状态到状态二元混合物动力学的Julia微分方程

我对朱莉娅是个新手，我正在考虑以下问题。 我想解（可能是刚性的）ODE系统，它描述了激波后流动的弛豫，根据一种状态到状态的方法，这意味着分子物种的每个振动能级都被视为一个具有连续性方程的伪物种。 在这里，我考虑了N2/N的二元混合物（实际上n＝0的浓度）。 我已将julia代码拆分为几个.jl文件。大体上，我调用ODE解算器，如下所示：

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，我在rpart.jl文件中定义了系统：

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，看起来好像什么都没有发生，所有的剖面都是相等和平坦的。事实上，每个时间步的解都是相等的。所以，我想我在u和du的更新中犯了一些错误，但我无法修复它。 在Matlab版本中，我获得了正确的进化

亲切问候,，

---

Lorenzo

您正在使用该版本对输出进行变异，但您正在创建一个数组，而不是对输出进行变异

du.=AA*B

谢谢，这就解决了问题。或者，我也可以把mul干掉！（du，AA，B）？是的，

mul会更好。