View Single Post
Dec17-12, 10:53 AM
P: 5,462
It does not really matter but I am going to put the c2 in its conventional place in what follows. Sorry that is what happens when you trust to an aging memory.

[tex]\frac{{{\partial ^2}y}}{{\partial {x^2}}} = \frac{1}{{{c^2}}}\frac{{{\partial ^2}y}}{{\partial {t^2}}}[/tex]

Now for fixed ends the boundary conditions are

[tex]y(0,t) = y(l,t) = 0[/tex]

If the ends are not 'free' but still participating in the wave then they can be attributed initial displacement and velocity conditions

y(x,0) = f(x) \\
{\left( {\frac{{\partial y}}{{\partial t}}} \right)_{t = 0}} = g(x) \\

The wave equation itself may be solved by the method of separating the variables

[tex]y(x,t) = F(x)G(t)[/tex]

F is a function of x only and G a function of t only.

Substituting and dividing through by y=FG

[tex]\frac{1}{F}\frac{{{d^2}F}}{{d{x^2}}} = \frac{1}{{G{c^2}}}\frac{{{d^2}G}}{{d{t^2}}}[/tex]

Both sides of this equation can only be equal if they are constant. Convention has this constant as -λ2.

Some algebra on the resultant pair of ordinary diffrential equations will lead to your required trigonometric solution ( not the one you offered )

where F has the form

[tex]{F_n}(x) = \sin \frac{{n\pi x}}{l}[/tex]

and G has the form

[tex]{G_n}(t) = {A_n}\cos {\omega _n}t + {B_n}\sin {\omega _n}t[/tex]


[tex]{y_n}(x,t) = {F_n}(x){G_n}(t) = \sin \frac{{n\pi x}}{l}\left[ {{A_n}\cos {\omega _n}t + {B_n}\sin {\omega _n}t} \right][/tex]

Where A and B are determined by the intial conditiions.
In general the solution above will not be complete since it depends upon n.

To obtain a complete solution you need to sum solutions over n from n=1 to ∞


y(x,0) = f(x) = \sum\limits_{n = 1}^\infty {{A_n}} \sin \frac{{n\pi x}}{l} \\
{\left( {\frac{{\partial y}}{{\partial t}}} \right)_{t = 0}} = g(x) = \sum\limits_{n = 1}^\infty {{B_n}} {\omega _n}\sin \frac{{n\pi x}}{l} \\