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What does this notation mean in a PDE: $(u_xu_t)_{xx}$? For example, I have checked that $(u_xu_t)_x=u_{xx}u_{xt}$ so the product rule does not apply. However, it applies in the former case. What's the general rule?

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    $\begingroup$ Actually, $(u_xu_t)_x=u_{xx}u_t+u_xu_{tx}$. $\endgroup$ – Pedro Feb 2 '18 at 14:12
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    $\begingroup$ Subzero-273K : You claim to have checked that $(u_xu_t)_x=u_{xx}u_{xt}$. Please, show your calculus for this supposed check. $\endgroup$ – JJacquelin Feb 2 '18 at 14:15
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    $\begingroup$ Where does the author of the linked paper use the result in your original post? $\endgroup$ – Kevin Feb 2 '18 at 14:46
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    $\begingroup$ @Bran In the linked paper, the expression $u_{xt}=u_{yy}+(uu_x)_{x}$ is a PDE. It is not obtained as a particular case of any general differentiation rule for products. It probably results from the modeling of some physical phenomenon. I suspect that the same is true for the equality in the post. $\endgroup$ – Pedro Feb 2 '18 at 15:21
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    $\begingroup$ @Bran: Don't confuse an equation with an identity. $(u_tu_x)_x=u_{tx}u_x+u_tu_{xx}$ is an identity which means that is is true any function $u(x,t)$. On the other hand $(u_tu_x)_x=u_{xx}u_{xt}$ which is an equation is not true for any $u(x,t)$, but it is true for some particular functions solutions of the equation. So, the relationship $(u_tu_x)_x=u_{xx}u_{xt}$ mut not be used everywhere for differents problems than the specific problem to which it is coming from. $\endgroup$ – JJacquelin Feb 2 '18 at 17:14
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It means that, for a function $u=u(x,t)$, you are computing the following quantity \begin{align*} \frac{\partial^2}{\partial x \partial x} \left( \frac{\partial u}{\partial x} \cdot \frac{\partial u}{\partial t} \right) = \frac{\partial}{\partial x} \left( \frac{\partial^2 u}{\partial x \partial x}\cdot \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} \frac{\partial^2 u}{\partial x \partial t} \right) \\ = \frac{\partial^3 u}{\partial x \partial x \partial x} \cdot \frac{\partial u}{\partial t} + 2\frac{\partial^2 u}{\partial x \partial x} \cdot \frac{\partial^2 u}{\partial x \partial t} + \frac{\partial u}{\partial x} \cdot \frac{\partial^3 u}{\partial x \partial x \partial t} \end{align*}

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