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Solving a linear PDE, I got the general solution $$f(x,t)=\frac{e^{-t/\tau}}x\sum^\infty_{n=1}a_n\sin(\gamma_n x)$$ where $\gamma_n$ is the nth positive root of $\tan x=x$.

To satisfy the initial condition we require $$\phi(x)=\frac1x\sum^\infty_{n=1}a_n\sin(\gamma_n x)$$

From a physical point of view, $\phi(x)=f(x,0)$ represents the initial temperature profile of the object, which can be arbitrarily defined. Hence, I assumed such a expansion form is possible for most sufficiently well-behaving $\phi(x)$, but I cannot prove it.

What are the necessary and sufficient conditions for expanding a function $\phi(x)$ into $\frac1x\sum^\infty_{n=1}a_n\sin(\gamma_n x)$?

Noteworthily, $\gamma_n\sim\frac{\pi}{2}+n\pi$ thus this series is ‘almost’ a Fourier sine series.

When this expansion turns out to be mathematically possible, how could one extract the $a_n$s? One possible way is performing fourier transform on both sides, so that $$\int^\infty_{-\infty}x\phi(x)e^{-i\omega x}dx = \sum^\infty_{n=1}a_n\delta(\omega-\gamma_n)$$ assuming $\omega>0$, but often does one have knowledge of $\phi(x)$ only on a physically meaningful region $[0,a]$.

Szeto
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  • It seems that the function must be able to be approximated using the sum of odd functions. An even function such as cosine can't be expanded this way – Moko19 Aug 13 '20 at 08:22
  • @Moko19 Yes, that's true. In the real setting, $\phi(x)$ is only known on the physically meaningful region $[0,a]$ so the problem of oddness is not a problem - we can oddly extend $x\phi(x)$ to the negatives. – Szeto Aug 13 '20 at 08:39

1 Answers1

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The functions $x$ and $\sin(\gamma_j x)$ with $\tan(\gamma_j) = \gamma_j$, are orthogonal on the interval $[0,1]$. Suitably normalized, I think we get an orthonormal basis of $L^2[0,1]$ (and I think this can be obtained from Sturm-Liouville theory). Thus for any square-integrable function $f$ on $[0,1]$, we should have the expansion $$ f(x) = c_0 x + \sum_{j=1}^\infty c_j \sin(\gamma_j x)$$ (converging in $L^2$) where $$ c_0 = 3 \int_0^1 f(x) x\; dx,\ c_j = \frac{2}{\sin^2(\gamma_j)} \int_0^1 f(x) \sin(\gamma_j x)\; dx$$

Robert Israel
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