The heat equation in 1D is: $$ \frac{\partial T(x,t)}{\partial t} = \alpha \,\frac{\partial^2T(x,t)}{\partial x^2} \, , \quad \alpha > 0 $$ My IC's and BC's \begin{cases} T'_x(0,t) = T'_x(L,t) = 0 \\ T(x,0) = f(x) \end{cases} I made the assumption that $$ T(x,t) = F(x)G(t) $$ Meaning that I get 2 dimensionally independent ODEs: \begin{cases} \frac{F''_x(x)}{F(x)} = \lambda\ \\[2mm] \alpha \, \frac{G'_t(t)}{G(t)} = \lambda \end{cases} Ending up with the solutions $$ T_n(x,t) = C_n\, \sin\left(\frac{n\pi x}{L} \right)\, e^{ \, - \frac{n^2\pi^2}{\alpha L^2}t} \, .$$ Which can be linearly superposed to have: $$T(x,t) = \sum_{n=1}^\infty C_n\, \sin\left(\frac{n\pi x}{L} \right)\, e^{ \, -\frac{n^2\pi^2}{\alpha L^2}t}\, $$
Can I do the same thing for an infinitely long rod by changing my BC's to (for example) \begin{cases} \lim_{n \to \infty} T'_x(-n,t) = \lim_{n \to \infty} T'_x(n,t) = 0 \\[3mm] T(x,0) = f(x) \end{cases} Or should I solve the heat equation using a different method?