- #1

- 3

- 0

_{e}(x) satisfies the wave equation? =(

i get that u

_{e}(x)=gx

^{2}/2c

^{2}+ ax + b where a and x are just constants but how does this satisfy the wave equation?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter tuan43
- Start date

- #1

- 3

- 0

i get that u

- #2

- 49

- 0

Generally to show that a function satisfies a DE, you'll need to show that its derivatives actually have the relationship in question. So I'd start by differentiating your definition of v twice with respect to each variable.

- #3

- 3

- 0

- #4

- 49

- 0

Ah, I see. :) So you've got a string with a uniform loading (which you've called Q) along the x-direction, sagging in the shape of a parabola because of that (note that in reality, if this loading were due to the weight of the string, the equilibrium shape would be a catenary, not a parabola).

And you're trying to prove that if u(x,t) is a solution to the wave equation on an equivalent (same boundary conditions) unloaded string, then:

[tex]u(x,t) - u_e(x)[/tex]

Will be a solution to the wave equation on the loaded string.

Did I understand the question correctly?

If so, then look up the superposition theorem for linear differential equations (such as the wave equation). This states that the sum of two solutions to a DE will also be a solution to the DE - So in this case, if both u

If you also need to show that it's the solution that you're looking for, then you'll need to check it satisfies the appropriate boundary/initial conditions.

- #5

- 9,557

- 767

Are we supposed to be psychic? Why don't you give us the equation you are trying to satisfy? How does Q enter in to it?

- #6

- 3

- 0

LCKurtz: Q(x,t) is just force acting on the string, so gravity in most cases. i got the answer now, sorry for not being more precise.

Share: