Second partial derivatives

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  • #1
CAF123
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Homework Statement


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?
 

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  • #2
CAF123
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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?
 
  • #3
HallsofIvy
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Homework Statement


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?
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]
 
  • #4
CAF123
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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]
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?
 
  • #5
SammyS
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Homework Statement


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?
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?
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]
 
  • #6
CAF123
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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]
Is this not the equation I wrote in my first post?
 
  • #7
SammyS
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Is this not the equation I wrote in my first post?
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]
 
  • #8
CAF123
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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.
 
  • #9
SammyS
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Yes, it tells me that, [tex] \frac{\partial x}{\partial s} = x \,\,\text{and}\,\,\frac{\partial y}{\partial s} =y [/tex]
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]
 
  • #10
CAF123
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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]
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]
 
  • #11
SammyS
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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]
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.​

Where do I go from here?
Use what HallsofIvy stated in post #3:
...

The derivative of anything with respect to s, assuming that f itself is a function of x and y, is
[tex]\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial s}[/tex]
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.
 
  • #12
CAF123
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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]
 
  • #13
SammyS
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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]
I failed to notice that you were computing the wrong partial derivative. :frown:

What you need is the following:
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]
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.
 

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