Use tensor methods to establish that grad1/2(u⋅u) = u × curl u + (u⋅grad)u. Now use this result and the general divergence theorem for tensors to show that, for a vector field A, ∫S [A(A⋅dS)−1/2A²dS] = ∫V[A divA−A×curl A]dV, where S is the surface enclosing the volume V.

Question

Use tensor methods to establish that

[latex]grad \frac { 1 }{ 2 } (u · u) = u × curl u + (u · grad)u[/latex].

Now use this result and the general divergence theorem for tensors to show that, for a vector field A,

[latex]\int_{ S }^{ }{ \left[ A(A\cdot dS)-\frac { 1 }{ 2 } { A }^{ 2 }dS ight] } =\int_{ V }^{ }{ \left[ A div A-A imes curl A ight] } dV[/latex],

where S is the surface enclosing the volume V.

Accepted Answer

We start with the most complicated of the terms in the identity:

[latex]\begin{aligned}{[ \mathbf{u} imes( abla imes \mathbf{u} )]_i } & =\epsilon_{i j k} u_j( abla imes \mathbf{u} )_k=\epsilon_{i j k} u_j \epsilon_{k l m} \frac{\partial \mathbf{u}_m}{\partial x_l} \& =\left(\delta_{i l} \delta_{j m}-\delta_{i m} \delta_{j l} ight) \mathbf{u}_j \frac{\partial\mathbf{u}_m}{\partial x_l}=\mathbf{u}_j \frac{\partial \mathbf{u}_j}{\partial x_i}-u_j \frac{\partial \mathbf{u}_i}{\partial x_j} \& =\frac{1}{2} \frac{\partial}{\partial x_i}\left(\mathbf{u}_j \mathbf{u}_j ight)-( \mathbf{u} \cdot abla) \mathbf{u}_i=\left[\frac{1}{2} abla(\mathbf{u} \cdot \mathbf{u} )-( \mathbf{u} \cdot abla) \mathbf{u} ight]_i,\end{aligned}[/latex]

which establishes the first result. To establish the second result we first note that

[latex]\frac{\partial}{\partial x_j}\left(A_i A_j ight)=A_i \frac{\partial A_j}{\partial x_j}+A_j \frac{\partial A_i}{\partial x_j}=[ \mathbf{A} abla \cdot \mathbf{A}+( \mathbf{A} \cdot abla) \mathbf{A} ]_i .[/latex] (∗)

Next we consider the ith component of the integrand on the RHS of the putative equation and use the first result to replace A × (∇ × A).

[latex]\begin{aligned}{[ RHS ]_i } & =A_i( abla \cdot \mathbf{A} )-[ \mathbf{A} imes( abla imes \mathbf{A} )]_i \& =A_i( abla \cdot \mathbf{A} )-\frac{1}{2} abla_i A^2+( \mathbf{A} \cdot abla) \mathbf{A} _i=\frac{\partial}{\partial x_j}\left(A_i A_j ight)-\frac{1}{2}\frac{\partial\left(A^2 ight)}{\partial x_i}\end{aligned}[/latex]

We can now integrate this equation over the volume V and apply the divergence theorem for tensors to both terms individually:

[latex]\begin{aligned}\int_V[ RHS ]_i d V & =\int_V \frac{\partial}{\partial x_j}\left(A_i A_j ight) d V-\frac{1}{2} \int_V \frac{\partial\left(A^2 ight)}{\partial x_i} d V \& =\int_S A_i A_j d S_j-\frac{1}{2} \int_S A^2 d S_i=\int_S\left[ \mathbf{A} ( \mathbf{A} \cdot d \mathbf{S})-\frac{1}{2} A^2 d \mathbf{S} ight]_i .\end{aligned}[/latex]

This concludes the proof.

Question 26.7: Use tensor methods to establish that grad1/2(u⋅u) = u × curl...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions