Prove that some function is the solution of some equation

Question

Show that $$x(t)=\sum_{n=0}^{\infty}\frac{(-1)^n(t/2)^{2n}}{(n!)^2}$$ is the solution of $$x*x=\int_{0}^t x(u)x(t-u)du=\sin t$$

My approach:

I suppose that I have to use the Laplace transform. I tried to calculate $$\mathcal{L}(\sin t)=\int_{0}^{\infty}e^{-\lambda t}\sin tdt=\frac{1}{\lambda^2+1}$$

But how do I continue?

Answer 1

$\mathcal{L}_{t\to\lambda}\left\{\int_0^tx(u)x(t-u)~du\right\}=\mathcal{L}_{t\to\lambda}\{\sin t\}$

$(X(\lambda))^2=\dfrac{1}{\lambda^2+1}$

$X(\lambda)=\pm\dfrac{1}{\sqrt{\lambda^2+1}}$

$x(t)=\mathcal{L}^{-1}_{\lambda\to t}\biggl\{\pm\dfrac{1}{\sqrt{\lambda^2+1}}\biggr\}=\mathcal{L}^{-1}_{\lambda\to t}\biggl\{\pm\dfrac{1}{\lambda\sqrt{1+\dfrac{1}{\lambda^2}}}\biggr\}=\mathcal{L}^{-1}_{\lambda\to t}\biggl\{\pm\dfrac{1}{\lambda}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2\lambda^{2n}}\biggr\}=\mathcal{L}^{-1}_{\lambda\to t}\biggl\{\pm\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2\lambda^{2n+1}}\biggr\}=\pm\sum\limits_{n=0}^\infty\dfrac{(-1)^nt^{2n}}{4^n(n!)^2}$

$x(t)=\sum\limits_{n=0}^\infty\dfrac{(-1)^nt^{2n}}{4^n(n!)^2}$ is one of the solutions.

Answer 2

By the uniqueness of the Laplace transform it suffices to show that: $$\mathcal{L}\left(\int_{0}^t x(u)x(t-u)du\right)=\frac{1}{\lambda^2+1} \tag{1}$$ To do so, insert the sum definition: $$x \left( t \right) =\sum _{n=0}^{\infty }{\frac { \left( -1 \right) ^{ n} \left(\dfrac{t}{2} \right) ^{2\,n}}{ \left( n! \right) ^{2}}} \tag{2}$$ into the integral and note that term by term integration is justified as the sum is absolutely convergent. Each term can be integrated by comparison with the Beta function: $$\begin{aligned} \int _{0}^{t}\! \left( \dfrac{u}{2} \right) ^{2\,n} \left( \dfrac{t}{2}-\dfrac{u}{2} \right) ^{2\,m}{du}&=2\, \left( \dfrac{t}{2} \right) ^{2\,m+1+2\,n}\int _{0} ^{1}\!{y}^{2\,n} \left( 1-y \right) ^{2\,m}{dy}\quad:y=ut\\ &=2\,{\frac { \left(\dfrac{t}{2}\right) ^{2\,m+1+2\,n}\Gamma \left( 2\,m+1 \right) \Gamma \left( 2\,n+1 \right) }{\Gamma \left( 2\,m+2+2\,n \right) }} \end{aligned} \tag{3} $$ and the Laplace transform can be applied to each term by comparison with the integral definition of the Gamma function: $$ \int _{0}^{\infty }\!2\, \left( \dfrac{t}{2}\right) ^{2\,m+1+2\,n}{{\rm e}^ {-\lambda\,t}}{dt}={\frac { \left(\dfrac{1}{2}\right) ^{2\,n+2\,m}\Gamma \left( 2\,m+2+2\,n \right) }{{\lambda}^{2\,m+2+2\,n}}} \tag{4}$$ and therefore: $$ \begin{aligned} \mathcal{L}\left(\int_{0}^t x(u)x(t-u)du\right)&=\sum _{m=0}^{\infty } \sum _{n=0}^{\infty }4\,{\frac { \left( - 1 \right) ^{m+n}\,\Gamma \left( 2\,n+1 \right) \Gamma \left( 2\,m+1 \right) }{ \left( \Gamma \left( n+1 \right) \right) ^{2} \left( \Gamma \left( 1+m \right) \right) ^{2}}} \left( \dfrac{\lambda}{2} \right) ^{-2\,m-2\,n-2} \\ &= \left( \sum _{m=0}^{\infty }{\frac { \left( -1 \right) ^{m}{2}^{-2\,m }{\lambda}^{-2\,m-1}\Gamma \left( 2\,m+1 \right) }{ \left( \Gamma \left( 1+m \right) \right) ^{2}}} \right) ^{2}\\ &=\left(\dfrac{1}{\lambda} \sum _{m=0}^{\infty }{-1/2\choose m}{\left(\lambda^2\right)}^{-m} \right) ^{2}\\ &=\left(\dfrac{1}{\lambda}\dfrac{1}{\sqrt{1+\dfrac{1}{\lambda^2}}}\right)^2\\ &=\frac{1}{\lambda^2+1} \end{aligned} $$