Partial Differentials and wave equation.

Click For Summary
To determine if a function is a solution to the wave equation, one must differentiate the function twice with respect to 'x' and 't', then check if the relationship f_{xx} = (1/c²) f_{tt} holds. For the function x² + c²t², the calculations confirm it satisfies the wave equation, resulting in 1=1. In contrast, for the function xt, the result is 0=0, indicating it does not satisfy the equation. The process involves substituting each function into the wave equation and verifying the equality of the second derivatives. This method effectively identifies whether a function meets the criteria of the wave equation.
hhhmortal
Messages
175
Reaction score
0

Homework Statement



How can I find out if a function is a solution of a wave equation such as:

(a) xt
(b) log(xt)
(c) x² + c²t²



The Attempt at a Solution



Is it simply differentiating the funtion with respect to 'x' twice and equating this to the product of 1/c² and differentiating the function twice with respect to 't'?

I tried doing it for part (c) and I got 1=1 which means its allowed I guess. As for part (a) I got 0=0 which i suppose it isn't?
 
Physics news on Phys.org
Yes, just plug each function into the wave equation and see if they satisfy it: for each function, f, find fxx and ftt and see if they satisfy f_{xx}= (1/c^2) f_{tt}
 
Last edited by a moderator:
Ok, thanks, most of the time they seem to be 1=1 or 0=0 hence satisfying and not satisfying the wave equation i suppose..thanks again!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
1K
Replies
19
Views
1K