The trend and influence of media information Propagation based on nonlinear Differential equation

Dongyu Yao; Enming Zhang; Yu Tang

Journal & Issue Details

Applied Mathematics and Nonlinear Sciences

Journal Details

Format: Journal
eISSN: 2444-8656
First Published: 01 Jan 2016
Publication timeframe: 2 times per year
Languages: English

Article

Solving process of nonlinear differential equations

When researchers explore phenomena in nature, they need to make use of the solving process of nonlinear differential equations for analysis. Therefore, this paper studies the trend and influence of media information transmission, and mainly selects the Solution method of Hirota bilinear derivative method, as shown below:

Definition 1

To study this solution method, we should first define the operator, the bilinear operator - D operator proposed by Royogo Hirota, as shown below: $D_{x}^{m} f . g = {(\frac{\partial}{\partial x} - \frac{\partial}{\partial x^{'}})}^{m} {f (x) g (x^{'}) |}_{x^{'} = x} = \frac{\partial^{m}}{\partial x^{' m}} {f (x + x^{'}) g (x - x^{'}) |}_{x^{'} = 0}$ D_x^mf.g = {\left({{\partial \over {\partial x}} - {\partial \over {\partial x^{'}}}} \right)^m}{\left. {f\left(x \right)g\left({x^{'}} \right)} \right|_{x^{'} = x}} = {{{\partial ^m}} \over {\partial {x^{^{'}m}}}}{\left. {f\left({x + x^{'}} \right)g\left({x - x^{'}} \right)} \right|_{x^{'} = 0}}

Can be converted to: $\begin{array}{l} D_{x}^{m} D_{t}^{n} f . g \\ = {(\frac{\partial}{\partial x} - \frac{\partial}{\partial x^{'}})}^{m} {(\frac{\partial}{\partial t} - \frac{\partial}{\partial t^{'}})}^{n} {f (x, t) g (x^{'} t^{'}) |}_{x^{'} = x, t^{'} = t} \\ = \frac{\partial^{m}}{\partial x^{' m}} \frac{\partial^{n}}{\partial t^{' n}} {f (x + x^{'}, t + t^{'}) g (x - x^{'}, t - t^{'}) |}_{x^{'} = 0, t^{'} = 0} \end{array}$ \eqalign{& D_x^mD_t^nf.g \cr & = {\left({{\partial \over {\partial x}} - {\partial \over {\partial x^{'}}}} \right)^m}{\left({{\partial \over {\partial t}} - {\partial \over {\partial t^{'}}}} \right)^n}{\left. {f\left({x,t} \right)g\left({x^{'}t^{'}} \right)} \right|_{x^{'} = x,t^{'} = t}} \cr & = {{{\partial ^m}} \over {\partial {x^{^{'}m}}}}{{{\partial ^n}} \over {\partial {t^{^{'}n}}}}{\left. {f\left({x + x^{'},t + t^{'}} \right)g\left({x - x^{'},t - t^{'}} \right)} \right|_{x^{'} = 0,t^{'} = 0}} \cr}

Abstract: Hirota bilinear method is used to study the LUMP solution, two kinds of reaction solution and respiration solution of Nizhnik equations. The actual equations are as follows: ${\begin{cases} u_{t} + u_{xxx} - 3 (uv) x = 0; \\ u_{x} = u_{y} \end{cases}$ \left\{\matrix{{u_t} + {u_{xxx}} - 3\left({uv} \right)x = 0; \hfill \cr {u_x} = {u_y} \hfill \cr} \right.

Can be converted to: ${\begin{cases} u = - 2 {(ln f)}_{xy}, \\ v = - 2 {(ln f)}_{xx} \end{cases}$ \left\{\matrix{u = - 2{\left({\ln f} \right)_{xy}}, \hfill \cr v = - 2{\left({\ln f} \right)_{xx}} \hfill \cr} \right.

Thus, the bilinear equation is obtained: $(D_{x} D_{t} + D_{y} D_{x}^{3}) f . f = 2 ({ff}_{yt} - f_{y} f_{t} + 3 f_{xy} f_{xx} - 3 f_{x} f_{xxy} - f_{y} f_{xxx} + f f_{xxxy}) = 0$ \left({{D_x}{D_t} + {D_y}D_x^3} \right)f.f = 2\left({f{f_{yt}} - {f_y}{f_t} + 3{f_{xy}}{f_{xx}} - 3{f_x}{f_{xxy}} - {f_y}{f_{xxx}} + f{f_{xxxy}}} \right) = 0

Theorem 1

To study these three types of solutions, we can assume: $f = α (x, y, t) + β (x, y, t) + γ (x, y, t) + δ (x, y, t) + c$ f = \alpha \left({x,y,t} \right) + \beta \left({x,y,t} \right) + \gamma \left({x,y,t} \right) + \delta \left({x,y,t} \right) + c

In the above formula, α, β, γ, δ represents different functions of x, y, and t, and c represents real parameters.

Assuming α = g² + h², β = 0, γ = 0, δ = 0, it can be obtained: $f = g^{2} + h^{2} + c$ f = {g^2} + {h^2} + c and ${\begin{cases} g = a_{1} x + a_{2} y + a_{3} t + a_{4} \\ h = a_{5} x + a_{6} y + a_{7} t + a_{8} \end{cases}$ \left\{\matrix{g = {a_1}x + {a_2}y + {a_3}t + {a_4} \hfill \cr h = {a_5}x + {a_6}y + {a_7}t + {a_8} \hfill \cr} \right.

Proposition 2

In the above formula, both a_i(i = 1, 2,..., 8) and C are real parameters, which can be obtained by calculation if this formula is substituted into the bilinear equation: $a_{1} = a_{1}, a_{2} = 0, a_{3} = a_{3}, a_{4} = a_{4}, a_{5} = a_{5}, a_{6} = 0, a_{7} = a_{7}, a_{8} = a_{8}, c = c$ {a_1} = {a_1},{a_2} = 0,{a_3} = {a_3},{a_4} = {a_4},{a_5} = {a_5},{a_6} = 0,{a_7} = {a_7},{a_8} = {a_8},c = c

Get lump solution: ${\begin{cases} u = 0 \\ v = \frac{8 {(a_{1} g + a_{5} h)}^{2} - 4 (a_{1}^{2} - a_{5}^{2}) f}{f^{2}} \end{cases}$ \left\{\matrix{u = 0 \hfill \cr v = {{8{{\left({{a_1}g + {a_5}h} \right)}^2} - 4\left({a_1^2 - a_5^2} \right)f} \over {{f^2}}} \hfill \cr} \right.

Among them: ${\begin{cases} f = {(a_{1} x + a_{3} t + a_{4})}^{2} + {(a_{5} x + a_{7} t + a_{8})}^{2} + c \\ g = a_{1} x + a_{3} t + a_{4} \\ h = a_{5} x + a_{7} t + a_{8} \end{cases}$ \left\{\matrix{f = {\left({{a_1}x + {a_3}t + {a_4}} \right)^2} + {\left({{a_5}x + {a_7}t + {a_8}} \right)^2} + c \hfill \cr g = {a_1}x + {a_3}t + {a_4} \hfill \cr h = {a_5}x + {a_7}t + {a_8} \hfill \cr} \right.

Assume that the parameters are: $a_{1} = 3, a_{2} = 0, a_{3} = - 2, a_{4} = 5, a_{5} = 2, a_{6} = 0, a_{7} = - 1, a_{8} = 2, c = 2$ {a_1} = 3,{a_2} = 0,{a_3} = - 2,{a_4} = 5,{a_5} = 2,{a_6} = 0,{a_7} = - 1,{a_8} = 2,c = 2

Then the contours of the corresponding solution are shown in Figure 1 below:

Lemma 3

At the same time, α = g² + h², β = coshl, γ = 0, δ = 0 assumes, then it can be obtained: $f = g^{2} + h^{2} + cos l + c$ f = {g^2} + {h^2} + \cos l + c and ${\begin{cases} g = a_{1} x + a_{2} y + a_{3} t + a_{4} \\ h = a_{5} x + a_{6} y + a_{7} t + a_{8} \\ l = a_{9} x + a_{10} y + a_{11} t + a_{12} \end{cases}$ \left\{\matrix{g = {a_1}x + {a_2}y + {a_3}t + {a_4} \hfill \cr h = {a_5}x + {a_6}y + {a_7}t + {a_8} \hfill \cr l = {a_9}x + {a_{10}}y + {a_{11}}t + {a_{12}} \hfill \cr} \right.

Corollary 4

In the above formula, a_i(i = 1, 2,..., 8) and C are real parameters, which are directly inserted into the bilinear equation for calculation. Four kinds of solutions can be obtained according to different parameter Settings:

The first: ${\begin{cases} a_{1} = a_{1}, a_{2} = 0, a_{3} = a_{3}, a_{4} = a_{4}, a_{5} = a_{5}, a_{6} = 0, a_{7} = a_{7} \\ a_{8} = a_{8}, a_{9} = a_{9}, a_{10} = 0, a_{11} = a_{11}, a_{12} = a_{12}, c = c \end{cases}$ \left\{\matrix{{a_1} = {a_1},{a_2} = 0,{a_3} = {a_3},{a_4} = {a_4},{a_5} = {a_5},{a_6} = 0,{a_7} = {a_7} \hfill \cr {a_8} = {a_8},{a_9} = {a_9},{a_{10}} = 0,{a_{11}} = {a_{11}},{a_{12}} = {a_{12}},c = c \hfill \cr} \right.

The lump-Soliton solution of the equation is: ${\begin{cases} u = 0 \\ v = \frac{2}{f^{2}} [{(2 a_{1} g + 2 a_{5} h + a_{9} sinh l)}^{2} - (2 a_{1}^{2} + 2 a_{5}^{2} + a_{9}^{2} cos l) f] \end{cases}$ \left\{\matrix{u = 0 \hfill \cr v = {2 \over {{f^2}}}\left[{{{\left({2{a_1}g + 2{a_5}h + {a_9}\sinh l} \right)}^2} - \left({2a_1^2 + 2a_5^2 + a_9^2\cos l} \right)f} \right] \hfill \cr} \right.

Among them: ${\begin{cases} f = {(a_{1} x + a_{3} t + a_{4})}^{2} + {(a_{5} x + a_{7} t + a_{8})}^{2} + \cosh (a_{9} x + a_{11} t + a_{12}) + c \\ g = a_{1} x + a_{3} t + a_{4} \\ h = a_{5} x + a_{7} t + a_{8} \\ l = a_{9} x + a_{11} t + a_{12} \end{cases}$ \left\{\matrix{f = {\left({{a_1}x + {a_3}t + {a_4}} \right)^2} + {\left({{a_5}x + {a_7}t + {a_8}} \right)^2} + \cosh \left({{a_9}x + {a_{11}}t + {a_{12}}} \right) + c \hfill \cr g = {a_1}x + {a_3}t + {a_4} \hfill \cr h = {a_5}x + {a_7}t + {a_8} \hfill \cr l = {a_9}x + {a_{11}}t + {a_{12}} \hfill \cr} \right.

The second: ${\begin{cases} a_{1} = 0, a_{2} = 0, a_{3} = 0, a_{4} = a_{4}, a_{5} = 0, a_{6} = 0, a_{7} = 0, a_{8} = a_{8} \\ a_{9} = a_{9}, a_{10} = a_{10}, a_{11} = - 4 a_{9}^{3}, a_{12} = a_{12}, c = - a_{4}^{2} - a_{8}^{2} \end{cases}$ \left\{\matrix{{a_1} = 0,{a_2} = 0,{a_3} = 0,{a_4} = {a_4},{a_5} = 0,{a_6} = 0,{a_7} = 0,{a_8} = {a_8} \hfill \cr {a_9} = {a_9},{a_{10}} = {a_{10}},{a_{11}} = - 4a_9^3,{a_{12}} = {a_{12}},c = - a_4^2 - a_8^2 \hfill \cr} \right.

Conjecture 5

The lump-Soliton solution of the equation is: ${\begin{cases} u = - 2 a_{9} a_{10} \sec h^{2} l \\ v = - 2 a_{9}^{2} \sec h^{2} l \end{cases}$ \left\{\matrix{u = - 2{a_9}{a_{10}}\sec {h^2}l \hfill \cr v = - 2a_9^2\sec {h^2}l \hfill \cr} \right.

Among them: ${\begin{cases} f = \cosh (a_{9} x + a_{10} y - 4 a_{9}^{3} t + a_{12}) \\ g = a_{4} \\ h = a_{8} \\ l = a_{9} x + a_{10} y - 4 a_{9}^{3} t + a_{12} \end{cases}$ \left\{\matrix{f = \cosh \left({{a_9}x + {a_{10}}y - 4a_9^3t + {a_{12}}} \right) \hfill \cr g = {a_4} \hfill \cr h = {a_8} \hfill \cr l = {a_9}x + {a_{10}}y - 4a_9^3t + {a_{12}} \hfill \cr} \right.

The third: ${\begin{cases} a_{1} = a_{1}, a_{2} = a_{2}, a_{3} = 0, a_{4} = a_{4}, a_{5} = - \frac{a_{1} a_{2}}{a_{6}}, a_{6} = a_{6}, a_{7} = 0 \\ a_{8} = a_{8}, a_{9} = a_{9}, a_{10} = 0, a_{11} = - a_{9}^{3}, a_{12} = a_{12}, c = c \end{cases}$ \left\{\matrix{{a_1} = {a_1},{a_2} = {a_2},{a_3} = 0,{a_4} = {a_4},{a_5} = - {{{a_1}{a_2}} \over {{a_6}}},{a_6} = {a_6},{a_7} = 0 \hfill \cr {a_8} = {a_8},{a_9} = {a_9},{a_{10}} = 0,{a_{11}} = - a_9^3,{a_{12}} = {a_{12}},c = c \hfill \cr} \right.

The lump-Soliton solution of the equation is: ${\begin{cases} u = \frac{4 (a_{2} g + a_{6} h) (2 a_{1} g - 2 \frac{a_{1} a_{2}}{a_{6}} h + a_{9} sinh l)}{f^{2}} \\ v = \frac{2}{f^{2}} [{(2 a_{1} g - 2 \frac{a_{1} a_{2}}{a_{6}} h + a_{9} sinh l)}^{2} - (2 a_{1}^{2} + a \frac{a_{1}^{2} a_{2}^{2}}{a_{6}^{2}} + a_{9}^{2} cosh l) f] \end{cases}$ \left\{\matrix{u = {{4\left({{a_2}g + {a_6}h} \right)\left({2{a_1}g - 2{{{a_1}{a_2}} \over {{a_6}}}h + {a_9}\sinh l} \right)} \over {{f^2}}} \hfill \cr v = {2 \over {{f^2}}}\left[{{{\left({2{a_1}g - 2{{{a_1}{a_2}} \over {{a_6}}}h + {a_9}\sinh l} \right)}^2} - \left({2a_1^2 + a{{a_1^2a_2^2} \over {a_6^2}} + a_9^2\cosh l} \right)f} \right] \hfill \cr} \right.

Among them: ${\begin{cases} f = {(a_{1} x + a_{2} y + a_{4})}^{2} + {(- \frac{a_{1} a_{2}}{a_{6}} x + a_{6} y + a_{8})}^{2} + cosh (a_{9} x + a_{9}^{3} t + a_{12}) + c \\ g = a_{1} x + a_{2} y + a_{4} \\ h = - \frac{a_{1} a_{2}}{a_{6}} x + a_{6} y + a_{8} \\ l = a_{9} x - a_{9}^{3} t + a_{12} \end{cases}$ \left\{\matrix{f = {\left({{a_1}x + {a_2}y + {a_4}} \right)^2} + {\left({- {{{a_1}{a_2}} \over {{a_6}}}x + {a_6}y + {a_8}} \right)^2} + \cosh \left({{a_9}x + a_9^3t + {a_{12}}} \right) + c \hfill \cr g = {a_1}x + {a_2}y + {a_4} \hfill \cr h = - {{{a_1}{a_2}} \over {{a_6}}}x + {a_6}y + {a_8} \hfill \cr l = {a_9}x - a_9^3t + {a_{12}} \hfill \cr} \right.

Fourth: ${\begin{cases} a_{1} = 0, a_{2} = a_{2}, a_{3} = 0, a_{4} = a_{4}, a_{5} = a_{5}, a_{6} = 0, a_{7} = 0 \\ a_{8} = a_{8}, a_{9} = a_{9}, a_{10} = 0, a_{11} = - a_{9}^{3}, a_{12} = a_{12}, c = c \end{cases}$ \left\{\matrix{{a_1} = 0,{a_2} = {a_2},{a_3} = 0,{a_4} = {a_4},{a_5} = {a_5},{a_6} = 0,{a_7} = 0 \hfill \cr {a_8} = {a_8},{a_9} = {a_9},{a_{10}} = 0,{a_{11}} = - a_9^3,{a_{12}} = {a_{12}},c = c \hfill \cr} \right.

The lump-Soliton solution of the equation is: ${\begin{cases} u = \frac{4 a_{2} g (2 a_{5} h + a_{9} sinh l)}{f^{2}} \\ v = 2 \frac{{(2 a_{5} h + a_{9} sinh l)}^{2} - (2 a_{5}^{2} + a_{9}^{2} cosh l) f}{f^{2}} \end{cases}$ \left\{\matrix{u = {{4{a_2}g\left({2{a_5}h + {a_9}\sinh l} \right)} \over {{f^2}}} \hfill \cr v = 2{{{{\left({2{a_5}h + {a_9}\sinh l} \right)}^2} - \left({2a_5^2 + a_9^2\cosh l} \right)f} \over {{f^2}}} \hfill \cr} \right.

Among them: ${\begin{cases} f = {(a_{2} y + a_{4})}^{2} + {(a_{5} x + a_{8})}^{2} + cosh (a_{9} x - a_{9}^{3} t + a_{12}) + c \\ g = a_{2} y + a_{4} \\ h = a_{5} x + a_{8} \\ l = a_{9} x + a_{9}^{3} t + a_{12} \end{cases}$ \left\{\matrix{f = {\left({{a_2}y + {a_4}} \right)^2} + {\left({{a_5}x + {a_8}} \right)^2} + \cosh \left({{a_9}x - a_9^3t + {a_{12}}} \right) + c \hfill \cr g = {a_2}y + {a_4} \hfill \cr h = {a_5}x + {a_8} \hfill \cr l = {a_9}x + a_9^3t + {a_{12}} \hfill \cr} \right.

Theorem 2

When obtaining the quadratically determined solution of the Nizhnik equations, assuming α = g² + h², β = 0, γ = e′, δ = 0, it can be concluded that: ${\begin{cases} a_{1} = a_{1}, a_{2} = 0, a_{3} = a_{3}, a_{4} = a_{4}, a_{5} = a_{5}, a_{6} = 0, a_{7} = a_{7} \\ a_{8} = a_{8}, a_{9} = a_{9}, a_{10} = 0, a_{11} = a_{11}, a_{12} = a_{12}, c = c \end{cases}$ \left\{\matrix{{a_1} = {a_1},{a_2} = 0,{a_3} = {a_3},{a_4} = {a_4},{a_5} = {a_5},{a_6} = 0,{a_7} = {a_7} \hfill \cr {a_8} = {a_8},{a_9} = {a_9},{a_{10}} = 0,{a_{11}} = {a_{11}},{a_{12}} = {a_{12}},c = c \hfill \cr} \right. and ${\begin{cases} g = a_{1} x + a_{2} y + a_{3} t + a_{4} \\ h = a_{5} x + a_{6} y + a_{7} t + a_{8} \\ l = a_{9} x + a_{10} y + a_{11} t + a_{12} \end{cases}$ \left\{\matrix{g = {a_1}x + {a_2}y + {a_3}t + {a_4} \hfill \cr h = {a_5}x + {a_6}y + {a_7}t + {a_8} \hfill \cr l = {a_9}x + {a_{10}}y + {a_{11}}t + {a_{12}} \hfill \cr} \right.

Theorem 3

In the above formula, a_i(i = 1, 2,..., 8) and C are real parameters, which can be directly substituted into the bilinear equations. Four kinds of solutions can be obtained according to different parameters:

The first: ${\begin{cases} a_{1} = 0, a_{2} = a_{2}, a_{3} = a_{3}, a_{4} = a_{4}, a_{5} = a_{5}, a_{6} = 0, a_{7} = a_{7} \\ a_{8} = a_{8}, a_{9} = a_{9}, a_{10} = 0, a_{11} = a_{11}, a_{12} = a_{12}, c = c \end{cases}$ \left\{\matrix{{a_1} = 0,{a_2} = {a_2},{a_3} = {a_3},{a_4} = {a_4},{a_5} = {a_5},{a_6} = 0,{a_7} = {a_7} \hfill \cr {a_8} = {a_8},{a_9} = {a_9},{a_{10}} = 0,{a_{11}} = {a_{11}},{a_{12}} = {a_{12}},c = c \hfill \cr} \right.

The quadrant-exponential solution of the equation is: ${\begin{cases} u = 0 \\ v = 2 \frac{{(2 a_{1} g + 2 a_{5} h + a_{9} e^{l})}^{2} - (2 a_{1}^{2} + 2 a_{5}^{2} + a_{9}^{2} e^{l}) f}{f^{2}} \end{cases}$ \left\{\matrix{u = 0 \hfill \cr v = 2{{{{\left({2{a_1}g + 2{a_5}h + {a_9}{e^l}} \right)}^2} - \left({2a_1^2 + 2a_5^2 + a_9^2{e^l}} \right)f} \over {{f^2}}} \hfill \cr} \right.

Among them: ${\begin{cases} f = {(a_{1} x + a_{3} t + a_{4})}^{2} + {(a_{5} x + a_{7} t + a_{8})}^{2} + e^{a_{9} x + a_{11} t + a_{12}} + c \\ g = a_{1} x + a_{3} t + a_{4} \\ h = a_{5} x + a_{7} t + a_{8} \\ l = a_{9} x + a_{11} t + a_{12} \end{cases}$ \left\{\matrix{f = {\left({{a_1}x + {a_3}t + {a_4}} \right)^2} + {\left({{a_5}x + {a_7}t + {a_8}} \right)^2} + {e^{{a_9}x + {a_{11}}t + {a_{12}}}} + c \hfill \cr g = {a_1}x + {a_3}t + {a_4} \hfill \cr h = {a_5}x + {a_7}t + {a_8} \hfill \cr l = {a_9}x + {a_{11}}t + {a_{12}} \hfill \cr} \right.

The second: ${\begin{cases} a_{1} = - \frac{a_{5} a_{6}}{a_{2}}, a_{2} = a_{2}, a_{3} = 0, a_{4} = a_{4}, a_{5} = a_{5}, a_{6} = a_{6}, \\ a_{7} = 0, a_{8} = a_{8}, a_{9} = a_{9}, a_{10} = 0, a_{11} = - a_{9}^{3}, a_{12} = a_{12}, c = c \end{cases}$ \left\{\matrix{{a_1} = - {{{a_5}{a_6}} \over {{a_2}}},{a_2} = {a_2},{a_3} = 0,{a_4} = {a_4},{a_5} = {a_5},{a_6} = {a_6}, \hfill \cr {a_7} = 0,{a_8} = {a_8},{a_9} = {a_9},{a_{10}} = 0,{a_{11}} = - a_9^3,{a_{12}} = {a_{12}},c = c \hfill \cr} \right.

The quadrant-exponential solution of the equation is: ${\begin{cases} u = \frac{4 (a_{2} g + a_{6} h) (- 2 \frac{a_{5} a_{6}}{a_{2}} g + 2 a_{5} h + a_{9} e^{l})}{f^{2}} \\ v = \frac{2}{f^{2}} [{(- 2 \frac{a_{5} a_{6}}{a_{2}} g + 2 a_{5} h + a_{9} e^{l})}^{2} - (2 \frac{a_{2}^{2} a_{6}^{2}}{a_{2}^{2}} + 2 a_{5}^{2} + a_{9}^{2} e^{l}) f] \end{cases}$ \left\{\matrix{u = {{4\left({{a_2}g + {a_6}h} \right)\left({- 2{{{a_5}{a_6}} \over {{a_2}}}g + 2{a_5}h + {a_9}{e^l}} \right)} \over {{f^2}}} \hfill \cr v = {2 \over {{f^2}}}\left[{{{\left({- 2{{{a_5}{a_6}} \over {{a_2}}}g + 2{a_5}h + {a_9}{e^l}} \right)}^2} - \left({2{{a_2^2a_6^2} \over {a_2^2}} + 2a_5^2 + a_9^2{e^l}} \right)f} \right] \hfill \cr} \right.

Among them: ${\begin{cases} f = {(- \frac{a_{5} a_{6}}{a_{2}} x + a_{2} y + a_{4})}^{2} + {(a_{5} x + a_{6} y + a_{8})}^{2} + e^{e_{9} x - a_{9}^{3} t + a_{12}} + c \\ g = - \frac{a_{5} a_{6}}{a_{2}} x + a_{2} y + a_{4} \\ h = a_{5} x + a_{6} y + a_{8} \\ l = a_{9} x - a_{9}^{3} t + a_{12} \end{cases}$ \left\{\matrix{f = {\left({- {{{a_5}{a_6}} \over {{a_2}}}x + {a_2}y + {a_4}} \right)^2} + {\left({{a_5}x + {a_6}y + {a_8}} \right)^2} + {e^{{e_9}x - a_9^3t + {a_{12}}}} + c \hfill \cr g = - {{{a_5}{a_6}} \over {{a_2}}}x + {a_2}y + {a_4} \hfill \cr h = {a_5}x + {a_6}y + {a_8} \hfill \cr l = {a_9}x - a_9^3t + {a_{12}} \hfill \cr} \right.

The third: ${\begin{cases} a_{1} = a_{1}, a_{2} = - \frac{a_{5} a_{6}}{a_{1}}, a_{3} = 0, a_{4} = a_{4}, a_{5} = a_{5}, a_{6} = a_{6}, \\ a_{7} = 0, a_{8} = a_{8}, a_{9} = a_{9}, a_{10} = 0, a_{11} = - a_{9}^{3}, a_{12} = a_{12}, c = c \end{cases}$ \left\{\matrix{{a_1} = {a_1},{a_2} = - {{{a_5}{a_6}} \over {{a_1}}},{a_3} = 0,{a_4} = {a_4},{a_5} = {a_5},{a_6} = {a_6}, \hfill \cr {a_7} = 0,{a_8} = {a_8},{a_9} = {a_9},{a_{10}} = 0,{a_{11}} = - a_9^3,{a_{12}} = {a_{12}},c = c \hfill \cr} \right.

The quadrant-exponential solution of the equation is: ${\begin{cases} u = \frac{4 a_{6} (a_{h} - a_{5} g) (2 a_{1} g + 2 a_{5} h + a_{9} e^{l})}{a_{1} f^{2}} \\ v = 2 \frac{{(2 a_{1} g + 2 a_{5} h + a_{9} e^{l})}^{2} - (2 a_{1}^{2} + 2 a_{5}^{2} + a_{9}^{2} e^{l}) f}{f^{2}} \end{cases}$ \left\{\matrix{u = {{4{a_6}\left({{a_h} - {a_5}g} \right)\left({2{a_1}g + 2{a_5}h + {a_9}{e^l}} \right)} \over {{a_1}{f^2}}} \hfill \cr v = 2{{{{\left({2{a_1}g + 2{a_5}h + {a_9}{e^l}} \right)}^2} - \left({2a_1^2 + 2a_5^2 + a_9^2{e^l}} \right)f} \over {{f^2}}} \hfill \cr} \right.

Among them: ${\begin{cases} f = {(a_{1} x - \frac{a_{5} a_{6}}{a_{1}} y + a_{4})}^{2} + {(a_{5} x + a_{6} y + a_{8})}^{2} + e^{e_{9} x - a_{9}^{3} t + a_{12}} + c \\ g = a_{1} x - \frac{a_{5} a_{6}}{a_{1}} y + a_{4} \\ h = a_{5} x + a_{6} y + a_{8} \\ l = a_{9} x - a_{9}^{3} t + a_{12} \end{cases}$ \left\{\matrix{f = {\left({{a_1}x - {{{a_5}{a_6}} \over {{a_1}}}y + {a_4}} \right)^2} + {\left({{a_5}x + {a_6}y + {a_8}} \right)^2} + {e^{{e_9}x - a_9^3t + {a_{12}}}} + c \hfill \cr g = {a_1}x - {{{a_5}{a_6}} \over {{a_1}}}y + {a_4} \hfill \cr h = {a_5}x + {a_6}y + {a_8} \hfill \cr l = {a_9}x - a_9^3t + {a_{12}} \hfill \cr} \right.

Fourth: ${\begin{cases} a_{1} = 0, a_{2} = a_{2}, a_{3} = 0, a_{4} = a_{4}, a_{5} = 0, a_{6} = 0, a_{7} = 0 \\ a_{8} = a_{8}, a_{9} = a_{9}, a_{10} = a_{10}, a_{11} = - a_{9}^{3}, a_{12} = a_{12}, c = c \end{cases}$ \left\{\matrix{{a_1} = 0,{a_2} = {a_2},{a_3} = 0,{a_4} = {a_4},{a_5} = 0,{a_6} = 0,{a_7} = 0 \hfill \cr {a_8} = {a_8},{a_9} = {a_9},{a_{10}} = {a_{10}},{a_{11}} = - a_9^3,{a_{12}} = {a_{12}},c = c \hfill \cr} \right.

The quadrant-exponential solution of the equation is: ${\begin{cases} u = 2 a_{9} e^{l} \frac{2 a_{2} g + 2 a_{6} h - a_{10} (g^{2} + h^{2} + c)}{f^{2}} \\ v = - 2 a_{9}^{2} e^{l} \frac{g^{2} + h^{2} + c}{f^{2}} \end{cases}$ \left\{\matrix{u = 2{a_9}{e^l}{{2{a_2}g + 2{a_6}h - {a_{10}}\left({{g^2} + {h^2} + c} \right)} \over {{f^2}}} \hfill \cr v = - 2a_9^2{e^l}{{{g^2} + {h^2} + c} \over {{f^2}}} \hfill \cr} \right.

Among them: ${\begin{cases} f = {(a_{2} y + a_{4})}^{2} + {(a_{6} y + a_{8})}^{2} + e^{e_{9} x + a_{10} y - a_{9}^{3} t + a_{12}} + c \\ g = a_{2} y + a_{4} \\ h = a_{6} y + a_{8} \\ l = a_{9} x + a_{10} y - a_{9}^{3} t + a_{12} \end{cases}$ \left\{\matrix{f = {\left({{a_2}y + {a_4}} \right)^2} + {\left({{a_6}y + {a_8}} \right)^2} + {e^{{e_9}x + {a_{10}}y - a_9^3t + {a_{12}}}} + c \hfill \cr g = {a_2}y + {a_4} \hfill \cr h = {a_6}y + {a_8} \hfill \cr l = {a_9}x + {a_{10}}y - a_9^3t + {a_{12}} \hfill \cr} \right.

When calculating and analyzing Nizhnik's respiratory solution, it is assumed that, then: $f = e^{- ξ} + cos η + e^{ξ} + c$ f = {e^{- \xi}} + \cos \eta + {e^\xi} + c and ${\begin{cases} ξ = a_{1} x + a_{2} y + a_{3} t + a_{4} \\ η = a_{5} x + a_{6} y + a_{7} t + a_{8} \end{cases}$ \left\{\matrix{\xi = {a_1}x + {a_2}y + {a_3}t + {a_4} \hfill \cr \eta = {a_5}x + {a_6}y + {a_7}t + {a_8} \hfill \cr} \right.

Example 6

In the above formula, both a_i(i = 1, 2,..., 8) and C are real parameters, which can be substituted into the bilinear equations. According to different parameters, the following four respiratory solutions can be obtained:

The first: ${a_{1} = 0, a_{2} = 0, a_{3} = a_{3}, a_{4} = a_{4}, a_{5} = a_{5}, a_{6} = 0, a_{7} = a_{7}, a_{8} = a_{8}, c = c$ \left\{{{a_1} = 0,{a_2} = 0,{a_3} = {a_3},{a_4} = {a_4},{a_5} = {a_5},{a_6} = 0,{a_7} = {a_7},{a_8} = {a_8},c = c} \right.

The respiratory solution of the equation is: ${\begin{cases} u = 0 \\ v = \frac{2}{f^{2}} [{(a_{1} (e^{ξ} - e^{- ξ}) - a_{5} \sin_{η})}^{2} - (a_{1}^{2} (e^{ξ} + e^{- ξ}) - a_{5}^{2} cos η) f] \end{cases}$ \left\{\matrix{u = 0 \hfill \cr v = {2 \over {{f^2}}}\left[{{{\left({{a_1}\left({{e^\xi} - {e^{- \xi}}} \right) - {a_5}{{\sin}_\eta}} \right)}^2} - \left({a_1^2\left({{e^\xi} + {e^{- \xi}}} \right) - a_5^2\cos \eta} \right)f} \right] \hfill \cr} \right.

Among them: ${\begin{cases} f = e^{- a_{1} x - a_{3} t - a_{4}} + cos (a_{5} x + a_{7} y + a_{8}) + e^{a_{1} x + a_{3} t + a_{4}} + c \\ ξ = a_{1} x + a_{3} t + a_{4} \\ η = a_{5} x + a_{7} t + a_{8} \end{cases}$ \left\{\matrix{f = {e^{- {a_1}x - {a_3}t - {a_4}}} + \cos \left({{a_5}x + {a_7}y + {a_8}} \right) + {e^{{a_1}x + {a_3}t + {a_4}}} + c \hfill \cr \xi = {a_1}x + {a_3}t + {a_4} \hfill \cr \eta = {a_5}x + {a_7}t + {a_8} \hfill \cr} \right.

The second: $a_{1} = 0, a_{2} = a_{2}, a_{3} = 0, a_{4} = a_{4}, a_{5} = a_{5}, a_{6} = 0, a_{7} = a_{5}^{3}, a_{8} = a_{8}, c = c$ {a_1} = 0,{a_2} = {a_2},{a_3} = 0,{a_4} = {a_4},{a_5} = {a_5},{a_6} = 0,{a_7} = a_5^3,{a_8} = {a_8},c = c

The respiratory solution of the equation is: ${\begin{cases} u = - \frac{2 a_{2} a_{5} e^{- ξ} (- 1 + e^{2 ξ}) \sin η}{f^{2}} \\ v = \frac{2 a_{5}^{2} (\sin^{2} η + cos η f)}{f^{2}} \end{cases}$ \left\{\matrix{u = - {{2{a_2}{a_5}{e^{- \xi}}\left({- 1 + {e^{2\xi}}} \right)\sin \eta} \over {{f^2}}} \hfill \cr v = {{2a_5^2\left({{{\sin}^2}\eta + \cos \eta f} \right)} \over {{f^2}}} \hfill \cr} \right.

Among them: ${\begin{cases} f = e^{- a_{2} y - a_{4}} + cos (a_{5} x + a_{5}^{3} t + a_{8}) + e^{a_{2} y + a_{4}} + c \\ ξ = a_{2} y + a_{4} \\ η = a_{5} x + a_{5}^{3} t + a_{8} \end{cases}$ \left\{\matrix{f = {e^{- {a_2}y - {a_4}}} + \cos \left({{a_5}x + a_5^3t + {a_8}} \right) + {e^{{a_2}y + {a_4}}} + c \hfill \cr \xi = {a_2}y + {a_4} \hfill \cr \eta = {a_5}x + a_5^3t + {a_8} \hfill \cr} \right.

The third: $a_{1} = a_{1}, a_{2} = 0, a_{3} = - a_{1}^{3}, a_{4} = a_{4}, a_{5} = 0, a_{6} = a_{6}, a_{7} = 0, a_{8} = a_{8}, c = c$ {a_1} = {a_1},{a_2} = 0,{a_3} = - a_1^3,{a_4} = {a_4},{a_5} = 0,{a_6} = {a_6},{a_7} = 0,{a_8} = {a_8},c = c

The respiratory solution of the equation is: ${\begin{cases} u = - \frac{2 a_{1} a_{6} (e^{ξ} - e^{- ξ}) \sin η}{f^{2}} \\ v = 2 a_{1}^{2} \frac{{(e^{ξ} - e^{- ξ})}^{2} - (e^{ξ} + e^{- ξ}) f}{f^{2}} \end{cases}$ \left\{\matrix{u = - {{2{a_1}{a_6}\left({{e^\xi} - {e^{- \xi}}} \right)\sin \eta} \over {{f^2}}} \hfill \cr v = 2a_1^2{{{{\left({{e^\xi} - {e^{- \xi}}} \right)}^2} - \left({{e^\xi} + {e^{- \xi}}} \right)f} \over {{f^2}}} \hfill \cr} \right.

Among them: ${\begin{cases} f = e^{- a_{1} x + a_{1}^{3} t - a_{4}} + cos (a_{6} y + a_{8}) + e^{a_{1} x - a_{1}^{3} t + a_{4}} + c \\ ξ = a_{1} x - a_{1}^{3} + a_{4} \\ η = a_{6} y + a_{8} \end{cases}$ \left\{\matrix{f = {e^{- {a_1}x + a_1^3t - {a_4}}} + \cos \left({{a_6}y + {a_8}} \right) + {e^{{a_1}x - a_1^3t + {a_4}}} + c \hfill \cr \xi = {a_1}x - a_1^3 + {a_4} \hfill \cr \eta = {a_6}y + {a_8} \hfill \cr} \right.

Fourth: ${\begin{cases} a_{1} = a_{1}, a_{2} = a_{2}, a_{3} = - a_{1}^{3} + 3 a_{1} a_{5}^{2}, a_{4} = a_{4}, a_{5} = a_{5}, a_{6} = - \frac{4 a_{1} a_{2}}{a_{5}} \\ a_{7} = - 3 a_{1}^{2} a_{5} + a_{5}^{3}, a_{8} = a_{8}, c = 0 \end{cases}$ \left\{\matrix{{a_1} = {a_1},{a_2} = {a_2},{a_3} = - a_1^3 + 3{a_1}a_5^2,{a_4} = {a_4},{a_5} = {a_5},{a_6} = - {{4{a_1}{a_2}} \over {{a_5}}} \hfill \cr {a_7} = - 3a_1^2{a_5} + a_5^3,{a_8} = {a_8},c = 0 \hfill \cr} \right.

The respiratory solution of the equation is: ${\begin{cases} u = 2 (\frac{a_{2} (e^{ξ} - e^{- ξ} + \frac{4 a_{1} \sin η}{a_{5}}) (a_{1} e^{ξ} - a_{1} e^{- ξ} + 4 cos η) f}{f^{2}} - \frac{a_{1} a_{2} (e^{ξ} + e^{- ξ} + 4 cos η) f}{f^{2}}) \\ v = \frac{2}{f^{2}} [{(a_{1} e^{ξ} - a_{1} e^{- ξ} - a_{5} \sin η)}^{2} - (a_{1}^{2} e^{ξ} + a_{1}^{2} e^{- ξ} - a_{5}^{2} cos η) f] \end{cases}$ \left\{\matrix{u = 2\left({{{{a_2}\left({{e^\xi} - {e^{- \xi}} + {{4{a_1}\sin \eta} \over {{a_5}}}} \right)\left({{a_1}{e^\xi} - {a_1}{e^{- \xi}} + 4\cos \eta} \right)f} \over {{f^2}}} - {{{a_1}{a_2}\left({{e^\xi} + {e^{- \xi}} + 4\cos \eta} \right)f} \over {{f^2}}}} \right) \hfill \cr v = {2 \over {{f^2}}}\left[{{{\left({{a_1}{e^\xi} - {a_1}{e^{- \xi}} - {a_5}\sin \eta} \right)}^2} - \left({a_1^2{e^\xi} + a_1^2{e^{- \xi}} - a_5^2\cos \eta} \right)f} \right] \hfill \cr} \right.

Among them: ${\begin{cases} f = e^{- a_{1} x - a_{2} y + (a_{1}^{3} - 3 a_{1} a_{5}^{2}) t - a_{4}} + cos (a_{5} x - \frac{4 a_{1} a_{2}}{a_{5}} y + (a_{5}^{3} - 3 a_{1}^{2} a_{5}) t + a_{8}) + e^{a_{1} x + a_{2} y + (3 a_{1} a_{5}^{2} - a_{1}^{3}) t + a_{4}} \\ ξ = a_{1} x + a_{2} y + (- a_{1}^{4} + 3 a_{1} a_{5}^{2}) t + a_{4} \\ η = a_{5} x - \frac{4 a_{1} a_{2}}{a_{5}} y + (- 3 a_{1}^{2} a_{5} + a_{5}^{3}) t + a_{8} \end{cases}$ \left\{\matrix{f = {e^{- {a_1}x - {a_2}y + \left({a_1^3 - 3{a_1}a_5^2} \right)t - {a_4}}} + \cos \left({{a_5}x - {{4{a_1}{a_2}} \over {{a_5}}}y + \left({a_5^3 - 3a_1^2{a_5}} \right)t + {a_8}} \right) + {e^{{a_1}x + {a_2}y + \left({3{a_1}a_5^2 - a_1^3} \right)t + {a_4}}} \hfill \cr \xi = {a_1}x + {a_2}y + \left({- a_1^4 + 3{a_1}a_5^2} \right)t + {a_4} \hfill \cr \eta = {a_5}x - {{4{a_1}{a_2}} \over {{a_5}}}y + \left({- 3a_1^2{a_5} + a_5^3} \right)t + {a_8} \hfill \cr} \right.

Analysis of media information transmission structure

Social network topology provides effective channel for information transmission, it can be thought of as the process of information through social networks of different community, the network structure and information dissemination is the coordination of influence each other, if the information transmission speed faster, so are likely to affect the user social relation network topology, and the information network structure also limits the diffusion process, This creates a feedback loop. Based on the online social network of Sina Weibo and the analysis of the structure of information transmission model, this paper makes an in-depth study of the actual communication trend and influence content[1.2].

2.1

Model design

According to the information transmission tree analysis shown in FIG. 2 below, it intuitively presents all transmission paths of microblog information, with nodes representing Sina Microblog users and directed edges representing information forwarding relations between users[3.4]:

Structure of information transmission tree

In this paper, the data set of information transmission tree required by the experiment is studied and collected. The network crawler is used to capture the microblog information published by users, and then the “forward” sign is extracted from it to identify the information source and forward information, so as to build the information transmission tree. This data set contains the complete propagation process of 340 different data sources, including 4469809 user nodes and 4469469 forwarding paths in total.

2.2

Research and analysis

Firstly, information dissemination breadth is mainly used to show the number of nodes in each layer of the information dissemination tree, which fully shows the once-spread classification of information released or forwarded by users. According to the empirical analysis of the above research data, the final result proves that, without studying the root node of information transmission, the lowest information transmission breadth is 1, while the maximum information transmission breadth is 67,352. It should be noted that in the information dissemination tree, about 99.3% of nodes are in the first 11 layers, which is in sharp contrast to the current scientific research results. Specific results are shown in the figure below:

Empirical analysis results of data set of information dissemination tree

According to the above analysis, (a) represents the distribution breadth of each information dissemination tree, (b) represents the distribution and fitting of all information dissemination data and breadth, among which the green node represents the distribution of original data.

Second, the depth of information transmission, which is mainly used to show the length of information transmission path, the level of information transmission tree, and the distribution of the number of information transmission termination nodes at each level. From the perspective of practice, in the information transmission tree studied in this paper, information transmission paths within three days account for 65% of the total, and most of the transmission paths are within nine days, which is in sharp contrast to the three-degree influence proposed by Twitter et al. The specific empirical analysis results are shown in Figure 4 below[5.6]:

Distribution of information transmission tree

According to the above analysis, (a) represents the depth distribution of all information dissemination trees, (b) represents the aggregation depth distribution and distribution fitting state of all information dissemination trees, and the green node represents the distribution of original data. According to the graph change analysis, it is found that all propagation depths have the characteristics of long tail. In other words, the propagation path length of all information propagation audiences has the characteristics of power-rate distribution and linear distribution, and the aggregation results of propagation path length of all information propagation trees also have these two characteristics. When the propagation depth is lower than or equal to 9, it conforms to the power distribution. When the propagation depth is more than 9, the data has the characteristics of linear distribution.

Third, node outgo distribution. It is found that all nodes except the root node have the same entry degree of 1 by studying the distribution of node exit degree of information dissemination tree. Therefore, based on the discussion of node outgo distribution of information transmission tree, the result graph as shown in FIG. 5 is obtained in this paper[7.8]:

Node exit degree distribution of information propagation tree

The analysis shown in the figure shows that the distribution of nodes is similar to that of most social networks, showing heavy tail characteristics during practical operation. Among them, scattered points at the bottom layer are obviously different from other distribution reasons. About 57% of nodes are in the first layer, and the output degree contributed by all nodes to the root node is 1. Therefore, the output degree of the root node occupies 57% of the input and output of all nodes in the information transmission tree. Since there are 340 root nodes in total and 4469809 nodes in all information transmission trees, the root nodes account for a relatively small proportion and are mainly at the lowest level.

Fourth, community characteristics. The research on community characteristics of information dissemination tree shows that community structure is one of the important features to present user relationship in online social networks. Information dissemination tree is a subgraph to present user relationship network, which also has community structure characteristics. Information transmission tree is a hierarchical nested tree structure. Users and their friends will form open triples, while friends cannot form closed triples because they cannot connect. It can be seen that the clustering coefficients of all nodes and the total information number are zero when analyzed in combination with the clustering coefficients, which also proves that the information propagation tree can present the unique characteristics of the user network complexgraph. It should be noted that it is difficult to present the community characteristics of information dissemination tree only by using clustering coefficient, so this paper also calculates the corresponding modularity. According to the analysis of the flow chart of information release as shown in Figure 6 below, it indirectly proves the law of information transmission, and the overall distribution of module degree is relatively balanced and has the characteristics of linear distribution. It can be seen that: on the one hand, information dissemination in social network media is characterized by linearity at macro level, which is in sharp contrast to randomness in the traditional sense. On the other hand, media information dissemination has limitations, which can only be transmitted among users' collective friends, thus forming the community structure of local clusters. Since users have similar concerns and preferences, they have similar adoption habits in the face of the same information, so forwarding will further spread the information.[6.7]

According to the above research results, information transmission tree not only has the characteristics of power distribution and community structure, but also has the characteristics of self-organization, and the information transmission tree as the user's social network is also a relatively complex network structure. Due to the different social media platforms, information dissemination size of the tree is different, so only calculate the number of information transmission path length, degree distribution, modules such as basic information, it is difficult to accurately judge the trend of network transmission, so in the future research trend of research scholars to explore the media information dissemination and the influence of the composition.[8.9.10]

Results

First, the centrality of media information disseminators has changed in the new era. In the traditional sense of mass communication, communicators not only influence the existence and development of communication process, but also determine the quantity and quality of information content, and some will change the role of social influence. Under this condition, everyone can use the Internet or mobile phone to disseminate media information in a peer-to-peer or peer-to-peer manner. This approach not only controls the distance between communicator and receiver, but also shifts the centrality of both during communication. Under the new media environment, interpersonal communication and information communication are gradually integrated, and the forms of information communication are gradually breaking through the limits of media boundaries. Under the influence of a variety of complex forces, communication personnel begin to rethink their communication responsibilities and communication process under the development of practice while changing their communication consciousness and behavior.

Second, change the role of the receiver of information. In the traditional media environment, the receiver of information is regarded as the audience, which pays more attention to the activeness of information. However, on new media platforms, both communicators and receivers belong to users of network or mobile clients. In the process of disseminating media information, users as disseminators have the right to disseminate information, and as receivers enjoy the right to obtain information. It can be seen that new media users have better functions than audiences to summarize the characteristics of current media information transmission.

According to the analysis of the operation structure chart of new media shown in Figure 7 below, users' roles and status have undergone comprehensive changes during information transmission, which also fully demonstrates their characteristics during information selection and communication reception.

First, information selection becomes more autonomous. Because of the new media platform combined with large quantities of image information, text, voice, digital video, etc, so the user can use technology to choose its own information content and form, this kind of personalized service technology can improve the user's right to choose, to make them according to their own values to build a personalized information environment; Secondly, information production presents the characteristics of socialization. Because new media can provide favorable conditions for individuals or organizations to release and disseminate information, and the recipients of information can directly participate in the production and dissemination of information, the trend of media information dissemination is more and more extensive, and it is affected by more and more factors. Thirdly, information exchange presents bidirectional characteristics. In the environment of new media technology, both communicators and receivers can carry out two-way communication. This form of communication is also the intuitive expression of anti-Runquan and the right of expression. Finally, the flow of information presents a trend of decentralization. Mass communication in the traditional sense is large-scale and unidirectional, and it will abide by most principles in the practice of communication development. However, in the era of new media, the flow of media information is more dispersed, and the actual communication audience will change from the masses to smaller or smaller groups. This requires information disseminators to customize professional services according to the personalized characteristics of receivers.

Structure diagram of new media operation

Third, the order of media communication has changed. Due to the spread of traditional media order was gradually break in new media environment, so the universal access to new media technology platform, to create a more broad and spread of space, forming a multi-level and complex environment, only in this way can guarantee the media information in effective communication, to improve its competitive advantage.

Conclusion

To sum up, in the social economy and technological innovation and development, the traditional communication environment has changed, which not only puts forward new requirements on the practice management system and operation form, but also systematically studies the trend and influencing factors of new media information communication. Only in this way can a better new media operation environment be built. Nowadays, the research scholars, according to media information dissemination form and the nonlinear differential equation has carried on the empirical research, the final results show that the new changes have taken place in the trend of media information transmission, the influence factors of facing more and more, so be combined with previous development experience, to build the new media environment has the rationality and the spread of open order. At the same time, it is necessary to strengthen the training of professional and technical personnel, actively apply advanced technical concepts, learn from the effective countermeasures of foreign media information dissemination, fully understand the personalized characteristics of information dissemination and reception, and build a high-quality new media information dissemination environment according to the needs of modern economic and cultural construction.

Figures & Tables

References

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Applied Mathematics and Nonlinear Sciences

The trend and influence of media information Propagation based on nonlinear Differential equation

Dongyu Yao,
Dongyu Yao,
Enming Zhang and
Enming Zhang and
Yu Tang
Yu Tang

Published Online: 15 Jul 2022

Volume & Issue: AHEAD OF PRINT

Page range: -

Received: 09 Mar 2022

Accepted: 14 May 2022

Journal & Issue Details

Applied Mathematics and Nonlinear Sciences

PDF Preview

Article

Figure 1

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Figures & Tables

Figure 1

FIG. 2

FIG. 3

FIG. 4

FIG. 5

Figure 6

FIG. 7

References

Recent Articles

Plan your remote conference with Sciendo

Applied Mathematics and Nonlinear Sciences

The trend and influence of media information Propagation based on nonlinear Differential equation

Dongyu Yao,Dongyu Yao,Enming Zhang andEnming Zhang andYu TangYu Tang

Published Online: 15 Jul 2022

Volume & Issue: AHEAD OF PRINT

Page range: -

Received: 09 Mar 2022

Accepted: 14 May 2022

Journal & Issue Details

Applied Mathematics and Nonlinear Sciences

PDF Preview

Article

Figure 1

FIG. 2

FIG. 3

FIG. 4

FIG. 5

Figure 6

FIG. 7

Figures & Tables

Figure 1

FIG. 2

FIG. 3

FIG. 4

FIG. 5

Figure 6

FIG. 7

References

Recent Articles

Plan your remote conference with Sciendo

Dongyu Yao,
Dongyu Yao,
Enming Zhang and
Enming Zhang and
Yu Tang
Yu Tang