of the system being modeled, as each individual function can be analyzed separately.

Are there any limitations to using the separation of variables method for the diffusion equation?

While the separation of variables method is a powerful tool, it does have some limitations. It can only be used for linear differential equations, and it may not always be possible to find closed-form solutions for all cases. In addition, this method can become more complicated for higher-dimensional problems.

How is the separation of variables method used in real-world applications?

The separation of variables method is used in a wide range of real-world applications, including modeling heat transfer in buildings, predicting the spread of contaminants in the environment, and analyzing diffusion processes in biological systems. It is also commonly used in engineering and physics to solve various differential equations and understand the behavior of complex systems.

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