Examination and Countermeasures of Network Education in Colleges and Universities Based on Ordinary Differential Equation Model

After social and economic development has entered the knowledge age, human development attaches more importance to educational innovation. [1]Using the theoretical method of differential equation, the researchers conducted a preliminary study based on the mathematical model under the condition that there were only two kinds of innovation generation relations, and mainly discussed whether the explanation with the initial value far from the equilibrium point could be inclined to the equilibrium point. The differential equation model of educational innovation competition is as follows[2]: {dxdt=k1x(t)(1−x(t)M1−b1y(t)M2)dydt=k2y(t)(1−b2x(t)M1−y(t)M2) \left\{\matrix{{{dx} \over {dt}} = {k_1}x\left(t \right)\left({1 - {{x\left(t \right)} \over {{M_1}}} - {b_1}{{y\left(t \right)} \over {{M_2}}}} \right) \hfill \cr {{dy} \over {dt}} = {k_2}y\left(t \right)\left({1 - {b_2}{{x\left(t \right)} \over {{M_1}}} - {{y\left(t \right)} \over {{M_2}}}} \right) \hfill \cr} \right.

In the above formula, K₁ and K₂ represent the penetration rate of educational innovation X and Y without considering the environmental impact; M₁ and M₂ represent the maximum possible number of X and Y; X (t) and y (t) represent the cumulative number of recipients of X and Y; B₁ >0 and B₂ >0 represent the interaction coefficients when X and Y coexist in the system.

According to the above formula, the positive equilibrium point P1(M1(1−b1)1−b1b2,M2(1−b2)1−b1b2) {P_1}\left({{{{M_1}\left({1 - {b_1}} \right)} \over {1 - {b_1}{b_2}}},{{{M_2}\left({1 - {b_2}} \right)} \over {1 - {b_1}{b_2}}}} \right) is globally uniformly asymptotic and tends to be stable if the conditions of 0<b2<M1M22−b1M1M2 0 < {b_2} < {{\root 2 \of {{M_1}{M_2}} - {b_1}{M_1}} \over {{M_2}}} and 0<M1M22−M2M1<b1<1 0 < {{\root 2 \of {{M_1}{M_2}} - {M_2}} \over {{M_1}}} < {b_1} < 1 are met in the case of coexistence of X and Y. It is proved that the equilibrium point of the above formula p1(M1(1−b1)1−b1b2,M2(1−b2)1−b1b2) {p_1}\left({{{{M_1}\left({1 - {b_1}} \right)} \over {1 - {b_1}{b_2}}},{{{M_2}\left({1 - {b_2}} \right)} \over {1 - {b_1}{b_2}}}} \right) , p₂(M₁,0), p₃(0, M₂), is p4 (0,0). In the case of coexistence of X and Y, research and analysis can be carried out under the condition of R+2={(x,y)|x>0,y>0} R_ + ^2 = \left\{{\left({x,y} \right)|x > 0,y > 0} \right\} , and only the equilibrium point P1 can be discussed.

Suppose p₁(x^*, y^*), where x*=M1(1−b1)1−b1b2>0 {x^*} = {{{M_1}\left({1 - {b_1}} \right)} \over {1 - {b_1}{b_2}}} > 0 , y*=M2(1−b2)1−b1b2>0 {y^*} = {{{M_2}\left({1 - {b_2}} \right)} \over {1 - {b_1}{b_2}}} > 0 , 0 < b₁ < b₂ < 1, then the following equation is satisfied: {1M1x*+b1M2y*=1b2M1x*+1M2y*=1 \left\{\matrix{{1 \over {{M_1}}}{x^*} + {{{b_1}} \over {{M_2}}}{y^*} = 1 \hfill \cr {{{b_2}} \over {{M_1}}}{x^*} + {1 \over {{M_2}}}{y^*} = 1 \hfill \cr} \right.

And the above model can be transformed into: {dxdt=−k1x(t)(x−x*M1+b1M2(y−y*))dydt=−k2y(t)(b2M1(x−x*)−1M2(y−y*)) \left\{\matrix{{{dx} \over {dt}} = - {k_1}x\left(t \right)\left({{{x - {x^*}} \over {{M_1}}} - {{{b_1}} \over {{M_2}}}\left({y - {y^*}} \right)} \right) \hfill \cr {{dy} \over {dt}} = - {k_2}y\left(t \right)\left({{{{b_2}} \over {{M_1}}}\left({x - {x^*}} \right) - {1 \over {{M_2}}}\left({y - {y^*}} \right)} \right) \hfill \cr} \right.

The corresponding function formula is: V(x,y)=1k1(x−x*−x*lnxx*)+1k2(y−y*−y*lnyy*) V\left({x,y} \right) = {1 \over {{k_1}}}\left({x - {x^*} - {x^*}\ln {x \over {{x^*}}}} \right) + {1 \over {{k_2}}}\left({y - {y^*} - {y^*}\ln {y \over {{y^*}}}} \right)

The final calculation is as follows: V(x*,y*)=0,∂V(x,y)∂x=1k1(1−x*x),∂V(x,y)∂y=1k2(1−y*y),∂2V(x,y)∂x2=x*k1x2,∂2V(x,y)∂y2=y*k2y2,∂2V(x,y)∂x∂y=0. \eqalign{& V\left({{x^*},{y^*}} \right) = 0,{{\partial V\left({x,y} \right)} \over {\partial x}} = {1 \over {{k_1}}}\left({1 - {{{x^*}} \over x}} \right),{{\partial V\left({x,y} \right)} \over {\partial y}} = {1 \over {{k_2}}}\left({1 - {{{y^*}} \over y}} \right), \cr & {{{\partial ^2}V\left({x,y} \right)} \over {\partial {x^2}}} = {{{x^*}} \over {{k_1}{x^2}}},{{{\partial ^2}V\left({x,y} \right)} \over {\partial {y^2}}} = {{{y^*}} \over {{k_2}{y^2}}},{{{\partial ^2}V\left({x,y} \right)} \over {\partial x\partial y}} = 0. \cr}

It follows that V (x, y) is a positive definite function inside R2, and as it approaches the x or y axis, it satisfies the condition V(x, y) → +∞, so it is a function with infinity inside. By analyzing it in accordance with the solution ideas of the above formula, the total derivative can be obtained as follows: V*(x,y)=−(x−x*)(x−x*M1+b1(y−y*)M2)−(y−y*)(b2(x−x*)M1+y−y*M2)=−[1M1(x−x*)2+(b2M1+b1M2)(x−x*)(y−y*)+1M2(y−y*)2] \eqalign{& \mathop V\limits^* \left({x,y} \right) = - \left({x - {x^*}} \right)\left({{{x - {x^*}} \over {{M_1}}} + {{{b_1}\left({y - {y^*}} \right)} \over {{M_2}}}} \right) - \left({y - {y^*}} \right)\left({{{{b_2}\left({x - {x^*}} \right)} \over {{M_1}}} + {{y - {y^*}} \over {{M_2}}}} \right) \cr & = - \left[{{1 \over {{M_1}}}{{\left({x - {x^*}} \right)}^2} + \left({{{{b_2}} \over {{M_1}}} + {{{b_1}} \over {{M_2}}}} \right)\left({x - {x^*}} \right)\left({y - {y^*}} \right) + {1 \over {{M_2}}}{{\left({y - {y^*}} \right)}^2}} \right] \cr}

V*(x,y) \mathop V\limits^* \left({x,y} \right) stands for quadratic homogeneous functions of (x-x*) and (y-y*). And in accordance with the condition of Δ=(b2M1+b1M2)2−4M1M2<0 \Delta = {\left({{{{b_2}} \over {{M_1}}} + {{{b_1}} \over {{M_2}}}} \right)^2} - {4 \over {{M_1}{M_2}}} < 0 , V*(x,y) \mathop V\limits^* \left({x,y} \right) is negative definite, and the simplified formula is as follows: Δ=(b1M+b2M2)2−4M1M2M12M22<0 \Delta = {{{{\left({{b_1}M + {b_2}{M_2}} \right)}^2} - 4{M_1}{M_2}} \over {M_1^2M_2^2}} < 0

Thus, it can be obtained: (b1M1+b2M2)2−4M1M2<0 {\left({{b_1}{M_1} + {b_2}{M_2}} \right)^2} - 4{M_1}{M_2} < 0

Since the corresponding symbols of X and y have been identified in the study, the following requirements should be met in order to ensure the invariable sign of D: {k1α+k2b2β+k1=0k1b1α+k2β+k2=0 \left\{\matrix{{k_1}\alpha + {k_2}{b_2}\beta + {k_1} = 0 \hfill \cr {k_1}{b_1}\alpha + {k_2}\beta + {k_2} = 0 \hfill \cr} \right.

From this, it can be calculated: α=k2b2−k1k1(1−b2b2),β=k1b1−k2k2(1−b1b2), \alpha = {{{k_2}{b_2} - {k_1}} \over {{k_1}\left({1 - {b_2}{b_2}} \right)}},\beta = {{{k_1}{b_1} - {k_2}} \over {{k_2}\left({1 - {b_1}{b_2}} \right)}},

And: D=k1(b1−1)+k2(b2−1)1−b1b2xα−1yβ−1 D = {{{k_1}\left({{b_1} - 1} \right) + {k_2}\left({{b_2} - 1} \right)} \over {1 - {b_1}{b_2}}}{x^{\alpha - 1}}{y^{\beta - 1}}

Assuming constant k₁(b₁−1)+k₂ (b₂−1) ≠ 0 meets this condition, it is proved that the system does not have closed orbits. If the constant k₁ (b₁ −1)+ k₂ (b₂ −1) = 0 satisfies this condition, then the proof system also has no limit cycles.

Researchers proposed a complementary differential equation model of educational innovation with the Logistic equation as the core, as shown below[3]: {dxdt=k1x(t)(1−x(t)M1+b1y(t)M2)dydt=k2y(t)(1−x(t)M2+b2y(t)M1) \left\{\matrix{{{dx} \over {dt}} = {k_1}x\left(t \right)\left({1 - {{x\left(t \right)} \over {{M_1}}} + {b_1}{{y\left(t \right)} \over {{M_2}}}} \right) \hfill \cr {{dy} \over {dt}} = {k_2}y\left(t \right)\left({1 - {{x\left(t \right)} \over {{M_2}}} + {b_2}{{y\left(t \right)} \over {{M_1}}}} \right) \hfill \cr} \right.

In the above model, the positive equilibrium point P1(M1(1+b1)1−b1b2,M2(1+b2)1−b1b2) {P_1}\left({{{{M_1}\left({1 + {b_1}} \right)} \over {1 - {b_1}{b_2}}},{{{M_2}\left({1 + {b_2}} \right)} \over {1 - {b_1}{b_2}}}} \right) is globally uniformly asymptotically stable where X and Y coexist and satisfy the conditions 0<b2<M1M22−b1M1M2 0 < {b_2} < {{\root 2 \of {{M_1}{M_2}} - {b_1}{M_1}} \over {{M_2}}} and 0<M1M22−M2M1<b1<1 0 < {{\root 2 \of {{M_1}{M_2}} - {M_2}} \over {{M_1}}} < {b_1} < 1 .

The equilibrium point of the above model is p1(M1(1+b1)1−b1b2,M2(1+b2)1−b1b2) {p_1}\left({{{{M_1}\left({1 + {b_1}} \right)} \over {1 - {b_1}{b_2}}},{{{M_2}\left({1 + {b_2}} \right)} \over {1 - {b_1}{b_2}}}} \right) , p₂ (M₁,0), p₃ (0, M₂) p4 (0,0). In the case of coexistence of X and Y, research and analysis can be carried out under the condition of R+2={(x,y)|x>0,y>0} R_ + ^2 = \left\{{\left({x,y} \right)|x > 0,y > 0} \right\} , and only the equilibrium point P1 can be discussed.

Suppose p₁(x* y*), where x*=M1(1−b1)1−b1b2>0 {x^*} = {{{M_1}\left({1 - {b_1}} \right)} \over {1 - {b_1}{b_2}}} > 0 , y*=M2(1−b2)1−b1b2>0 {y^*} = {{{M_2}\left({1 - {b_2}} \right)} \over {1 - {b_1}{b_2}}} > 0 , 0 < b₁, b₂ < 1, then the following equation is satisfied: {1M1x*−b1M2y*=1−b2M1x*+1M2y*=1 \left\{\matrix{{1 \over {{M_1}}}{x^*} - {{{b_1}} \over {{M_2}}}{y^*} = 1 \hfill \cr - {{{b_2}} \over {{M_1}}}{x^*} + {1 \over {{M_2}}}{y^*} = 1 \hfill \cr} \right.

Convert the above formula to: {dxdt=−k1x(t)(x−x*M1−b1M2(y−y*))dydt=−k2y(t)(−b2M1(x−x*)+1M2(y−y*)) \left\{\matrix{{{dx} \over {dt}} = - {k_1}x\left(t \right)\left({{{x - {x^*}} \over {{M_1}}} + {{{b_1}} \over {{M_2}}}\left({y - {y^*}} \right)} \right) \hfill \cr {{dy} \over {dt}} = - {k_2}y\left(t \right)\left({{{{b_2}} \over {{M_1}}}\left({x - {x^*}} \right) - {1 \over {{M_2}}}\left({y - {y^*}} \right)} \right) \hfill \cr} \right.

And get: V(x,y)=1k1(x−x*−x*lnxx*)+1k2(y−y*−y*lnyy*) V\left({x,y} \right) = {1 \over {{k_1}}}\left({x - {x^*} - {x^*}\ln {x \over {{x^*}}}} \right) + {1 \over {{k_2}}}\left({y - {y^*} - {y^*}\ln {y \over {{y^*}}}} \right)

The actual calculation results are:

So V of x, y is a positive definite function inside R2, and as (x, y) approaches the x and y axes, we get V(x, y) → +∞.

The total derivative of V (x, y) according to the above model is: V*(x,y)=−(x−x*)(x−x*M1−b1(y−y*)M2)−(y−y*)(−b2(x+x*)M1+y−y*M2)=−[1M1(x−x*)2−(b2M1+b1M2)(x−x*)(y−y*)+1M2(y−y*)2] \eqalign{& \mathop V\limits^* \left({x,y} \right) = - \left({x - {x^*}} \right)\left({{{x - {x^*}} \over {{M_1}}} - {{{b_1}\left({y - {y^*}} \right)} \over {{M_2}}}} \right) - \left({y - {y^*}} \right)\left({- {{{b_2}\left({x - {x^*}} \right)} \over {{M_1}}} + {{y - {y^*}} \over {{M_2}}}} \right) \cr & = - \left[{{1 \over {{M_1}}}{{\left({x - {x^*}} \right)}^2} - \left({{{{b_2}} \over {{M_1}}} + {{{b_1}} \over {{M_2}}}} \right)\left({x - {x^*}} \right)\left({y - {y^*}} \right) + {1 \over {{M_2}}}{{\left({y - {y^*}} \right)}^2}} \right] \cr}

Under the condition of b₁b₂ ≠ 1, limit cycles will not appear outside the equilibrium point p1(M1(1−b1)1−b1b2,M2(1−b2)1−b1b2) {p_1}\left({{{{M_1}\left({1 - {b_1}} \right)} \over {1 - {b_1}{b_2}}},{{{M_2}\left({1 - {b_2}} \right)} \over {1 - {b_1}{b_2}}}} \right) inside the system.

After the symbols of x and y are specified, in order to keep D unchanged, the following requirements must be met: {k1α−k2b2β+k1=0k2β−k1b1α+k2=0 \left\{\matrix{{k_1}\alpha - {k_2}{b_2}\beta + {k_1} = 0 \hfill \cr {k_2}\beta - {k_1}{b_1}\alpha + {k_2} = 0 \hfill \cr} \right.

It can be calculated as follows: α=k1+k2b2k1(b1b2−1),β=k1b1+k2k2(b1b2−1) \alpha = {{{k_1} + {k_2}{b_2}} \over {{k_1}\left({{b_1}{b_2} - 1} \right)}},\beta = {{{k_1}{b_1} + {k_2}} \over {{k_2}\left({{b_1}{b_2} - 1} \right)}}

And: D=k1(1+b1)+k2(1+b2)1−b1b2xα−1yβ−1 D = {{{k_1}\left({1 + {b_1}} \right) + {k_2}\left({1 + {b_2}} \right)} \over {1 - {b_1}{b_2}}}{x^{\alpha - 1}}{y^{\beta - 1}}

Assuming constant k₁(b₁ −1)+ k₂ (b₂ −1) ≠ 0 meets this condition, combined with the above studies, it is proved that the system does not have closed orbits.