Dangelo Rosario

2022-10-02

Proof of: ${u}_{j}\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}{u}_{j}=\frac{1}{2}\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}({u}_{j}^{2})$

I'm reading a proof of:

$u\times \omega =\mathrm{\nabla}\text{}(\frac{u\cdot \text{}u}{2})-u\cdot \mathrm{\nabla}\text{}u$

In this proof, it says:

${u}_{j}\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}{u}_{j}=\frac{1}{2}\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}({u}_{j}^{2})$

It might be trivial, but I cannot get my head around this later equality. Does anyone know how to prove/motivate it (using Einstein notation)?

I'm reading a proof of:

$u\times \omega =\mathrm{\nabla}\text{}(\frac{u\cdot \text{}u}{2})-u\cdot \mathrm{\nabla}\text{}u$

In this proof, it says:

${u}_{j}\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}{u}_{j}=\frac{1}{2}\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}({u}_{j}^{2})$

It might be trivial, but I cannot get my head around this later equality. Does anyone know how to prove/motivate it (using Einstein notation)?

Marcel Mccullough

Beginner2022-10-03Added 11 answers

You can do this for fixed j by either the chain rule or the product rule, and then sum over j. It may be easier to start on the right-hand side and evaluate it. By the chain rule:

$\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}({u}_{j}^{2})=\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}({u}_{j}{u}_{j})=\frac{\mathrm{\partial}}{\mathrm{\partial}{u}_{k}}({u}_{j}{u}_{j})\frac{\mathrm{\partial}{u}_{k}}{\mathrm{\partial}{x}_{i}}=2{u}_{j}{\delta}_{jk}\frac{\mathrm{\partial}{u}_{k}}{\mathrm{\partial}{x}_{i}}=2{u}_{j}\frac{\mathrm{\partial}{u}_{j}}{\mathrm{\partial}{x}_{i}}.$

By the product rule:

$\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}({u}_{j}^{2})=\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}({u}_{j}{u}_{j})=\frac{\mathrm{\partial}{u}_{j}}{\mathrm{\partial}{x}_{i}}{u}_{j}+{u}_{j}\frac{\mathrm{\partial}{u}_{j}}{\mathrm{\partial}{x}_{i}}=2{u}_{j}\frac{\mathrm{\partial}{u}_{j}}{\mathrm{\partial}{x}_{i}}.$

$\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}({u}_{j}^{2})=\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}({u}_{j}{u}_{j})=\frac{\mathrm{\partial}}{\mathrm{\partial}{u}_{k}}({u}_{j}{u}_{j})\frac{\mathrm{\partial}{u}_{k}}{\mathrm{\partial}{x}_{i}}=2{u}_{j}{\delta}_{jk}\frac{\mathrm{\partial}{u}_{k}}{\mathrm{\partial}{x}_{i}}=2{u}_{j}\frac{\mathrm{\partial}{u}_{j}}{\mathrm{\partial}{x}_{i}}.$

By the product rule:

$\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}({u}_{j}^{2})=\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}({u}_{j}{u}_{j})=\frac{\mathrm{\partial}{u}_{j}}{\mathrm{\partial}{x}_{i}}{u}_{j}+{u}_{j}\frac{\mathrm{\partial}{u}_{j}}{\mathrm{\partial}{x}_{i}}=2{u}_{j}\frac{\mathrm{\partial}{u}_{j}}{\mathrm{\partial}{x}_{i}}.$

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