wave equation and Imposing ICs

[john]

Hello everyone.

I'm having an awful time understanding the following. It's from the
website,

http://math.etsu.edu/multicalc/Chap2/Chap2-4/part4.htm

Given a wave equations, the separation yields the following
solutions .

1). T( t) = A1 cos( awt) +B1 sin(awt)
2). Phi( x) = A2 cos( wx) +B2 sin( wx)

let w = n pi / l

Then the solution to the wave equation is

3). u (x,t) = [ A1 cos ( a*n*pi*t / l ) +B1 sin ( a*n*pi*t / l ) ] B2
sin ( n*pi*x/ l )

Let's apply the following ICs to the solution above:

4). u (x, 0) = f (x)
5). du/dt (x, 0) = 0

Second IC implies that Phi (x) T ' (0) = 0.
Given the T above, T ' = - a w A1 sin ( awt) + a w B1 sin (awt)

Choose T ' (0) = 0 to avoid trivial solution.

Then

6). 0 = - a w A1 sin ( 0) + a w B1 cos(0) = a w B1 = B2

(below is the part I don't understand)

As a result, we must have

7). T(t) = A1 cos ( a*n*pi*t / l )

(how does one know this? how do I go from T' to T being simplified to
above? )

then define bn = A1 B2

So subject to the ICs, we get

8). u(x,t) = bn cos (a*n*pi*t / l) sin ( n*pi*x / l)

Thanks for the help.

[Ray Vickson]

On Jan 4, 1:18 am, john <johnboy98...@yahoo.com> wrote:
> Hello everyone.
>
> I'm having an awful time understanding the following. It's from the
> website,
>
> http://math.etsu.edu/multicalc/Chap2/Chap2-4/part4.htm
>
> Given a wave equations, the separation yields the following
> solutions .
>
> 1). T( t) = A1 cos( awt) +B1 sin(awt)
> 2). Phi( x) = A2 cos( wx) +B2 sin( wx)
>
> let  w = n pi / l
>
> Then the solution to the wave equation is
>
> 3). u (x,t) = [ A1 cos ( a*n*pi*t / l ) +B1 sin ( a*n*pi*t / l ) ] B2
> sin ( n*pi*x/ l )
>
> Let's apply the following ICs to the solution above:
>
> 4). u (x, 0) = f (x)
> 5). du/dt (x, 0) = 0
>
> Second IC implies that Phi (x) T ' (0) = 0.
> Given the T above, T ' = - a w A1 sin ( awt) + a w B1 sin (awt)
>
> Choose T ' (0) = 0 to avoid trivial solution.

So, using T'(0) = 0, sin(0) = 0 and cos(0) = 1 we get 0 = a*w*B1,
hence B1 = 0.
Thus u = A1*B2*cos(a*n*pi*t/L)*sin(n*pi*x/L).

By the way: the link you posted is broken: I get a "page not found"
error.

R.G. Vickson

[Han de Bruijn]

Typical. It worked yesterday ..

> > Then
>
> > 6). 0 = - a w A1 sin ( 0) + a w B1 cos(0) = a w B1 = B2
>
> > (below is the part I don't understand)
>
> > As a result, we must have
>
> > 7). T(t) = A1 cos ( a*n*pi*t / l )
>
> > (how does one know this? how do I go from T' to T being simplified to
> > above? )
>
> > then define bn = A1 B2
>
> > So subject to the ICs, we get
>
> > 8). u(x,t) = bn cos (a*n*pi*t / l) sin ( n*pi*x / l)
>
> > Thanks for the help.

Han de Bruijn