Multivariate Higher Order Derivatives

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
Yagoda
Messages
45
Reaction score
0

Homework Statement


Let [itex]h(u,v) = f(u+v, u-v)[/itex]. Show that [itex]f_{xx} - f_{yy} = h_{uv}[/itex] and [itex]f_{xx} + f_{yy} = \frac12(h_{uu}+h_{vv})[/itex].


Homework Equations





The Attempt at a Solution


I'm always confused on how to tackle these types of questions because there isn't an actual function to differentiate.
So I am assuming here that x = u+v and y = u-v and going from there. So I need to find the first partial with respect to x, which might be something like [itex]f_x = \frac{\partial f}{\partial(u+v)}[/itex] since I need to get it in terms of u and v, but this doesn't seem right. What sort of approach do I need to use on these questions?

Edit: What I've done now is write [itex]h_{uv} = \frac{\partial}{\partial v}\frac{\partial h}{\partial u} = \frac{\partial}{\partial v} ( \frac{\partial f}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial u}) =\frac{\partial}{\partial v} (\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y})[/itex]. I'm having trouble converting from u's and v's to x's and y's.
 
Last edited:
Physics news on Phys.org
[itex]u=\frac{x+y}{2}[/itex] and [itex]v=\frac{x-y}{2}[/itex] so [itex]f(x,y)=h(\frac{x+y}{2},\frac{x-y}{2})[/itex]

So we have:

[itex] f_x=h_u u_x+h_v v_x \Rightarrow f_x=\frac{1}{2}h_u+\frac{1}{2}h_v[/itex]

[itex] f_{xx}=\frac{1}{2}(h_{uu}u_x+h_{uv}v_x+h_{vu}u_x+h_{vv}v_x)[/itex]

I think you can continue yourself.
 
Yagoda said:

Homework Statement


Let [itex]h(u,v) = f(u+v, u-v)[/itex]. Show that [itex]f_{xx} - f_{yy} = h_{uv}[/itex] and [itex]f_{xx} + f_{yy} = \frac12(h_{uu}+h_{vv})[/itex].

Homework Equations


The Attempt at a Solution


I'm always confused on how to tackle these types of questions because there isn't an actual function to differentiate.
So I am assuming here that x = u+v and y = u-v and going from there. So I need to find the first partial with respect to x, which might be something like [itex]f_x = \frac{\partial f}{\partial(u+v)}[/itex] since I need to get it in terms of u and v, but this doesn't seem right. What sort of approach do I need to use on these questions?

Edit: What I've done now is write [itex]h_{uv} = \frac{\partial}{\partial v}\frac{\partial h}{\partial u} = \frac{\partial}{\partial v} ( \frac{\partial f}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial u}) =\frac{\partial}{\partial v} (\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y})[/itex]. I'm having trouble converting from u's and v's to x's and y's.

For thinking about these things it's sometimes good to think about the operators without worrying about the functions. Define, for example, ##\partial_u(F)=\frac{\partial F}{\partial u}##. Then wouldn't it be true that ##\partial_u=\partial_x+\partial_y## and ##\partial_v=\partial_x-\partial_y## from the chain rule? It can save you a lot of texing and even spare some confusion. It's kind of the same as your underscore notation for partial derivatives.
 
Last edited: