Do I have to find separately T_1, T_3 and T_n for n = 1,3?

[mathmari]

Hey!

I have to solve the following problem:
$$u_t=u_{xx}+f(x,t), 0<x<L, t>0$$
$$u(0,t)=u(L,t)=0, t>0$$
$$u(x,0)=0, 0<x<L$$
$$f(0,t)=f(L,t)=0, t>0$$
$$f(x,t)=e^t \sin{(\frac{\pi x }{L})}+2t \sin{(\frac{3 \pi x }{L})}$$

I have done the following:

I write $f$ as a Fourier series:
$$f(x,t)=\sum_{n=1}^{\infty}{F_n(t) \sin{(\frac{n \pi x}{L})}}$$
$$n \neq 1,3: F_n(t)=\frac{2}{L} \int_0^L \sin{(\frac{n \pi x}{L})}dx=0$$
$$F_1(t)=\frac{2}{L} \int_0^L (e^t \sin{(\frac{ \pi x}{L})}+2t \sin{(\frac{3 \pi x}{L})} ) \sin{(\frac{\pi x}{L})}dx=e^t$$
$$F_3(t)=\frac{2}{L} \int_0^L (e^t \sin{(\frac{ \pi x}{L})}+2t \sin{(\frac{3 \pi x}{L})} ) \sin{(\frac{3 \pi x}{L})}dx=2t$$
So $$f(x,t)=F_1(t) \sin{(\frac{\pi x}{L})}+F_3(t) \sin{(\frac{3 \pi x}{L})}=e^t \sin{(\frac{\pi x}{L})}+2t \sin{(\frac{3 \pi x}{L})}$$

I write $u$ as a Fourier series:
$$u(x,t)=\sum_{n=1}^{\infty}{T_n(t) \sin{(\frac{n \pi x}{L})}}$$
$$u_t=\sum_{n=1}^{\infty}{T_n'(t) \sin{(\frac{n \pi x}{L})}}$$
$$u_{xx}=\sum_{n=1}^{\infty}{-\frac{n^2 \pi^2}{L^2}T_n(t) \sin{(\frac{n \pi x}{L})}}$$
Replacing at the problem, we get:
$$\sum_{n=1}^{\infty}{T_n'(t) \sin{(\frac{n \pi x}{L})}}=\sum_{n=1}^{\infty}{(-\frac{n^2 \pi^2}{L^2}T_n(t) +F_n(t))\sin{(\frac{n \pi x}{L})}}$$
$$T_n'(t) + \frac{n^2 \pi^2}{L^2}T_n(t)=F_n(t)$$

How can I continue??
Do I have to find separately $T_1$, $T_3$ and $T_n \text{ for } n\neq 1,3$?? (Wondering)

 

[I like Serena]

Hey!

Hi! (Happy)

How can I continue??
Do I have to find separately $T_1$, $T_3$ and $T_n \text{ for } n\neq 1,3$?? (Wondering)

Sounds like a good plan to me... (Mmm)

 

[mathmari]

Hi! (Happy)
Sounds like a good plan to me... (Mmm)

I tried to solve it in that way.

I got the following finding first the solution of the homogeneous equation and then the particular solution:

$$T_1(t)=A_n(t) e^{-\frac{\pi^2 t}{L^2}}, \text{ where } A_n(t)=\frac{L^2}{L^2+\pi^2} (e^{(1+\frac{\pi^2}{L^2})t}-1)$$

$$T_3(t)=B_n(t) e^{-\frac{9 \pi^2 t}{L^2}}, \text{ where } B_n(t)=\frac{2 t L^2}{9 \pi^2} e^{\frac{9 \pi^2 t}{L^2}}-\frac{2 L^4}{81 \pi^4} (e^{\frac{ 9 \pi^2 t}{L^2}}-1)$$

$$T_n=0, \text{ for } n \neq 0$$

So the solution would be:
$$u(x,t)=T_1(t) \sin{(\frac{\pi x }{L})}+T_3(t) \sin{(\frac{3 \pi x}{L})}$$

In my notes there is the following:
$$T_1(t)=\frac{1}{3}(e^t-e^{-2t})$$
$$T_3(t)=\frac{1}{9}(t+\frac{1}{18}(e^{-18 t}-1))$$
$$u(x,t)=T_1(t) \sin{(\frac{\pi x }{L})}+T_3(t) \sin{(\frac{3 \pi x}{L})}$$

It's different from what I've found...What have I done wrong? (Wondering)

 

[mathmari]

I tried to solve it in that way.

I got the following finding first the solution of the homogeneous equation and then the particular solution:

$$T_1(t)=A_n(t) e^{-\frac{\pi^2 t}{L^2}}, \text{ where } A_n(t)=\frac{L^2}{L^2+\pi^2} (e^{(1+\frac{\pi^2}{L^2})t}-1)$$

$$T_3(t)=B_n(t) e^{-\frac{9 \pi^2 t}{L^2}}, \text{ where } B_n(t)=\frac{2 t L^2}{9 \pi^2} e^{\frac{9 \pi^2 t}{L^2}}-\frac{2 L^4}{81 \pi^4} (e^{\frac{ 9 \pi^2 t}{L^2}}-1)$$

$$T_n=0, \text{ for } n \neq 0$$

So the solution would be:
$$u(x,t)=T_1(t) \sin{(\frac{\pi x }{L})}+T_3(t) \sin{(\frac{3 \pi x}{L})}$$

In my notes there is the following:
$$T_1(t)=\frac{1}{3}(e^t-e^{-2t})$$
$$T_3(t)=\frac{1}{9}(t+\frac{1}{18}(e^{-18 t}-1))$$
$$u(x,t)=T_1(t) \sin{(\frac{\pi x }{L})}+T_3(t) \sin{(\frac{3 \pi x}{L})}$$

It's different from what I've found...What have I done wrong? (Wondering)

I replaced the result I found at the problem and it satisfies it...

I also replaced the result from my notes at the problem, but it doesn't satisfy it, does it? (Wondering)

So is the my result correct?? (Thinking)

 

[mathmari]

I replaced the result I found at the problem and it satisfies it...

I also replaced the result from my notes at the problem, but it doesn't satisfy it, does it? (Wondering)

So is the my result correct?? (Thinking)

Or isn't it correct?? (Wondering) (Tauri)

 

[mathmari]

$$u_t=u_{xx}+f(x,t), 0<x<L, t>0$$
$$u(0,t)=u(L,t)=0, t>0$$
$$u(x,0)=0, 0<x<L$$
$$f(0,t)=f(L,t)=0, t>0$$
$$f(x,t)=e^t \sin{(\frac{\pi x }{L})}+2t \sin{(\frac{3 \pi x }{L})}$$

I have done the following:

I write $f$ as a Fourier series:
$$f(x,t)=\sum_{n=1}^{\infty}{F_n(t) \sin{(\frac{n \pi x}{L})}}$$

Do we expand the function $f$ in an odd way due to the conditions:
$u(0,t)=u(L,t)=0, t>0$
or due to the conditions:
$f(0,t)=f(L,t)=0, t>0$ ?? (Thinking)