P18

Incompressible, Viscous Navier-Stokes Equations

P19

Equation Fomulation

✅ 这是一个描述了速度场的公式，它可以告许你速度如何更新。公式1限制不可压，公式2 diffusion 的目的是粘滞。

Method of Characteristics: solving a long partial differential equation (PDE) in steps

• Step 1: Update $\mathbf{u}$ by solving $∂\mathbf{u}∕∂t=\mathbf{g}$
• Step 2: Update $\mathbf{u}$ by solving $∂\mathbf{u}∕∂t=−(\mathbf{u}\cdot ∇)\mathbf{u}$
• Step 3: Update $\mathbf{u}$ by solving $∂\mathbf{u}∕∂t=υ∆\mathbf{u}$
• Step 4: Update $\mathbf{u}$ by solving $∂\mathbf{u}∕∂t=−∇\mathbf{p}$

✅ 把偏微分方程分解几个小块，依次轮流优化每一小块。
❓ 这种方法为什么可行？

P20

Step 1: External Acceleration

The Update of $\mathbf{u}$ by $∂\mathbf{u}∕∂t=\mathbf{g}$ is straightforward, just add acceleration to $u$ and $v$.

✅ $v_{i,j}$代表向下的速度，对所有格子更新$v_{i,j}$.
✅ 其它外部速度同理。

P21

Step 2: Advection

✅ Advection,代表流动。即速度会跟着粒子移动，基于欧拉的方法才需要考虑这个问题。因为固定的格子无法描述水的流动。
✅ 基于拉格朗日的方法，变量定义在粒子上，天然满足这个特点。

数学模型

Next we need to update $\mathbf{u}$ by solving $∂\mathbf{u}∕∂t=−(\mathbf{u}\cdot ∇)\mathbf{u}$.

Solving this in an Eulerian way can be a source of instability.

✅ Eulerian way： $\mathbf{u}^{\mathrm{new} }=\frac{\partial u}{\partial t} ·Δt＋\mathbf{u}$ 不稳定

To solve this problem, we come to realize that advection means to carry physical quantities by velocity.

P22

Solution: Semi-Lagrangian Method

The solution is to trace a virtual particle backward over time.

✅ 例如要求$\mathbf{x}_0$的速度，倒推哪个粒子会运动到$\mathbf{x}_0$处;因此找到$\mathbf{x}_1$，从$\mathbf{x}_1$的下一刻速度来更新$\mathbf{x}_0$的速度。

• Define $\mathbf{x}_0←(i−0.5, j)$
• Compute $\mathbf{u}(\mathbf{x}_0)$
• $\mathbf{x}_1←\mathbf{x}_0−∆t \mathbf{u}(\mathbf{x}_0)$

✅ 假设短时间内速度不变，根据当前速度猜测上一帧的位置。

• Compute $\mathbf{u}(\mathbf{x}_1)$
• $u_{i,j}^{new}←u(\mathbf{x}_1)$

Note that if the velocities are staggered, we need to do staggered bilinear interpolation.

P23

✅ 对每个墙上的速度都以相同的方式更新。

P24

We could also subdivided the time step for better tracing.

✅ 反推找$\mathbf{x}_1$时 step 细一点，这样能找得准一点
✅ 怎么计算每个$\mathbf{x}$的$\mathbf{u}$?答：双线性插值方法、
✅ 做模拟通常更在乎稳定而不是误差，此方法更稳定，但会有模糊的 artifacts.

P25

Step 3: Diffusion

Next we need to update $\mathbf{u}$ by solving $∂\mathbf{u}∕∂t=\upsilon ∆\mathbf{u}$.

✅ 分别对$u$和 $v$ 做 laplacian.
✅ 注意公式中$v$和$\nu$的不同，后者为粘滞系数。

We could also use even smaller sub-steps…

P27

Step 4: Pressure Projection

Finally, we need to update $\mathbf{u}$ by solving $∂\mathbf{u}∕∂t=−∇\mathbf{p}$.

Staggering makes this very straightforward:

$$u_{i,j}^{new}←u_{i,j}−\frac{∆t}{ℎ}(p_{i,j}−p_{i−1,j})$$

$$v_{i,j}^{new}←v_{i,j}−\frac{∆t}{ℎ}(p_{i,j}−p_{i,j−1})$$

But what is $\mathbf{p}$?

✅ 公式第二项离散化后在特定方向上的压强。
✅ $u$和$v$分别为两个方向上的速度。

P28

压强的来源

The pressure is caused by incompressibility.

✅ 压强的原因：由于流体不可压缩、对于流体的压力会传导到每个点上。
✅ 每个点都有压强，虽然压强未知，但可以根据不可压条件构造方程组。
✅ 不可压的表现为有压强，产生的效果是散度为0．

In other words, after this update by pressure, we should achieve:

which means

$$\Downarrow$$

P29

压强的数学模型

The pressure is caused by incompressibility. Eventually, we get a Poisson equation:

Eventually, we get a Poisson equation:

$$4p_{i,j}−p_{i−1,j}−p_{i+1,j}−p_{i,j−1}−p_{i,j+1}= \\ \\ ℎ(−u_{i+1,j}−v_{i,j+1}+u_{i,j}+v_{i,j})$$

with boundary conditions:

$$\text{Dirichlet boundary (open) } p_{i−1,j}=P \\ \text{Neumann boundary (close) } p_{i−1,j}=p_{i,j}$$

Once we solve $\mathbf{p}$, we update $\mathbf{u}$ and done.

P30

After-Class Reading

Jos Stam. 1999. Stable Fluids. TOG (SIGGRAPH).

✅ 这篇论文主要讨论了step2，但也包含了全部过程

本文出自CaterpillarStudyGroup，转载请注明出处。

https://caterpillarstudygroup.github.io/GAMES103_mdbook/