t*****r
发帖数: 42 | 1 假设a1,a2,b1,b2都大于零，如何计算下面这个二元常微分方程组的显式解？
dx/dt=a1(1-x-y)x-b1*x,
dy/dt=a2(1-x-y)y-b2*y,
需要显式解，不要数值解。非常感谢！ | o*******w
发帖数: 349 | 2 You can use coordinate transformation on {x,y} to change the original
equation to
dx/dt = a(1-x-y)x - b1*x,
dy/dt = a(1-x-y)y - b2*y,
(i.e. we can only consider the case of a1=a2)
Dividing by x (and y respectively) both hand sides yields
(dx/dt)/x + b1 = a(1-x-y)
(dy/dt)/y + b2 = a(1-x-y)
Thus you have
(dx/dt)/x + b1 = (dy/dt)/y + b2
i.e.
d/dt (ln x + b1*t) = d/dt (ln y + b2*t)
=> d/dt (ln x + b1*t) - d/dt(ln y + b2*t) = 0
ln(x/y) + [b1 - b2]*t = arbitrary constant
x/y = C*exp(b2 - b1)t
So
x = C*y* exp(b2 - b1]*t or y = C*x*exp(b1 - b2)t
so a(1 - x - y)*x = a(1 - C*y* exp(b2 - b1)*t - y)*C*y*exp(b2 - b1]*t
From dx/dt = a(1-x-y)x - b1*x, you now have
dy/dt * C exp(b2-b1)*t + (b1-b2)C*y*exp(b1-b2)t
= a(1 - C*y* exp(b2 - b1)*t - y)*C*y* exp(b2 - b1]*t
=>
dy/dt + (b1-b2)y = a(1 - C*y* exp(b2 - b1)*t - y) y
Divinding by y^2 on both sides, you have
(dy/dt)/y^2 + (b1-b2)/y = a(1/y - C* exp(b2 - b1)*t - 1)
(dy/dt)/y^2 + (b1-b2)/y = a(1/y - C* exp(b2 - b1)*t - 1)
- d(1/y)/dt + (b1-b2)/y = a(1/y - C* exp(b2 - b1)*t - 1)
Let 1/y be a new variable u and you get this form of differential equation,
-du/dt + A*u + B*exp(b2 - b1)t + C = 0
This is a perfectly solvable equation. You can look at website eqWorld for
the solution formula.
【在 t*****r 的大作中提到】
: 假设a1,a2,b1,b2都大于零，如何计算下面这个二元常微分方程组的显式解？
: dx/dt=a1(1-x-y)x-b1*x,
: dy/dt=a2(1-x-y)y-b2*y,
: 需要显式解，不要数值解。非常感谢！
| t*****r
发帖数: 42 | 3 dy/dt * C exp(b2-b1)*t + (b1-b2)C*y*exp(b1-b2)t
= a(1 - C*y* exp(b2 - b1)*t - y)*C*y* exp(b2 - b1]*t
这步有点问题，应该是
dy/dt * C exp(b2-b1)*t + b2*C*y*exp(b1-b2)t
= a(1 - C*y* exp(b2 - b1)*t - y)*C*y* exp(b2 - b1]*t
谢谢你的帮助。
【在 o*******w 的大作中提到】
: You can use coordinate transformation on {x,y} to change the original
: equation to
: dx/dt = a(1-x-y)x - b1*x,
: dy/dt = a(1-x-y)y - b2*y,
: (i.e. we can only consider the case of a1=a2)
: Dividing by x (and y respectively) both hand sides yields
: (dx/dt)/x + b1 = a(1-x-y)
: (dy/dt)/y + b2 = a(1-x-y)
: Thus you have
: (dx/dt)/x + b1 = (dy/dt)/y + b2
| o*******w
发帖数: 349 | 4 Right, should be b2-b1 in all
【在 t*****r 的大作中提到】
: dy/dt * C exp(b2-b1)*t + (b1-b2)C*y*exp(b1-b2)t
: = a(1 - C*y* exp(b2 - b1)*t - y)*C*y* exp(b2 - b1]*t
: 这步有点问题，应该是
: dy/dt * C exp(b2-b1)*t + b2*C*y*exp(b1-b2)t
: = a(1 - C*y* exp(b2 - b1)*t - y)*C*y* exp(b2 - b1]*t
: 谢谢你的帮助。
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