Laplace Transform Methods for Bessel and Modified Bessel Equations: Theory and Applications

Abstract

This paper revisits the use of the Laplace transform in the study of Bessel and modified Bessel differential equations. Rather than presenting the method as a new alternative to the classical Frobenius approach, the paper is organized as a pedagogical review of how standard operational rules for the Laplace transform lead to transform-domain differential equations for the regular solutions of these variable-coefficient problems. For Bessel’s equation, the transformed equation yields the classical Laplace transform of $J_{\nu }\left(x\right)$; for the modified Bessel equation, the same procedure yields the corresponding transform of $I_{\nu }\left(x\right)$. These formulas are then used to recover standard recurrence and derivative identities and to derive several illustrative transform evaluations. The paper also explains why the singular companion solutions $Y_{\nu }\left(x\right)$ and $K_{\nu }\left(x\right)$ are not obtained directly from the regular Laplace-transform framework and must instead be introduced through classical connection formulas. Particular attention is given to placing the calculations in the context of the earlier literature, especially the classical treatise of Watson and later work on Laplace and Lipschitz–Hankel integrals. In this form, the paper is intended as a self-contained review and tutorial on a useful operational approach to Bessel-type equations.

1. Introduction

Bessel and modified Bessel functions occupy a central place in analysis, applied mathematics, and mathematical physics. Their differential equations, recurrence relations, asymptotic expansions, and integral representations are classical material; standard references include Watson’s treatise and the modern handbook literature [1,2,3]. In most presentations, Bessel’s equation and the modified Bessel equation are introduced through the Frobenius method, since both equations have a regular singular point at the origin.

The aim of the present paper is to provide a pedagogical review of how the operational rules of the Laplace transform may be used to recover classical results for Bessel-type equations. The emphasis is on exposition, organization, and the transparent derivation of standard transform formulas and identities directly from the differential equations.

This point of view has clear precedents. The classical theory of Bessel functions is already developed in Watson [1]. Tables and handbooks record many Laplace transforms involving Bessel and modified Bessel functions; see, for example, the compendia cited in [1,2,3]. Particularly relevant to the present discussion is the paper of Eason, Noble, and Sneddon on Lipschitz–Hankel-type integrals, which contains systematic evaluations of Laplace-type integrals involving Bessel functions [4]. Rathie also considered related operational-calculus integrals involving Bessel and other special functions [5]. Later work by Ragab and Srivastava gives explicit formulas for Laplace transforms of families of modified Bessel functions [6,7], and Maier revisited complete Lipschitz–Hankel integrals and corrected a number of tabulated formulas [8]. A related transform-based treatment using the ${\mathcal{L}}_2$-transform was given by Yürekli and Wilson [9]. Accordingly, the contribution of the present paper is chiefly expository: it gathers a coherent Laplace-transform treatment of the regular solutions of the two differential equations and organizes several classical consequences in one place.

Recent work confirms that Laplace-type and related integral-transform methods remain active in the analysis of Bessel-type functions and equations. For example, Shah and Mubeen expressed Laplace transforms of several special functions in generalized Meijer–G form [10]; Khammash studied integral transforms associated with generalized k-Bessel functions [11]; distributional solutions of equations related to the modified Bessel equation were obtained using right-sided distributional Laplace transforms [12]; Aghili used joint Laplace–Hankel transforms for fractional diffusion equations [13]; and recent applied work continues to use Laplace-domain modified Bessel functions in semidiscrete and boundary-value problems [14,15]. These papers have different aims from the present review: they primarily develop generalized transforms, distributional methods, fractional-diffusion applications, or numerical inversion techniques. The present paper is narrower and more classical, concentrating on the ordinary Laplace transform and on a transparent route from the Bessel and modified Bessel equations to the familiar transforms of $J_{\nu }$ and $I_{\nu }$.

The main idea is straightforward. Starting from Bessel’s equation or the modified Bessel equation, one applies the standard derivative and multiplication rules for the Laplace transform to obtain a second-order differential equation in the transform variable. Solving that transformed equation yields the classical Laplace transform of the regular solution. The inverse transform can then be identified with $J_{\nu }$ or $I_{\nu }$ by appeal to standard formulas from the literature or, equivalently, by comparison with their classical series expansions. In this sense, the Laplace transform does not replace the standard theory; rather, it provides a useful operational route to familiar formulas.

The paper is organized as follows. Section 2 records the Laplace-transform facts used throughout, with conditions stated explicitly and with references to standard sources. Section 3 applies the transform method to Bessel’s equation and the modified Bessel equation, derives the corresponding transform-domain equations, and discusses the regular and singular solutions. Section 4 presents illustrative consequences, including recurrence and derivative identities and a few classical transform evaluations. The final section summarizes the expository role and limitations of the method.

2. Preliminaries

In this section, we collect the Laplace-transform facts needed in the sequel. The statements are standard; see, for example, Widder and Doetsch [16,17]. We record them here only to fix notation and hypotheses.

Throughout, f is assumed to be piecewise continuous on every finite interval of $[0,\infty )$. When we say that f is of exponential order a, we mean that there exist constants $M>0$ and $x_0\ge 0$ such that

$$\left|f\left(x\right)\right|\le M{e}^{ax},x\ge x_0.$$

Under this assumption, the Laplace integral converges absolutely for $\Re \left(y\right)>a$.

Definition 1 

(Laplace transform). Let $f\left(x\right)$ be a function defined on $[0,\infty )$. Its Laplace transform is

$$\mathcal{L}\{f\left(x\right);y\}={\int }_0^{\infty }{e}^{-xy}f\left(x\right)dx,$$

(1)

whenever the integral converges.

Theorem 1 

(Linearity [16] (Ch. I)). If $\mathcal{L}\left\{f\right(x);y\}$ and $\mathcal{L}\left\{g\right(x);y\}$ exist, then for any constants $\alpha ,\beta \in \mathbb{C}$,

$$\mathcal{L}\left\{\alpha f\right(x)+\beta g(x);y\}=\alpha \mathcal{L}\left\{f\right(x);y\}+\beta \mathcal{L}\left\{g\right(x);y\}.$$

(2)

Theorem 2 

(Derivative formulas [16] (Ch. II)). Assume that $f\in C^n\left([0,\infty )\right)$, that each derivative $f,f^{\prime },\dots ,f^{\left(n\right)}$ is piecewise continuous on every finite interval, and that each of these functions is of exponential order. Then, for $\Re \left(y\right)$ sufficiently large,

$$\mathcal{L}\{f^{\left(n\right)}\left(x\right);y\}=y^n\mathcal{L}\{f\left(x\right);y\}-\sum _{k=0}^{n-1}{y}^{n-1-k}{f}^{\left(k\right)}\left(0\right).$$

(3)

In particular,

$$\begin{array}{c}\mathcal{L}\{f^{\prime }\left(x\right);y\}=y\mathcal{L}\left\{f\right(x);y\}-f\left(0\right),\end{array}$$

(4)

$$\begin{array}{c}\mathcal{L}\{f^{″}\left(x\right);y\}=y^2\mathcal{L}\{f\left(x\right);y\}-yf\left(0\right)-f^{\prime }\left(0\right).\end{array}$$

(5)

Theorem 3 

(Differentiation with respect to the transform variable [16] (Ch. II)). Let $F\left(y\right)=\mathcal{L}\left\{f\right(x);y\}$. Assume that $x^{n}f\left(x\right)$ is Laplace transformable for the relevant n. Then

$$\mathcal{L}\{x^{n}f\left(x\right);y\}={(-1)}^n{\displaystyle \frac{d^n}{d{y}^n}}F\left(y\right),n=0,1,2,\dots .$$

(6)

In particular,

$$\begin{array}{cc}\hfill \mathcal{L}\left\{xf\right(x);y\}& =-{\displaystyle \frac{d}{dy}}F\left(y\right),\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill \mathcal{L}\{x^{2}f\left(x\right);y\}& ={\displaystyle \frac{d^2}{d{y}^2}}F\left(y\right).\hfill \end{array}$$

(8)

As immediate consequences of Theorems 2 and 3, if $U\left(y\right)=\mathcal{L}\left\{u\right(x);y\}$, then

$$\begin{array}{cc}\hfill \mathcal{L}\{x{u}^{\prime }\left(x\right);y\}& =-{\displaystyle \frac{d}{dy}}\left(yU\left(y\right)-u\left(0\right)\right)=-\left(U\left(y\right)+y{U}^{\prime }\left(y\right)\right),\hfill \\ \hfill \mathcal{L}\{x^2{u}^{″}\left(x\right);y\}& ={\displaystyle \frac{d^2}{d{y}^2}}\left(y^{2}U\left(y\right)-yu\left(0\right)-u^{\prime }\left(0\right)\right)\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill & =2U\left(y\right)+4y{U}^{\prime }\left(y\right)+y^2{U}^{″}\left(y\right).\hfill \end{array}$$

(10)

The following elementary solution bases for the transformed equations are used only to identify the branches selected by the Laplace transform. Similar changes of variables are standard in the classical treatment of Bessel-type equations; see, for example, Watson and the handbook accounts [1,2,3].

Proposition 1 

(Transform-domain solution bases). Let $\nu \in \mathbb{C}$, $\nu \ne 0$. On a simply connected domain where the square roots and powers are defined, a fundamental set of solutions of

$$(1+y^2)U^{″}\left(y\right)+3y{U}^{\prime }\left(y\right)+(1-{\nu }^2)U\left(y\right)=0$$

(11)

is

$${\displaystyle \frac{{\left(y+\sqrt{1+y^2}\right)}^{\nu }}{\sqrt{1+y^2}}},{\displaystyle \frac{{\left(y+\sqrt{1+y^2}\right)}^{-\nu }}{\sqrt{1+y^2}}}.$$

(12)

Similarly, a fundamental set of solutions of

$$(y^2-1)U^{″}\left(y\right)+3y{U}^{\prime }\left(y\right)+(1-{\nu }^2)U\left(y\right)=0$$

(13)

is

$${\displaystyle \frac{{\left(y+\sqrt{y^2-1}\right)}^{\nu }}{\sqrt{y^2-1}}},{\displaystyle \frac{{\left(y+\sqrt{y^2-1}\right)}^{-\nu }}{\sqrt{y^2-1}}}.$$

(14)

Proof. 

For (11), put $y=\sinh t$ and write $V\left(t\right)=\cosh tU\left(\sinh t\right)$. A direct calculation gives

$$(1+y^2)U^{″}\left(y\right)+3y{U}^{\prime }\left(y\right)+U\left(y\right)={\displaystyle \frac{1}{\cosh t}}{V}^{″}\left(t\right).$$

Hence (11) is equivalent to

$$V^{″}\left(t\right)-{\nu }^{2}V\left(t\right)=0.$$

Thus $V\left(t\right)=A{e}^{\nu t}+B{e}^{-\nu t}$. Since $e^t=y+\sqrt{1+y^2}$ and $\cosh t=\sqrt{1+y^2}$, this gives (12).

For (13), put $y=\cosh t$ and $V\left(t\right)=\sinh tU\left(\cosh t\right)$. Then

$$(y^2-1)U^{″}\left(y\right)+3y{U}^{\prime }\left(y\right)+U\left(y\right)={\displaystyle \frac{1}{\sinh t}}{V}^{″}\left(t\right),$$

so the transformed equation is again $V^{″}\left(t\right)-{\nu }^{2}V\left(t\right)=0$. Since $e^t=y+\sqrt{y^2-1}$ and $\sinh t=\sqrt{y^2-1}$, the basis (14) follows. When $\nu =0$, the second independent solution is obtained by the usual limiting process, equivalently by differentiating the displayed power solution with respect to $\nu$ and then setting $\nu =0$. □

We shall also use the Gamma function and the classical transform of a power.

Definition 2 

(Gamma function). For $\Re \left(\nu \right)>0$, the Gamma function is defined by

$$\Gamma \left(\nu \right)={\int }_0^{\infty }{x}^{\nu -1}{e}^{-x}dx.$$

(15)

$$\mathcal{L}\{x^{\nu -1};y\}={\displaystyle \frac{\Gamma \left(\nu \right)}{y^{\nu }}},\Re \left(\nu \right)>0,\Re \left(y\right)>0,$$

(16)

and the duplication formula

$$\Gamma (2\nu +1)={\displaystyle \frac{2^{2\nu }}{\sqrt{\pi }}}\Gamma \left(\nu +\frac{1}{2}\right)\Gamma (\nu +1)$$

(17)

will be used to simplify constants.

3. Main Results

We begin with Bessel’s differential equation and show how the Laplace transform leads to the classical transform of its regular solution. Since the relevant formulas are classical, the point here is expository: the transformed differential equation is derived directly, and the resulting transform is then identified with the known transform of $J_{\nu }$.

Definition 3 

(Bessel function of the first kind). Let $\nu \in \mathbb{C}$. The Bessel function of the first kind of order ν, denoted by $J_{\nu }\left(x\right)$, is the regular solution of

$$x^2{u}^{″}\left(x\right)+x{u}^{\prime }\left(x\right)+\left(x^2-{\nu }^2\right)u\left(x\right)=0,$$

(18)

normalized by

$$J_{\nu }\left(x\right)\sim {\displaystyle \frac{1}{\Gamma (\nu +1)}}{\left({\displaystyle \frac{x}{2}}\right)}^{\nu }(x\to 0),$$

for $\Re \left(\nu \right)>-1$. Equivalently,

$$J_{\nu }\left(x\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k}{k!\Gamma (k+\nu +1)}}{\left({\displaystyle \frac{x}{2}}\right)}^{2k+\nu },$$

(19)

which converges for every $x\in \mathbb{C}$; see [1,2,3].

The next statement is the classical Laplace-transform formula for the regular Bessel solution, derived here from the transformed differential equation; the transform pair itself is tabulated in the standard references [1,2,3] (§13.31).

Theorem 4 

(Classical Laplace transform of $J_{\nu }$). Let u be a regular solution of (18) with $\Re \left(\nu \right)>-1$, and let $U\left(y\right)=\mathcal{L}\left\{u\right(x);y\}$. Then U satisfies

$$(1+y^2)U^{″}\left(y\right)+3y{U}^{\prime }\left(y\right)+(1-{\nu }^2)U\left(y\right)=0.$$

(20)

For the regular branch,

$$\mathcal{L}\{J_{\nu }\left(x\right);y\}={\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{\nu }}{\sqrt{1+y^2}}},\Re \left(y\right)>0.$$

(21)

Proof. 

Let $U\left(y\right)=\mathcal{L}\left\{u\right(x);y\}$. Applying the Laplace transform to (18) and using (8)–(10), we obtain

$$\mathcal{L}\{x^2{u}^{″}\left(x\right);y\}+\mathcal{L}\{x{u}^{\prime }\left(x\right);y\}+\mathcal{L}\{x^{2}u\left(x\right);y\}-{\nu }^{2}U\left(y\right)=0.$$

Hence

$$\left(2U+4y{U}^{\prime }+y^2{U}^{″}\right)-\left(U+y{U}^{\prime }\right)+U^{″}-{\nu }^{2}U=0,$$

which simplifies to (20).

By Proposition 1, a basis of solutions of (20) is

$$U\left(y\right)={\displaystyle \frac{A{\left(y+\sqrt{1+y^2}\right)}^{\nu }+B{\left(\sqrt{1+y^2}-y\right)}^{\nu }}{\sqrt{1+y^2}}}.$$

Since

$$\sqrt{1+y^2}-y={\displaystyle \frac{1}{y+\sqrt{1+y^2}}},$$

the first term grows like $y^{\nu -1}$ as $y\to \infty$, whereas the second term decays like $y^{-\nu -1}$. For the Laplace transform of the regular solution, the large-y behavior must agree with the small-x normalization

$$u\left(x\right)\sim {\displaystyle \frac{1}{\Gamma (\nu +1)}}{\left({\displaystyle \frac{x}{2}}\right)}^{\nu },x\to 0,$$

which implies

$$U\left(y\right)\sim {\displaystyle \frac{1}{2^{\nu }{y}^{\nu +1}}}(y\to \infty ).$$

Thus the growing branch must be absent, so $A=0$. Matching the remaining leading term gives $B=1$. Therefore

$$U\left(y\right)={\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{\nu }}{\sqrt{1+y^2}}}.$$

Formula (21) is the classical Laplace transform of $J_{\nu }$; see [1] (§13.31) and the standard tables discussed in [2,3]. Hence the inverse transform is $J_{\nu }\left(x\right)$, i.e., the regular solution is identified with the Bessel function of the first kind. □

Remark 1. 

The preceding argument derives the transform-domain differential equation and selects the regular branch from its asymptotic behavior. The actual inversion is classical and should be viewed here as an identification step rather than as a claim of a new transform pair.

The next examples are retained only as illustrations of how the transform Formula (21) may be used in practice. The first is a classical Laplace transform involving $x^{\nu }{J}_{\nu }\left(ax\right)$, closely related to the Bessel-transform entries collected in Watson and standard tables [1,2,3].

Example 1 

(Classical transform of $x^{\nu }{J}_{\nu }\left(ax\right)$). Let $a>0$. Then

$$\mathcal{L}\{x^{\nu }{J}_{\nu }\left(ax\right);y\}={\displaystyle \frac{2^{\nu }\Gamma \left(\nu +\frac{1}{2}\right)a^{\nu }}{\sqrt{\pi }{(y^2+a^2)}^{\nu +\frac{1}{2}}}},\Re \left(y\right)>0.$$

(22)

Proof. 

Let

$$f\left(x\right)=x^{\nu }{J}_{\nu }\left(ax\right),F\left(y\right)=\mathcal{L}\{f\left(x\right);y\}.$$

Since $u\left(x\right)=J_{\nu }\left(ax\right)$ satisfies

$$x^2{u}^{″}\left(x\right)+x{u}^{\prime }\left(x\right)+(a^2{x}^2-{\nu }^2)u\left(x\right)=0,$$

(23)

it follows, after writing $u\left(x\right)=x^{-\nu }f\left(x\right)$, that f satisfies

$$x{f}^{″}\left(x\right)+(1-2\nu )f^{\prime }\left(x\right)+a^{2}xf\left(x\right)=0.$$

(24)

Applying the Laplace transform to (24), and using

$$\mathcal{L}\{f^{\prime }\left(x\right);y\}=yF\left(y\right)-f\left(0\right),\mathcal{L}\{xf\left(x\right);y\}=-F^{\prime }\left(y\right),$$

we obtain

$$(y^2+a^2)F^{\prime }\left(y\right)+(2\nu +1)yF\left(y\right)=2\nu f\left(0\right).$$

Since $f\left(0\right)=0$ for $\Re \left(\nu \right)>0$, we get

$$(y^2+a^2)F^{\prime }\left(y\right)+(2\nu +1)yF\left(y\right)=0.$$

Thus

$$F\left(y\right)=C{(y^2+a^2)}^{-\nu -{\displaystyle \frac{1}{2}}},$$

(25)

where C is constant. Matching the large-y asymptotics determined by the small-x behavior of $x^{\nu }{J}_{\nu }\left(ax\right)$ gives

$$C={\displaystyle \frac{a^{\nu }\Gamma (2\nu +1)}{2^{\nu }\Gamma (\nu +1)}}={\displaystyle \frac{2^{\nu }\Gamma \left(\nu +\frac{1}{2}\right)a^{\nu }}{\sqrt{\pi }}},$$

by (17). Substituting into (25) proves (22). □

Example 2 

(Second linearly independent solution of Bessel’s equation). A second linearly independent solution of Bessel’s equation is given, up to normalization, by the standard reduction-of-order formula for Bessel’s equation [1,2]

$$Y_{\nu }\left(x\right)={\displaystyle \frac{2}{\pi }}{J}_{\nu }\left(x\right){\int }^x{\displaystyle \frac{1}{t{J}_{\nu }{\left(t\right)}^2}}dt.$$

(26)

Proof. 

Dividing (18) by $x^2$, we obtain

$$y^{″}+{\displaystyle \frac{1}{x}}{y}^{\prime }+\left(1-{\displaystyle \frac{{\nu }^2}{x^2}}\right)y=0.$$

(27)

From Theorem 4, one nontrivial solution is

$$y_1\left(x\right)=J_{\nu }\left(x\right).$$

For an equation of the form

$$y^{″}+P\left(x\right)y^{\prime }+Q\left(x\right)y=0,$$

a second solution is

$$y_2\left(x\right)=y_1\left(x\right){\int }^x{\displaystyle \frac{e^{-\int P\left(t\right)dt}}{y_1{\left(t\right)}^2}}dt.$$

Here $P\left(x\right)=1/x$, so $e^{-\int P\left(t\right)dt}=1/t$, and therefore

$$y_2\left(x\right)=J_{\nu }\left(x\right){\int }^x{\displaystyle \frac{1}{t{J}_{\nu }{\left(t\right)}^2}}dt.$$

If $W(f,g)=f{g}^{\prime }-f^{\prime }g$ denotes the Wronskian, then the standard normalization $W(J_{\nu },Y_{\nu })\left(x\right)=2/\left(\pi x\right)$ yields (26). □

Remark 2. 

The standard Laplace-transform framework is naturally adapted to the regular solution at the origin. For this reason, it yields $J_{\nu }\left(x\right)$ directly but not the singular solution $Y_{\nu }\left(x\right)$, which must be introduced by reduction of order or by the classical connection formula below.

Theorem 5 

(Connection formula). Let $\nu \notin \mathbb{Z}$ and $-1<\Re \left(\nu \right)<1$. Then, for $x>0$,

$$Y_{\nu }\left(x\right)={\displaystyle \frac{\cos \left(\pi \nu \right)J_{\nu }\left(x\right)-J_{-\nu }\left(x\right)}{\sin \left(\pi \nu \right)}}.$$

(28)

Proof. 

The formula is classical; see [1] (§3.63) or [2] (§10.4). It is included here only to complete the discussion of the singular companion solution. □

Corollary 1 

(Laplace transform of the singular companion solution). Let $\nu \notin \mathbb{Z}$ and $-1<\Re \left(\nu \right)<1$. Then, for $\Re \left(y\right)>0$,

$$\begin{array}{cc}\hfill \mathcal{L}\{Y_{\nu }\left(x\right);y\}& ={\displaystyle \frac{1}{\sin \left(\pi \nu \right)}}\left[\cos \left(\pi \nu \right){\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{\nu }}{\sqrt{1+y^2}}}-{\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{-\nu }}{\sqrt{1+y^2}}}\right].\hfill \end{array}$$

(29)

Moreover, this transform satisfies the same transform-domain Equation (20).

Proof. 

Under the stated restriction, both $J_{\nu }\left(x\right)$ and $J_{-\nu }\left(x\right)$ are locally integrable at the origin, and the factor $e^{-xy}$ gives convergence at infinity for $\Re \left(y\right)>0$. Applying the Laplace transform to the connection Formula (28) gives (29). Since (20) contains ${\nu }^2$, the transforms of $J_{\nu }$ and $J_{-\nu }$ both satisfy it. Their linear combination in (29) therefore also satisfies (20). This clarifies that the singular solution can be represented in the transform-domain equation, although it is not selected by the regularity condition used in Theorem 4. □

We now turn to the modified Bessel equation.

Definition 4 

(Modified Bessel functions). The modified Bessel functions of order ν are associated with

$$x^2{u}^{″}\left(x\right)+x{u}^{\prime }\left(x\right)-\left(x^2+{\nu }^2\right)u\left(x\right)=0,x>0.$$

(30)

The regular solution is the modified Bessel function of the first kind, $I_{\nu }\left(x\right)$, characterized by

$$I_{\nu }\left(x\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{1}{k!\Gamma (k+\nu +1)}}{\left({\displaystyle \frac{x}{2}}\right)}^{2k+\nu },$$

(31)

and by the local behavior

$$I_{\nu }\left(x\right)\sim {\displaystyle \frac{1}{\Gamma (\nu +1)}}{\left({\displaystyle \frac{x}{2}}\right)}^{\nu }(x\to 0),$$

for $\Re \left(\nu \right)>-1$. A second linearly independent solution is $K_{\nu }\left(x\right)$; for $\nu \notin \mathbb{Z}$,

$$K_{\nu }\left(x\right)={\displaystyle \frac{\pi }{2}}{\displaystyle \frac{I_{-\nu }\left(x\right)-I_{\nu }\left(x\right)}{\sin \left(\pi \nu \right)}}.$$

(32)

See [1,2,3].

The corresponding formula for the regular modified Bessel solution is also classical and is included here in the same operational form; compare [1,2,3].

Theorem 6 

(Classical Laplace transform of $I_{\nu }$). Let u be a regular solution of (30) with $\Re \left(\nu \right)>-1$, and let $U\left(y\right)=\mathcal{L}\left\{u\right(x);y\}$. Then U satisfies

$$(y^2-1)U^{″}\left(y\right)+3y{U}^{\prime }\left(y\right)+(1-{\nu }^2)U\left(y\right)=0.$$

(33)

For the regular branch,

$$\mathcal{L}\{I_{\nu }\left(x\right);y\}={\displaystyle \frac{{\left(y-\sqrt{y^2-1}\right)}^{\nu }}{\sqrt{y^2-1}}},\Re \left(y\right)>1.$$

(34)

Proof. 

Applying the Laplace transform to (30) and using the same operational rules as before gives

$$\mathcal{L}\{x^2{u}^{″}\left(x\right);y\}+\mathcal{L}\{x{u}^{\prime }\left(x\right);y\}-\mathcal{L}\{x^{2}u\left(x\right);y\}-{\nu }^{2}U\left(y\right)=0.$$

Hence

$$\left(2U+4y{U}^{\prime }+y^2{U}^{″}\right)-\left(U+y{U}^{\prime }\right)-U^{″}-{\nu }^{2}U=0,$$

which reduces to (33).

By Proposition 1, a basis of solutions of (33) is

$$U\left(y\right)={\displaystyle \frac{A{\left(y+\sqrt{y^2-1}\right)}^{\nu }+B{\left(y+\sqrt{y^2-1}\right)}^{-\nu }}{\sqrt{y^2-1}}}.$$

For $\Re \left(y\right)\to +\infty$, the first term behaves like $y^{\nu -1}$, whereas the second behaves like $y^{-\nu -1}$. Since the regular solution satisfies the same local normalization at the origin as $J_{\nu }$, namely

$$u\left(x\right)\sim {\displaystyle \frac{1}{\Gamma (\nu +1)}}{\left({\displaystyle \frac{x}{2}}\right)}^{\nu },x\to 0,$$

its Laplace transform must again behave like $2^{-\nu }{y}^{-\nu -1}$ for large y. Thus $A=0$, and matching the leading term gives $B=1$. Therefore

$$U\left(y\right)={\displaystyle \frac{{\left(y+\sqrt{y^2-1}\right)}^{-\nu }}{\sqrt{y^2-1}}}={\displaystyle \frac{{\left(y-\sqrt{y^2-1}\right)}^{\nu }}{\sqrt{y^2-1}}},$$

since

$$\left(y+\sqrt{y^2-1}\right)\left(y-\sqrt{y^2-1}\right)=1.$$

This is the classical Laplace transform of $I_{\nu }\left(x\right)$; see the standard tables and handbook formulas in [1,2,3]. Hence the regular solution is $I_{\nu }\left(x\right)$. □

The following consequence uses the classical connection formula for $K_{\nu }$; see Watson and the DLMF account of modified Bessel functions [1,2].

Corollary 2 

(Laplace transform of $K_{\nu }$). Let $\nu \notin \mathbb{Z}$ and $-1<\Re \left(\nu \right)<1$. Then, for $\Re \left(y\right)>1$,

$$\begin{array}{cc}\hfill \mathcal{L}\{K_{\nu }\left(x\right);y\}& ={\displaystyle \frac{\pi }{2\sin \left(\pi \nu \right)}}\left[{\displaystyle \frac{{\left(y-\sqrt{y^2-1}\right)}^{-\nu }}{\sqrt{y^2-1}}}-{\displaystyle \frac{{\left(y-\sqrt{y^2-1}\right)}^{\nu }}{\sqrt{y^2-1}}}\right].\hfill \end{array}$$

(35)

Moreover, this transform satisfies the modified transform-domain Equation (33).

Proof. 

For $-1<\Re \left(\nu \right)<1$, both $I_{\nu }\left(x\right)$ and $I_{-\nu }\left(x\right)$ are locally integrable at the origin, and their Laplace transforms converge for $\Re \left(y\right)>1$. Applying the Laplace transform to (32) gives (35). Since the transformed equation depends on ${\nu }^2$, the transforms of $I_{\nu }$ and $I_{-\nu }$ both satisfy (33); hence their linear combination does as well. □

Remark 3. 

As in the ordinary Bessel case, the Laplace transform naturally selects the solution regular at the origin. The companion solution $K_{\nu }$ is therefore introduced through the standard connection Formula (32), rather than obtained directly from the regular Laplace-transform calculation.

Example 3 

(Complete Lipschitz–Hankel integral). For $a>0$,

$$\mathcal{L}\{u^{\nu /2}{J}_{\nu }\left(a{u}^{1/2}\right);y\}={\displaystyle \frac{a^{\nu }}{2^{\nu }{y}^{\nu +1}}}{e}^{-a^2/\left(4y\right)},\Re \left(y\right)>0.$$

(36)

Proof. 

Using the definition of the Laplace transform, we have

$$\mathcal{L}\{u^{\nu /2}{J}_{\nu }\left(a{u}^{1/2}\right);y\}={\int }_0^{\infty }{e}^{-yu}{u}^{\nu /2}{J}_{\nu }\left(a{u}^{1/2}\right)du.$$

Set $u=x^2$. Then $du=2xdx$, and therefore

$$u^{\nu /2}{J}_{\nu }\left(a{u}^{1/2}\right)du=x^{\nu }{J}_{\nu }\left(ax\right)\left(2xdx\right)=2x^{\nu +1}{J}_{\nu }\left(ax\right)dx.$$

Hence

$$\mathcal{L}\{u^{\nu /2}{J}_{\nu }\left(a{u}^{1/2}\right);y\}=2{\int }_0^{\infty }{e}^{-y{x}^2}{x}^{\nu +1}{J}_{\nu }\left(ax\right)dx.$$

This is a complete Lipschitz–Hankel integral; see [4,8]. For completeness, we derive the evaluation used here. Expanding $J_{\nu }$ by (19) and integrating term by term gives

$$\begin{array}{cc}\hfill {\int }_0^{\infty }{e}^{-y{x}^2}{x}^{\nu +1}{J}_{\nu }\left(ax\right)dx& =\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k{(a/2)}^{2k+\nu }}{k!\Gamma (k+\nu +1)}}{\int }_0^{\infty }{e}^{-y{x}^2}{x}^{2k+2\nu +1}dx\hfill \\ \hfill & =\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k{(a/2)}^{2k+\nu }}{k!\Gamma (k+\nu +1)}}{\displaystyle \frac{\Gamma (k+\nu +1)}{2y^{k+\nu +1}}}\hfill \\ \hfill & ={\displaystyle \frac{a^{\nu }}{{\left(2y\right)}^{\nu +1}}}\sum _{k=0}^{\infty }{\displaystyle \frac{1}{k!}}{\left(-{\displaystyle \frac{a^2}{4y}}\right)}^k\hfill \\ \hfill & ={\displaystyle \frac{a^{\nu }}{{\left(2y\right)}^{\nu +1}}}{e}^{-a^2/\left(4y\right)},\Re \left(y\right)>0.\hfill \end{array}$$

Substitution into the preceding expression proves (36). □

4. Operational Identities for Bessel and Modified Bessel Functions via the Laplace Transform

The Laplace transform representations obtained in Section 3 provide a compact operational framework for recovering several classical recurrence and derivative identities. The results in this section are included as illustrative consequences of the transform formulas rather than as new identities; the identities themselves are standard in the classical theory of Bessel functions [1,2,3]. We begin with the Bessel function of the first kind.

Theorem 7 

(Classical derivative identities for $J_{\nu }$ via the Laplace transform). Let $\Re \left(\nu \right)>-1$. Then

$$\begin{array}{c}{J}_{\nu }^{\prime }\left(x\right)={\displaystyle \frac{1}{2}}\left(J_{\nu -1}\left(x\right)-J_{\nu +1}\left(x\right)\right),\end{array}$$

(37)

$$\begin{array}{c}x{J}_{\nu }^{\prime }\left(x\right)=\nu J_{\nu }\left(x\right)-x{J}_{\nu +1}\left(x\right).\end{array}$$

(38)

Proof. 

Using (21), we have

$$\mathcal{L}\{J_{\nu }\left(x\right);y\}={\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{\nu }}{\sqrt{1+y^2}}},\Re \left(y\right)>0.$$

Applying (4) and noting that $J_{\nu }\left(0\right)=0$ for $\Re \left(\nu \right)>0$, we obtain

$$\mathcal{L}\{J_{\nu }^{\prime }\left(x\right);y\}=y\mathcal{L}\{J_{\nu }\left(x\right);y\}.$$

Now

$$\begin{array}{cc}\hfill y\mathcal{L}\{J_{\nu }\left(x\right);y\}& =y{\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{\nu }}{\sqrt{1+y^2}}}\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{\nu -1}}{\sqrt{1+y^2}}}-{\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{\nu +1}}{\sqrt{1+y^2}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left(\mathcal{L}\{J_{\nu -1}\left(x\right);y\}-\mathcal{L}\{J_{\nu +1}\left(x\right);y\}\right).\hfill \end{array}$$

Taking inverse Laplace transforms proves (37).

The second identity is proved independently in the transform domain, avoiding any use of the recurrence relation. Since multiplication by x corresponds to $-d/dy$, it is enough to verify

$$-{\displaystyle \frac{d}{dy}}\left(y\mathcal{L}\{J_{\nu }\left(x\right);y\}\right)=\nu \mathcal{L}\{J_{\nu }\left(x\right);y\}+{\displaystyle \frac{d}{dy}}\mathcal{L}\{J_{\nu +1}\left(x\right);y\}.$$

Substituting (21) into the two sides gives the same expression,

$${\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{\nu }}{{(1+y^2)}^{3/2}}}\left(\nu y\sqrt{1+y^2}-1\right).$$

Taking inverse Laplace transforms gives (38). The endpoint cases are obtained by the usual limiting argument in the order parameter. □

The following half-order formulas are standard elementary special cases of Bessel functions [1,2,3].

Example 4 

(Half-order Bessel functions). Using (21), verify that

$$\begin{array}{c}{J}_{1/2}\left(x\right)=\sqrt{{\displaystyle \frac{2}{\pi x}}}\sin x,\end{array}$$

(39)

$$\begin{array}{c}{J}_{-1/2}\left(x\right)=\sqrt{{\displaystyle \frac{2}{\pi x}}}\cos x.\end{array}$$

(40)

Proof. 

Setting $\nu =\frac{1}{2}$ in (21), we obtain

$$\mathcal{L}\{J_{1/2}\left(x\right);y\}={\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{1/2}}{\sqrt{1+y^2}}}.$$

On the other hand, the previously established Laplace transform formula for $\sqrt{2/\left(\pi x\right)}\sin x$ gives

$$\mathcal{L}\left\{\sqrt{{\displaystyle \frac{2}{\pi x}}}\sin x;y\right\}={\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{1/2}}{\sqrt{1+y^2}}}.$$

Hence, by uniqueness of the Laplace transform, (39) follows.

The verification of (40) is similar and is left as an exercise. □

Theorem 8 

(Classical recurrence relation for $J_{\nu }$). Let $\nu \in \mathbb{C}$ with $\Re \left(\nu \right)>-1$. Then, for $x>0$,

$$x{J}_{\nu +1}\left(x\right)=2\nu J_{\nu }\left(x\right)-x{J}_{\nu -1}\left(x\right).$$

(41)

Proof. 

Applying the Laplace transform to (41), it suffices to verify

$$-{\displaystyle \frac{d}{dy}}\mathcal{L}\{J_{\nu +1}\left(x\right);y\}=2\nu \mathcal{L}\{J_{\nu }\left(x\right);y\}+{\displaystyle \frac{d}{dy}}\mathcal{L}\{J_{\nu -1}\left(x\right);y\}.$$

(42)

Using (21), we write

$$\mathcal{L}\{J_{\mu }\left(x\right);y\}={\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{\mu }}{\sqrt{1+y^2}}},\Re \left(y\right)>0.$$

Differentiating with respect to y, we obtain

$${\displaystyle \frac{d}{dy}}\mathcal{L}\{J_{\mu }\left(x\right);y\}=-\mu {\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{\mu }}{1+y^2}}-y{\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{\mu }}{{(1+y^2)}^{3/2}}}.$$

(43)

Applying (43) with $\mu =\nu +1$ and $\mu =\nu -1$, and using (21) for $\mu =\nu$, we reduce (42) to a straightforward algebraic identity in y. Hence (42) holds, and uniqueness of the Laplace transform yields (41). □

Example 5 

(Three-half-order Bessel functions). Using (39)–(41), verify that

$$\begin{array}{c}{J}_{3/2}\left(x\right)=\sqrt{{\displaystyle \frac{2}{\pi x}}}\left({\displaystyle \frac{\sin x}{x}}-\cos x\right),\end{array}$$

(44)

$$\begin{array}{c}{J}_{-3/2}\left(x\right)=\sqrt{{\displaystyle \frac{2}{\pi x}}}\left(-{\displaystyle \frac{\cos x}{x}}-\sin x\right).\end{array}$$

(45)

Proof. 

Setting $\nu =\frac{1}{2}$ in (41), we obtain

$$x{J}_{3/2}\left(x\right)=J_{1/2}\left(x\right)-x{J}_{-1/2}\left(x\right).$$

Substituting (39) and (40) gives

$$x{J}_{3/2}\left(x\right)=\sqrt{{\displaystyle \frac{2}{\pi x}}}\sin x-x\sqrt{{\displaystyle \frac{2}{\pi x}}}\cos x,$$

and therefore

$$J_{3/2}\left(x\right)=\sqrt{{\displaystyle \frac{2}{\pi x}}}\left({\displaystyle \frac{\sin x}{x}}-\cos x\right).$$

This proves (44).

The verification of (45) is similar and is left as an exercise. □

This last example illustrates a standard type of Bessel-series summation; such series identities are closely related to the classical generating-function approach to Bessel functions [1,2].

Example 6 

(Alternating odd-order Bessel sum). For all $x\in \mathbb{R}$,

$$J_1\left(x\right)-J_3\left(x\right)+J_5\left(x\right)-J_7\left(x\right)+\dots ={\displaystyle \frac{\sin x}{2}}.$$

(46)

Proof. 

Let

$$S\left(x\right)=\sum _{k=0}^{\infty }{(-1)}^k{J}_{2k+1}\left(x\right).$$

Using (21), we obtain

$$\mathcal{L}\{J_n\left(x\right);y\}={\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^n}{\sqrt{1+y^2}}},\Re \left(y\right)>0.$$

Therefore,

$$\begin{array}{cc}\hfill \mathcal{L}\left\{S\right(x);y\}& =\sum _{k=0}^{\infty }{(-1)}^k{\displaystyle \frac{{\left(\sqrt{1+y^2}-y\right)}^{2k+1}}{\sqrt{1+y^2}}}\hfill \\ \hfill & ={\displaystyle \frac{\sqrt{1+y^2}-y}{\sqrt{1+y^2}}}\sum _{k=0}^{\infty }{\left[-{\left(\sqrt{1+y^2}-y\right)}^2\right]}^k.\hfill \end{array}$$

Summing the geometric series gives

$$\mathcal{L}\{S\left(x\right);y\}={\displaystyle \frac{\sqrt{1+y^2}-y}{\sqrt{1+y^2}}}·{\displaystyle \frac{1}{1+{\left(\sqrt{1+y^2}-y\right)}^2}}.$$

Since

$$1+{\left(\sqrt{1+y^2}-y\right)}^2=2\sqrt{1+y^2}(\sqrt{1+y^2}-y),$$

it follows that

$$\mathcal{L}\{S\left(x\right);y\}={\displaystyle \frac{1}{2(1+y^2)}}.$$

But

$$\mathcal{L}\left\{{\displaystyle \frac{\sin x}{2}};y\right\}={\displaystyle \frac{1}{2(1+y^2)}}.$$

Hence, by uniqueness of the Laplace transform,

$$S\left(x\right)={\displaystyle \frac{\sin x}{2}},$$

which proves (46). □

We now record the analogous operational identities for modified Bessel functions.

Theorem 9 

(Classical derivative identities for $I_{\nu }$ via the Laplace transform). Let $\Re \left(\nu \right)>-1$. Then

$$\begin{array}{c}{I}_{\nu }^{\prime }\left(x\right)={\displaystyle \frac{1}{2}}\left(I_{\nu -1}\left(x\right)+I_{\nu +1}\left(x\right)\right),\end{array}$$

(47)

$$\begin{array}{c}x{I}_{\nu }^{\prime }\left(x\right)=\nu I_{\nu }\left(x\right)+x{I}_{\nu +1}\left(x\right).\end{array}$$

(48)

Proof. 

Using (34), we have

$$\mathcal{L}\{I_{\nu }\left(x\right);y\}={\displaystyle \frac{{\left(y-\sqrt{y^2-1}\right)}^{\nu }}{\sqrt{y^2-1}}},\Re \left(y\right)>1.$$

Applying (4) and simplifying, we obtain

$$y\mathcal{L}\{I_{\nu }\left(x\right);y\}={\displaystyle \frac{1}{2}}\left(\mathcal{L}\{I_{\nu -1}\left(x\right);y\}+\mathcal{L}\{I_{\nu +1}\left(x\right);y\}\right),$$

which yields (47).

The second identity is again obtained directly in the transform domain. It is enough to show that

$$-{\displaystyle \frac{d}{dy}}\left(y\mathcal{L}\{I_{\nu }\left(x\right);y\}\right)=\nu \mathcal{L}\{I_{\nu }\left(x\right);y\}-{\displaystyle \frac{d}{dy}}\mathcal{L}\{I_{\nu +1}\left(x\right);y\}.$$

Using (34), both sides reduce to

$${\displaystyle \frac{{\left(y-\sqrt{y^2-1}\right)}^{\nu }}{{(y^2-1)}^{3/2}}}\left(\nu y\sqrt{y^2-1}+1\right).$$

Taking inverse Laplace transforms gives (48). □

Theorem 10 

(Classical recurrence relation for $I_{\nu }$). Let $\nu \in \mathbb{C}$ with $\Re \left(\nu \right)>-1$. Then, for $x>0$,

$$x{I}_{\nu +1}\left(x\right)=x{I}_{\nu -1}\left(x\right)-2\nu I_{\nu }\left(x\right).$$

(49)

Proof. 

The result follows immediately by combining (47) and (48). Indeed, from (47),

$$2x{I}_{\nu }^{\prime }\left(x\right)=x{I}_{\nu -1}\left(x\right)+x{I}_{\nu +1}\left(x\right),$$

while (48) gives

$$x{I}_{\nu }^{\prime }\left(x\right)=\nu I_{\nu }\left(x\right)+x{I}_{\nu +1}\left(x\right).$$

Eliminating $x{I}_{\nu }^{\prime }\left(x\right)$ yields

$$x{I}_{\nu -1}\left(x\right)-x{I}_{\nu +1}\left(x\right)=2\nu I_{\nu }\left(x\right),$$

which is equivalent to (49). □

5. Conclusions

This paper presents a pedagogical review of how standard Laplace-transform rules may be used to analyze Bessel’s equation and the modified Bessel equation. Its main purpose is to show, in a self-contained way, how these operational methods convert the original equations into differential equations in the transform variable, and how the regular branches of the transformed equations reproduce the classical Laplace transforms of $J_{\nu }$ and $I_{\nu }$.

From that starting point, a number of familiar consequences follow naturally: explicit transform evaluations, recurrence relations, derivative identities, and several half-integer specializations. At the same time, the discussion also makes clear a limitation of the method. The ordinary Laplace transform is adapted to solutions that are regular at the origin, so the singular companion solutions $Y_{\nu }$ and $K_{\nu }$ are not obtained directly from the regular Laplace-transform calculation and must instead be introduced through classical connection formulas or by reduction of order.

The broader value of the method is therefore expository. It complements, but does not replace, the classical Frobenius and special-function approaches developed in Watson and the handbook literature [1,2,3]. It also sits naturally alongside the literature on operational-calculus, Laplace-type, and Lipschitz–Hankel integrals, including the work of Eason, Noble, and Sneddon [4], Rathie’s treatment of related special-function integrals [5], later transform formulas for modified Bessel functions [6,7], modern corrections and extensions by Maier [8], and the related ${\mathcal{L}}_2$ treatment in [9]. In this form, the paper is best read as a concise tutorial on an operational viewpoint that may be useful to readers encountering Bessel-type equations for the first time.