📂Odinary Differential Equations

Bessel Functions as Solutions to Differential Equations

Theorem¹

Theorem 1

Given a differential equation slightly different from the Bessel equation as follows:

$$ \begin{equation} \begin{aligned} && y^{\prime \prime}+\frac{1-2a}{x}y^{\prime}+\left[ (bcx^{c-1})^{2}+\frac{a^{2}-\nu^{2}c^{2}}{x^{2}} \right]y =&\ 0 \\ \text{or} && x^{2}y^{\prime \prime}+(1-2a)xy^{\prime}+\left[ b^{2}c^{2}x^{2c}+(a^{2}-\nu^{2}c^{2}) \right]y =&\ 0 \end{aligned} \label{1} \end{equation} $$

And let $Z_{\nu}(x)$ be any linear combination of $J_{\nu}(x)$ and $N_{\nu}(x)$. Then, the solution to the given differential equation is as follows:

$$ y=x^{a}Z_{\nu}(bx^{c})=x^{a}[AJ_{\nu}(bx^{c})+BN_{\nu}(bx^{c})] $$

$\nu$, $a$, $b$, $c$, $A$, $B$ are constants.

Theorem 2

$$ x^{2}y^{\prime \prime} + xy^{\prime}+(K^{2}x^{2}-\nu^{2})y=0 $$

The general solution to the differential equation is as follows:

$$ y=AJ_{\nu}(Kx)+BN_{\nu}(Kx) $$

Explanation

Bessel’s Equation

$$ \begin{align*} x^2 y^{\prime \prime} +xy^{\prime} +(x^2-\nu^2)y =&\ 0 \\ x(xy^{\prime})^{\prime}+(x^2- \nu ^2) y =&\ 0 \\ y^{\prime \prime}+\frac{1}{x} y^{\prime} + \left( 1-\frac{\nu^{2}}{x^{2}} \right)y =&\ 0 \end{align*} $$

Even if a differential equation is not exactly like the Bessel equation, the fact that its solutions can be expressed using Bessel functions is of significant help in solving various differential equations. For example,

$$ \begin{align*} 3xy^{\prime \prime}+y^{\prime}+12y =&\ 0 \\ xy^{\prime \prime}+2y^{\prime}+4y =&\ 0 \\ y^{\prime \prime}+9xy =&\ 0 \end{align*} $$

Although this differential equation is not exactly in the form of Bessel’s equation, its solution can still be expressed by Bessel functions. The solutions are as follows:

$$ \begin{align*} y =&\ x^{1/3}Z_{2/3}(4x^{1/2} )=x^{1/3}\left[A J_{2/3}(4x^{1/2} )+BN_{2/3}(4x^{1/2} ) \right] \\ y =&\ x^{-1/2}Z_{1}(4x^{1/2} )=x^{-1/2}\left[A J_{1}(4x^{1/2} )+BN_{1}(4x^{1/2} ) \right] \\ y =&\ x^{1/2}Z_{1/3}(2x^{3/2} )=x^{1/2}\left[A J_{1/3}(2x^{3/2} )+BN_{1/3}(2x^{3/2} ) \right] \end{align*} $$

Proof

Proof 1

It is sufficient to show that it holds for $y=x^{a}J_{\nu}(bx^{c})$. First, obtaining $y^{\prime}$, $y^{\prime \prime}$ gives the following:

$$ \begin{align*} y =&\ x^{a}J_{\nu}(bx^{c}) \\ y^{\prime} =&\ ax^{a-1}J_{\nu}(bx^{c})+bcx^{a+c-1}J_{\nu}^{\prime}(bx^{c}) \\ y^{\prime \prime} =&\ a(a-1)x^{a-2}J_{\nu}(bx^{c}) +abcx^{a+c-2}J_{\nu}^{\prime}(bx^{c}) \\ & +(a+c-1)bcx^{a+c-2}J_{\nu}^{\prime} (bx^{c})+b^{2}c^{2}x^{a+2c-2}J_{\nu}^{\prime \prime}(bx^{c}) \end{align*} $$

Substituting this into $\eqref{1}$ results in:

$$ \begin{align*} & \Big[a(a-1)x^{a}J_{\nu}(bx^{c}) +abcx^{a+c}J_{\nu}^{\prime}(bx^{c})+(a+c-1)bcx^{a+c}J_{\nu}^{\prime} (bx^{c}) +b^{2}c^{2}x^{a+2c}J_{\nu}^{\prime \prime}(bx^{c}) \Big] \\ & +\Big[(1-2a)ax^{a}J_{\nu}(bx^{c})+(1-2a)bcx^{a+c}J_{\nu}^{\prime}(bx^{c}) \Big] \\ & +\Big[ b^{2}c^{2}x^{a+2c}J_{\nu}(bx^{c})+(a^{2}-\nu^{2}c^{2})x^{a}J_{\nu}(bx^{c})\Big] = 0 \end{align*} $$

After organizing according to the differential coefficients:

$$ \begin{align*} & (a^{2}-a+a-2a^{2}+b^{2}c^{2}x^{2c}+a^{2}-\nu^{2}c^{2})x^{a}J_{\nu}(bx^{c}) \\ & +(abc+abc+bc^{2}-bc+bc-2abc)x^{a+c}J_{\nu}^{\prime}(bx^{c}) \\ & +(b^{2}c^{2}x^{2c})x^{a}J_{\nu}^{\prime \prime}(bx^{c}) = 0 \end{align*} $$

Multiplying both sides by $\dfrac{1}{x^{a}}$ and organizing the coefficients gives:

$$ \begin{align*} && c^{2}(b^{2}x^{2c}-\nu^{2})J_{\nu}(bx^{c}) + c^{2}(bx^{c})J_{\nu}^{\prime}(bx^{c})+c^{2}(b^{2}x^{2c})J_{\nu}^{\prime \prime}(bx^{c}) =&\ 0 \\ \implies && (b^{2}x^{2c}-\nu^{2})J_{\nu}(bx^{c}) + (bx^{c})J_{\nu}^{\prime}(bx^{c})+(b^{2}x^{2c})J_{\nu}^{\prime \prime}(bx^{c}) =&\ 0 \end{align*} $$

Let’s denote by $ bx^{c}=z$, $J_{\nu}(z)=y$, then from the above equation, we obtain the following:

$$ \begin{align*} && (z^{2}-\nu^{2})y+zy^{\prime}+z^{2}y^{\prime \prime}=&\ 0 \\ \implies && x^{2}y^{\prime \prime}+xy^{\prime}+(x^{2}-\nu^{2})y =&\ 0 \end{align*} $$

This is a Bessel equation, and since Bessel functions are solutions to the Bessel equation, the above statement is valid.

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Proof 2

$$ x^{2}\frac{ d ^{2}y}{ dx^{2} }+x\frac{ d y}{ d x }+(x^{2}-\nu^{2})y=0 $$ The general solution to the above Bessel differential equation is $y=AJ_{\nu}(Kx)+BN_{\nu}(Kx)$. By substituting $x=Kx$ into the equation and general solution, we obtain the following:

$$ K^{2}x^{2}\frac{ d ^{2}y }{ d (Kx)^{2} } +Kx\frac{ d y}{ d(Kx) } +(K^{2}x^{2}-\nu^{2})y=0 $$

The general solution to this differential equation is $y=AJ_{\nu}(Kx)+BN_{\nu}(x)$. By organizing the constants of the differential equation, we get:

$$ x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{ d y}{ dx }+(K^{2}x^{2}-\nu^{2})y=0 $$

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