Question

In Section 6.4 we saw that$\mathbf{ty}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{ty}\mathbf{=}\mathbf{0}$is Bessel’s equation of order$\mathit{v}\mathbf{=}\mathbf{0}$. In view of$\left(24\right)$of that section $\mathbf{ty}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{ty}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{}\mathit{i}\mathit{s} \mathbf{y}\mathbf{=}{\mathbf{J}}_{\mathbf{0}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$and Table 6.4.1 a solution of the initial-value problem. Use this result and the procedure outlined in the instructions to Problems$\mathbf{17}$and$\mathbf{18}$to show that.

$\mathbf{L}\left\{{\mathbf{J}}_{\mathbf{u}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right\}\mathbf{=}\frac{\mathbf{1}}{\sqrt{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}}$

Short answer

Hence, by using the theorem derivatives of transforms, it is proved that$\mathcal{L}\left\{J_{\text{u}}\left(t\right)\right\}=\frac{1}{\sqrt{s^2+1}}$

Step-by-step solution

Define derivatives of transforms theorem:

If $\mathbf{F}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}\mathbf{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right\}}$and$\mathbf{n}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{\dots }$then $\mathbf{L}\left\{{\mathbf{t}}^{\mathbf{n}}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right\}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{n}}\frac{{\mathbf{d}}^{\mathbf{n}}}{{\mathbf{ds}}^{\mathbf{\pi }}}\mathbf{F}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{.}$

Step 2:Solve the given equation by taking Laplace transform:

Consider the equations,

$\left(*\right)\left\{\begin{array}{l}\mathcal{L}\left\{y\text{'}\text{'}\right\}=s^2\mathcal{L}\left\{y\right\}-sy\left(0\right)-y\text{'}\left(0\right)\\ \mathcal{L}\left\{y\text{'}\right\}=s\mathcal{L}\left\{y\right\}-y\left(0\right)\end{array}\right.$

Hence, take Laplace transform on both sides,

$\mathcal{L}\left\{0\right\}=\mathcal{L}\left\{ty\text{'}\text{'}\right\}+\mathcal{L}\left\{y\text{'}\right\}+\mathcal{L}\left\{ty\right\}$

Use the theorem,

$=-\frac{\text{d}}{\text{d}s}\left[\mathcal{L}\left\{y\text{'}\text{'}\right\}\right]+\mathcal{L}\left\{y\text{'}\right\}-\frac{\text{d}}{\text{d}s}[\mathcal{L}\{y\left\}\right]$

Evaluate,

$\stackrel{*}{=}-\frac{\text{d}}{\text{d}s}\left[s^2\mathcal{L}\left\{y\right\}-sy\left(0\right)-y\text{'}\left(0\right)\right]+s\mathcal{L}\left\{y\right\}-y\left(0\right)-\frac{\text{d}}{\text{d}s}[\mathcal{L}\{y\left\}\right]\left(y\left(0\right)=0y\text{'}\left(0\right)=0\right)$

$=-\frac{\text{d}}{\text{d}s}\left[s^2\mathcal{L}\left\{y\right\}\right]+s\mathcal{L}\left\{y\right\}-\frac{\text{d}}{\text{d}s}[\mathcal{L}\{y\left\}\right]$

Assume $\mathcal{L}\left\{y\right\}=Y\left(s\right)$and substitute in the equation

$0=-\frac{\text{d}}{\text{d}s}\left[s^{2}Y\left(s\right)\right]+sY\left(s\right)-\frac{\text{d}}{\text{d}s}\left[Y\right(s\left)\right]$

$=-2sY\left(s\right)-s^{2}Y\text{'}\left(s\right)+sY\left(s\right)-Y\text{'}\left(s\right)$

$=Y\text{'}\left(s\right)\left(-s^2-1\right)-sY\left(s\right)$

$=Y\text{'}\left(s\right)\left(s^2+1\right)+sY\left(s\right)$

Rewrite the equation as,

$\frac{\text{d}Y\left(s\right)}{\text{d}s}\left(s^2+1\right)=-sY\left(s\right)$

$\frac{\text{d}Y\left(s\right)}{Y\left(s\right)}=-\frac{s}{s^2+1}\text{d}s$

Derive the equation by taking integration on both sides:

Thus, the equation can be written as,

$\int \frac{\text{d}Y\left(s\right)}{Y\left(s\right)}=-\int \frac{s}{s^2+1}\text{d}s$

$\ln \left|Y\right(s\left)\right|=-\frac{1}{2}\ln \left|s^2+1\right|+\ln \left|c\right|$

$\ln \left|Y\right(s\left)\right|=\ln {\left|s^2+1\right|}^{-\frac{1}{2}}+\ln \left|c\right|$

$\ln \left|Y\right(s\left)\right|=\ln \left|\right|s^2+{\left.1\right|}^{-\frac{1}{2}}·c\mid$

$Y\left(s\right)={\left|s^2+1\right|}^{-\frac{1}{2}}·c$

$Y\left(s\right)=\frac{c}{\sqrt{s^2+1}}$

Substitute$\mathcal{L}\left\{y\right(t\left)\right\}=Y\left(s\right)$,and hence we know that,$\mathcal{L}\left\{y\right(t\left)\right\}=Y\left(s\right)$

$\mathcal{L}\left\{J_0\left(t\right)\right\}=\frac{c}{\sqrt{s^2+1}}---\left(1\right)$

Prove the equation by defining the function:

$J_0\left(0\right)=c$Consider,$f\left(0\right)=\underset{s\to \infty }{limsF\left(s\right)}$

Substitute in the equation,

$J_0\left(0\right)=\underset{s\to \infty }{lim}\frac{sc}{\sqrt{s^2+1}}$

$=\underset{s\to \infty }{lim}\frac{sc}{\sqrt{s^2\left(1+\frac{1}{s^2}\right)}}$

Solve,

$=\underset{s\to \infty }{lim}\frac{c}{\sqrt{1+\frac{1}{s^2}}}$

$=\frac{c}{\sqrt{1+0}}$

$c=1$

Substitute $c=1$in equation (1)

$\mathcal{L}\left\{J_0\left(t\right)\right\}=\frac{1}{\sqrt{s^2+1}}$

Therefore, by solving the equation it is proved that$\mathcal{L}\left\{J_0\left(t\right)\right\}=\frac{1}{\sqrt{s^2+1}}$