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Nonhomogeneous linear differential equations can often be solved by Fourier transforming the differential equation into an algebraic equation which is easier to solve. However, this only gives one particular solution, while the general solution of a linear differential equation forms an affine space of the schematic form (particular solution) + (all solutions to homogeneous part). The usual story I've heard to explain this discrepancy is that typically only one particular solution is Lebesgue-integrable and therefore has as well-defined Fourier transform, and only this solution can be found by the Fourier transform method. (Fourier transforming a homogeneous differential equation typically just yields the trivial zero solution, or possibly a set of plane waves with discrete wave vectors.)

Is it true that linear differential equations generically only have one integrable solution? That's not at all obvious to me. If so, is there an easy way to see why? If not, which particular solution does the Fourier transform method find?

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    $\begingroup$ Are we talking about DEs with constant coefficients? Because if the coefficients aren't constant the Fourier transform is not going to give you an algebraic equation. $\endgroup$ – eyeballfrog May 9 '17 at 6:12
  • $\begingroup$ @eyeballfrog Good point. And for DEs with constant coefficients, the homogeneous term always has solutions that are (non-integrable) plane waves, right? So it makes sense that at most one solution (the one with all the homogeneous-solution plane waves "turned off") can be integrable, and that's the one that the FT finds. Is that line of reasoning correct? $\endgroup$ – tparker May 9 '17 at 6:16
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    $\begingroup$ The homogeneous term could also have exponentially growing or decaying parts, but they aren't integrable either. Your point is correct, though it's worth noting that there may be no integrable solutions. $y'' + y = 1$ is an obvious example of such a scenario. $\endgroup$ – eyeballfrog May 9 '17 at 6:25
  • $\begingroup$ @eyeballfrog Yes, by "plane wave" I mean $\exp(i {\bf k \cdot x})$ with ${\bf k}$ possibly complex, so including exponentially decaying/growing solutions. Thanks! I'll give you the check if you want to convert your comments to answers. $\endgroup$ – tparker May 9 '17 at 6:36
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For differential equations with constant coefficients (the ones that transform into algebraic equations), the nonzero homogeneous solutions are never integrable. They all take the form $$ \sum_i A_i x^{n_i}e^{k_ix}, $$ where the $k_i$ are some complex numbers. The $n_i$ are usually zero, but they are needed for cases where the characteristic polynomial has roots with multiplicity greater than one. Such functions either diverge to infinity in one direction or are pure sines and cosines, which have Fourier transforms in a sense but not what we're looking for here. Thus, there can be at most one solution in $L^1$, and if such a solution exists that's what the Fourier transform should find.

With nonconstant coefficients, the differential equation may indeed have integrable homogeneous solutions, such as $$ y'' - (x^2 -1) y = 0, $$ which has a Gaussian as a solution. These tend to be eigenfunctions of differential operators, and thus can be seen as somewhat special cases.

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  • $\begingroup$ It is not true that the solutions of a constant coefficient linear ODE are of that form: that is only true if the root of the characteristic polynomial are simple. $\endgroup$ – Mariano Suárez-Álvarez May 9 '17 at 16:41
  • $\begingroup$ oh darn it you're right. fixing. $\endgroup$ – eyeballfrog May 9 '17 at 16:53

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