Partial Differential Equations Help

Click For Summary

Homework Help Overview

The problem involves demonstrating that a specific function form satisfies a given partial differential equation (PDE). The function is expressed in terms of two arbitrary functions, f and g, which are twice differentiable. The context is within the subject area of partial differential equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the computation of second derivatives of the function u, questioning the application of the chain rule and product rule in their calculations. There is confusion regarding the derivatives of f and g, particularly in relation to their dependence on the variables x and y.

Discussion Status

Some participants are clarifying their understanding of the derivatives involved, while others are providing corrections and guidance on the application of differentiation rules. There is an ongoing exploration of the correct forms of the derivatives and how to substitute them into the PDE.

Contextual Notes

Participants express a desire to understand the underlying concepts rather than seeking a direct solution. There is an emphasis on ensuring the correct application of differentiation techniques without providing complete solutions.

walter9459
Messages
20
Reaction score
0

Homework Statement


Show that u=f(2x+y^2)+g(2x-y^2) satisfies the equation y^2 d^2u/dx^2 + (1/y) du/dy - d^2u/dy^2=0 where f and g are arbitrary (twice differentiable) functions.



Homework Equations





The Attempt at a Solution


I came up with fxx=0 fyy=2 gxx=0 gyy= 2. But didn't know if this was correct and if so what to do next. I do not want the solution but how to go about solving this problem. Thanks!
 
Physics news on Phys.org
You want to find things like uxx, uy and uyy to substitute into your equation. What are those? Oh, and fxx is not 0. E.g. fx=f'(2x+y^2)*2. Use the chain rule.
 
Last edited:
I am still confused, I do not see where the chain rule comes into play, f(x,y) = 2x + y2. So why isn't fx equal to 2? I do not see how you are coming up with (2x + y2)*2. I desperately need to understand this whole concept! Thanks!
 
Chain rule. f(x,y) isn't equal to 2x+y^2. It's equal to f(2x+y^2). The x derivative is f'(2x+y^2)*2.
 
Sorry to be a pain but I think I am finally getting the idea and want to make sure this is true. ux = f'(2x+y2)*2 + g'(2x-y2)*2
uy = f'(2x+y2)*2y + g'(2x-y2)*-2y
uxx = f''(2x+y2)*0 + g''(2x-y2)*0
uyy = f''(2x+y2)*2 + g''(2x-y2)*-2

then I just plug these results into the equation y2uxx + (1/y) uy - uyy = 0. Is that correct?

Thanks!
 
ux and uy are ok. Not uxx and uyy. fxx=(f'(2x+y^2)*2)'=2*(f'(2x+y^2))'=4*f''(2x+y^2). The '2' is a multiplicative constant. You need to use the product rule on 2y*f'(2x+y^2), since it's the product of two functions of y. Yes, then just plug them into your equation.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K