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I'm looking for some documentation about solving DEs using Fourier Transformation (not Fourier Series).

In particular, I have this one DE that I solved using another technique. Somebody mentioned that it can be solved using FT as well:

$$\varepsilon - \frac{1}{2}l^2 \frac{d^2 \varepsilon}{dx^2} = \delta(x)$$

Where $\varepsilon$ is a function of $x$ only, $l$ is a scalar constant and $\delta(x)$ is the Dirac "function".

The solution for the homogeneous case is

$$\varepsilon = A \; exp \left[ -\frac{|x|\sqrt2}{l} \right]$$

If I remember correctly, taking some derivatives resulted in the final solution of $A = \dfrac{1}{l \sqrt{2}}$.

Ok, now with FT. I found this site, the example looks a little like my DE -- with in my case $g(t) = \delta(x)$, which is convenient considering the convolution in the end.

Without the constant $\frac{1}{2}l^2$ I get

$$Y(f) - (2\pi i f)^2 Y(f) = \left( 1 - (2 \pi i f)^2 \right)\;Y(f) = G(f)$$


$$Y(f) = \frac{G(f)}{1+4 \pi^2 f^2}$$.

According to the site (would like to know how to compute it myself),

$$F^{-1}\left( \frac{1}{1+4 \pi^2 f^2} \right) = \frac{e^{-|t|}}{2}$$

Taking the convolution with the Dirac "function" is again the same function, $\dfrac{e^{-|t|}}{2} = exp \left[ \dfrac{-|t|}{2} \right]$.

So to conclude:

  1. Any documentation about solving DEs using FT is more than welcome
  2. How to do the same, now with the constant $\frac{1}{2}l^2$
  3. How to compute the inverse Fourier Transform of expressions like above (I know it involves an integral to infinity, but I never learned how to solve integrals like that)
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Well, since Fourier and Laplace transforms are more or less the same thing... – J. M. Sep 29 '11 at 17:24
This is a large question, and many books treat this method. If you can describe your background, people here will be able to give some useful references. – Gerben Sep 29 '11 at 18:58
@J.M. I thought that Laplace Transforms only worked with BVPs or IVPs, not just plain DEs? – Ailurus Sep 29 '11 at 19:35
If you stare for quite a while at the expressions for the Laplace and Fourier transformations side-by-side, you'll soon see a resemblance... – J. M. Sep 29 '11 at 19:39
@Gerben, my background is (bachelor) mechanical engineering. Currently finishing my master, focused on numerical methods in engineering (e.g. FEA). – Ailurus Sep 29 '11 at 19:40
up vote 0 down vote accepted

You probably know how a function and its Fourier transform are related. Choosing a certain convention, we can write

$$f(x) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} e^{ikx} \hat{f}(k) = \mathfrak{F}^{-1}[\hat{f}].$$

The RHS only depends on $x$ through the exponential factor; in particular,

$$\frac{\mathrm{d}f}{\mathrm{d}x} = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} e^{ikx} (ik) \hat{f}(k) = \mathfrak{F}^{-1}[ik\hat{f}].$$

Similarly, for the second derivative, we find

$$\frac{\mathrm{d}^2f}{\mathrm{d}x^2} = \mathfrak{F}^{-1}[-k^2\hat{f}].$$

You probably also know that

$$\delta(x) = \mathfrak{F}[\frac{1}{\sqrt{2\pi}}].$$

You know apply an FT to your entire differential equation, which yields:

$$\hat{e}(k) + \frac{1}{2}l^2 k^2 \hat{e}(k) = \frac{1}{\sqrt{2\pi}},$$ which directly implies that

$$\hat{e}(k) = \frac{1}{\sqrt{2\pi} (1 + \frac{1}{2}l^2 k^2)}.$$

Applying the inverse FT yields your answer. However, it's non-trivial. You'll need to calculate the integral

$$\int_{\mathbb{R}} \mathrm{d}k \frac{e^{-ikx}}{{1 + \frac{1}{2}l^2 k^2}}.$$

The way most people would solve this integral is using complex analysis: you'd interpret $k$ as a complex variable, identify the poles of the integrand (at $k = \pm i \sqrt{2}/l$) and use Cauchy's theorem in combination with the residue theorem in order to find its value. If you haven't studied complex analysis, it's wisest to look up the Fourier transform of the above function in a textbook on Fourier transforms. It's a Fourier transform that comes up often, so you won't have trouble finding it (here, for example).

share|cite|improve this answer
Ok so the $\frac{1}{2}l^2$ is a constant, and doesn't affect the Fourier Transformation, fairly logical. Indeed, I don't have knowledge about complex analysis, but it definitely sounds interesting. Thanks! By the way, your $k$ is $2 \pi f$ in my case, right? – Ailurus Sep 29 '11 at 19:46
It's indeed the same thing, just renamed. – Gerben Sep 29 '11 at 20:13

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