Mathematica 中的非线性 PDE 求解

Question

我正在尝试使用 NDsolve 求解以下与 ginzburg Landau 相关的非线性耦合 PDE 方程。我是 Mathematica 的新手。我收到以下错误。我在做什么？

pde = {D[u[t, x, y], t] == 
D[u[t, x, y], {x, x}] + 
 D[u[t, x, y], {y, 
   y}] - (1/u[t, x, y])^3*(D[v[t, x, y], y]^2 + 
    D[v[t, x, y], x]^2) - u[t, x, y] + u[t, x, y]^3, 
   D[v[t, x, y], t] == 
D[v[t, x, y], {x, x}] + D[v[t, x, y], {y, y}] - 
 v[t, x, y]*u[t, x, y] + 
       (2/u[t, x, y])*(D[u[t, x, y], x]*D[v[t, x, y], x] - 
    D[u[t, x, y], y]*D[v[t, x, y], y])};bc = {u[0, x, y] == 0, v[0, x, y]== 0, u[t, 5, y] == 1, u[t, x, 5] == 1, D[v[t, 0, y], x] == 0, D[v[t, x, 0], y] == 0};
NDSolve[{pde, bc}, {u, v}, {x, 0, 5}, {y, 0, 5}, {t, 0, 2}]

'Error: NDSolve::deqn: Equation or list of equations expected instead of True in the first argument {{(u^(1,0,0))[t,x,y]==-u[t,x,y]+u[t,x,y]^3+(u^(0,0,y))[t,x,y]-((<<1>>^(<<3>>))[<<3>>]^2+(<<1>>^(<<3>>))[<<3>>]^2)/u[t,x,y]^3+(u^(0,x,0))[t,x,y],(v^(1,0,0))[t,x,y]==-u[t,x,y] v[t,x,y]+(v^(0,0,y))[t,x,y]+(2 (-(<<1>>^(<<3>>))[<<3>>] (<<1>>^(<<3>>))[<<3>>]+(<<1>>^(<<3>>))[<<3>>] (<<1>>^(<<3>>))[<<3>>]))/u[t,x,y]+(v^(0,x,0))[t,x,y]},{u[0,x,y]==0,v[0,x,y]==0,u[t,5,y]==1,u[t,x,5]==1,True,True}}.

NDSolve[{{Derivative[1, 0, 0][u][t, x, y] == -u[t, x, y] + 
     u[t, x, y]^3 + Derivative[0, 0, y][u][t, x, y] - 
              (Derivative[0, 0, 1][v][t, x, y]^2 + 
     Derivative[0, 1, 0][v][t, x, y]^2)/u[t, x, y]^3 + 
     Derivative[0, x, 0][u][t, x, y], 
       Derivative[1, 0, 0][v][t, x, y] == (-u[t, x, y])*v[t, x, y] + 
     Derivative[0, 0, y][v][t, x, y] + 
           (2*((-Derivative[0, 0, 1][u][t, x, y])*
           Derivative[0, 0, 1][v][t, x, y] + 
          Derivative[0, 1, 0][u][t, x, y]*
           Derivative[0, 1, 0][v][t, x, y]))/u[t, x, y] + 
           Derivative[0, x, 0][v][t, x, y]}, {u[0, x, y] == 0, 
   v[0, x, y] == 0, u[t, 5, y] == 1, u[t, x, 5] == 1, True, 
   True}}, {u, v}, {x, 0, 5}, {y, 0, 5}, {t, 0, 2}]

Answer 1

如果您查看bc 的值，您会看到

bc = {u[0, x, y] == 0, v[0, x, y] == 0, u[t, 5, y] == 1, 
u[t, x, 5] == 1, D[v[t, 0, y], x] == 0, D[v[t, x, 0], y] == 0}

给你

{u[0, x, y] == 0, v[0, x, y] == 0, u[t, 5, y] == 1, u[t, x, 5] == 1, True, True}

这就是您关于 True 的错误消息的来源。

您所做的是对表达式进行微分，但表达式中没有 x，因此结果为零。并且 0==0 始终为真。同样与 y。所以让我们改变你试图告诉它边界条件的方式。

bc = {u[0, x, y] == 0, v[0, x, y] == 0, u[t, 5, y] == 1, u[t, x, 5] == 1,
Derivative[0, 1, 0][v][t, 0, y] == 0, Derivative[0, 0, 1][v][t, x, 0] == 0}

或

bc = {u[0, x, y] == 0, v[0, x, y] == 0, u[t, 5, y] == 1, u[t, x, 5] == 1,
D[v[t, x, y], x] == 0/.x->0, D[v[t, x, y], y] == 0/.y->0}

我认为其中任何一个都可以满足您的需求。

一旦你修复了这些，你就会得到一个关于导数顺序和非负整数的不同错误。

我相信你可以通过将 pde 从 {x,x} 和 {y,y} 更改为 {x,2} 和 {y,2} 来解决这个问题

pde = {D[u[t, x, y], t] == D[u[t, x, y], {x, 2}] + D[u[t, x, y], {y, 2}] -
  (1/u[t, x, y])^3*(D[v[t, x, y], y]^2 + D[v[t, x, y], x]^2) - u[t, x, y] +
  u[t, x, y]^3,
  D[v[t, x, y], t] == D[v[t, x, y], {x, 2}] + D[v[t, x, y], {y, 2}] - v[t, x, y]*
  u[t, x, y] + (2/u[t, x, y])*(D[u[t, x, y], x]*D[v[t, x, y], x] - D[u[t, x, y], y]*
  D[v[t, x, y], y])};

这会让错误消失。

一旦你解决了这个问题并尝试你的NDSolve，那么你的分母中的零就会开始咬你。

修复这些看起来不仅仅是理解 MMA 语法。这可能需要了解您的问题并查看是否可以消除那些零分母。