# Completely missing the idea of this solution (finding an arbitrary function in integrand)

I'm hoping that someone can explain this (partial) solution to me.

In my textbook (Haberman PDEs book, q. 10.2.1), we're asked to find (complex) $c(\omega)$ so that the following are equivalent (with $A(\omega), B(\omega) \in \mathbb{R}$):

$$u(x,t)=\int^\infty_0\left[A\cos{(\omega x)}e^{-k\omega^2t}+B\sin{(\omega x)}e^{-k\omega^2t}\right]\mathrm{d}\omega$$ $$u(x,t)=\int^\infty_{-\infty}c(\omega)e^{-i\omega x}e^{-k\omega^2t}\,\mathrm{d}\omega$$

My first question - without any boundary or initial (time) conditions, is it possible to find $c(\omega)$ from the integral itself?

Moving on, the second integral can be rewritten as the sum of two integrals over the halves of the interval, but there's something in the solution I didn't understand:

$$\int_{-\infty}^0c(\omega)e^{-i\omega x}e^{-k\omega^2t}\,\mathrm{d}\omega +\int^\infty_0c(-\omega)e^{-i\omega x}e^{-k\omega^2t}\,\mathrm{d}\omega$$

Second question - why is the arbitary function in the second term now a function of $-\omega$?

It then replaces $-\omega$ with the dummy variable $\omega$ in the first term and rewrites the expression as:

$$\int^\infty_0c(\omega)e^{i\omega x}e^{-k\omega^2t}\,\mathrm{d}\omega+\int^\infty_0c(-\omega)e^{-i\omega x}e^{-k\omega^2t}\,\mathrm{d}\omega$$

Third question - How did this occur, that by using this (to me, unintuitive) dummy variable, we're now dealing with two integrals over the same interval?

Finally, using Euler's formula ($e^{ix}=\cos x + i\sin x$), everything is rewritten:

$$\int^\infty_0[c(\omega)+c(-\omega)]\cos{(\omega x)}e^{-k\omega^2t}\,\mathrm{d}\omega+\int^\infty_0i[-c(\omega)+c(-\omega)]\sin{(\omega x)}e^{-k\omega^2t}\,\mathrm{d}\omega$$

Last question(s?) - I'm not sure I follow on how they got to this form at all. How did the cosine and sine functions end up in separate integrands? How did $c(\omega)+c(-\omega)$ and $-c(\omega)+c(-\omega)$ show up?

1rst question: you surely know the relation between $sin(w),cos(w)$ and $exp^{iw}$, the symetries of cos and sin, and how to factor $A cos(w)+B sin(w)$ into $C cos(w+\phi)$
3rd question: same. + back to question 1: when you have variable changes like $a=\frac{c+d}2, b=\frac{c-d}2$, then you also have $c=a+b, d=a-b$
• Oh, well this is discouraging... Some of this is less than obvious to me, and I would never have guessed to use it (e.g. $A\cos(\omega)+B\cos(\omega)=C\cos(\omega+\phi)$) – galois Nov 8 '15 at 6:03