Question

Solve the boundary value problem:du/dt=βd^2u/dx^2,0<x<l,t>0
U(0,t)=0,u(l,t)=0,t>0
U(x,0)=f(x)0<x<l

Answer 1

$u = u(x,t)=a(x)b(t)\neq0.\\ a(x)b'(t)=\beta a''(x)b(t)\Rightarrow \\ \frac{b'(t)}{\beta b(t)}=\frac{a''(x)}{a(x)}=-(\frac{\pi n}{l})^2 \Rightarrow\\ b(t)=e^{-\beta \cdot (\frac{\pi n}{l})^2 \cdot t},\quad a(x)=c_1sin(\frac{\pi n}{l}x)+c_2cos(\frac{\pi n}{l}x)\\ u(0,t)=u(l,t)=0 \Rightarrow c_2=0 \Rightarrow u_n(x,t)=e^{-\beta \cdot (\frac{\pi n}{l})^2 \cdot t}\cdot sin(\frac{\pi n}{l}x).\\ U(x,t)=\sum\limits_{n=0}^\infty a_nu_n(x,t)=\sum\limits_{n=0}^\infty a_n\cdot e^{-\beta \cdot (\frac{\pi n}{l})^2 \cdot t}\cdot sin(\frac{\pi n}{l}x).\\ U(x,0)=f(x)=\sum\limits_{n=0}^\infty a_n\cdot sin(\frac{\pi n}{l}x) \Rightarrow a_n=\frac{2}{l}\int\limits_0^lf(t)sin(\frac{\pi n}{l}t)dt\Rightarrow\\ U(x,t)=\sum\limits_{n=0}^\infty a_nu_n(x,t)=\sum\limits_{n=0}^\infty e^{-\beta \cdot (\frac{\pi n}{l})^2 \cdot t}\cdot sin(\frac{\pi n}{l}x) \cdot \frac{2}{l}\int\limits_0^lf(t)sin(\frac{\pi n}{l}t)dt.$