微分方程BVP [英] differential equations BVP
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问题描述
#包括微分方程和参数
def模型(x,dydx,p):
s = p [0]#第一个参数
a = p [1]#第二个参数
dydx [0] = -2 *(s + a)* y [0] + 2 * s * y [1] + s / 2 * y [2]
dydx [1] = + 2 *(s + a)* y [1] -2 * s * y [0] -s / 2 * y [ 2]
dydx [2] =-(s + a)* y [2]
return np.vstack(dydx [0],dydx [1],dydx [2])
#边界条件
def bc(ya,yb,yc,p):
s = p [0]
a = p [1]
I0 = 1
返回np.array(([ya [0],yb [0],yc [0],a,s]))
#x值
x = np.array([1、2、3、4、5、6、7、8、9])
y = np.zeros((3,x.size))
# p = np.zeros((3,d.size))
#y初始值
y [0] = 3
y [1] = 3
y [2] = 2
I0 = 1
sol = solve_bvp(模型,bc,x,y,p = [1,1])$ b $ b
我不知道如何写边界条件来求解三个微分方程
I w ant求解方程并具有 y
值和参数值
解决方案
对于其他问题,您有固定的参数和3个边界条件。这需要被编码为
#边界条件
def bc(y0,yd ,I0):
返回np.array([y0 [0],y0 [2] -I0,yd [1]])
然后,初始 x
数组需要以边界点为边界
<前置类= lang-py prettyprint-override>
x = np.linspace(0,d,9)
y = np.zeros((3,x.size))
#y初始值
y [0] = 3
y [1] = 3
y [2] = 2
,并且需要在不使用可变参数的情况下调用求解器,并通过包装器/局部求值将它们全部固定为常数值
sol = solve_bvp(lambda t,y:model(t,y,[s,a]),lambda y0,yd:bc(y0,yd,I0) ,x,y)
#including the differential equations and parameters
def model(x,dydx,p):
s=p[0] #first parameter
a=p[1] #second parameter
dydx[0] = -2*(s+a)*y[0]+2*s*y[1]+s/2*y[2]
dydx[1] = +2*(s+a)*y[1]-2*s*y[0]-s/2*y[2]
dydx[2] = -(s+a)*y[2]
return np.vstack(dydx[0],dydx[1],dydx[2])
# boundary conditions
def bc(ya, yb,yc, p):
s=p[0]
a=p[1]
I0=1
return np.array(([ya[0], yb[0],yc[0],a,s]))
#x values
x = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9])
y = np.zeros((3, x.size))
#p= np.zeros((3, d.size))
#y initial values
y[0]=3
y[1]=3
y[2]=2
I0 = 1
sol = solve_bvp(model, bc,x,y, p=[1,1])
I do not know how can I write the boundary conditions to solve the three differential equations
I want to solve the equations and have the y
values and parameters values
解决方案
Per the other question you have fixed parameters and 3 boundary conditions. This needs to be encoded as
# boundary conditions
def bc(y0, yd, I0):
return np.array([y0[0], y0[2]-I0, yd[1]])
Then the initial x
array needs to be bounded by the boundary points
x = np.linspace(0,d,9)
y = np.zeros((3, x.size))
#y initial values
y[0]=3
y[1]=3
y[2]=2
and the solver needs to be called without variable parameters, fixing them all to their constant values via wrappers/partial evaluation
sol = solve_bvp(lambda t,y:model(t,y,[s,a]), lambda y0,yd: bc(y0,yd,I0), x,y)
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