Paritial derivative.

  • Thread starter yungman
  • Start date
  • #1
5,588
209

Homework Statement



Verify if ##t=\lambda x## then ##x^2\frac{\partial^2 y}{\partial x^2} = t^2\frac{\partial^2 y}{\partial t^2}##

The Attempt at a Solution



[tex]t=\lambda x\;\Rightarrow\; \frac{\partial t}{\partial x}=\lambda[/tex]
[tex]\frac{\partial y}{\partial x} = \frac{\partial y}{\partial t}\frac{\partial t}{\partial x}= \lambda\;\frac{\partial y}{\partial t}[/tex]
[tex]\frac{\partial^2 y}{\partial x^2}=\lambda \frac{\partial^2 y}{\partial t^2} \frac{\partial t}{\partial x}=\lambda^2\frac{\partial^2 y}{\partial t^2}[/tex]

[tex]x^2\frac{\partial^2 y}{\partial x^2}=\frac{t^2}{\lambda^2}\lambda^2\frac{\partial^2 y}{\partial t^2}\;\Rightarrow\;x^2\frac{\partial^2 y}{\partial x^2} = t^2\frac{\partial^2 y}{\partial t^2}[/tex]

Am I correct?

Thanks
 

Answers and Replies

  • #2
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
135
Yes, with just a minor quibble:
"t" is only dependent on "x", and not in addition dependent on other variables.
Thus, you should use dt/dx, rather than the symbol for the partial derivative here.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
969
Your differential equation has dependent variable "y" so it makes no sense to ask if "[itex]t= \lambda x[/itex]" satisfies the equation.

I suspect you are asked to show that [itex]f(t-\lambda x)[/itex] satisfies the equation for f any twice differentiable function of a single variable.
 
  • #4
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
135
I don't agree HallsofIvy. Here, it is about how a simple linear scaling will affect the shape of a particular derivative from the "unscaled" x as independent variable, to its shape in its "scaled" variable "t". That is, we are dealing with the correct IDENTITY, not an equation!! It is not about an "equation" as such, but meant as an intermediate step in transforming an equation from the "x"-formulation to its logically equivalent "t"-formulation.

TECHNICALLY, we are looking at the relation between the shapes of the derivatives of functions y and Y, respectively, fulfilling the relation
y(x,....)=Y(t(x),...)

making an abuse of notation with replacing "Y" with "y" everywhere.
 
Last edited:
  • #5
5,588
209
Thanks everyone. I meant exactly what I asked, it's only about the steps to perform the differential equation.
 

Related Threads on Paritial derivative.

Replies
8
Views
4K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
773
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
1K
Replies
2
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
4
Views
768
  • Last Post
Replies
2
Views
2K
Replies
3
Views
2K
Top