**Example 2.**

*(i) δ*-(*t*), *ϕ*(*t*) = − *δ*(*t*), *ϕ*-(*t*) = −*ϕ*-(0)*; (ii) δ*(*k*)(*t*), *ϕ*(*t*) = (−<sup>1</sup>)*<sup>k</sup> δ*(*t*), *ϕ*(*k*)(*t*) = (−<sup>1</sup>)*<sup>k</sup>ϕ*(*k*)(0)*.*

**Definition 4.** *Let α*(*t*) *be an infinitely differentiable function. We define the* product *of α*(*t*) *with any distribution T in* D- *by α*(*t*)*<sup>T</sup>*, *ϕ*(*t*) = *<sup>T</sup>*, *α*(*t*)*ϕ*(*t*) *for all ϕ*(*t*) ∈ D*.*

**Definition 5.** *If y*(*t*) *is a singular distribution and satisfies the equation*

$$\sum\_{m=0}^{n} a\_m(t) y^{(n)}(t) = f(t),\tag{4}$$

*where am*(*t*) *is an infinitely differentiable function and f*(*t*) *is an arbitrary known distribution, in the sense of distribution, and is called a distributional solution of* (4)*.*

**Definition 6.** *Let M* ∈ R *and f*(*t*) *be a locally integrable function satisfying the following conditions:*


*The Laplace transform of f*(*t*) *is defined by*

$$F(s) = \mathcal{L}\left\{f(t)\right\} = \int\_M^\infty f(t)e^{-st}dt,\tag{5}$$

*where s is a complex variable.*

It is well known that if *f*(*t*) is continuous, then *<sup>F</sup>*(*s*) is an analytic function on the half-plane (*s*) > *σa*, where *σa* is an abscissa of absolute convergence for L { *f*(*t*)}.

Recall that the Laplace transform *<sup>G</sup>*(*s*) of a locally integrable function *g*(*t*) satisfying the conditions of definition 6, that is,

$$G(\mathbf{s}) = \mathcal{L}\left\{\mathbf{g}(t)\right\} = \int\_{M}^{\infty} \mathbf{g}(t)e^{-st}dt,\tag{6}$$

where (*s*) > *σa*, can be written in the form *<sup>G</sup>*(*s*) = *g*(*t*),*e*<sup>−</sup>*st*.

**Definition 7.** *Let S be the space of test functions of* rapid decay *containing the complex-valued functions φ*(*t*) *having the following properties:*


$$|t^p \phi^{(k)}(t)| < \mathcal{C}\_{pk\prime}$$

*where Cpk is a constant depending on p*, *k, and φ*(*t*)*. Then φ*(*t*) *is called a test function in the space S.* **Definition 8.** *A distribution of slow growth or tempered distribution T is a continuous linear functional over the space S of test function of rapid decay and contains the complex-valued functions—i.e., there is assigned a complex number <sup>T</sup>*, *φ*(*t*) *with properties:*


*We shall let S*- *denote the set of all distributions of slow growth.*

**Definition 9.** *Let f*(*t*) *be a distribution satisfying the following properties:*


*The Laplace transform of a right-sided distribution f*(*t*) *satisfying (ii) is defined by*

$$F(s) = \mathcal{L}\left\{ f(t) \right\} = \left\langle e^{-\varepsilon t} f(t), X(t)e^{-(s-\varepsilon)t} \right\rangle,\tag{7}$$

*where X*(*t*) *is an infinitely differentiable function with support bounded on the left, which equals* 1 *over a neighbourhood of the support of f*(*t*)*.*

For (*s*) > *c*, the function *<sup>X</sup>*(*t*)*e*<sup>−</sup>(*<sup>s</sup>*−*<sup>c</sup>*)*<sup>t</sup>* is a testing function in the space *S* and *e*<sup>−</sup>*ct f*(*t*) is in the space *S*-. Equation (7) can be reduced to

$$F(s) = \mathcal{L}\left\{f(t)\right\} = \left\langle f(t), e^{-st} \right\rangle. \tag{8}$$

Now *<sup>F</sup>*(*s*) is a function of *s* defined over the right half-plane (*s*) > *c*. Zemanian [19] proved that *<sup>F</sup>*(*s*) is an analytic function in the region of convergence (*s*) > *σ*1, where *σ*1 is the abscissa of convergence and *e*<sup>−</sup>*ct f*(*t*) ∈ *S*- for some real number *c* > *σ*1.

**Example 3.** *Let δ*(*t*) *be the Dirac delta function, H*(*t*) *be the Heaviside function, and f*(*t*) *be a Laplace-transformable distribution in* D-*R. If k is a positive integer, then the following holds:*

*(i)* L{(*tk*−1*H*(*t*))/(*<sup>k</sup>* − 1)!} = 1/*sk,* (*s*) > 0*;*

$$\{\text{ii)}\quad \mathcal{L}\left\{\delta(t)\right\} = 1, \quad -\infty < \Re(s) < 1$$

$$\text{(iii)}\quad \mathcal{L}\left\{\delta^{(k)}(t)\right\} = \mathbf{s}^k, \quad -\infty < \Re(\mathbf{s}) < \infty;$$


The proof of following lemma 1 is given in [14].

**Lemma 1.** *Let ψ*(*t*) *be an infinitely differentiable function. Then*

∞*;*

$$\begin{split} \psi(t)\delta^{(m)}(t) &= (-1)^{m}\psi^{(m)}(0)\delta(t) + (-1)^{m-1}m\psi^{(m-1)}(0)\delta'(t) \\ &+ (-1)^{m-2}\frac{m(m-1)}{2!}\psi^{(m-1)}(0)\delta''(t) + \cdots + \psi(0)\delta^{(m)}(t). \end{split} \tag{9}$$

A useful formula that follows from (9), for any monomial *ψ*(*t*) = *t<sup>n</sup>*, is that

$$t^n \delta^{(m)}(t) = \begin{cases} 0, & \text{if } m < n; \\ (-1)^n \frac{m!}{(m-n)!} \delta^{(m-n)}(t), & \text{if } m \ge n. \end{cases} \tag{10}$$

#### **3. Main Results**

In this section, we will state our main results and give their proofs.

**Theorem 1.** *Consider the differential equation of the form*

$$t^2y''(t) + 2ty'(t) - [t^2 + \nu(\nu + 1)]y(t) = 0,\tag{11}$$

*where ν* ∈ N ∪ {0} *and t* ∈ R*. The distributional solutions of* (11) *are given by*

$$y(t) = P\_\mathcal{V}(D)\delta(t),\tag{12}$$

*where*

$$P\_{\boldsymbol{\nu}}(D) = \frac{1}{2^{\nu}} \sum\_{k=0}^{\lfloor \nu/2 \rfloor} (-1)^k \frac{(2\nu - 2k)!}{k!(\nu - k)!(\nu - 2k)!} D^{\boldsymbol{\nu} - 2k} \boldsymbol{\nu}$$

*is a Legendre polynomial of distributional derivative operator D* = *d*/*dt.*

**Proof.** Applying the Laplace transform to both sides of (11) with L{*y*(*t*)} = *<sup>F</sup>*(*s*), and using Example 3(iv), (v), we obtain

$$\frac{1}{2}(1-s^2)F''(s) - 2sF'(s) + \nu(\nu+1)F(s) = 0. \tag{13}$$

Suppose that a solution of (13) is of the form *<sup>F</sup>*(*s*) = ∑∞*<sup>n</sup>*=<sup>0</sup> *ans<sup>n</sup>*. Differentiating term by term, we obtain

$$F'(s) = \sum\_{n=1}^{\infty} n a\_n s^{n-1}$$

and

$$F''(s) = \sum\_{n=2}^{\infty} n(n-1)a\_n s^{n-2}.$$

Substituting these terms into (13), we have

$$\begin{aligned} \left[2a\_2 + \nu(\nu+1)a\_0\right] + \left[(3\cdot2)a\_3 - (2-\nu(\nu+1)a\_1)s \\ + \sum\_{n=2}^{\infty} \left[ (n+2)(n+1)a\_{n+2} - n(n-1)a\_{\mathbb{II}} - 2na\_{\mathbb{II}} + \nu(\nu+1)a\_{\mathbb{II}} \right] s^n = 0. \end{aligned}$$

Since *s<sup>n</sup>* = 0, it follows that

$$\begin{aligned} 2a\_2 + \nu(\nu+1)a\_0 &= 0, \quad (3\cdot 2)a\_3 - (2-\nu(\nu+1)a\_1 = 0, \\ (n+2)(n+1)a\_{n+2} - n(n-1)a\_n - 2na\_{\nu} + \nu(\nu+1)a\_{\nu} &= 0, \quad n \ge 2, \end{aligned}$$

which leads to a recurrence relation

$$a\_{n+2} = -\frac{(\nu - n)(\nu + n + 1)}{(n+1)(n+2)} a\_n. \tag{14}$$

> Thus, we obtain

$$\begin{aligned} a\_2 &= -\frac{\nu(\nu+1)}{2!} a\_0 \\ a\_4 &= (-1)^2 \frac{\nu(\nu-2)(\nu+1)(\nu+3)}{4!} a\_0 \\ &\vdots \\ a\_{2n} &= (-1)^n \frac{\nu(\nu-2)\cdots(\nu-2n+2)(\nu+1)(\nu+3)\cdots(\nu+2n-1)}{(2n)!} a\_0 \end{aligned}$$

Similarly,

$$\begin{aligned} a\_3 &= -\frac{(\nu - 1)(\nu + 2)}{2 \cdot 3} a\_1 \\ a\_5 &= (-1)^2 \frac{(\nu - 1)(\nu - 3)(\nu + 2)(\nu + 4)}{5!} a\_1 \\ &\vdots \\ a\_{2n+1} &= (-1)^n \frac{(\nu - 1)(\nu - 3) \cdots (\nu - 2n + 1)(\nu + 2)(\nu + 4) \cdots (\nu + 2n)}{(2n + 1)!} a\_1. \end{aligned}$$

Letting *a*0 = *a*1 = 1, we ge<sup>t</sup> the two solutions of (13) in the forms

$$F\_t(s) = 1 + \sum\_{n=1}^{\infty} (-1)^n \frac{\nu(\nu - 2) \cdots (\nu - 2n + 2)(\nu + 1)(\nu + 3) \cdots (\nu + 2n - 1)}{(2n)!} s^{2n}$$

and

$$F\_{\nu}(s) = s + \sum\_{n=1}^{\infty} (-1)^n \frac{(\nu - 1)(\nu - 3) \cdots (\nu - 2n + 1)(\nu + 2)(\nu + 4) \cdots (\nu + 2n)}{(2n + 1)!} s^{2n + 1}.$$

If *ν* is even, letting *ν* = 2*m*, *m* ∈ N ∪ {0}, we note that

$$\nu(\nu-2)\cdots(\nu-2n+2) = 2m(2m-2)\cdots(2m-2n+2) = \begin{cases} 0, & m=0; \\ 2^n m! & m>0, \ n \le m; \\ \frac{(m-n)!}{(m-n)!}, & n>m>0. \end{cases}$$

and

$$(\nu+1)(\nu+3)\cdots(\nu+2n-1) = (2m+1)(2m+3)\cdots(2m+2n-1) = \frac{(2m+2n)!m!}{2^n(2m)!(m+n)!}.$$

Then, in this case, *Fe*(*s*) only becomes the finite series of the form

$$F\_t(s) = \frac{(m!)^2}{(2m)!} \sum\_{k=0}^m (-1)^k \frac{(2m+2k)! s^{2k}}{(m-k)!(m+k)!(2k)!}.\tag{15}$$

If *ν* is odd, letting *ν* = 2*m* + 1, *m* ∈ N ∪ {0}, we note that

$$(\nu - 1)(\nu - 3) \cdots (\nu - 2n + 1) = 2m(2m - 2) \cdot \cdots (2m - 2n + 2) = \begin{cases} 0, & m = 0; \\ 2^n m! & m > 0, \ n \le m; \\ \frac{2^n m!}{(m - n)!}, & n > m > 0, \end{cases}$$

and

$$(\nu+2)(\nu+4)\cdots(\nu+2n) = (2m+3)(2m+5)\cdots(2m+2n+1) = \frac{(2m+2n+1)!m!}{2^n(2m+1)!(m+n)!},$$

Then, in this case, *Fo*(*s*) only becomes the finite series of the form

$$F\_o(s) = \frac{(m!)^2}{(2m+1)!} \sum\_{k=0}^m (-1)^k \frac{(2m+2k+1)! s^{2k+1}}{(m-k)!(m+k)!(2k+1)!}.\tag{16}$$

For *ν* = 0, 1, 2, . . ., we have *<sup>F</sup>ν*(*s*), as follows:

$$\begin{aligned} F\_0(s) &= 1 = P\_0(s), \\ F\_1(s) &= s = P\_1(s), \\ F\_2(s) &= 1 - 3s^2 = -2P\_2(s), \\ F\_3(s) &= s - (5/3)s^3 = -(2/3)P\_3(s), \\ F\_4(s) &= 1 - 10s^2 + (35/3)s^4 = (8/3)P\_4(s), \\ F\_5(s) &= s - (14/3)s^3 + (21/5)s^5 = (8/15)P\_5(s), \\ F\_6(s) &= 1 - 21s^2 + 63s^4 - (231/5)s^6 = -(16/5)P\_6(s), \\ F\_7(s) &= s - 9s^3 + (99/5)s^5 - (429/35)s^7 = -(16/35)P\_7(s). \end{aligned}$$

and so on, where *Pn*(*s*) is the Legendre polynomial of *s* for *n* = 0, 1, 2, .... Since (13) is linear, *Pν* is also its solution for all non-negative integer *ν*. Taking the inverse Laplace transform to *<sup>P</sup>ν*(*s*), and using Example 3(ii),(iii), we obtain the solutions of (11),

$$\begin{split} y(t) &= \frac{1}{2^{\nu}} \sum\_{k=0}^{\lfloor \nu/2 \rfloor} (-1)^{k} \frac{(2\nu - 2k)!}{k!(\nu - k)!(\nu - 2k)!} D^{\nu - 2k} \delta(t), \quad (D \equiv \frac{d}{dt} \text{ distribution derivative}) \\ &= P\_{\boldsymbol{\nu}}(D) \delta(t), \end{split} \tag{17}$$

which are the distributional solutions of the form (12).

**Example 4.** *Letting ν* = 1*,* (11) *becomes*

$$t^2y''(t) + 2ty'(t) - (t^2 + 2)y(t) = 0. \tag{18}$$

*From Theorem 1,* (18) *has a solution*

$$y(t) = \delta'(t). \tag{19}$$

*Letting ν* = 4*,* (11) *becomes*

$$t^2y''(t) + 2ty'(t) - (t^2 + 20)y(t) = 0.\tag{20}$$

*From Theorem 1,* (20) *has a solution*

$$y(t) = \frac{35}{8}\delta^{(4)}(t) - \frac{15}{4}\delta''(t) + \frac{3}{8}\delta(t). \tag{21}$$

*By applying* (10)*, it is not difficult to verify that* (19) *and* (21) *satisfy* (18) *and* (20)*, respectively.*

**Theorem 2.** *Consider the equation of the form*

$$t^2y''(t) + 3ty'(t) - (t^2 + \nu^2 - 1)y(t) = 0,\tag{22}$$

*where ν* ∈ N ∪ {0} *and t* ∈ R*. The distributional solutions of* (22) *are given by*

$$y(t) = T\_\vee(D)\delta(t),\tag{23}$$

*where*

$$T\_{\nu}(D) = \frac{\nu}{2} \sum\_{k=0}^{\lfloor \nu/2 \rfloor} (-1)^k \frac{2^{\nu-2k} (n-k-1)!}{k! (\nu-2k)!} D^{\nu-2k} \zeta$$

*is a Chebyshev polynomial of the first kind of distributional derivative operator D* = *d*/*dt.*

**Proof.** Applying the Laplace transform L{*y*(*t*)} = *<sup>F</sup>*(*s*) to (22) and using Example 3(iv), (v), we obtain

$$s\left(1-s^{2}\right)F''(s) - sF'(s) + \nu^{2}F(s) = 0.\tag{24}$$

Suppose that a solution of (24) is of the form *<sup>F</sup>*(*s*) = ∑∞*<sup>n</sup>*=<sup>0</sup> *ans<sup>n</sup>*. Differentiating *<sup>F</sup>*(*s*) term by term, we obtain

$$F'(s) = \sum\_{n=1}^{\infty} n a\_n s^{n-1}$$

and

$$F^{\prime\prime}(\mathbf{s}) = \sum\_{n=2}^{\infty} n(n-1)a\_n \mathbf{s}^{n-2}.$$

Substituting these terms into (24), we ge<sup>t</sup>

$$2a\_2 + \nu^2 a\_0 + \left(6a\_3 - a\_1 + \nu^2 a\_1\right)s + \sum\_{n=2}^{\infty} \left\{ (n+2)(n+1)a\_{n+2} - [n(n-1) + n - \nu^2]a\_n \right\} s^n = 0.5$$

Since *s<sup>n</sup>* = 0, it follows that

$$\begin{aligned} 2a\_2 + \nu^2 a\_0 &= 0, \\ 6a\_3 - a\_1 + \nu^2 a\_1 &= 0, \end{aligned}$$

and for *n* = 2, 3, . . .,

$$(n+2)(n+1)a\_{n+2} - (n^2 - \nu^2)a\_n = 0.$$

Hence,

$$a\_2 = -\frac{\nu^2}{2} a\_{0\nu}$$

$$a\_3 = \frac{1 - \nu^2}{3!} a\_{1\nu}$$

and for *n* = 2, 3, . . .,

$$a\_{n+2} = \frac{n^2 - \nu^2}{(n+2)(n+1)} a\_n. \tag{25}$$

Thus, we obtain

$$\begin{aligned} a\_2 &= -\frac{\nu^2}{2} a\_{0\prime} \\ a\_4 &= \frac{(2^2 - \nu^2)(-\nu^2)}{4!} a\_{0\prime} \end{aligned}$$

and so on. Similarly,

$$\begin{aligned} a\_3 &= \frac{1 - \nu^2}{3!} a\_1 \\ a\_5 &= \frac{(3^2 - \nu^2)(1 - \nu^2)}{5!} a\_1 \end{aligned}$$

and so on. A pattern clearly emerges:

$$a\_{2n} = \frac{[(2n-2)^2 - \nu^2][(2n-4)^2 - \nu^2] \cdot \cdots \cdot (2^2 - \nu^2)(-\nu^2)}{(2n)!} a\_0$$

and

$$a\_{2n+1} = \frac{[(2n-1)^2 - \nu^2][(2n-3)^2 - \nu^2] \cdots (3^2 - \nu^2)(1 - \nu^2)}{(2n+1)!} a\_1.$$

Hence, we ge<sup>t</sup> two solutions of (24) in the forms

$$F\_{\varepsilon}(s) = a\_0 + \sum\_{n=1}^{\infty} \frac{[(2n-2)^2 - \nu^2][(2n-4)^2 - \nu^2] \cdots (2^2 - \nu^2)(-\nu^2)}{(2n)!} a\_0 s^{2n}$$

and

$$F\_{\sigma}(s) = a\_1 s + \sum\_{n=1}^{\infty} \frac{[(2n-1)^2 - \nu^2][(2n-3)^2 - \nu^2] \cdots (3^2 - \nu^2)(1 - \nu^2)}{(2n+1)!} a\_1 s^{2n+1}.$$

If *ν* is even, letting *ν* = 2*m*, *m* ∈ N ∪ {0}, then *<sup>a</sup>*2*m*+2 = 0, so that *<sup>a</sup>*2*n*+2 = 0 for *n* ≥ *m*. Hence,

$$F\_{\ell}(s) = 1 + \sum\_{n=1}^{m} \frac{[(2n-2)^2 - (2m)^2][(2n-4)^2 - (2m)^2] \cdots [2^2 - (2m)^2][-(2m)^2]}{(2n)!} s^{2n} \,, s$$

equivalently to

$$F\_t(s) = 1 + \sum\_{n=1}^{m} \frac{\prod\_{k=1}^{n} 4(n - k + m)(n - k - m)}{(2n)!} s^{2n}.$$

If *ν* is odd, letting *ν* = 2*m* + 1, *m* ∈ N ∪ {0}, then *<sup>a</sup>*2*m*+3 = 0, so that *<sup>a</sup>*2*n*+3 = 0 for *n* ≥ *m*. Hence,

$$F\_{\theta}(s) = s + \sum\_{n=1}^{m} \frac{[(2n-1)^2 - (2m+1)^2][(2n-3)^2 - (2m+1)^2] \cdots [3^2 - (2m+1)^2][(1-(2m+1)^2)]}{(2n+1)!} s^{2n+1}, \quad s \in (0,1)$$

equivalently to

$$F\_o(s) = s + \sum\_{n=1}^{m} \frac{\prod\_{k=1}^{n} 4(n - k + m + 1)(n - k - m)}{(2n + 1)!} s^{2n + 1}.$$

For *ν* = 0, 1, 2, . . ., we have *<sup>F</sup>ν*(*s*), as follows:

$$\begin{aligned} F\_0(s) &= 1 = T\_0(s), \\ F\_1(s) &= s = T\_1(s), \\ F\_2(s) &= 1 - 2s^2 = -T\_2(s), \\ F\_3(s) &= s - (4/3)s^3 = -(1/3)T\_3(s), \\ F\_4(s) &= 1 - 8s^2 + 8s^4 = T\_4(s), \\ F\_5(s) &= s - 4s^3 + (16/5)s^5 = (1/5)T\_5(s), \\ F\_6(s) &= 1 - 18s^2 + 48s^4 - 32s^6 = -T\_6(s), \\ F\_7(s) &= s - 8s^3 + 16s^5 - (64/7)s^7 = -(1/7)T\_7(s). \end{aligned}$$

and so on, where *Tn*(*s*) is the Chebyshev polynomial of the first kind of *s* for *n* = 0, 1, 2, .... Since (24) is linear, *Tn*(*s*) is also its solution. Taking the inverse Laplace transform to *Tn*(*s*), and using Example 3(ii),(iii), we obtain the solutions of (22),

$$\begin{split} y\_{\boldsymbol{n}}(t) &= \frac{\nu}{2} \sum\_{k=0}^{\lfloor \nu/2 \rfloor} (-1)^{k} \frac{2^{\nu-2k} (n-k-1)!}{k! (\nu-2k)!} D^{\nu-2k} \delta(t), \\ &= T\_{\boldsymbol{n}}(\boldsymbol{D}) \delta(t), \end{split} \tag{26}$$

which are the distributional solutions of the forms (23).

**Example 5.** *Letting ν* = 1*,* (22) *becomes*

$$t^2y''(t) + 3ty'(t) - t^2y(t) = 0.\tag{27}$$

*From Theorem 2,* (27) *has a solution*

$$
\delta y(t) = \delta'(t). \tag{28}
$$

*Letting ν* = 4*,* (22) *becomes*

$$t^2y''(t) + 2ty'(t) - (t^2 + 15)y(t) = 0.\tag{29}$$

*From Theorem 2,* (29) *has a solution*

$$y(t) = 8\delta^{(4)}(t) - 8\delta^{\prime\prime}(t) + \delta(t). \tag{30}$$

*By applying* (10)*, it is not difficult to verify that* (28) *and* (30) *satisfy* (27) *and* (29)*, respectively.*
