What is the Fourier Sine Series Coefficient in this case?

Question

Using the wave equation, $$u_{tt}=u_{xx}$$and initial conditions $$u(t,0)=u_{x}(t,1)=0$$ $$u(0,x)=1$$ $$u_{t}(0,x)=0$$ on the interval $0 \leq x \leq 1$, I found the only nontrivial solution to be in the case where $\lambda = -\omega^2 <0$, and found $$u(t,x)=\sum_{n=1}^{\infty}\Bigg[b_{n} \cos\Big(\frac{\pi(2n-1)t}{2}\Big)+d_{n} \sin\Big(\frac{\pi(2n-1)t}{2}\Big)\Bigg] \sin\Big(\frac{\pi(2n-1)x}{2}\Big)$$ From here, $$u(0,x)=\sum_{n=1}^{\infty} b_{n} \sin\Big(\frac{\pi(2n-1)x}{2}\Big) =1$$ Here is the problem. I have never dealt with a situation where the sine term is not of the form $\sin(n\pi x)$, so I don't know if $b_{n}$ will be of the form $$b_{n}=\int_{0}^{1}1 \cdot \sin\Big(\frac{\pi(2n-1)x}{2}\Big)dx$$ or $$b_{n}=\int_{0}^{1}1 \cdot \sin(n\pi x)dx$$ Please let me know which of the two is correct, or if I am completely wrong about both.

Answer 1

If we separate variables and write $u(x)=X(x)T(t)$, then we find that

$$\frac{X''(x)}{X(x)}=\frac{T''(t)}{T(t)}=-\lambda^2$$

Hence, $X(x)=A\sin(\lambda x)+B\cos(\lambda x)$ and $T(t)=C\sin(\lambda t)+D\cos(\lambda t)$.

Applying the boundary conditions, we find that $B=0$ and $\lambda = (2n-1)\pi/2$. Then, we can write

$$u(x,t)=\sum_{n=1}^\infty \sin((2n-1)\pi x/2)\left(a_n\sin((2n-1)\pi t/2)+b_n\cos((2n-1)\pi t/2)\right)$$

Applying the initial conditions, we find that $a_n=0$ and

$$1=\sum_{n=1}^\infty b_n \sin((2n-1)\pi x/2)$$

where $b_n=2\int_0^1 \sin((2n-1)\pi x/2)\,dx=4\left(\frac{1}{(2n-1)\pi}\right)$.

Finally, we have

$$u(x,t)=\frac4\pi \sum_{n=1}^\infty \frac{\sin((2n-1)\pi x/2)\cos((2n-1)\pi t/2)}{2n-1}$$