Partial Differential Equations Help

In summary, the conversation discusses finding the values of uxx, uy, uyy, and other derivatives in order to solve the equation y^2 d^2u/dx^2 + (1/y) du/dy - d^2u/dy^2=0. The use of the chain rule and product rule is necessary to determine the correct values of these derivatives, which can then be substituted into the equation to solve for u.
  • #1
walter9459
20
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
  • #2
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:
  • #3
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!
 
  • #4
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.
 
  • #5
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!
 
  • #6
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.
 

Related to Partial Differential Equations Help

1. What is a partial differential equation (PDE)?

A PDE is a mathematical equation that involves multiple variables and their partial derivatives. It is used to describe the relationship between a function and its partial derivatives in multiple dimensions.

2. What are the applications of PDEs in science and engineering?

PDEs are used to model and study a wide range of phenomena in physics, engineering, and other scientific fields. They are commonly used in fluid mechanics, electromagnetism, quantum mechanics, and many other areas.

3. What are the different types of PDEs?

There are several types of PDEs, including elliptic, parabolic, and hyperbolic equations. These types differ in their characteristics and the types of problems they are used to solve.

4. How are PDEs solved?

PDEs can be solved analytically using mathematical techniques or numerically using computational methods. The choice of method depends on the complexity of the equation and the desired level of accuracy.

5. What are the challenges in solving PDEs?

Solving PDEs can be challenging due to their complexity and the need for advanced mathematical and computational skills. Additionally, PDEs often have no exact solution and require approximation methods to obtain a solution.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
539
  • Calculus and Beyond Homework Help
Replies
5
Views
790
  • Calculus and Beyond Homework Help
Replies
3
Views
367
  • Calculus and Beyond Homework Help
Replies
6
Views
983
  • Calculus and Beyond Homework Help
Replies
18
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
921
  • Calculus and Beyond Homework Help
Replies
1
Views
581
  • Calculus and Beyond Homework Help
Replies
2
Views
880
  • Calculus and Beyond Homework Help
Replies
14
Views
575
  • Calculus and Beyond Homework Help
Replies
21
Views
1K
Back
Top