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10.8: Solving DEs using the Fourier transform

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    51139
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    Let

    \[D = \dfrac{d}{dt}. \nonumber \]

    Our goal is to see how to use the Fourier transform to solve differential equations like

    \[P(D) y = f(t). \nonumber \]

    Here \(P(D)\) is a polynomial operator, e.g.

    \[D^2 + 8D + 7I. \nonumber \]

    We first note the following formula:

    \[\widehat{Df} (\omega) = i \omega \hat{f}. \nonumber \]

    \(Proof\). This is just integration by parts:

    \[\begin{array} {rcl} {\widehat{Df} (\omega)} & = & {\int_{-\infty}^{\infty} f'(t) e^{-i \omega t}\ dt} \\ {} & = & {f(t) e^{i\omega t} \vert_{-\infty}^{\infty} - \int_{-\infty}^{\infty} f(t) (-i \omega e^{-i \omega t})\ dt} \\ {} & = & {i \omega \int_{-\infty}^{\infty} f(t) e^{-i \omega t} \ dt} \\ {} & = & {i \omega \hat{f} (\omega) \ \ \ \text{QED}} \end{array} \nonumber \]

    In the third line we assumed that \(f\) decays so that \(f(\infty) = f(-\infty) = 0\).

    It is a simple extension of Equation 10.9.4 to see

    \[(\widehat{P(D)f}) (\omega) = P(i \omega) \hat{f}. \nonumber \]

    We can now use this to solve some differential equations.

    Example \(\PageIndex{1}\)

    Solve the equation

    \[y''(t) + 8y'(t) + 7y(t) = f(t) = \begin{cases} e^{-at} & \text{ if } t > 0 \\ 0 & \text{ if } t < 0 \end{cases} \nonumber \]

    Solution

    In this case, we have

    \[P(D) = D^2 + 8D + 7I, \nonumber \]

    so

    \[P(s) = s^2 + 8s + 7 = (s + 7)(s + 1). \nonumber \]

    The DE

    \[P(D)y = f(t) \nonumber \]

    transforms to

    \[P(iw) \hat{y} = \hat{f}. \nonumber \]

    Using the Fourier transform of \(f\) found in Example 10.8.1 we have

    \[\hat{y} (\omega) = \dfrac{\hat{f}}{P(i\omega)} = \dfrac{1}{(a + i \omega)(7 + i \omega)(1 + i \omega)}. \nonumber \]

    Fourier inversion says that

    \[y(t) = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} \hat{y} (\omega) e^{i \omega t} \ d \omega \nonumber \]

    As always, we want to extend \(\hat{y}\) to be function of a complex variable \(z\). Let's call it \(g(z)\):

    \[g(z) = \dfrac{1}{(a + iz)(7 + iz)(1 + iz)}. \nonumber \]

    Now we can proceed exactly as in Example 10.8.1. We know \(|g(z)| < M/|z|^3\) for some constant \(M\). Thus, the conditions of Theorem 10.2.2 are easily met. So, just as in Example 10.8.1, we have:

    For \(t > 0\), \(e^{izt}\) is bounded in the upper half-plane, so we use the contour in Figure \(\PageIndex{1}\) on the left.

    \[\begin{array} {rcl} {y(t) = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} \hat{y} (\omega) e^{i \omega t} \ d \omega} & = & {\dfrac{1}{2\pi} \lim_{x_1 \to \infty, x_2 \to \infty} \int_{C_4} g(z) e^{izt}\ dz} \\ {} & = & {\dfrac{1}{2\pi} \lim_{x_1 \to \infty, x_2 \to \infty} \int_{C_1 + C_2 + C_3 + C_4} g(z) e^{izt}\ dz} \\ {} & = & {i \sum \text{ residues of } e^{izt} g(z) \text{ in the upper half-plane}} \end{array} \nonumber \]

    The poles of \(e^{izt} g(z)\) are at

    \(ia\), \(7i\), \(i\).

    These are all in the upper half-plane. The residues are respectively,

    \[\dfrac{e^{-at}}{i(7 - a)(1 - a)}, \ \ \dfrac{e^{-7t}}{i(a - 7)(6)}, \ \ \dfrac{e^{-t}}{i(a - 1)(6)} \nonumber \]

    Thus, for \(t > 0\) we have

    \[y(t) = \dfrac{e^{-at}}{(7 - a)(1 - a)} - \dfrac{e^{-7t}}{(a - 7)(6)} + \dfrac{e^{-t}}{(a - 1)(6)}. \nonumber \]

    014 - (Example 10.9.1).svg
    Figure \(\PageIndex{1}\): (CC BY-NC; Ümit Kaya)

    More briefly, when \(t < 0\) we use the contour above on the right. We get the exact same string of equalities except the sum is over the residues of \(e^{izt} g(z)\) in the lower half-plane. Since there are no poles in the lower half-plane, we find that

    \[\hat{y} (t) = 0 \nonumber \]

    when \(t < 0\).

    Conclusion (reorganizing the signs and order of the terms):

    \[y(t) = \begin{cases} 0 & \text{ for } t < 0 \\ \dfrac{e^{-at}}{(7 - a)(1 - a)} + \dfrac{e^{-7t}}{(7 - a)(6)} - \dfrac{e^{-t}}{(1 - a)(6)} & \text{ for } t > 0. \end{cases} \nonumber \]

    Note

    Because \(|g(z)| < M/|z|^3\), we could replace the rectangular contours by semicircles to compute the Fourier inversion integral.

    Example \(\PageIndex{2}\)

    Consider

    \[y'' + y = f(t) = \begin{cases} e^{-at} & \text{ if } t > 0 \\ 0 & \text{ if } t < 0. \end{cases} \nonumber \]

    Find a solution for \(t > 0\).

    Solution

    We work a little more quickly than in the previous example.

    Taking the Fourier transform we get

    \[\hat{y} (\omega) = \dfrac{\hat{f} (\omega)}{P(i \omega)} = \dfrac{\hat{f} (\omega)}{1 - \omega ^2} = \dfrac{1}{(a + i \omega)(1 - \omega ^2)}. \nonumber \]

    (In the last expression, we used the known Fourier transform of \(f\).)

    As usual, we extend \(\hat{y} (\omega)\) to a function of \(z\):

    \[g(z) = \dfrac{1}{(a + iz) (1 - z^2)}. \nonumber \]

    This has simple poles at

    -1, 1, \(ai\).

    Since some of the poles are on the real axis, we will need to use an indented contour along the real axis and use principal value to compute the integral.

    The contour is shown in Figure \(\PageIndex{2}\). We assume each of the small indents is a semicircle with radius \(r\). The big rectangular path from \((R, 0)\) to \((-R, 0)\) is called \(C_R\).

    015 - (Example 10.9.2).svg
    Figure \(\PageIndex{2}\): (CC BY-NC; Ümit Kaya)

    For \(t > 0\) the function \(e^{izt} g(z) , M/|z|^3\) in the upper half-plane. Thus, we get the following limits:

    \(\begin{array} {ccl} {\lim_{R \to \infty} \int_{C_R} e^{izt} g(z)\ dz = 0} & \ \ \ \ \ \ & {\text{(Theorem 10.2.2(b))}} \\ {\lim_{R \to \infty, r \to 0} \int_{C_2} e^{izt} g(z) \ dz = \pi i \text{Res} (e^{izt} g(z), -1)} & \ \ \ \ \ \ & {\text{(Theorem 10.7.2)}} \\ {\lim_{R \to \infty, r \to 0} \int_{C_4} e^{izt} g(z)\ dz = \pi i \text{Res} (e^{izt} g(z), 1)} & \ \ \ \ \ \ & {\text{(Theorem 10.7.2)}} \\ {\lim_{R \to \infty, r \to 0} \int_{C_1 + C_3 + C_5} e^{izt} g(z) \ dz = \text{p.v.} \hat{y} (t) e^{i \omega t}\ dt} & \ \ \ \ \ \ & {} \end{array}\)

    Putting this together with the residue theorem we have

    \[\begin{array} {rcl} {\lim_{R \to \infty, r \to 0} \int_{C_1 - C_2 + C_3 - C_4 + C_5 + C_R} e^{izt} g(z)\ dz} & = & {\text{p.v.} \int_{-\infty}^{\infty} \hat{y} (t) e^{i \omega t} \ dt - \pi i \text{Res} (e^{izt} g(z), - 1) - \pi i \text{Res} (e^{izt} g(z), 1)} \\ {} & = & {2\pi i \text{Res} (e^{izt}, ai).} \end{array} \nonumber \]

    All that’s left is to compute the residues and do some arithmetic. We don’t show the calculations, but give the results

    \[\begin{array} {l} {\text{Res} (e^{izt} g(z), -1) = \dfrac{e^{-it}}{2(a - i)}} \\ {\text{Res} (e^{izt} g(z), 1) = -\dfrac{e^{it}}{2(a + i)}} \\ {\text{Res} (e^{izt} g(z), ai) = -\dfrac{e^{-at}}{i(1 + a^2)}} \end{array} \nonumber \]

    We get, for \(t > 0\),

    \[\begin{array} {rcl} {y(t)} & = & {\dfrac{1}{2\pi} \text{p.v.} \int_{-\infty}^{\infty} \hat{y} (t) e^{i \omega t} \ dt} \\ {} & = & {\dfrac{i}{2} \text{Res} (e^{izt} g(z), -1) + \dfrac{i}{2} \text{Res} (e^{izt} g(z), 1) + i \text{Res} (e^{izt} g(z), ai)} \\ {} & = & {\dfrac{e^{-at}}{1 + a^2} + \dfrac{a}{1 + a^2} \sin (t) - \dfrac{1}{1 + a^2} \cos (t).} \end{array} \nonumber \]


    This page titled 10.8: Solving DEs using the Fourier transform is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jeremy Orloff (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.