1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Second partial derivatives

  1. Oct 9, 2012 #1

    CAF123

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    I have an expression for the partial derivative of u with respect to s, which is [tex] \frac{\partial\,u}{\partial\,s} = \frac{\partial\,u}{\partial\,x}x + \frac{\partial\,u}{\partial\,y}y [/tex]
    How do I compute [itex] \frac{\partial^2u}{\partial\,s^2} [/itex] from this?
     
  2. jcsd
  3. Oct 10, 2012 #2

    CAF123

    User Avatar
    Gold Member

    Is something missing from the question that is required ?( I have the explicit functions x = x(s,t) and y=y(s,t)). Anyone any ideas?
     
  4. Oct 10, 2012 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Use the chain rule:
    [tex]\frac{\partial^2 u}{\partial s^2}= \frac{\partial}{\partial u}\left(\frac{\partial u}{\partial x}x\right)+ \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial y}y\right)[/tex]

    The derivative of anything with respect to s, assuming that s itself is a function of x and y, is
    [tex]\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+ \frac{\partial f}{\partal y}\frac{\partial y}{\partial s}[/tex]
     
  5. Oct 10, 2012 #4

    CAF123

    User Avatar
    Gold Member

    Should that be an s above?
    Given this, I have [tex] \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\right)x + \frac{\partial}{\partial s}\left(x\right) \frac{\partial u}{\partial x} [/tex] for the first term.
    What do I do from here?
     
  6. Oct 10, 2012 #5

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes, it's important to know that x = x(s,t) and y=y(s,t)

    What is the general form of [itex]\displaystyle \frac{\partial\,u}{\partial\,s}\,,\ \ \text{ if } \ \ x = x(s,t)\ \text{ and }\ y=y(s,t)\ ?[/itex]
     
  7. Oct 10, 2012 #6

    CAF123

    User Avatar
    Gold Member

    Is this not the equation I wrote in my first post?
     
  8. Oct 10, 2012 #7

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The equation in your first post is for some particular x = x(s,t) and y = y(s,t).

    In general, [itex]\displaystyle \frac{\partial u}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}\ .[/itex]

    This along with the equation in your first post tells you something about [itex]\displaystyle \frac{\partial x}{\partial s}[/itex] and [itex]\displaystyle \frac{\partial y}{\partial s}\ .[/itex]
     
  9. Oct 11, 2012 #8

    CAF123

    User Avatar
    Gold Member

    Yes, it tells me that, [tex] \frac{\partial x}{\partial s} = x \,\,\text{and}\,\,\frac{\partial y}{\partial s} =y [/tex] I.e the partial derivatives with respect to s are the actual functions.
     
  10. Oct 11, 2012 #9

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes. Now proceed with what you said in post #4 regarding: [itex]\displaystyle \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\,x\right)\ ,[/itex]

    [itex]\displaystyle \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\,x\right)=\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\right)x + \frac{\partial}{\partial s}\left(x\right) \frac{\partial u}{\partial x}\ ,[/itex]

    which can be written as:

    [itex]\displaystyle \left(\frac{\partial^2 u}{\partial s\,\partial x}\right)x + \frac{\partial x}{\partial s}\, \frac{\partial u}{\partial x}\ .[/itex]
     
  11. Oct 11, 2012 #10

    CAF123

    User Avatar
    Gold Member

    Where do I go from here?
    I have written the above as [tex] \frac{\partial u}{\partial x} x + \left(\frac{\partial u}{\partial s}\right)\left(\frac{\partial u}{\partial x}\right) x = \frac{\partial u}{\partial x} x \left( 1 + \frac{\partial u}{\partial s}\right) [/tex]
     
  12. Oct 11, 2012 #11

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Those are not correct .

    [itex]\displaystyle \left(\frac{\partial^2 u}{\partial s\,\partial x}\right)x + \frac{\partial x}{\partial s}\, \frac{\partial u}{\partial x}[/itex]
    [itex]\displaystyle
    =x\frac{\partial u}{\partial x} + x \left(\frac{\partial^2 u}{\partial s\,\partial x}\right)[/itex]

    [itex]\displaystyle =x\frac{\partial }{\partial x}\left( u + \frac{\partial u}{\partial s}\right)\,, \ \ \ [/itex] but I wouldn't put it into this last form. Perhaps only factor out x.​

    Use what HallsofIvy stated in post #3:
    where he's giving the the result of using the chain rule on, [itex]\displaystyle \frac{\partial f}{\partial s}\ .[/itex]

    Use [itex]\displaystyle \frac{\partial u}{\partial s}\ [/itex] in place of the function, f.
     
  13. Oct 11, 2012 #12

    CAF123

    User Avatar
    Gold Member

    Ok, so [tex] \frac{\partial}{\partial s} (f) = \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial s} \right) = \frac{\partial}{\partial x} \left(\frac{\partial u}{\partial s}\right) \frac{\partial x}{\partial s} + \frac{\partial}{\partial y}\left(\frac{\partial u}{\partial s} \right) \frac{\partial y}{\partial s} [/tex]
    But before I computed [tex] \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial s}\right),\,\, \text{and not} \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial s}\right) ?[/tex]
     
  14. Oct 11, 2012 #13

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I failed to notice that you were computing the wrong partial derivative. :frown:

    What you need is the following:
    Now, plug-in what you were given for [itex]\displaystyle \frac{\partial\,u}{\partial\,s}[/itex] in the statement of the problem.

    The resulting expression will have some occurrences of [itex]\displaystyle \frac{\partial x}{\partial s}[/itex] and[itex]\displaystyle \frac{\partial y}{\partial s}\ .[/itex] Don't forget to change these in accordance with your previous results.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Second partial derivatives
Loading...