Solving 2nd ODE and Multivariable Calculus for Wave Equation

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weakness66
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Hello guys,
I would like to ask some questions regarding my coursework, which is about 2nd ODE and multivariable calculus.
Since we have the one-dimensional wave equation and values for the string stretched between x=0 and L=2: 0≤x≤L, t≥0
The string is fixed at both ends so we have :
u(_,t)=u(_,t) = 0 , t≥0
now we assume string is raised a distance μ at x=b and released from rest at t=0, Zero velocity implies: ∂u/∂t (_,_)=0 . 0≤x≤L

thanks beforehand!
 
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Welcome to PF;
I would like to ask some questions regarding my coursework...
That's what we are here for. However, you didn't ask any questions. Instead, you appear to have presented a situation.

Since we have the one-dimensional wave equation and values for the string stretched between x=0 and L=2: 0≤x≤L, t≥0
The string is fixed at both ends so we have :
u(_,t)=u(_,t) = 0 , t≥0
now we assume string is raised a distance μ at x=b and released from rest at t=0, Zero velocity implies: ∂u/∂t (_,_)=0 . 0≤x≤L
... what are the underscore characters supposed to represent in the above?

Are these blanks in the question you are supplied with?
 
weakness66 said:
Hello guys,
I would like to ask some questions regarding my coursework, which is about 2nd ODE and multivariable calculus.
Since we have the one-dimensional wave equation and values for the string stretched between x=0 and L=2: 0≤x≤L, t≥0
The string is fixed at both ends so we have :
u(0,t)=u(2,t) = 0 , t≥0
now we assume string is raised a distance μ at x=b and released from rest at t=0, Zero velocity implies: ∂u/∂t (x,0)=0 . 0≤x≤L

thanks beforehand!

And the initial position is u(x,0) = ? Or is that what you are asking?
 
I'm concerned that this is a homework question of the "fill in the blanks" style.
If that is the case then LCKurtz may just have done your homework for you - well done!
You should be able to figure the initial function though.
 
Simon Bridge said:
I'm concerned that this is a homework question of the "fill in the blanks" style...

Perhaps you are correct, although I wouldn't expect something that trivial here.