- #1
Wuberdall
- 34
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Hi PH.
Let ##u_i(\mathbf{x},t)## be the velocity field in a periodic box of linear size ##2\pi##. The spectral representation of ##u_i(\mathbf{x},t)## is then
$$u_i(\mathbf{x},t) = \sum_{\mathbf{k}\in\mathbb{Z}^3}\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j}$$ where ι denotes the usual imaginary unit.
A flow is incompressible iff ##u_i(\mathbf{x})## is a solenodial verctor field, that is
$$\nabla_iu_i(\mathbf{x},t) = 0$$
Combining the above, I get
$$\begin{aligned}0 &= \nabla_iu_i(\mathbf{x},t) \\
&= \sum_{\mathbf{k}\in\mathbb{Z}^3}\hat{u}_i(\mathbf{k},t)\nabla_ie^{\iota k_jx_j} \\
&= \sum_{\mathbf{k}\in\mathbb{Z}^3}\iota k_j\delta_{ij}\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j} \\
&= \iota\sum_{\mathbf{k}\in\mathbb{Z}^3}k_j\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j}\end{aligned}$$
If ##u_i(\mathbf{x},t)## is to represent a physical flow velocity field, then ##u_i(\mathbf{x},t)## must be real function. That is ##\mathbf{u}(\mathbf{x}):[0,2\pi]^3\times\mathbb{R}\rightarrow\mathbb{R}^3## and thus ##\hat{u}_i(\mathbf{k}) = \hat{u}^\ast_i(-\mathbf{k})##. Therefore
$$\begin{aligned} \iota\sum_{\mathbf{k}\in\mathbb{Z}^3}k_j\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j} &= -\iota\sum_{\mathbf{k}\in\mathbb{N}^3}k_j\hat{u}_i(-\mathbf{k},t)e^{-\iota k_jx_j} + 0 + \iota\sum_{\mathbf{k}\in\mathbb{N}^3}k_j\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j} \\
&= \iota\sum_{\mathbf{k}\in\mathbb{N}^3}k_j\big[\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j} - \hat{u}_i(-\mathbf{k},t)e^{-\iota k_jx_j}\big] \\
&= \iota\sum_{\mathbf{k}\in\mathbb{N}^3}k_j\big[\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j} - \hat{u}^\ast_i(\mathbf{k},t)\big(e^{\iota k_jx_j}\big)^\ast\big] \\
&= 2\iota\sum_{\mathbf{k}\in\mathbb{N}^3}\Im\big[k_j\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j}\big] \end{aligned}$$
From the harmonic analysis I would expect the incompressibility condition to be $$k_i\hat{u}_i(\mathbf{k},t) = 0$$
Clearly, if this is true the incompressibility condition would be true. But I can't see how (unless I assume a sort of detailed balance in the above sum) the incompressibility condition imply the orthogonal relation ##k_i\hat{u}_i(\mathbf{k},t) = 0##.
Any help is appreciated :-)
Let ##u_i(\mathbf{x},t)## be the velocity field in a periodic box of linear size ##2\pi##. The spectral representation of ##u_i(\mathbf{x},t)## is then
$$u_i(\mathbf{x},t) = \sum_{\mathbf{k}\in\mathbb{Z}^3}\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j}$$ where ι denotes the usual imaginary unit.
A flow is incompressible iff ##u_i(\mathbf{x})## is a solenodial verctor field, that is
$$\nabla_iu_i(\mathbf{x},t) = 0$$
Combining the above, I get
$$\begin{aligned}0 &= \nabla_iu_i(\mathbf{x},t) \\
&= \sum_{\mathbf{k}\in\mathbb{Z}^3}\hat{u}_i(\mathbf{k},t)\nabla_ie^{\iota k_jx_j} \\
&= \sum_{\mathbf{k}\in\mathbb{Z}^3}\iota k_j\delta_{ij}\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j} \\
&= \iota\sum_{\mathbf{k}\in\mathbb{Z}^3}k_j\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j}\end{aligned}$$
If ##u_i(\mathbf{x},t)## is to represent a physical flow velocity field, then ##u_i(\mathbf{x},t)## must be real function. That is ##\mathbf{u}(\mathbf{x}):[0,2\pi]^3\times\mathbb{R}\rightarrow\mathbb{R}^3## and thus ##\hat{u}_i(\mathbf{k}) = \hat{u}^\ast_i(-\mathbf{k})##. Therefore
$$\begin{aligned} \iota\sum_{\mathbf{k}\in\mathbb{Z}^3}k_j\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j} &= -\iota\sum_{\mathbf{k}\in\mathbb{N}^3}k_j\hat{u}_i(-\mathbf{k},t)e^{-\iota k_jx_j} + 0 + \iota\sum_{\mathbf{k}\in\mathbb{N}^3}k_j\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j} \\
&= \iota\sum_{\mathbf{k}\in\mathbb{N}^3}k_j\big[\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j} - \hat{u}_i(-\mathbf{k},t)e^{-\iota k_jx_j}\big] \\
&= \iota\sum_{\mathbf{k}\in\mathbb{N}^3}k_j\big[\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j} - \hat{u}^\ast_i(\mathbf{k},t)\big(e^{\iota k_jx_j}\big)^\ast\big] \\
&= 2\iota\sum_{\mathbf{k}\in\mathbb{N}^3}\Im\big[k_j\hat{u}_i(\mathbf{k},t)e^{\iota k_jx_j}\big] \end{aligned}$$
From the harmonic analysis I would expect the incompressibility condition to be $$k_i\hat{u}_i(\mathbf{k},t) = 0$$
Clearly, if this is true the incompressibility condition would be true. But I can't see how (unless I assume a sort of detailed balance in the above sum) the incompressibility condition imply the orthogonal relation ##k_i\hat{u}_i(\mathbf{k},t) = 0##.
Any help is appreciated :-)
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