Question

Solve the heat equation
\[\alpha^2\frac{\partial^2u}{\partial x^2}=\frac{\partial u}{\partial t}\], 00 subject to the conditions u(0,t)=0, u (L,t)=0 u(x,0) =x (L-x) Please help

Answer 1

I am sorry, I don't know

Answer 2

Can you help me check mine? I just imitate what my prof do without understanding a word

Answer 3

\[\alpha^2\frac{\partial^2u}{\partial x^2}=\frac{\partial u}{\partial t}\]

Answer 4

yes

Answer 5

What do you mean, "check" did you already solve it?

Answer 6

I told you, I just imitate my prof's stuff, I didn't understand

Answer 7

I guess you are using Fourier series right..

Answer 8

I really don't know

Answer 9

I see neumann conditions. So shouldn't it be fourier cosine series?

Answer 10

no no, wait you do have fourier sine series! Sorry, you're right!

Answer 11

I am looking at your 1, but i gotta step out for a bit.

Answer 12

your boundaries should be like this, do you think you have typo\[U_{x}(0,t)=0 \\U_{x}(L,t)=0\\ U_{x}(x,0)=x(L-x)\]

Answer 13

nope, there is no x as sub in U (0,t) or U(L,t)

Answer 14

ok this is different problem..

Answer 15

I got Cn using by mathematica..

Answer 16

It cannot have sin there, right? because sin ... pi =0

Answer 17

yes

Answer 18

when we simplify we get \[\frac{4L^2}{n^3\pi^3}-\frac{4L^2\cos n \pi}{n^3\pi^3}\]

Answer 19

you forgot 2 when you calculate c_1

Answer 20

I don't get why at out(4) , numerator says 2L^2 (2+ (-2+ n^2 pi^2...) they don't cancel out 2 and -2 inside the bracket?

Answer 21

yeah, you are right, thank you

Answer 22

why my denominator just n^2pi^2 , theirs is n^3pi^3:(

Answer 23

your C_1 is right, let me check C_2

Answer 24

I see my mistake at denominator from page 3, the second term should be n^3 pi^3 at denominator, ha!!

Answer 25

that's right!!

Answer 26

\[C_{1}=-\frac{2L^2\cos n \pi}{n \pi}\\C_{2}=\frac{4L^2\cos n \pi}{n \pi}-\frac{4L^2\cos n \pi}{n^3 \pi^3}\]

Answer 27

c2 has another term without cos term

Answer 28

nope, I combine C_1 and C_2 , cancel out some like term to get
\[C_n= -\dfrac{4L^2 cos(n\pi)}{n^3\pi^3}-\dfrac{2L^2 sin(n\pi)}{n^2\pi^2}+\dfrac{4L^2}{n^3\pi^3}\]

Answer 29

\[C_{1}=-\frac{2L^2\cos n \pi}{n \pi}\\C_{2}=\frac{4L^2\cos n \pi}{n \pi}-\frac{4L^2\cos n \pi}{n^3 \pi^3}+\frac{4L^2}{n^3\pi^3}\]

Answer 30

well you can cancel sin term in c_1

Answer 31

from then, I can construct U(x,t) , right?

Answer 32

\[C_{1}=-\frac{2L^2\cos n \pi}{n \pi}\\C_{2}=\frac{2L^2\cos n \pi}{n \pi}-\frac{4L^2\cos n \pi}{n^3 \pi^3}+\frac{4L^2}{n^3\pi^3}\]

Answer 33

yes, you can

Answer 34

It's too late here. I will redo and repost it, just the term I messed up (page 3 and 4) , Can you help me check mine tomorrow?

Answer 35

I'll try (:

Answer 36

One more question, that is the way I have to do, right? that 's fourier series, right?

Answer 37

hey, (: means???

Answer 38

I saw everybody use :) which stands for smiling face and :( for crying one, yours is (: I don't know

Answer 39

Anyway, thanks, thanks, thanks a ton.

Answer 40

smiling (:

Answer 41

is that the final answer? ok, let me try to get it.

Answer 42

where \(\alpha^2\)?

Answer 43

I used k

Answer 44

I forgot t, this is the correct one

Answer 45

my book uses different letter, don't be confused, use your letter..

Answer 46

http://tutorial.math.lamar.edu/Classes/DE/SolvingHeatEquation.aspx

Answer 47

aha, yours is k, got you. thanks again

Answer 48

@cinar I know you are offline, just post in case you came online, check mine, please. Compare to wolframalpha, my \(C_n\) is ok, but the last step when plug everything into the formula, I am not sure. My argument is whatever n is, sin npi =0 and cos npi =-1, therefore, the last result should be as mine.

Answer 49

when n is odd cosnpi=-1 n is even cosnpi=1

so you can replace cosnpi by (-1)^n

I can't see your last sol.

Answer 50

\[ \Large C_{n}=\frac{-4(-1)^nL^2}{n^3 \pi ^3}+\frac{4L^2}{n^3 \pi^3}\\

\Large U(x,t)=\sum_{n=1}^{\infty}(\frac{-4(-1)^nL^2}{n^3\pi ^3}+\frac{4L^2}{n^3 \pi^3})e^{\frac{-n^2\pi^2\alpha^2}{L^2}t}sin{\frac{n\pi x}{L}}\]

Answer 51

Thank you, cinar