Solving Heat Equation by Fourier Transform

[HAMJOOP]

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 ?

 

[Jolb]

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 e^i(kx±ωt). Using this you can easily deduce what the coefficients should be for the sine and cosine terms, using the identity e^iθ=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.]

 

[HAMJOOP]

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 ?