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?


bag of hints:

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 $C cos(w+\phi)$

2nd question: again, look at symetries in the integrand, and play spliting the infinite range into the positive vs negative parts.

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 $c=a+b, d=a-b$

  • $\begingroup$ 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. $A\cos(\omega)+B\cos(\omega)=C\cos(\omega+\phi)$) $\endgroup$ – galois Nov 8 '15 at 6:03
  • $\begingroup$ but is it better now ? what do you get ? $\endgroup$ – Fabrice NEYRET Nov 8 '15 at 8:36

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