0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Actually, $(u_xu_t)_x=u_{xx}u_t+u_xu_{tx}$. $\endgroup$ – Pedro Feb 2 '18 at 14:12
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Where does the author of the linked paper use the result in your original post? $\endgroup$ – Kevin Feb 2 '18 at 14:46
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
0
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.