Q14 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Verify that the formulas for the Bessel functions ${\mathrm{J}}_{\frac{1}{2}}\left(\mathrm{t}\right),{\mathrm{Y}}_{\frac{1}{2}}\left(\mathrm{t}\right)$ do indeed solve equation (16).

Step-by-Step Solution

Answer

Therefore, we verified that the given statement is true. That is, ${\mathrm{J}}_{\frac{1}{2}}\left(\mathrm{t}\right),{\mathrm{Y}}_{\frac{1}{2}}\left(\mathrm{t}\right)$ are solutions to Bessel’s equation.

Step 1: General form

The Energy Integral Lemma: 

 

Let y(t) be a solution to the differential equation $\mathrm{y}=\mathrm{f}\left(\mathrm{y}\right)$, where f(y) is a continuous function that does not depend on y’ or the independent variable t. Let F(y) is an indefinite integral of $\mathrm{f}\left(\mathrm{y}\right)$, that is, $\mathrm{f}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{F}\left(\mathrm{y}\right)$. Then the quantity $\mathrm{E}\left(\mathrm{t}\right):=\frac{1}{2}\mathrm{y}\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{y}\left(\mathrm{t}\right)\right)$ is constant, i.e., $\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{E}\left(\mathrm{t}\right)=0.$

 

Bessel’s equation: (16)

  $\mathrm{y}+\frac{1}{\mathrm{t}}\mathrm{y}\text{'}+\left(1-\frac{{\mathrm{n}}^2}{{\mathrm{t}}^2}\right)\mathrm{y}=0$ … (1)

The mass–spring oscillator equation: 

 

  $$\begin{gathered} {\mathrm{F}}_{\mathrm{ext}}=\left[\mathrm{inertia}\right]\mathrm{y}+\left[\mathrm{damping}\right]\mathrm{y}\text{'}+\left[\mathrm{stiffness}\right]\mathrm{y} \\ =\mathrm{my}+\mathrm{by}\text{'}+\mathrm{ky} \end{gathered}$$ .....… (2)

Step 2: Find the Equation

Bessel’s equation is:

 

(16) $\mathrm{y}+\frac{1}{\mathrm{t}}\mathrm{y}\text{'}+\left(1-\frac{{\mathrm{n}}^2}{{\mathrm{t}}^2}\right)\mathrm{y}=0$ .

To verify: ${\mathrm{J}}_{\frac{1}{2}}\left(\mathrm{t}\right),{\mathrm{Y}}_{\frac{1}{2}}\left(\mathrm{t}\right)$.

Let’s take $\mathrm{n}=\frac{1}{2}$ and we know that,

$$\begin{gathered} {\mathrm{J}}_{\frac{1}{2}}\left(\mathrm{t}\right)=\sqrt{\frac{2}{\mathrm{\pi t}}}\mathrm{sint} \\ {\mathrm{Y}}_{\frac{1}{2}}\left(\mathrm{t}\right)=\sqrt{\frac{2}{\mathrm{\pi t}}}\mathrm{sint} \end{gathered}$$

Step 3: Verify the equation

Let us check whether both are solutions to the equation (16) or not.

 

Case (1):

 

If ${\mathrm{J}}_{\frac{1}{2}}\left(\mathrm{t}\right)=\sqrt{\frac{2}{\mathrm{\pi t}}}\mathrm{sint}$. Then, differentiate two times with respect to t.

 $$\begin{gathered} J_{\frac{1}{2}}^{\text{'}}\left(\mathrm{t}\right)=\frac{-2\mathrm{sint}-2\mathrm{tcost}}{\sqrt{2\mathrm{\pi }}{\mathrm{t}}^{\frac{3}{2}}} \\ {J}_{\frac{{\displaystyle 1}}{{\displaystyle 2}}}^{\text{'}}\left(\mathrm{t}\right)=\frac{\left(4{\mathrm{t}}^2+3\right)\mathrm{sint}}{2^{\frac{3}{2}}{\mathrm{t}}^{\frac{5}{2}}\sqrt{\mathrm{\pi }}} \end{gathered}$$

Now substitute the values in equation (1).

$$\begin{gathered} \mathrm{y}+\frac{1}{\mathrm{t}}\mathrm{y}\text{'}+\left(1-\frac{{\mathrm{n}}^2}{{\mathrm{t}}^2}\right)\mathrm{y}=\frac{\left(4{\mathrm{t}}^2+3\right)\mathrm{sint}}{2^{\frac{3}{2}}{\mathrm{t}}^{\frac{5}{2}}\sqrt{\mathrm{\pi }}}+\frac{1}{\mathrm{t}}\left(\frac{-2\mathrm{sint}-2\mathrm{tcost}}{\sqrt{2\mathrm{\pi }}{\mathrm{t}}^{\frac{3}{2}}}\right)+\left(1-\frac{1}{4{\mathrm{t}}^2}\right)\sqrt{\frac{2}{\mathrm{\pi t}}}\mathrm{sint} \\ =\frac{\left(4{\mathrm{t}}^2+3\right)\mathrm{sint}}{2^{\frac{3}{2}}{\mathrm{t}}^{\frac{5}{2}}\sqrt{\mathrm{\pi }}}-\frac{2\left(\mathrm{sint}+\mathrm{tcost}\right)}{2^{\frac{1}{2}}{\mathrm{t}}^{\frac{5}{2}}\sqrt{\mathrm{\pi }}}+\left(1-\frac{1}{4{\mathrm{t}}^2}\right)\sqrt{\frac{2}{\mathrm{\pi t}}}\mathrm{sint} \\ =0 \end{gathered}$$

Step 4: Verify the equation

Case (1):

${\mathrm{Y}}_{\frac{1}{2}}\left(\mathrm{t}\right)=\sqrt{\frac{2}{\mathrm{\pi t}}}\mathrm{sint}$If. Then, differentiate two times with respect to t.

$$\begin{gathered} Y_{\frac{1}{2}}^{\text{'}}\left(\mathrm{t}\right)=\frac{-2\mathrm{sint}-2\mathrm{tcost}}{\sqrt{2\mathrm{\pi }}{\mathrm{t}}^{\frac{3}{2}}} \\ {Y}_{\frac{{\displaystyle 1}}{{\displaystyle 2}}}^{\text{'}}\left(\mathrm{t}\right)=\frac{\left(4{\mathrm{t}}^2+3\right)\mathrm{sint}}{2^{\frac{3}{2}}{\mathrm{t}}^{\frac{5}{2}}\sqrt{\mathrm{\pi }}} \end{gathered}$$

Now substitute the values in equation (1).

$$\begin{gathered} \mathrm{y}+\frac{1}{\mathrm{t}}\mathrm{y}\text{'}+\left(1-\frac{{\mathrm{n}}^2}{{\mathrm{t}}^2}\right)\mathrm{y}=\frac{\left(4{\mathrm{t}}^2+3\right)\mathrm{sint}}{2^{\frac{3}{2}}{\mathrm{t}}^{\frac{5}{2}}\sqrt{\mathrm{\pi }}}+\frac{1}{\mathrm{t}}\left(\frac{-2\mathrm{sint}-2\mathrm{tcost}}{\sqrt{2\mathrm{\pi }}{\mathrm{t}}^{\frac{3}{2}}}\right)+\left(1-\frac{1}{4{\mathrm{t}}^2}\right)\sqrt{\frac{2}{\mathrm{\pi t}}}\mathrm{sint} \\ =\frac{\left(4{\mathrm{t}}^2+3\right)\mathrm{sint}}{2^{\frac{3}{2}}{\mathrm{t}}^{\frac{5}{2}}\sqrt{\mathrm{\pi }}}-\frac{2\left(\mathrm{sint}+\mathrm{tcost}\right)}{2^{\frac{1}{2}}{\mathrm{t}}^{\frac{5}{2}}\sqrt{\mathrm{\pi }}}+\left(1-\frac{1}{4{\mathrm{t}}^2}\right)\sqrt{\frac{2}{\mathrm{\pi t}}}\mathrm{sint} \\ =0 \end{gathered}$$

 

Therefore, ${\mathrm{Y}}_{\frac{1}{2}}\left(\mathrm{t}\right)$is a solution.