Solve the boundary value problem \begin{cases} &u_{t} - 3u_{xx} = 0 && \text{in} \ \pi_{T} = (0,\pi) \times (0,t], \\ &u(x,0) = 4\sin\frac{x}{2} - \frac43 \sin \frac{3x}{2} && \text{in} \ [0,\pi],\\ &u(0,t)= 0, \ \ \ u_{x}(\pi,t) && \text{in} \ (0,T]. \end{cases}
Attempt at solution
Since the initial condition is sinusoidal, the solution must be of the form $$ u(x,t) = e^{-k\lambda^2 t} (A\cos (\lambda x) + B\sin(\lambda x)),$$ for which $k=3.$ Starting with the boundary conditions : \begin{align} u(0,t) &= e^{-3 \lambda^2 t}A = 0 \implies A = 0.\\ u(\pi,t) &= e^{-3 \lambda^2 t}B\sin(\lambda \pi) \implies \sin(\lambda\pi)=0 \\ &\implies \lambda = n \quad \text{for} \ \ n\in \mathbb{Z} \tag{a} \end{align} So we have that \begin{align} u(\pi,t) &= e^{-3 \lambda^2 t} B\sin(n \pi) \\\implies u(x,t) &= e^{-3 \lambda^2 t}B\sin(nx) \\ \implies u(x,t) &= \sum_{n=0}^{N}e^{-3 \lambda^2 t}B_{n}\sin(nx) \tag{1} \end{align} Finally, I equate $(1)$ with the initial condition of the BVP to find the constants $B_n$ : \begin{gather} u(x,0) = \underbrace{\sum_{n=0}^{N}B_{n}\sin( nx)}_{= 0 + B_{1}\sin(x)+B_{2}\sin(2x)+\dots} = 4\sin\frac{x}{2} - \frac43 \sin \frac{3x}{2} \end{gather}
I can not seem to find the constants $B_{n}$ from the last equality, suggesting I'm missing a division by 2 in the argument of $\sin(nx)$. Perhaps in $(\text{a})$ I should have $\lambda = n/2$ but that doesn't seem to make sense. Anyone know where I'm going wrong ?