常密度无粘流体：如何从欧拉运动方程推导压力表达式

Alright, let's break down how to derive the pressure expression from the given momentum equation (Euler equation) for an inviscid, constant-density fluid. We’ll rely on vector calculus identities and the continuity equation to simplify things step by step:

• Step 1: Start with the continuity equation (key simplification)
  For a constant-density fluid, the continuity equation reduces to:
  $$\nabla \cdot \mathbf{u} = 0$$
  This tells us the velocity field is solenoidal, which will eliminate several terms later on.

• Step 2: Take the divergence of the Euler equation
  Begin with the given equation:
  $$\frac{\partial \mathbf{u}}{\partial t} + \left(\mathbf{u} \cdot \nabla\right)\mathbf{u} = -\frac{1}{\rho}\nabla P + \mathbf{f}$$
  Apply the divergence operator $\nabla \cdot$ to both sides. This transforms the vector momentum equation into a scalar equation involving pressure:
  $$\nabla \cdot \frac{\partial \mathbf{u}}{\partial t} + \nabla \cdot \left[\left(\mathbf{u} \cdot \nabla\right)\mathbf{u}\right] = -\frac{1}{\rho}\nabla^2 P + \nabla \cdot \mathbf{f}$$

• Step 3: Simplify the left-hand side (LHS)
  First, swap the order of time differentiation and divergence (they commute, so this is valid):
  $$\frac{\partial}{\partial t}\left(\nabla \cdot \mathbf{u}\right) + \nabla \cdot \left[\left(\mathbf{u} \cdot \nabla\right)\mathbf{u}\right] = -\frac{1}{\rho}\nabla^2 P + \nabla \cdot \mathbf{f}$$
  From the continuity equation, $\nabla \cdot \mathbf{u} = 0$, so the first term disappears entirely. Next, use the vector identity for convective acceleration to rewrite the remaining term:
  $$\left(\mathbf{u} \cdot \nabla\right)\mathbf{u} = \frac{1}{2}\nabla\left(|\mathbf{u}|^2\right) - \mathbf{u} \times \left(\nabla \times \mathbf{u}\right)$$
  Take the divergence of this identity. The divergence of a cross product is always zero (you can verify this using the vector identity for divergence of cross products), so we’re left with:
  $$\nabla \cdot \left[\left(\mathbf{u} \cdot \nabla\right)\mathbf{u}\right] = \frac{1}{2}\nabla^{2\left(|\mathbf{u}|}2\right)$$

• Step 4: Rearrange to get the Poisson equation for pressure
  Substitute the simplified LHS back into our equation:
  $$\frac{1}{2}\nabla^{2\left(|\mathbf{u}|}2\right) = -\frac{1}{\rho}\nabla^2 P + \nabla \cdot \mathbf{f}$$
  Rearrange terms to isolate the Laplacian of pressure:
  $$\nabla^2 P = -\rho\left[\frac{1}{2}\nabla^{2\left(|\mathbf{u}|}2\right) - \nabla \cdot \mathbf{f}\right]$$
  Alternatively, using the original convective acceleration term, this can also be written as:
  $$\nabla^2 P = -\rho \nabla \cdot \left(\left(\mathbf{u} \cdot \nabla\right)\mathbf{u}\right) + \rho \nabla \cdot \mathbf{f}$$

• Step 5: Solve for the explicit pressure expression
  The equation above is a Poisson equation for pressure. To get the explicit form of $P(\mathbf{x}, t)$, we use the fundamental solution (Green’s function) for the 3D Laplacian. For a fluid domain with appropriate boundary conditions, the solution is:
  $$P(\mathbf{x}, t) = \frac{\rho}{4\pi} \int \frac{\nabla' \cdot \left(\left(\mathbf{u}(\mathbf{x}', t) \cdot \nabla'\right)\mathbf{u}(\mathbf{x}', t)\right) - \nabla' \cdot \mathbf{f}(\mathbf{x}', t)}{|\mathbf{x} - \mathbf{x}'|} dV' + \text{boundary contribution}$$
  The boundary contribution depends on the specific problem’s constraints—for example, prescribed pressure at a boundary or normal velocity conditions.

内容的提问来源于stack exchange，提问作者Sam Plant