Questão 2. Seja f(x)=e^{i x^{2}}=\cos \left(x^{2}\right)+i \operatorname{sen}\left(x^{2}\right).
(a) Mostre que f(x) satisfaz a equação diferencial f^{\prime}(x)-i 2 x f(x)=0.
(b) Aplique a transformada de Fourier na equação diferencial do item anterior obtendo
\hat{f}^{\prime}(\omega)-\frac{\omega}{2 i} \hat{f}(\omega)=0 .
(c) Resolva a equação diferencial do item anterior e encontre que
\hat{f}(\omega)=\hat{f}(0) e^{-i \frac{\omega^{2}}{4}}
(d) Usando o fato de que
\hat{f}(0)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x) d x=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{i x^{2}} d x=\frac{1}{2}+\frac{i}{2}
mostre que
\hat{f}(\omega)=\frac{1}{\sqrt{2 \pi}}\left[\cos \left(\frac{\omega^{2}}{4}-\frac{\pi}{4}\right)+i \cos \left(\frac{\omega^{2}}{4}+\frac{\pi}{4}\right)\right] .
(e) Use o item anterior para mostrar que
\begin{aligned} & \mathcal{F}\left(\cos \left(x^{2}\right)\right)(\omega)=\frac{1}{\sqrt{2}} \cos \left(\frac{\omega^{2}}{4}-\frac{\pi}{4}\right) \\\ & \mathcal{F}\left(\operatorname{sen}\left(x^{2}\right)\right)(\omega)=\frac{1}{\sqrt{2}} \cos \left(\frac{\omega^{2}}{4}+\frac{\pi}{4}\right) \end{aligned}
(f) Usando o item anterior encontre \mathcal{F}\left(\cos \left(a x^{2}\right)\right)(\omega) e \mathcal{F}\left(\operatorname{sen}\left(a x^{2}\right)\right)(\omega), para $a>0.

Passo 1

O enunciado define f(x)=e^{i x^{2}}=\cos\bigl(x^{2}\bigr)+i\,\operatorname{sen}\bigl(x^{2}\bigr) e pede, em sequência:
– (a) mostrar que f satisfaz a EDO
f'(x)-i\,2x\,f(x)=0;
– (b) aplicar transformada de Fourier nessa EDO e obter
\hat f'(\omega)-\frac{\omega}{2i}\,\hat f(\omega)=0;
– (c) resolver essa EDO em \omega e achar
\hat f(\omega)=\hat f(0)\,e^{-i\omega^{2}/4};
– (d) usar \hat f(0)=\tfrac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{i x^{2}}dx=\tfrac12+\tfrac{i}{2} para reescrever
\hat f(\omega)=\frac1{\sqrt{2\pi}}\biggl[\cos\Bigl(\tfrac{\omega^{2}}{4}-\tfrac\pi4\Bigr)+i\,\cos\Bigl(\tfrac{\omega^{2}}{4}+\tfrac\pi4\Bigr)\biggr];
– (e) separar parte real e imaginária para obter
\mathcal F\bigl(\cos(x^{2})\bigr)(\omega) =\tfrac1{\sqrt2}\cos\Bigl(\tfrac{\omega^{2}}{4}-\tfrac\pi4\Bigr), \quad \mathcal F\bigl(\operatorname{sen}(x^{2})\bigr)(\omega) =\tfrac1{\sqrt2}\cos\Bigl(\tfrac{\omega^{2}}{4}+\tfrac\pi4\Bigr);
– (f) finalmente usar propriedade de escala para generalizar a qualquer a>0.
Conteúdos envolvidos: EDO de primeira ordem, propriedades da transformada de Fourier (derivação e escala) e integrais de Fresnel.

Passo 2

(a) Calculamos f'(x). Como f(x)=e^{i x^{2}}, pela regra da cadeia,
f'(x) =\frac{d}{dx}\bigl(i x^{2}\bigr)\,e^{i x^{2}} =(i\cdot 2x)\,e^{i x^{2}} =i\,2x\,f(x).
Logo
f'(x)-i\,2x\,f(x)=i\,2x\,f(x)-i\,2x\,f(x)=0,
como queríamos.

Passo 3

(b) Aplicamos a transformada de Fourier (definida como \hat f(\omega)=\tfrac1{\sqrt{2\pi}}\int e^{-i\omega x}f(x)\,dx) em cada termo de
f'(x)-i\,2x\,f(x)=0.
Usamos que
\mathcal F\{f'\}(\omega)=i\omega\,\hat f(\omega), \quad \mathcal F\{x\,f(x)\}(\omega)=i\frac{d}{d\omega}\hat f(\omega).
Então
\mathcal F\{f'\}-i\,2\,\mathcal F\{x f\} =i\omega\,\hat f(\omega)-i\,2\Bigl(i\hat f'(\omega)\Bigr) =i\omega\,\hat f(\omega)+2\,\hat f'(\omega)=0.
Reescrevendo, 2\,\hat f'(\omega)=-\,i\omega\,\hat f(\omega), ou seja
\hat f'(\omega)-\frac{\omega}{2i}\,\hat f(\omega)=0.

Passo 4

(c) A EDO em \omega é de primeira ordem e separável:
\frac{d\hat f}{d\omega} =\frac{\omega}{2i}\,\hat f \quad\Longrightarrow\quad \frac{d\hat f}{\hat f} =\frac{\omega}{2i}\,d\omega.
Integrando,
\ln\hat f(\omega) =\frac{\omega^{2}}{4i}+C \quad\Longrightarrow\quad \hat f(\omega) =A\,e^{\omega^{2}/(4i)} =A\,e^{-\,i\,\omega^{2}/4},
onde A=\hat f(0).

Passo 5

(d) Pelo enunciado,
\hat f(0) =\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{i x^{2}}dx =\frac12+\frac{i}{2}.
Logo
\hat f(\omega) =\Bigl(\tfrac12+\tfrac{i}{2}\Bigr)e^{-i\,\omega^{2}/4} =\frac1{\sqrt{2\pi}}\Bigl(\tfrac12+\tfrac{i}{2}\Bigr)\bigl(\cos\tfrac{\omega^{2}}{4}-i\sin\tfrac{\omega^{2}}{4}\bigr).
Distribuindo e usando
\cos\tfrac{\omega^{2}}{4}+\sin\tfrac{\omega^{2}}{4} =\sqrt2\,\cos\!\Bigl(\tfrac{\omega^{2}}{4}-\tfrac\pi4\Bigr), \quad \cos\tfrac{\omega^{2}}{4}-\sin\tfrac{\omega^{2}}{4} =\sqrt2\,\cos\!\Bigl(\tfrac{\omega^{2}}{4}+\tfrac\pi4\Bigr),
chegamos a
\hat f(\omega) =\frac1{\sqrt{2\pi}}\Bigl[\cos\!\Bigl(\tfrac{\omega^{2}}{4}-\tfrac\pi4\Bigr) +i\,\cos\!\Bigl(\tfrac{\omega^{2}}{4}+\tfrac\pi4\Bigr)\Bigr].

Passo 6

(e) Escrevendo \hat f(\omega)=\mathcal F\bigl(\cos(x^{2})\bigr)(\omega)+i\,\mathcal F\bigl(\operatorname{sen}(x^{2})\bigr)(\omega) e comparando parte real e imaginária, obtemos
\mathcal F\bigl(\cos(x^{2})\bigr)(\omega) =\frac1{\sqrt2}\,\cos\!\Bigl(\tfrac{\omega^{2}}{4}-\tfrac\pi4\Bigr), \quad \mathcal F\bigl(\operatorname{sen}(x^{2})\bigr)(\omega) =\frac1{\sqrt2}\,\cos\!\Bigl(\tfrac{\omega^{2}}{4}+\tfrac\pi4\Bigr).

Passo 7

(f) Para a>0, usamos que \cos(a x^{2})=g(bx) com g(x)=\cos(x^{2}) e b=\sqrt{a}. A propriedade de escala diz
\mathcal F\{g(bx)\}(\omega)=\frac1{|b|}\,\hat g\!\Bigl(\frac{\omega}{b}\Bigr).
Assim
\mathcal F\bigl(\cos(a x^{2})\bigr)(\omega) =\frac1{\sqrt{a}}\, \mathcal F\bigl(\cos(x^{2})\bigr)\!\Bigl(\frac{\omega}{\sqrt{a}}\Bigr) =\frac1{\sqrt{2a}} \cos\!\Bigl(\frac{\omega^{2}}{4a}-\frac\pi4\Bigr),
e analogamente
\mathcal F\bigl(\operatorname{sen}(a x^{2})\bigr)(\omega) =\frac1{\sqrt{2a}} \cos\!\Bigl(\frac{\omega^{2}}{4a}+\frac\pi4\Bigr).

Resposta

(a) f'(x)-i\,2x\,f(x)=0.
(b) \hat f'(\omega)-\tfrac{\omega}{2i}\,\hat f(\omega)=0.
(c) \hat f(\omega)=\hat f(0)e^{-i\,\omega^{2}/4}.
(d) \hat f(\omega)=\tfrac1{\sqrt{2\pi}}\bigl[\cos(\tfrac{\omega^{2}}{4}-\tfrac\pi4)+i\cos(\tfrac{\omega^{2}}{4}+\tfrac\pi4)\bigr].
(e) \mathcal F(\cos(x^{2}))=\tfrac1{\sqrt2}\cos(\tfrac{\omega^{2}}{4}-\tfrac\pi4), \mathcal F(\operatorname{sen}(x^{2}))=\tfrac1{\sqrt2}\cos(\tfrac{\omega^{2}}{4}+\tfrac\pi4).
(f) \mathcal F(\cos(a x^{2}))=\tfrac1{\sqrt{2a}}\cos(\tfrac{\omega^{2}}{4a}-\tfrac\pi4), \mathcal F(\operatorname{sen}(a x^{2}))=\tfrac1{\sqrt{2a}}\cos(\tfrac{\omega^{2}}{4a}+\tfrac\pi4).

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