> [!NOTE] > Consider the problem the 1D [[Heat Equation|heat equation]] on the bounded domain $(0,l)$ together with homogeneous Dirichlet boundary conditions $\begin{align} > \partial_{t}u(x,t) &= k\partial_{xx}(x,t) & (x,t)\in (0,l)\times(0\times \infty) \\ > u(0,t)&= u(l,t) &t\in (0,\infty). > \end{align}$Assuming the solution has the form $u(x,t)=X(x)T(t)$ where $X:(0,l)\to \mathbb{R}$ and $T :(0,\infty)\to \mathbb{R}$, then $u(x,t) =\sum_{j\in \mathbb{N}}T_{j}(t)X_{j}(x) = \sum_{j\in \mathbb{N}}e^{-k\beta_{j}^2t}D_{j}\sin(\beta_{j}x).$ ###### Proof Substituting in the heat equation yields $X(x)T'(t)=kX''(x)T(t)$ and assuming we can divide we get $\frac{X''(x)}{X(x)}=\frac{T'(t)}{kT(t)}:=\lambda.$We must have that $\lambda$ is a constant since it neither depends on $x$ nor $t$. We get the following equations $T'(t)-k\lambda T(t)=0 \quad \quad \text{and} \quad \quad X''(x) -\lambda X(x)= 0.$The boundary conditions yields $X(0)=0=X(l).$ We can argue that $\lambda<0$ so by [[Solution to homogenous 2nd order linear scalar ODE with real coefficients]], we get $X(x)=D_{j}\sin(\beta_{j}x), \quad D_{j}\in \mathbb{R},j\in \mathbb{N},$where $\beta_{j}=j\pi/L$ (since $X(0)=0$ implies coefficients of $\cos$ terms are $0$ and $X(l)=0$ implies $\sin(\beta_{j}L)=0$ for all $j\in \mathbb{N}$). With $\lambda=-\beta_{j}^2$ the $T$ ODE then has the corresponding solutions $T_{j}(t) =A_{j}e^{-k\beta_{j}^2t},\quad A_{j}\in \mathbb{R},j\in \mathbb{N}.$Hence $u(x,t) =\sum_{j\in \mathbb{N}}T_{j}(t)X_{j}(x) = \sum_{j\in \mathbb{N}}e^{-k\beta_{j}^2t}D_{j}\sin(\beta_{j}x).$