Roo2 said:
Thanks for your reply. I'm starting to get it, but I still need some help (it's been quite a while since I've done math; I'm a bit slow on the uptake). I think you can drop the integrals because if ∫x dk = ∫y dk, then x = y. However, I'm not terribly sure about that; is it not possible to have two different functions with the same area under the curve?
Hey, sorry I hadn't replied yet. I've been out of town for a few days.
In general, ∫x dk = ∫y dk does
not imply x = y. For example, ##\int_0^{2\pi} dk \cos(k) = \int_0^{2\pi} dk \sin(k) = 0##, but we know that ##\cos(k) \neq \sin(k) \neq 0##. The cases in which you can deduce that ∫x dk = ∫y dk implies x = y typically have some sort of arbitrary parameter in both integrals, such that the equality has to hold for any value of the parameter. For example, in the Fourier case you are considering in the notes, the variable x is arbitrary. The equation has to hold true for every possible value of x. This generally implies that the two integrands have to be equal. You can easily show this to be the case for Fourier integrals. If
$$\int_{-\infty}^\infty dk~f(k)e^{ikx} = \int_{-\infty}^\infty dk~g(k)e^{ikx},$$
then you can conclude that f(k) = g(k) by multiplying both equations by ##\exp(iqx)## and integrating over x. Try it out. (You will need the identity ##\int_{-\infty}^\infty dx \exp(i(k-q)x) = \delta(k-q)##).
Also, could you please show me the process of differentiating a more simple function via Fourier transform? I'm not seeing how you can apply a derivative through an integral (I think that I just never learned the operation or the rules for it).
The general rule for differentiating an integral is
$$\frac{\partial}{\partial x} \left[ \int_{\alpha(x)}^{\beta(x)} dt~f(x,t)\right] = f(x,\beta(x))\frac{\partial \beta}{\partial x} - f(x,\alpha(x))\frac{\partial \alpha}{\partial x} + \int_{\alpha(x)}^{\beta(x)} dt~\frac{\partial f(x,t)}{\partial x}.$$
In your case, ##\alpha## and ##\beta## are constants (infinite, but constants), so only the last term remains. Basically, for constant limits of integration, a derivative of an integral (with respect to a variable that is
not the integration variable) will pass through the integral.
Finally, why is U(k,t) an ODE while u(x,t) is a PDE, when they're both functions of two variables?
Just to be clear on terminology: you
have an ODE for ##\hat{u}(k,t)## and you
have a PDE for u(x,t). The functions themselves are not ODEs or PDEs.
When insert the Fourier transform of u(x,t) into ##\partial u(x,t)/\partial x##, you will find it gives you ##\hat{u}(k,t)## times some k-dependent factor - no derivatives leftoever. Since your PDE only had x or t derivatives, after plugging in the Fourier representation of u(x,t) and dropping the integrals, only t derivatives will remain. Since you only have one kind of derivative, you have an ODE.
Similarly, say you had an ODE for u(x) - only x, no t. If you insert the Fourier transform of u(x), all of the x derivatives would go away, leaving you with a purely algebraic equation for the Fourier transform ##\hat{u}(k)##.
Also, for completeness, I should point out that you can also derive the ODE for ##\hat{u}(k,t)## by Fourier transforming the PDE directly.