Fourier Transform to solve heat equation in infinite domain

[SUDOnym]

I'm having trouble following a step in my notes:

first off the heat equation is given by:

$$\frac{\partial u}{\partial t}=k^{2}\frac{\partial^{2}u}{\partial x^{2}}$$

then take the Fourier transform of this w.r.t.x, where in this notation the Ftransform of u(x,t) is denoted by U(alpha,t):

$$\frac{\partial U}{\partial t}=-\alpha^{2}k^{2}U$$

***it is this step I don't quite follow**** integrate with respect to t:

$$U=c(\alpha)e^{-k^{2}\alpha^{2}t}$$

Can someone please explain this step ie. how do you know what limits to integrate between and how do you know that the rights side of the equation ends up in the form?:

$$c(\alpha)e^{-k^{2}\alpha^{2}t}$$

obviously an intermediate step is:

$$\partial U=-k^{2}\alpha^{2}Udt$$

 

[Jory]

$$\partial U=-k^{2}\alpha^{2}Udt$$

Here, the terms concerning U need to be brought on to the left hand side, and then both sides integrated. You will end up with x as a function of ln(y), then exponentiating both sides will give you the form you have

 

[SUDOnym]

thanks.

and the c(alpha) term multiplying it...that's got nothing to do with the integration itself, right? it's added in there to make the solution more general...?

 

[Jory]

Well alpha is presumably the variable you use for the Fourier transform of your function, so it is to say, since when solving the resulting ODE you integrated the RHS with respect to t, it is entirely possible that the standard integration constant is instead a function of alpha. (Since d/dt (alpha) = 0)

 

[SUDOnym]

and how do you know d/dt(alpha)=0?

 

[Jory]

C(alpha) is a function of alpha only, so it's derivative with respect to t = 0

So instead of adding a constant, you add an arbitrary function of the non integrating variables.

 

[SUDOnym]

thanks. I have not seen that before, does that "function of integration" have a name that I could look up in a calculus book?

 

[Jory]

It should be mentioned when you solved basic PDEs.

For example if U is a function of (t,x)

Ut = f(t,x)

You would integrate wrt t, and add an arbitrary function of x.

When I was taught this, he simply called them arbitrary functions. I'm not sure if they have a special name.