Julia DifferentialEquations for state-to-state binary mixture kinetics

Question

我对 Julia 很陌生，我正在考虑以下问题。我想解决（可能是刚性的）ODE 系统，它根据状态到状态的方法描述冲击波后面的流动的松弛，这意味着分子物种的每个振动水平都被视为伪物种及其连续性方程。这里，我考虑的是N2/N的二元混合（实际上是N=0的浓度）。

我已将 julia 代码拆分为多个 .jl 文件。大体上，我调用 ODE 求解器如下：

prob = ODEProblem(rpart!,Y0_bar,xspan, 1.)                                                                                          
sol  = DifferentialEquations.solve(prob, Tsit5(), reltol=1e-8, abstol=1e-8, save_everystep=true, progress=true)

其中 Y0_bar 和 xspan 已在前面定义，在 rpart.jl 文件中我定义了系统：

function rpart!(du,u,p,t)

  ni_b = zeros(l);

  ni_b[1:l] = u[1:l]; print("ni_b = ", ni_b, "\n")
  na_b = u[l+1];      print("na_b = ", na_b, "\n")
  v_b  = u[l+2];      print("v_b = ",  v_b,  "\n")
  T_b  = u[l+3];      print("T_b = ",  T_b,  "\n")

  nm_b = sum(ni_b);   #print("nm_b = ", nm_b, "\n")
  Lmax = l-1;         #println("Lmax = ", Lmax, "\n")
  temp = T_b*T0;      #print("T = ", temp, "\n")

  ef_b = 0.5*D/T0;    #println("ef_b = ", ef_b, "\n")

  ei_b = e_i./(k*T0); #println("ei_b = ", ei_b, "\n")
  e0_b = e_0/(k*T0);  #println("e0_b = ", e0_b, "\n")

  sigma   = 2.;                     #println("sigma = ", sigma, "\n")
  Theta_r = Be*h*c/k;               #println("Theta_r = ", Theta_r, "\n")
  Z_rot   = temp./(sigma.*Theta_r); #println("Z_rot = ", Z_rot, "\n")

  M  = sum(m); #println("M = ", M, "\n")
  mb = m/M;    #println("mb = ", mb, "\n")

  A = zeros(l+3,l+3)

  for i = 1:l
    A[i,i]   = v_b
    A[i,l+2] = ni_b[i]
  end

  A[l+1,l+1] = v_b
  A[l+1,l+2] = na_b

  for i = 1:l+1
    A[l+2,i] = T_b
  end
  A[l+2,l+2] = M*v0^2/k/T0*(mb[1]*nm_b+mb[2]*na_b)*v_b
  A[l+2,l+3] = nm_b+na_b

  for i = 1:l
    A[l+3,i] = 2.5*T_b+ei_b[i]+e0_b
  end
  A[l+3,l+1] = 1.5*T_b+ef_b
  A[l+3,l+2] = 1/v_b*(3.5*nm_b*T_b+2.5*na_b*T_b+sum((ei_b.+e0_b).*ni_b)+ef_b*na_b)
  A[l+3,l+3] = 2.5*nm_b+1.5*na_b

  AA = inv(A); println("AA = ", AA, "\n", size(AA), "\n")

  # Equilibrium constant for DR processes
  Kdr = (m[1]*h^2/(m[2]*m[2]*2*pi*k*temp))^(3/2)*Z_rot*exp.(-e_i/(k*temp))*exp(D/temp); println("Kdr = ", Kdr, "\n")

  # Equilibrium constant for VT processes
  Kvt = exp.((e_i[1:end-1]-e_i[2:end])/(k*temp)); println("Kvt = ", Kvt, "\n")

  # Dissociation processes
  kd = zeros(2,l)
  kd = kdis(temp) * Delta*n0/v0;
  println("kd = ", kd, "\n", size(kd), "\n")

  # Recombination processes
  kr = zeros(2,l)
  for iM = 1:2
    kr[iM,:] = kd[iM,:] .* Kdr * n0
  end
  println("kr = ", kr, "\n", size(kr), "\n")

  RD  = zeros(l)

  for i1 = 1:l

    RD[i1] = nm_b*(na_b*na_b*kr[1,i1]-ni_b[i1]*kd[1,i1]) + na_b*(na_b*na_b*kr[2,i1]-ni_b[i1]*kd[2,i1])

  end

  println("RD = ",  RD,  "\n", size(RD))

  B      = zeros(l+3)
  for i  = 1:l
    B[i] = RD[i]
  end
  B[l+1] = - 2*sum(RD)

  du     = AA*B

end

问题是，当我运行模拟并绘制解决方案时，它看起来什么也没发生，所有配置文件都相等且平坦。事实上，每个时间步的解都等于它自己。所以，我想我在更新 u 和 du 时犯了一些错误，但我无法修复它。在Matlab版本中我得到了正确的进化。

亲切的问候，洛伦佐

Answer 1

您正在使用版本来改变输出，但您正在创建一个数组而不是改变输出。 du .= AA*B

Julia DifferentialEquations for state-to-state binary mixture kinetics

Julia DifferentialEquations for state-to-state binary mixture kinetics

ode

julia

differentialequations.jl