Mostrar que $L^{-1}\{(s^2+a^2)^{-\frac{3}{2}}\}=\frac{t}{a}J_1(ax)$

Question

Pregunta:

Suponga que $J_\alpha(x)$ se define de la siguiente manera:
$$J_\alpha(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{n!\, \Gamma(n+\alpha+1)}\left(\frac{x}{2}\right)^{2n+\alpha}.$$

(a) Probar que $\frac{d}{dx}(x^{-\alpha} J_\alpha(x))=-x^{-\alpha}J_{\alpha+1}(x)$;

(b) Si $L\{J_0(ax)\}=\frac{1}{\sqrt {s^2+a^2}}$, muestran que $L^{-1}\{(s^2+a^2)^{-\frac{3}{2}}\}=\frac{x}{a}J_1(ax)$ .

He demostrado la parte (a). Pero acerca de la parte (b), lo que tengo es:

$$L^{-1}\{(s^2+a^2)^{-\frac{3}{2}}\}=L^{-1}\{({(s^2+a^2)^{-\frac{1}{2}}})^{3}\}=J_0(x)*J_0(x)*J_0(x).$$

¿Cómo debo llegar a $\frac{t}{a}J_1(ax)$$J_0(x)*J_0(x)*J_0(x)$ ?

Answer 1

Bien, vamos a escribir:

$$\mathscr{L}_x\left[\frac{x}{\text{a}}\cdot\mathscr{J}_\alpha\left(\text{b}\cdot x\right)\right]_{\left(\text{s}\right)}=\frac{1}{\text{a}}\cdot\mathscr{L}_x\left[x\cdot\mathscr{J}_\alpha\left(\text{b}\cdot x\right)\right]_{\left(\text{s}\right)}\tag1$$

Utilizando el 'dominio de la frecuencia derivado de lapropiedad de la transformada de Laplace:

$$\frac{1}{\text{a}}\cdot\mathscr{L}_x\left[x\cdot\mathscr{J}_\alpha\left(\text{b}\cdot x\right)\right]_{\left(\text{s}\right)}=-\frac{1}{\text{a}}\cdot\frac{\partial}{\partial\text{s}}\left\{\mathscr{L}_x\left[\mathscr{J}_\alpha\left(\text{b}\cdot x\right)\right]_{\left(\text{s}\right)}\right\}\tag2$$

Utilizando la"escala de tiempo"la propiedad de la transformada de Laplace:

$$-\frac{1}{\text{a}}\cdot\frac{\partial}{\partial\text{s}}\left\{\mathscr{L}_x\left[\mathscr{J}_\alpha\left(\text{b}\cdot x\right)\right]_{\left(\text{s}\right)}\right\}=-\frac{1}{\text{a}}\cdot\frac{\partial}{\partial\text{s}}\left\{\frac{1}{\text{b}}\cdot\mathscr{L}_x\left[\mathscr{J}_\alpha\left(x\right)\right]_{\left(\frac{\text{s}}{\text{b}}\right)}\right\}=$$ $$-\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\frac{\partial}{\partial\text{s}}\left\{\mathscr{L}_x\left[\mathscr{J}_\alpha\left(x\right)\right]_{\left(\frac{\text{s}}{\text{b}}\right)}\right\}\tag3$$

Utilizando la definición de la función de Bessel:

$$-\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\frac{\partial}{\partial\text{s}}\left\{\mathscr{L}_x\left[\mathscr{J}_\alpha\left(x\right)\right]_{\left(\frac{\text{s}}{\text{b}}\right)}\right\}=$$ $$-\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\frac{\partial}{\partial\text{s}}\left\{\mathscr{L}_x\left[\left(\frac{x}{2}\right)^{\alpha+2\text{n}}\right]_{\left(\frac{\text{s}}{\text{b}}\right)}\right\}=$$ $$-\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\frac{\partial}{\partial\text{s}}\left\{\frac{1}{2^{\alpha+2\text{n}}}\cdot\frac{\Gamma\left(1+\alpha+2\text{n}\right)}{\left(\frac{\text{s}}{\text{b}}\right)^{1+\alpha+2\text{n}}}\right\}=$$ $$-\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\sum_{\text{n}=0}^\infty\frac{1}{2^{\alpha+2\text{n}}}\cdot\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{\Gamma\left(1+\alpha+2\text{n}\right)}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\frac{\partial}{\partial\text{s}}\left\{\frac{1}{\left(\frac{\text{s}}{\text{b}}\right)^{1+\alpha+2\text{n}}}\right\}=$$ $$-\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\sum_{\text{n}=0}^\infty\frac{1}{2^{\alpha+2\text{n}}}\cdot\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{\Gamma\left(1+\alpha+2\text{n}\right)}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\left(-\frac{1+\alpha+2\text{n}}{\text{s}}\cdot\frac{1}{\left(\frac{\text{s}}{\text{b}}\right)^{1+\alpha+2\text{n}}}\right)=$$ $$\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\sum_{\text{n}=0}^\infty\frac{1}{2^{\alpha+2\text{n}}}\cdot\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{\Gamma\left(2+\alpha+2\text{n}\right)}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\frac{1}{\text{s}}\cdot\frac{1}{\left(\frac{\text{s}}{\text{b}}\right)^{1+\alpha+2\text{n}}}\tag4$$

Al $\Re\left(\alpha+2\text{n}\right)>-1\space\wedge\space\Re\left(\text{s}\right)>0$

Ahora, cuando $\alpha=1$:

$$\mathscr{L}_x\left[\frac{x}{\text{a}}\cdot\mathscr{J}_1\left(\text{b}\cdot x\right)\right]_{\left(\text{s}\right)}=$$ $$\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\sum_{\text{n}=0}^\infty\frac{1}{2^{1+2\text{n}}}\cdot\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{\Gamma\left(2+1+2\text{n}\right)}{\Gamma\left(1+1+\text{n}\right)}\cdot\frac{1}{\text{s}}\cdot\frac{1}{\left(\frac{\text{s}}{\text{b}}\right)^{1+1+2\text{n}}}=$$ $$\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\sum_{\text{n}=0}^\infty\frac{1}{2^{1+2\text{n}}}\cdot\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{\Gamma\left(3+2\text{n}\right)}{\Gamma\left(2+\text{n}\right)}\cdot\frac{1}{\text{s}}\cdot\frac{1}{\left(\frac{\text{s}}{\text{b}}\right)^{2+2\text{n}}}\tag5$$

Ahora, cuando $\text{a}=\text{b}$:

$$\mathscr{L}_x\left[\frac{x}{\text{a}}\cdot\mathscr{J}_1\left(\text{a}\cdot x\right)\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}^3}\cdot\frac{1}{\left(1+\left(\frac{\text{a}}{\text{s}}\right)^2\right)^\frac{3}{2}}\tag6$$

Ahora, que te deje probar:

$$\frac{1}{\text{s}^3}\cdot\frac{1}{\left(1+\left(\frac{\text{a}}{\text{s}}\right)^2\right)^\frac{3}{2}}=\frac{1}{\left(\text{s}^2+\text{a}^2\right)^\frac{3}{2}}\tag7$$