- #1
jaejoon89
- 187
- 0
Hi, I'm trying to understand this.
The given equation is y_tt = 4 y_xx
0 < x < pi, t>0
where y_tt is the 2nd derivative with respect to t, y_xx is 2nd wrt x
Boundary conditions
y(0,t) = 0 and y(pi,t) = 0
And initial conditions
y_t (x,0) = 0 = g(x)
y(x,0) = sin^2 x = f(x)
---
My teacher wrote that F(x) is the odd periodic extension of f(x), and then wrote
F(x) = sign(sinx)sin^2 x
1) I assume this is to make it odd but why wouldn't he just write sign(x)sin^2 x?
2) Also, there was a similar question in class but -infinity < x < infinity and no boundary conditions given with one of the initial conditions y(x,0) = 1/(1+x^2) = f(x). In that case, since f(x) is even why isn't it necessary to use the sign function?
Thanks for your help!
The given equation is y_tt = 4 y_xx
0 < x < pi, t>0
where y_tt is the 2nd derivative with respect to t, y_xx is 2nd wrt x
Boundary conditions
y(0,t) = 0 and y(pi,t) = 0
And initial conditions
y_t (x,0) = 0 = g(x)
y(x,0) = sin^2 x = f(x)
---
My teacher wrote that F(x) is the odd periodic extension of f(x), and then wrote
F(x) = sign(sinx)sin^2 x
1) I assume this is to make it odd but why wouldn't he just write sign(x)sin^2 x?
2) Also, there was a similar question in class but -infinity < x < infinity and no boundary conditions given with one of the initial conditions y(x,0) = 1/(1+x^2) = f(x). In that case, since f(x) is even why isn't it necessary to use the sign function?
Thanks for your help!