- The problem is to solve the diferential equation
\[ a _ {n} \frac {d ^ {n}}{d t ^ {n}} u (t) + a _ {n - 1} \frac {d ^ {n - 1}}{d t ^ {n - 1}} u (t) + \dots + a _ {0} u (t) = f (t), \]
where \(a _ { 0 } , a _ { 1 } , \ldots , a _ { n }\) are complex constants, and \(f\) is a given function. Here we suppose that \(f\) has bounded support and is smooth (say of class \(C ^ { 2 } )\)
- Let
\[ \hat {f} (z) = \int_ {- \infty} ^ {\infty} f (t) e ^ {- 2 \pi i z t} d t. \]
Observe that \(\hat { f }\) is an entire function, and using integration by parts show that
\[ | \hat {f} (x + i y) | \leq \frac {A}{1 + x ^ {2}} \]
if \(| y | \le a\) for any fixed \(a \geq 0\)
- Write
\[ P (z) = a _ {n} (2 \pi i z) ^ {n} + a _ {n - 1} (2 \pi i z) ^ {n - 1} + \dots + a _ {0}. \]
Find a real number c so that \(P ( z )\) does not vanish on the line
\[ L = \{z: z = x + i c, x \in \mathbb {R} \}. \]
- Set
\[ u (t) = \int_ {L} \frac {e ^ {2 \pi i z t}}{P (z)} \hat {f} (z) d z. \]
Check that
\[ \sum_{j=0}^n a_j\left(\frac{d}{dt}\right)^j u(t) =\int_L e^{2\pi i zt}\hat f(z)\,dz \]
and
\[ \int_L e^{2\pi i zt}\hat f(z)\,dz =\int_{-\infty}^{\infty}e^{2\pi ixt}\hat f(x)\,dx. \]
Conclude by the Fourier inversion theorem that
\[ \sum_{j=0}^n a_j\left(\frac{d}{dt}\right)^j u(t)=f(t). \]
Solution
Let the support of \(f\) be contained in \([-R,R]\).
For (a), since the defining integral is over a compact interval and its integrand is entire in \(z\), differentiation under the integral sign shows that \(\widehat f\) is entire. Because \(f\) is \(C^2\) with compact support, integration by parts twice gives, for \(z\ne0\), \[ \widehat f(z) =\frac1{(2\pi iz)^2} \int_{-R}^{R}f''(t)e^{-2\pi izt}\,dt. \tag{1} \] If \(z=x+iy\) and \(|y|\le a\), then \[ |e^{-2\pi izt}|=e^{2\pi yt}\le e^{2\pi aR}. \] Hence for \(|x|\ge1\), \[ |\widehat f(x+iy)| \le \frac{e^{2\pi aR}\|f''\|_1}{4\pi^2|x+iy|^2} \le \frac{C_a}{x^2}. \] For \(|x|\le1\), the original integral gives the uniform bound \[ |\widehat f(x+iy)|\le e^{2\pi aR}\|f\|_1. \] Combining the two estimates yields \[ |\widehat f(x+iy)|\le \frac{A_a}{1+x^2} \qquad(|y|\le a). \tag{2} \]
For (b), assuming the differential operator is nonzero, \(P\) is a nonzero polynomial and therefore has finitely many roots. Choose any real number \(c\) that is not the imaginary part of any root of \(P\). Then \[ L=\{x+ic:x\in\mathbb R\} \] contains no zero of \(P\).
For (c), parametrize \(L\) by \(z=x+ic\). The estimate (2), together with polynomial growth of \(P\), shows that \[ u(t)=\int_L\frac{e^{2\pi izt}}{P(z)}\widehat f(z)\,dz \] converges absolutely. Moreover, differentiating \(j\le n\) times multiplies the integrand by \((2\pi iz)^j\); after division by the degree-\(n\) polynomial \(P(z)\) and multiplication by the \(O(x^{-2})\) factor \(\widehat f(z)\), the resulting integrand is still absolutely integrable. Thus differentiation under the integral sign is justified and \[ \begin{aligned} \sum_{j=0}^n a_j\left(\frac d{dt}\right)^j u(t) &=\int_L \frac{\sum_{j=0}^n a_j(2\pi iz)^j}{P(z)} e^{2\pi izt}\widehat f(z)\,dz\\ &=\int_L e^{2\pi izt}\widehat f(z)\,dz. \end{aligned} \tag{3} \]
It remains to move \(L\) to the real axis. The integrand in (3) is entire. On the vertical sides of the rectangle joining \([-N,N]\) to \([-N+ic,N+ic]\), estimate (2) uniformly for imaginary parts between \(0\) and \(c\); for fixed \(t\) the exponential factor is also uniformly bounded there. Hence each vertical-side integral is \(O(N^{-2})\) and tends to \(0\). Cauchy’s theorem therefore gives \[ \int_L e^{2\pi izt}\widehat f(z)\,dz =\int_{-\infty}^{\infty}e^{2\pi ixt}\widehat f(x)\,dx. \tag{4} \] Since \(f\) is compactly supported and \(C^2\), Fourier inversion applies to the right side of (4), giving \(f(t)\). Combining (3) and (4), \[ \boxed{ \sum_{j=0}^n a_j\left(\frac d{dt}\right)^j u(t)=f(t)}. \] Thus the displayed contour integral gives a solution of the original constant-coefficient ODE. Its form depends on the chosen horizontal line \(L\), equivalently on \(c\).