Dangelo Rosario

2022-10-02

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

In this proof, it says:
${u}_{j}\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_{i}}{u}_{j}=\frac{1}{2}\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_{i}}\left({u}_{j}^{2}\right)$
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

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:
$\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_{i}}\left({u}_{j}^{2}\right)=\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_{i}}\left({u}_{j}{u}_{j}\right)=\frac{\mathrm{\partial }}{\mathrm{\partial }{u}_{k}}\left({u}_{j}{u}_{j}\right)\frac{\mathrm{\partial }{u}_{k}}{\mathrm{\partial }{x}_{i}}=2{u}_{j}{\delta }_{jk}\frac{\mathrm{\partial }{u}_{k}}{\mathrm{\partial }{x}_{i}}=2{u}_{j}\frac{\mathrm{\partial }{u}_{j}}{\mathrm{\partial }{x}_{i}}.$
By the product rule:
$\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_{i}}\left({u}_{j}^{2}\right)=\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_{i}}\left({u}_{j}{u}_{j}\right)=\frac{\mathrm{\partial }{u}_{j}}{\mathrm{\partial }{x}_{i}}{u}_{j}+{u}_{j}\frac{\mathrm{\partial }{u}_{j}}{\mathrm{\partial }{x}_{i}}=2{u}_{j}\frac{\mathrm{\partial }{u}_{j}}{\mathrm{\partial }{x}_{i}}.$

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