Understanding the Fourier Transform in Solving the Heat Equation

[member 428835]

Hi PF! I was wondering if you could clarify something for me. Specifically, I am solving the heat equation ##u_t = u_{xx}## subject to ##| u(\pm \infty , t ) | < \infty##. Now this implies a solution of sines and cosines times an exponential. Since we have a linear PDE, we may superimpose each solution (we have infinitely many since ##\lambda##, the separation constant, need only be positive or negative, depending on how we define it).

Since ##\lambda## is continuous we may sum via the Reimann integral, as $$\int_0^\infty c_1(\lambda ) \sin (\sqrt{\lambda} x) \exp (- \lambda k t) d \lambda$$. There would also be a cosine expression. My question is, where does the ##d \lambda## come from? Can someone please explain?

Thanks so much!

Josh

 

[Orodruin]

It is part of the expansion coefficient. If you compare with the ##1/L## you get when you normalise the basis of a Fourier series, you cannot do that here and the normalisation condition is that you get a delta function rather than one, the ##1/L## essentially becomes your ##d\lambda##.

 

[member 428835]

so it comes from determining the coefficients?

 

[Orodruin]

If you look at the finite case, you can always change your coefficients by changing the normalisation of the basis. This is the case also in the infinite case, but in order to have a normalisation that makes sense, you need to add the infinitesimal ##d\lambda## (i.e., the coefficient of ##\sin(\ldots)\exp(\ldots)## is really ##c(\lambda) d\lambda## in order for the normalisation of the basis to make sense).

 

[member 428835]

If you look at the finite case, you can always change your coefficients by changing the normalisation of the basis.

Can you elaborate on this?

 

[Orodruin]

The general idea is to write the solution as
$$
\sum c_k f_k(x,t)
$$
where the ##f_k(x,t)## each satisfy the differential equation. If you define ##g_k(x,t) = f_k(x,t) \sqrt{L}##, you will still be able to write the solution as
$$
\sum c_k' g_k(x,t) = \sum \frac{c_k}{\sqrt L} g_k(x,t)
$$
and ##g_k(x,t)## also satisfies the differential equation, but also an additional normalisation condition. When you go to the infinite case, the sum goes to an integral.

 

[member 428835]

When you go to the infinite case, the sum goes to an integral.

When you say infinite case are you referring to ##L \to \infty##? I'm sure you're referring to ##k \to \infty## but then where is the ##\Delta k##?

 

[Orodruin]

Yes. But I think you are reading too much into the appearance of the ##d\lambda## in the integral. I suggest simply seeing it as a linear combination of your (uncountable) set of solutions. The coefficient of each is ##c(\lambda) d\lambda##.

 

[member 428835]

Thanks for your help!