Solve d'Alembert BVP with Boundary Conditions & f(x)=2x,g(x)=x

[erok81]

Homework Statement

Use d'Alembert's formula to solve the boundary value problem for a string of unit length, subject to the given conditions.

I can't get this to work with LaTeX so I hope this makes sense.

f(x) = 2x if 0 ≤ x ≤ 1/2
f(x) = 2(1-x) if 1/2 < x ≤ 1

g(x) = x
c=1

Homework Equations

$$u(x,t)=\frac{1}{2}[f^{*}(x-ct)+f^{*}(x+ct)]+\frac{1}{2c}\int^{x+ct}_{x-ct}g^{*}(s)ds$$

The Attempt at a Solution

I've been able to do previous problems up until this one. Previous problems were all a single function that was 2 periodic. This one I am not sure where to start. I at least thought the g^* wouldn't be too hard since it's only x, but there is an example that shows the same thing and they have completely different integration limits.

I think my whole problem is even/odd extensions. I am not the best at them and plan to go back and study them again since they are a couple sections back.

Any tips to get this one even started?

 

[erok81]

Actually. I can write out/graph the odd extension of g(x)=x. But if I integrate that over a 2p-period I would get zero.

My text has an example where g^*=x but it is on a interval 0 < x < 1.

So their integral comes out to be...

$$G(x)=\int^{x}_{-1}g^{*}(s) ds$$

Since I don't have a range, I don't see how to integrate the odd extension and not get zero.

 

[erok81]

So I drew this out and it makes slightly more sense.

Here is where I am confused. g(x)=x. So is g^*(x) only the odd extension of the entire thing? Because if it is only the odd extension then it makes sense it isn't zero. But do the integration limits change?

As for the f^*(x) parts, I am still clueless. I don't see how the two intervals play into the solution form.
Edit...

So for the odd extension of g(x) I could set my limits as [-1,0]. But that still doesn't explain why the text has theirs [-1,x]. Unless I am just integrating from -1 to the function itself. That would give me the correct range then. Hmm...