Solving Heat Equation by Fourier Transform

In summary, the transform is used to solve the heat equation when the boundary conditions are even or odd, depending on the symmetry of the problem.
  • #1
HAMJOOP
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When the rod is infinite or semi-infinite, I was taught to use Fourier transform.

But I don't know when should the full Fourier transform or sine/cosine transform be used.

how's the B.C. related to the choice of the transform ?
 
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  • #2
In general, to solve the heat equation, you should use a full Fourier transform--i.e. the one where you find the Fourier coefficients associated with plane waves ei(kx±ωt). Using this you can easily deduce what the coefficients should be for the sine and cosine terms, using the identity e=cos(θ) + i sin(θ).

If you use this prescription you can deduce the following helpful rule of thumb: when you are decomposing an odd function, you will only get nonzero contributions from the sin terms, but when you decompose an even function, you will only get nonzero contributions from the cos terms. If the function is neither even nor odd, you're forced to include both cosine and sine terms, which as I suggested before can be packaged together with plane waves.

This suggests a way to attack heat equation problems based on the symmetry of the boundary conditions. If you have boundary conditions which are odd, you expect the solution to be odd, and thus you decompose in sines, and if the boundary conditions are even, decompose in cosines.

For example, for a finite rod extending from -l/2 to l/2, initially at T=0, and then both ends are heated to T=T* at both ends, the solution should be even since the boundary conditions are even, so use cosines. On the other hand, if one end is held at T=T* and the other is held at T=-T*, the solution should be odd so use sines.

Things are not quite as obvious when dealing with a semi-infinite rod, but if you use a little imagation, then if you set the end of the rod to x=0 and apply your boundary condition there, you can convince yourself you must need to use cosines. [One way to see this is that you can imagine the semi-infinite rod to be one half of the problem where two semi infinite rods are joined at the origin, where we put the boundary condition T=T* at the origin for both rods, then the solution should by even.]
 
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  • #3
Solving infinite rod, full Fourier transform(complex form) is used in my note.
But the (natural) boundary conditions are even, should I use cosine transform ?
 

1. What is the heat equation?

The heat equation is a partial differential equation that describes the flow of heat through a given medium over time. It is often used in physics and engineering to model heat distribution and transfer in various systems.

2. How is the heat equation solved using Fourier transform?

The heat equation can be solved using Fourier transform by converting it into an ordinary differential equation in terms of the Fourier coefficients. The solution is then obtained by using the inverse Fourier transform to recover the original function.

3. What are the advantages of using Fourier transform to solve the heat equation?

Using Fourier transform to solve the heat equation allows for a simpler and more efficient solution compared to other methods. It also provides a way to analyze the behavior of the system in the frequency domain.

4. Are there any limitations to using Fourier transform for solving the heat equation?

One limitation of using Fourier transform is that it assumes the boundary conditions of the system are periodic. This may not always be the case in real-world applications, leading to inaccuracies in the solution.

5. How is the solution to the heat equation using Fourier transform validated?

The solution obtained using Fourier transform can be validated by comparing it to other known solutions or by using numerical methods to approximate the solution. Additionally, physical experiments can be conducted to test the validity of the solution.

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