#### 3.1. Exact Solutions of the (1 + 1)-Dimensional Cauchy Problem

In case (i), one easily construct the ansatz:

Substituting ansatz (

9) into Equation (

4), the (1 + 1)-dimensional Cauchy problem,

is obtained. Thus, the same problem is derived, however, in the case of a single spacial variable. Let us reduce the governing system to a single equation, extracting

$\rho =-{S}_{xx}$ from the second equation of Equation (

10) and substituting it into the first. Hence, the 4th order differential equation:

is obtained, which is equivalent to the 3rd order equation:

where

$\theta (t)$ is an arbitrary function. The obvious substitution:

transforms the last equation to the nonlinear 2nd order equation:

It can be noted that the substitution:

where

$j(t)={\int}_{0}^{t}({\int}_{0}^{\tau}\phantom{\rule{-0.166667em}{0ex}}\theta ({\tau}_{1})\phantom{\rule{0.166667em}{0ex}}\mathrm{d}{\tau}_{1})\mathrm{d}\tau $, reduces Equation (

11) to the Burgers equation:

The Burgers equation is linearizable via the famous Cole–Hopf substitution [

19,

20]:

to the linear heat equation:

All the substitutions mentioned above could be combined as follows:

It should be stressed that the Substitution (

13) reduces the nonlinear Cauchy Problem (

10) to the linear problem for the heat Equation (

12), which can be exactly solved. In fact, having the specified initial profiles in (

10), we find the initial condition for Equation (

12) as follows:

i.e.,

Obviously, the exact solution of Cauchy Problems (

12) and (

14) is the Poisson integral:

Thus, calculating the derivatives

${V}_{z}(t,z)$ and

${V}_{zz}(t,z)$, and using Substitution (

13), one can construct the solution of the Cauchy Problem (

10):

Because the Cole–Hopf substitution in non-local transform (i.e., involves derivatives), we need to examine the behavior of the solution as

$t\to 0$:

Hence, having chosen the function $A(t)$ with the property $A(0)=0$, we obtain ${lim}_{t\to 0}S(t,x)=S(0,x)$, i.e., the function $S(t,x)$ is continuous at $t=0$.

In order to prove that

${lim}_{t\to 0}\rho (t,x)=\rho (0,x)$, one needs to show that

${lim}_{t\to 0}{V}_{z}(t,z)={V}_{z}(0,z)$ and

${lim}_{t\to 0}{V}_{zz}(t,z)={V}_{zz}(0,z)$. The proof of the first equality can be found in [

19]. We have proved that

${lim}_{t\to 0}{V}_{zz}(t,z)={V}_{zz}(0,z)$ under the restriction

$\varphi (x)=o({x}^{2}),x\to \infty $ (here, the relevant calculations are omitted). Thus, the following statement can be formulated.

**Theorem** **2.** The classical solution of the Cauchy Problem (10), in the case when $\varphi (x)$ is differentiable twice and $\varphi (x)=o({x}^{2}),x\to \infty $ can be presented as:where $p(\xi )={e}^{-\frac{1}{2}\varphi (\xi )}\phantom{\rule{-0.166667em}{0ex}},\phantom{\rule{0.166667em}{0ex}}A(0)=0.$ Obviously, the exact Solution (

16) is not unique because one contains two arbitrary functions

$\theta (t)$ and

$A(t)$. To specify these functions, one needs additional biologically motivated restrictions. We remind the reader that function

$S(t,x)$ describing the density of chemicals should be bounded in space. Therefore, functions

$\theta (t)$ must vanish (in this case, function

$j(t)$, which depends on

$\theta (t)$, also vanishes). In order to specify both functions

$\theta (t)$ and

$A(t)$ one needs, for example, to assume that the quantity of the chemical

$S(t,x)$ is finite in space and time, i.e.,

${\int}_{-\infty}^{+\infty}\left|S(t,x)\right|dx<0$ for

$\forall t>0$. This assumption immediately leads to the unique solution of Cauchy Problem (

10) in the form:

In order to present a non-trivial exact Solution of (

10) in terms of elementary functions, we consider the initial profiles:

where

γ is an arbitrary constant.

Using Formula (

17), one obtains:

To calculate the integral

$I(t,x)={\int}_{-\infty}^{+\infty}sin(\gamma \xi )\phantom{\rule{0.166667em}{0ex}}{e}^{-\frac{1}{4t}{\xi}^{2}+\frac{x}{2t}\xi}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}\xi $, the Mellin transformation (see, e.g., [

21]) has been used:

Here,

${D}_{-1}(\xb7)$ is a parabolic cylinder function. Using the known properties of such functions, the integral in question can be explicitly calculated:

Now, we find the function:

Because

$\rho (t,x)=-{S}_{xx}(t,x)$, one also finds the function

$\rho (t,x)$. Thus, the exact solution of the Cauchy Problems (

10) and (

18) has the form:

Plots of this solution with

$\gamma =1$ are presented in

Figure 1 and

Figure 2. Both functions,

$\rho (t,x)$ and

$S(t,x)$, have attractive properties. For example, they are periodic, bounded and tend to some constants if

$t\to +\infty $. However, one notes that the function

$\rho (t,x)$, which usually describes the densities of cells or species (see Introduction), are not non-negative for

$\forall x\in \mathbb{R}$. It turns out that this unrealistic behavior (for real world applications) is a natural property of each non-constant solution of the (1 + 1)-dimensional Cauchy Problem (

10) with an arbitrary non-negative function

$\varphi (x)\ne const$.

Let us show this assuming that there is a Solution of (

10) with a non-negative function

$\varphi (x)\ne const$, such that the functions

$\rho (t,x)$ and

$S(t,x)$ are non-negative for

$\forall (t,x)\in {\mathbb{R}}^{+}\times \mathbb{R}$. Let us fix an arbitrary

$t={t}_{0}>0$. Because

$0\le \rho ({t}_{0},x)=-{S}_{xx}({t}_{0},x)$, the second derivative

${S}_{xx}({t}_{0},x)$ is non-positive for all

x. It means the continues function

$S({t}_{0},x)$ is convex upwards for all

x (otherwise, one is a constant). Now, one realizes that any function with such property (like

$-{x}^{2}+c(t)$,

$-{e}^{-x}+c(t)$, etc.) cannot be non-negative for all

x.

#### 3.2. Reduction and Exact Solutions of (1 + 2)-Dimensional Cauchy Problem

Now, we apply the operator

${G}_{2}^{\infty}$ with an arbitrary non-constant function

${f}_{2}(t)$. In order to reduce System (

4), we need to construct an ansatz by solving the corresponding system of characteristic equations for this operator. After rather standard calculations, the ansatz:

is obtained. Ansatz (

20) reduces Cauchy Problem (

4) to the (1 + 1)-dimensional Cauchy problem:

The function

$\mathrm{\Psi}(t,x)$ can be easily found by using the second equation:

Substituting this expression into the first Equation of (

21), one arrives at the fourth order partial differential equation:

which can be integrated twice w.r.t. the variable

x. Thus, the second order nonlinear equation:

is obtained (

${c}_{1}(t)$ and

${c}_{0}(t)$ are arbitrary smooth functions).

In contrast to the nonlinear Equation (

11), we were unable to linearize Equation (

22) (we remind the reader that

${f}_{2}^{\prime}(t)\ne 0$). To find particular solutions, we used a non-Lie ansatz (see [

22] for an example):

in order to reduce the partial differential Equation (

22) to a system of ODEs. In fact, Substituting (

23) into Equation (

22), the system of three ODEs:

is obtained to find unknown functions

${\psi}_{0}(t),{\psi}_{1}(t)$ and

${\psi}_{2}(t)$. Because System (

24) has the same structure as (38) [

22], one is integrable. In fact, the first equation can be solved for arbitrary function

${f}_{2}(t)$. By substituting the known function

${\psi}_{2}(t)$ into the second equation, one obtains the first order linear ODE to find the function

${\psi}_{1}(t)$. Finally, having the known functions

${\psi}_{1}(t)$ and

${\psi}_{2}(t)$, the third equation of System (

24) can be easily solved. As a result, the general Solution of (

24) has the form:

Taking into account that

${c}_{1}(t)$ and

${c}_{0}(t)$ are arbitrary smooth functions, we may simplify the general Solution (

25) to the form:

by introducing new notations

${c}_{1}^{*}(t)$ and

${c}_{0}^{*}(t)$, which are arbitrary smooth functions, while

${t}_{0}$ is an arbitrary parameter.

Thus, using Equations (

20), (

23) and (

26), we construct the exact solution:

of Cauchy Problem (

4) with the correctly-specified initial profile

$\varphi (x)=-\frac{1}{2{t}_{0}}{x}^{2}+{c}_{1}x+{c}_{0}$.

Obviously, this solution blows up for the finite time

${t}_{0}>0$, which makes its immediate biological interpretation unlikely. On the other hand, it is in agreement with [

18] (see also the references cited therein) because the global solution of this problem requires the constraint

${\int}_{{\mathbb{R}}^{2}}\rho (0,x,y)\mathrm{d}x\mathrm{d}y<8\pi $, which does not take place for

$\rho (0,x,y)=\frac{1}{{t}_{0}}$. Notably, blow-up solutions occur in some physical models, and they have been intensively studied since the 1980s (see, e.g., [

23] and references therein).