differential equations BVP

三世轮回 提交于 2020-05-24 07:54:05

问题


#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


回答1:


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) 


来源:https://stackoverflow.com/questions/60697878/differential-equations-bvp

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