This section introduces the product of distributions defined in [21, 22].

Let \(\mathcal{D}\) be the space of compactly supported infinitely differentiable complex-valued functions defined on *R*, let \(\mathcal{D}'\) be the space of Schwartz distributions, and let \(\alpha\in\mathcal{D}\) be even with \(\int_{-\infty}^{\infty}\alpha= 1\). In the theory of products in [21, 22], for computing the *α*-product \(T_{\dot{\alpha }} S \), they arrive at a relation of the form

$$ T_{\dot{\alpha}} S = T \beta+ ( T \ast\alpha) f $$

(3.1)

for \(T\in\mathcal{D}'\) and \(S=\beta+f\in C^{p}\oplus\mathcal {D}^{\prime }_{\mu}\), where \(p\in{0,1,2,\ldots,\infty}\), \(\mathcal{D}^{\prime p}\) is the space of distributions of order *p* in the sense of Schwartz (\({\mathcal{D}}^{\prime \infty}\) means \(\mathcal{D}'\)), \(\mathcal{D}^{\prime }_{\mu}\) is the space of distributions whose support has measure zero in the sense of Lebesgue, and *Tβ* is the usual Schwartz product of a \(\mathcal{D}^{\prime p}\) distribution by a \(C^{p}\)-function.

### Remark 3.1

The *α*-product is a generalization of the classical product of functions in the distribution sense. Therefore, the weak solution of the nonlinear PDE is related to this product. It is clear to see that, in (3.1), if the functions *T* and *S* are classical functions, then \(S=\beta+f\), \(f=0\) and \(T_{\dot{\alpha}} S = T \beta+ ( T \ast \alpha) f=T\beta\). Hence the *α*-product is equivalent to the classical product.

For instance, if *δ* stands for the Dirac measure, we have

$$\begin{aligned}& \delta_{\dot{\alpha}} \delta=\delta_{\dot{\alpha}} (0+\delta )=(\delta\ast \alpha)\delta=\alpha\delta= \alpha(0)\delta, \end{aligned}$$

(3.2)

$$\begin{aligned}& \delta_{\dot{\alpha}} (D\delta)=(\delta\ast\alpha) (D\delta )=\alpha(0) (D \delta)-\alpha'(0)\delta=\alpha(0) (D\delta), \end{aligned}$$

(3.3)

$$\begin{aligned}& (D\delta)_{\dot{\alpha}} \delta= \bigl((D\delta)\ast\alpha \bigr)\delta = \bigl( \delta\ast\alpha' \bigr)\delta=\alpha'(0)\delta= 0, \end{aligned}$$

(3.4)

where *D* denotes the generalized derivative.

It is easy to define the product of a distribution with a smooth function. A limitation of the theory of distributions is that there is no associative product of two distributions extending the product of a distribution by a smooth function, as has been proved by Laurent Schwartz in the 1950s [42, 43]. So about the properties of this kind of product of distributions, it is quite different from the pointwise product of classical functions.

This *α*-product is bilinear, has unit element(the constant function taking the value 1 viewed as a distribution), is invariant under translations and also under the action of the transformation \(t \to-t\) from *R* onto *R*. In general, this product is neither associative nor commutative; however,

$$ \int_{R}T _{\dot{\alpha}} S= \int_{R}S_{\dot{\alpha}} T $$

(3.5)

for any *α* if \(T,S \in\mathcal{D}'_{\mu}\) and *T* or *S* is compactly supported. In general, *α*-products cannot be completely localized. This becomes clear by noticing that \(\operatorname{supp}(T _{\dot{\alpha}} S) \subset \operatorname{supp} S\) (as for ordinary functions), but it can happen that \(\operatorname{supp}(T _{\dot{\alpha}} S) \subset \operatorname{supp} T\). Thus, in the following, *α*-product is regarded as a global product, and when we apply the product to differential equations, the solutions are naturally viewed as global solutions. Product (3.1) is consistent with the Schwartz product of \(\mathcal {D}^{\prime p}\)-distributions by \(C^{p}\)-functions (if these functions are placed on the right-hand side) and satisfy the standard differential rules.

In general, *α*-product cannot be completely localized. Thus, in the following, *α*-product is regarded as a global product. The Leibniz formula must be represented in the form

$$ D ( T _{\dot{\alpha}} S ) = ( D T ) _{\dot{\alpha}} S + T _{\dot {\alpha}} ( D S ) , $$

(3.6)

where *D* is the derivative operator in the distributional sense.

Besides, we can use *α*-products (3.1) to define powers of some distributions. Thus, if \(T=\beta+f\in C^{p}\oplus\mathcal {D}'_{\mu}\cap\mathcal{D}'{p}\), then

$$ T_{\dot{\alpha}} T = \beta^{2}+ \bigl[\beta+(\beta * \alpha)+(f * \alpha) \bigr]f, $$

(3.7)

because \(T\in\mathcal{D}'{p}\cap(C^{p}\oplus\mathcal{D}'_{\mu})\). Since \(T_{\dot{\alpha}} T \in C^{p}\oplus\mathcal{D}'_{\mu}\cap \mathcal{D}'{p}\), we can define the *α*-powers \(T_{\alpha}^{n}\) (\(n\geq0\) is an integer) by the recurrence formula

$$\begin{aligned}& T_{\alpha}^{0}=1, \end{aligned}$$

(3.8)

$$\begin{aligned}& T_{\alpha}^{n}= \bigl(T_{\alpha}^{n-1} \bigr)_{\dot{\alpha}} T. \end{aligned}$$

(3.9)

Since the distributional products (3.1) are consistent with the Schwartz products of distributions by functions (when functions are placed on the right-hand side), we have \(\beta_{\alpha}^{n}=\beta^{n}\) for all \(\beta\in C^{p}\), and the consistency of this definition with the ordinary powers of \(C^{p}\)-functions is proved. For instance, if \(m \in C\), then \((m\delta)^{0}_{\alpha}= 1\) and \((m\delta)^{n}_{\alpha}= m^{n}[\alpha(0)]^{n-1}\) for \(n\geq2\), which can readily be seen by induction.

We also have \((\tau_{a}T)_{\alpha}^{n}=\tau_{a}(T)_{\alpha}^{n}\) in the distributional sense, where \(\tau_{a}\) is the translation operator defined by \(a \in R\). Thus, in what follows, we shall write \(T^{n}\) instead of \(T_{\alpha}^{n}\) (supposing that *α* is fixed), which will also simplify the notation.

Notice that, under the definition of this kind product of distributions, if \(\phi(u)\) is an entire function of *u*, then \(\phi \circ u\) is well defined, here \(\phi\circ u\) is used to denote the expression of \(\phi(u)\) involving the product of distributions, and we have the following result.

### Lemma 3.1

([22])

*If*
\(\phi(u)\)*is an entire function of**u*, *then*

$$ \phi\circ(m\delta)= \textstyle\begin{cases} \phi(0)+\phi'(0)m\delta &\textit{if }\alpha(0)=0,\\ \phi(0)+\frac{\phi[m\alpha(0)]-\phi(0)}{\alpha(0)}\delta& \textit{if }\alpha(0)\neq0. \end{cases} $$

(3.10)

### Proof

If \(\phi(u)\) is an entire function of *u*, then we have

$$ \phi(u)=a_{0}+a_{1}u+a_{2}u^{2}+ \cdots, $$

(3.11)

where \(a_{n}=\frac{\phi^{n}(0)}{n!}\) for \(n=0,1,2,\dots\). For \(T\in C^{p}\oplus(\mathcal{D}^{\prime p}\cup\mathcal{D}'_{\mu})\), we define the composition \(\phi\circ T\) as follows:

$$ \phi\circ T=a_{0}+a_{1}T+a_{2}T^{2}+ \cdots $$

(3.12)

provided this series converges in \(\mathcal{D}'\). This is clearly a consistent definition, and we have \(\tau_{a}(\phi \circ T) = \phi\circ(\tau_{a}T)\) if \(\phi\circ T\) or \(\phi\circ (\tau_{a}T)\) is well defined. Recall that \(\phi\circ T\) depends on *α* in general. Now we shall show that \(\phi\circ(m\delta)\) is a distribution for all \(m\in\mathbb{C}\). We have \((m\delta)^{0} =1\) and \((m\delta)^{1} =m\delta\) and, for \(n\geqslant2\),

$$ (m\delta)^{n}=m^{n} \bigl[\alpha (0) \bigr]^{n-1}\delta, $$

(3.13)

as we have already seen. Then, according to (3.12),

$$ \phi\circ(m\delta)=a_{0}+a_{1}m\delta+a_{2}(m \delta)^{2}+\cdots, $$

(3.14)

because, as we shall see, this series is convergent in \(\mathcal{D}'\). Indeed, by (3.13), we have

$$ \phi\circ(m\delta)=a_{0}+a_{1}m\delta+a_{2}m^{2} \alpha (0)\delta+a_{3}m^{3} \bigl[\alpha (0) \bigr]^{2} \delta+ \cdots, $$

(3.15)

and thus, if \(\alpha (0) = 0\), then \(\phi\circ(m\delta)= a_{0} + a_{1}m\delta\), while if \(\alpha (0)\neq0\), then

$$ \alpha (0) \bigl[\phi\circ(m\delta)-a_{0} \bigr]=a_{1}\alpha (0)m \delta+a_{2}m^{2} \bigl[\alpha (0) \bigr]^{2} \delta+a_{3}m^{3} \bigl[\alpha (0) \bigr]^{3}\delta+ \cdots, $$

(3.16)

which is equivalent to

$$ \alpha (0) \bigl[\phi\circ(m\delta)-a_{0} \bigr]= \bigl[{a_{1} \alpha (0)m+a_{2}m^{2} \bigl[\alpha (0) \bigr]^{2}+a_{3}m^{3} \bigl[\alpha (0) \bigr]^{3}+\cdots} \bigr]\delta, $$

(3.17)

because, by (3.11), the series \(\{\cdots\} \) converges to \(\phi (m\alpha (0))-a_{0}\). In this case,

$$ \alpha (0) \bigl[\phi\circ(m\delta)-a_{0} \bigr]= \bigl[{\phi \bigl(m\alpha (0) \bigr)-a_{0}} \bigr]\delta, $$

(3.18)

from the above equation, we have

$$ \phi\circ(m\delta)=\phi(0)+\frac{\phi[m\alpha(0)]-\phi (0)}{\alpha(0)}\delta. $$

(3.19)

This completes the proof. □