求解数学中的双调和方程

发布于 2024-12-10 17:30:37 字数 487 浏览 1 评论 0原文

我正在尝试使用 DSolve 求解 mathematica 中的线性双调和方程。我认为这个问题不仅限于双调和方程，而且当我试图求解它时，数学只是吐出方程。

我尝试过求解其他偏微分方程，没有遇到任何问题。

双调和方程就是：

Laplacian^2[f]=0

这是我的方程：

DSolve[
 D[f[x, y], {x, 4}] + 2 D[D[f[x, y], {x, 2}, {y, 2}]] + 
   D[f[x, y], {y, 4}] == 0,
 f,
 {x, y}]

解被吐出为

DSolve[(f^(0,4))[x,y]+2 (f^(2,2))[x,y]+(f^(4,0))[x,y]==0,f,{x,y}]

这显然不是解。什么给？我缺少什么？我已经在没有边界条件的情况下解决了其他偏微分方程。

原文

I am attempting to solve the linear biharmonic equation in mathematica using DSolve. I think this issue is not just limited to the biharmonic equation but MATHEMATICA just spits out the equation when I attempt to solve it.

I've tried solving other partial differential equations and there was no trouble.

The biharmonic equation is just:

Laplacian^2[f]=0

Here is my equation:

DSolve[
 D[f[x, y], {x, 4}] + 2 D[D[f[x, y], {x, 2}, {y, 2}]] + 
   D[f[x, y], {y, 4}] == 0,
 f,
 {x, y}]

The solution is spit out as

DSolve[(f^(0,4))[x,y]+2 (f^(2,2))[x,y]+(f^(4,0))[x,y]==0,f,{x,y}]

That is obviously not the solution. What gives? What am I missing? I've solved other PDEs without boundary conditions.

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宫墨修音 2024-12-17 17:30:37

在极坐标下试试怎么样？如果 f(r, \[Theta]) 相对于方位角 \[Theta] 对称，则双调和方程可简化为 Mathematca 可以符号求解的值（参见 http://mathworld.wolfram.com/BiharmonicEquation.html）：

In[22]:= eq = D[r D[D[r D[f[r],r],r]/r,r],r]/r;
eq//FullSimplify//TraditionalForm

Out[23]//TraditionalForm= f^(4)(r) + (2 r^2 f^(3)(r) - r f''(r)
                           + f'(r))/r^3

In[24]:= DSolve[eq==0,f,r]
Out[24]= {{f -> Function[{r}, 
                 1/2 r^2 C[2] - 1/4 r^2 C[3] + C[4] + C[1] Log[r] 
                   + 1/2 r^2 C[3] Log[r]
                ]}}

In[25]:= ReplaceAll[
    1/2 r^2 C[2]-1/4 r^2 C[3]+C[4]+C[1] Log[r]+1/2 r^2 C[3] Log[r],
    r->Sqrt[x^2+y^2]
]
Out[25]= 1/2 (x^2+y^2) C[2]-1/4 (x^2+y^2) C[3]+C[4]+C[1] Log[Sqrt[x^2+y^2]]+ 
1/2 (x^2+y^2) C[3] Log[Sqrt[x^2+y^2]]

How about try it in polar coordinates? If f(r, \[Theta]) is symmetric with respect to azimuth \[Theta], the biharmonic equation reduces to something Mathematca can solve symbolically (c.f. http://mathworld.wolfram.com/BiharmonicEquation.html):

In[22]:= eq = D[r D[D[r D[f[r],r],r]/r,r],r]/r;
eq//FullSimplify//TraditionalForm

Out[23]//TraditionalForm= f^(4)(r) + (2 r^2 f^(3)(r) - r f''(r)
                           + f'(r))/r^3

In[24]:= DSolve[eq==0,f,r]
Out[24]= {{f -> Function[{r}, 
                 1/2 r^2 C[2] - 1/4 r^2 C[3] + C[4] + C[1] Log[r] 
                   + 1/2 r^2 C[3] Log[r]
                ]}}

In[25]:= ReplaceAll[
    1/2 r^2 C[2]-1/4 r^2 C[3]+C[4]+C[1] Log[r]+1/2 r^2 C[3] Log[r],
    r->Sqrt[x^2+y^2]
]
Out[25]= 1/2 (x^2+y^2) C[2]-1/4 (x^2+y^2) C[3]+C[4]+C[1] Log[Sqrt[x^2+y^2]]+ 
1/2 (x^2+y^2) C[3] Log[Sqrt[x^2+y^2]]

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