- #1
mathmari
Gold Member
MHB
- 5,049
- 7
Hey! 
Having the following problem:
$$(1): u_t=u_{xx}+f(x,t), 0<x<L, t>0$$
$$u(0,t)=u(L,t)=0, t>0$$
$$u(x,0)=0, 0<x<L$$
$$f(0,t)=f(L,t)=0, t>0$$
we do the following to find the general solution:
We write the function $f(x,t)$ as a Fourier series:
$$(2): f(x,t)=\sum_{n=1}^{\infty}{F_n(t) \sin{(\frac{n \pi x}{L})}}, F_n(t)=\frac{2}{L} \int_0^L f(x,t) \in{(\frac{n \pi x}{L})}dx$$
We write also $u(x,t)$ as a Fourier series:
$$(3): u(x,t)=\sum_{n=1}^{\infty}{T_n(t) \sin{(\frac{n \pi x}{L})}}, T_n(t)=\frac{2}{L} \int_0^L u(x,t) \sin{(\frac{n \pi x}{L})}dx$$
Replacing the relations $(2),(3)$ in $(1)$ we get:
$$(4):T_n'(t)+(\frac{n \pi}{L})^2 T_n(t)=F_n(t), n=1,2, \dots$$
The solution of the homogeneous $T_n'(t)+(\frac{n \pi}{L})^2 T_n(t)=0$ is:
$$T_n(t)=e^{-(\frac{n \pi}{L})^2t}=Q(t)$$
$$T_n(t)=Q_n(t)A_n(t)$$
So replacing this at the relation $(4)$ we get:
$$Q_n'A_n+Q_nA_n'+(\frac{n \pi}{L})^2Q_nA_n=F_n \Rightarrow Q_nA_n'=F_n \Rightarrow A_n'=\frac{F_n}{Q_n} \Rightarrow A_n'(t)=F_n(t) e^{(\frac{n \pi}{L})^2t} $$
So $$A_n(t)=\int_0^t F_n(s) e^{(\frac{n \pi}{L})^2s}ds$$
Therefore, $$(9): T_n(t)=\int_0^t F_n(s) e^{-(\frac{n \pi}{L})^2(t-s)}ds$$
So the solution is the relation $(3)$ and the coefficient $T_n(t)$ are given from the relation $(9)$.That is what I have in my notes.
When I solve the homogeneous $T_n'(t)+(\frac{n \pi}{L})^2 T_n(t)=0$, I get the following solution:
$$T_n(t)=D_n e^{-(\frac{n \pi}{L})^2t}$$
$D_n$ does not depend from $t$, does it?
How do I get the relation $T_n(t)=Q_n(t)A_n(t)$ ?? (Wondering)
Having the following problem:
$$(1): u_t=u_{xx}+f(x,t), 0<x<L, t>0$$
$$u(0,t)=u(L,t)=0, t>0$$
$$u(x,0)=0, 0<x<L$$
$$f(0,t)=f(L,t)=0, t>0$$
we do the following to find the general solution:
We write the function $f(x,t)$ as a Fourier series:
$$(2): f(x,t)=\sum_{n=1}^{\infty}{F_n(t) \sin{(\frac{n \pi x}{L})}}, F_n(t)=\frac{2}{L} \int_0^L f(x,t) \in{(\frac{n \pi x}{L})}dx$$
We write also $u(x,t)$ as a Fourier series:
$$(3): u(x,t)=\sum_{n=1}^{\infty}{T_n(t) \sin{(\frac{n \pi x}{L})}}, T_n(t)=\frac{2}{L} \int_0^L u(x,t) \sin{(\frac{n \pi x}{L})}dx$$
Replacing the relations $(2),(3)$ in $(1)$ we get:
$$(4):T_n'(t)+(\frac{n \pi}{L})^2 T_n(t)=F_n(t), n=1,2, \dots$$
The solution of the homogeneous $T_n'(t)+(\frac{n \pi}{L})^2 T_n(t)=0$ is:
$$T_n(t)=e^{-(\frac{n \pi}{L})^2t}=Q(t)$$
$$T_n(t)=Q_n(t)A_n(t)$$
So replacing this at the relation $(4)$ we get:
$$Q_n'A_n+Q_nA_n'+(\frac{n \pi}{L})^2Q_nA_n=F_n \Rightarrow Q_nA_n'=F_n \Rightarrow A_n'=\frac{F_n}{Q_n} \Rightarrow A_n'(t)=F_n(t) e^{(\frac{n \pi}{L})^2t} $$
So $$A_n(t)=\int_0^t F_n(s) e^{(\frac{n \pi}{L})^2s}ds$$
Therefore, $$(9): T_n(t)=\int_0^t F_n(s) e^{-(\frac{n \pi}{L})^2(t-s)}ds$$
So the solution is the relation $(3)$ and the coefficient $T_n(t)$ are given from the relation $(9)$.That is what I have in my notes.
When I solve the homogeneous $T_n'(t)+(\frac{n \pi}{L})^2 T_n(t)=0$, I get the following solution:
$$T_n(t)=D_n e^{-(\frac{n \pi}{L})^2t}$$
$D_n$ does not depend from $t$, does it?
How do I get the relation $T_n(t)=Q_n(t)A_n(t)$ ?? (Wondering)