Heat Equation - bar, length L, heated at one end, cooled at the other end

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No. v(x, t) = u(x, t) - s(x, t).

Sorry mistype I meant v(x,0)=u(x,o)-s(x)

So in my case v(x,0) = sin(xpi/L)-1+2x/L

I think this is right just want to double check, I know it may seem "dumb" but I just want to double check everything so I don't learn it wrong

EDIT: because I am working through it using v(x,0) as stated above and it looks like it is going really weird, and rather too longwinded.

(edited 12 years ago)

Reply 22

DFranklin

Well, I think that's the correct expression. Note that you can "write down" the Fourier expansion of sin(x pi / L), because it's just sin(x pi / L). So you just need to deal with the 2x/L - 1 term by writing it as a Fourier Series.

Reply 23

MimuiSnoopy_2922

Original post by DFranklin

I'm sorry I am confused. I thought I was supposed to be solving

c_n = \frac{2}{L} \int_0^L v(x,0) \cos( \frac{n \pi x}{L} ) dx

To put in this equation:

u(x,t) = \sum c_n e^\frac{-n^2 x^2 k t}{L^2} \cos ( \frac{n \pi x}{L} ) + 1 - \frac{2x}{L}

I'm not sure if I understand what you mean in your reply.

Reply 24

DFranklin

It should be

\sin\frac{n\pi x}{L}

in the integral, which is what you've been writing in every post up until the one just above - I don't know where cos(n pi x / L) has come from all of a sudden.

So you need to find

\int \left[ \sin \frac{x \pi}{L} -1 + \dfrac{2x}{L}\right] \sin \frac{n\pi x}{L}\,dx

.

You can split this into

\int \sin \frac{x \pi}{L} \sin \frac{n\pi x}{L}\,dx

and

\int \left[-1 + \dfrac{2x}{L}\right] \sin \frac{n\pi x}{L}\,dx

, and you should be able to write down the value of the first of these by considering the Fourier sin series for

\sin \frac{x \pi}{L}

.

[Although if I'm completely honest, given your other postings in this thread, I think it's pretty likely you won't be able to do this and will just get confused. In which case just work out the integral the hard way].

Reply 25

MimuiSnoopy_2922

Original post by DFranklin

It should be

\sin\frac{n\pi x}{L}

in the integral, which is what you've been writing in every post up until the one just above - I don't know where cos(n pi x / L) has come from all of a sudden.

So you need to find

\int \left[ \sin \frac{x \pi}{L} -1 + \dfrac{2x}{L}\right] \sin \frac{n\pi x}{L}\,dx

.

You can split this into

\int \sin \frac{x \pi}{L} \sin \frac{n\pi x}{L}\,dx

and

\int \left[-1 + \dfrac{2x}{L}\right] \sin \frac{n\pi x}{L}\,dx

, and you should be able to write down the value of the first of these by considering the Fourier sin series for

\sin \frac{x \pi}{L}

I'm trying to do it without working out the integral the hard way

would

\int \sin \frac{x \pi}{L} \sin \frac{n\pi x}{L}\,dx

be equal to 0?

Reply 26

around

For n not equal to 1, yes.

For n=1 you have to use a trig identity.

Reply 27

MimuiSnoopy_2922

Original post by around

For n not equal to 1, yes.

For n=1 you have to use a trig identity.

Thank you

Reply 28

DFranklin

Original post by around

For n not equal to 1, yes.

For n=1 you have to use a trig identity.

Or, as I said, observe that the Fourier sin series for sin(pi x / L) is going to be, um, sin(pi x / L)...

Reply 29

MimuiSnoopy_2922

Original post by DFranklin

Or, as I said, observe that the Fourier sin series for sin(pi x / L) is going to be, um, sin(pi x / L)...

I do really appreciate the help you (and others) have given me with these questions I have been stuck on.
However, I have to say that I sometimes find your tone to me (and others) comes across as condescending. I am not sure whether you mean it to come across that way or not. But it can be a little hurtful to someone who is just trying to learn.