# integral transforms: why do roots in frequency domain correspond to eigenvalues in time domain (and how does it help solve differential equations)?

In Wikipedia you can read about integral transforms, esp. the Laplace transform which maps a differential equation in the time domain into a polynomial equation in the complex frequency domain:

roots of the polynomial equations in the complex frequency domain correspond to eigenvalues in the time domain

Can anybody give a simple example and foremost give an intuition why this is so - and why does this help solving the differential equation? I don't see the big picture.

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(Linear, homogeneous, constant-coefficient) differential equations are already polynomial equations. That is, let $D$ denote the differential operator $\frac{d}{dx}$. Then linear homogeneous constant-coefficient differential equations are nothing more than equations of the form $p(D) y = 0$, where $y$ is some function of $x$ and $p$ is some polynomial. Writing $p(x) = \prod_{i=1}^n (x - r_i)^{e_i}$ where the roots $r_i$ are distinct, it follows that we can write the equation as

$$(D - r_1)^{e_1} (D - r_2)^{e_2} ... (D - r_n)^{e_n} y = 0.$$

The point here is that $D$ is a linear operator, and it generates a commutative algebra of differential operators. Now, from the above it follows that any solution to $(D - r_i)^{e_i} y = 0$ is a solution to the above, and it turns out that linear combinations of these give all solutions. (This is a general fact about linear operators on finite-dimensional vector spaces; see, for example, Axler's Linear Algebra Done Right.)

In the special case that $e_i = 1$ for all $i$ things are particularly simple: $(D - r_i) y = 0$ just means $Dy = r_i y$, or $y = c_i e^{r_i x}$ for some constant $c_i$. This gives $n$ linearly independent solutions, which is the correct number, hence the general solution has the form

$$y = \sum_{i=1}^n c_i e^{r_i x}.$$

If the multiplicities $e_i$ are higher then we must consider solutions to $(D - r_i)^{e_i} y = 0$. Let $w(x) = e^{-r_i x} y(x)$. Then $D w = e^{-r_i x} (D - r_i) y$, from which it follows that the above holds if and only if $D^{e_i} w = 0$. But this says precisely that $w$ is a polynomial of degree at most $e_i - 1$. Hence

$$y = c_i(x) e^{r_i x}$$

where $c_i(x)$ is a polynomial of degree at most $e_i - 1$. This gives $e_i$ linearly independent solutions, so again taking linear combinations we have found all of them.

None of this requires that you know anything about integral transforms. From the abstract point of view, integral transforms are just a convenient way to diagonalize $D$. The Laplace transform (more or less) takes $D$ to multiplication by $s$, hence turns $p(D)$ into $p(s)$ (the roots of which are the eigenvalues $r_i$ above), but this is just a more concrete way of realizing the abstract structure already evident in the problem: that the set of solutions to $p(D) y = 0$ is naturally acted on by $\mathbb{C}[D]$.

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I'm going to give an example of how it works. I don't know why it works. The example is the simplest case from electric circuit theory, the discharge of a capacitor. You know the solution f(t) is just an exponential decay, truncated to the left at t=0. The Fourier Transform of f(t) is F(w) = 1/1+jw. Now observe:

F(w) + jw*F(w) = 1/1+jw + jw/1+jw = 1

Now remember some properties of the Fourier transform: multiplication by jw in the frequency domain is equivalent to differentiation in the time domain. And the Fourier transform of a constant function in one domain is the delta function in the other domain. Using only these properties we can translate the above equation into words:

"A function plus its derivative is equal to the delta function".

This sentence is exactly the differential equation of a capactor which has a unit of charge dumped into it at t=0.

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