Solution of Heat Equation Through Fourier Series.

[Je m'appelle]

OK, so I was trying to solve the Heat Equation with Inhomogeneous boundary conditions for a rod through Fourier Series when I got stuck at the solution for the coefficient c_n, the part where I'm stuck is highlighted in red.

The following is just a step-by-step solution of how I got to c_n.

Heat Equation

\alpha^2 \frac{\partial^2}{\partial x^2}u(x,t) = \frac{\partial}{\partial t}u(x,t)

Initial Conditions


1. u(0,t) = T_1 = 20,\ u(30,t) = T_2 = 50,\ \forall \ t > 0

2. u(x,0) = 60 - 2x \ \forall \ 0 < x < 30

3. \alpha^2 = 1


Solution

So the general solution would be the steady-state temperature v(x) plus the transient temperature w(x,t) such as

u(x,t) = v(x) + w(x,t)

I found out that

v(x) = 20 + x

f(x) = u(x,0) + v(x) = w(x,0) + v(x) = 80 - x

Then by evaluating the general expression for the nonhomogeneous heat equation I got to

u(x,t) = T_1 + (T_2 - T_1)\frac{x}{L} + \sum_{n=1}^{+\infty} c_n e^{-\frac{n^2 \pi^2 \alpha^2}{L^2}t} sin(\frac{n \pi x}{L})

Where

c_n = \frac{2}{L}\int_{0}^{L} [f(x) - (T_2 - T_1)\frac{x}{L} - T_1] sin(\frac{n \pi x}{L}) dx

My struggling begins here, at c_n, let's evaluate it.

By using the initial conditions it can be rewritten as

c_n = \frac{1}{15}\int_{0}^{30} (60 - 2x) sin(\frac{n \pi x}{30}) dx

I'll break it into two integrals as follows

c_n = \frac{1}{15}\int_{0}^{30} 60 sin(\frac{n \pi x}{30}) dx - \frac{1}{15}\int_{0}^{30} 2x sin(\frac{n \pi x}{30})dx

c_n = (I) + (II)

Let's solve the integrals separately, solving (I) first

(I) = 4 \int_{0}^{30} sin(\frac{n \pi x}{30}) dx

(I) = \frac{120}{n \pi}(-cos(n \pi) + 1)

Now here, I can consider two solutions, (I) = 0 for n ''even'' or (I) = \frac{240}{n \pi} for n ''odd''.

Now, solving (II)

(II) = \frac{2}{15}\int_{0}^{30} x sin(\frac{n \pi x}{30}) dx

(II) = \frac{120}{n \pi}(\frac{sin(n\pi)}{\pi} - cos(n \pi))

Now, sin(n\pi) = 0 for n ''odd'' or ''even'', therefore the solution of (II) becomes

(II) = \frac{120}{n \pi}( - cos(n \pi))

Which will be \frac{120}{n \pi} for n ''odd'' or -\frac{120}{n \pi}for n ''even''.

So as you can see, I have two possible solutions for c_n which happens when either n is ''odd'' or ''even'', my question is, which one should I consider? The one for all n ''odd'' or the one for all n ''even''?

Like, if I carry on I would eventually get to

For all n ''odd'':

c_n = (I)_{odd} + (II)_{odd}

c_n = \frac{120}{n\pi}

u(x,t) = (20 + x) + \frac{120}{\pi}\sum_{n=1,3,5,...}^{+\infty} \frac{1}{n} e^{-\frac{n^2 \pi^2}{900}t} sin(\frac{n \pi x}{30})

For all n ''even'':

c_n = (I)_{even} + (II)_{even}

c_n = -\frac{120}{n\pi}

u(x,t) = (20 + x) - \frac{120}{\pi}\sum_{n=2,4,6,...}^{+\infty} \frac{1}{n} e^{-\frac{n^2 \pi^2}{900}t} sin(\frac{n \pi x}{30})

So which solution should I use? The one for the n ''odd'' or the one for n ''even''?

 

[arkajad]

How it is that your two different conditions give one u(0,0)=20 and the second one u(0,0)=60?

 

[Je m'appelle]

How it is that your two different conditions give one u(0,0)=20 and the second one u(0,0)=60?

They don't.

For u(x,0) = 60 - 2x

we have 0 < x < 30

and for u(0,t) = 20

we have t > 0.

I guess I haven't specified it very clear in the initial conditions on the original post, I'll edit that.

Here's the problem:

[PLAIN]http://img3.imageshack.us/img3/4238/ff212.jpg

 

[arkajad]

From (19) you have discontinuity at x=0? For t=0 you have u(0+,0)=60-0=60
And for t>0 you have u(0,t)=20. Just checking ...

 

[arkajad]

As for your odd-even problem: why don't you use odd solution for odd n and even solution for even n? I do not get it. After all you have to sum over n from 1 to infinity...

 

[Je m'appelle]

Hey, I think I've got it, listen

For all n ''odd'':

c_n = (I)_{odd} + (II)_{odd}

c_n = \frac{120}{n\pi}

u(x,t) = (20 + x) + \frac{120}{\pi}\sum_{n=1,3,5,...}^{+\infty} \frac{1}{n} e^{-\frac{n^2 \pi^2}{900}t} sin(\frac{n \pi x}{30})

For all n ''even'':

c_n = (I)_{even} + (II)_{even}

c_n = -\frac{120}{n\pi}

u(x,t) = (20 + x) - \frac{120}{\pi}\sum_{n=2,4,6,...}^{+\infty} \frac{1}{n} e^{-\frac{n^2 \pi^2}{900}t} sin(\frac{n \pi x}{30})

So which solution should I use? The one for the n ''odd'' or the one for n ''even''?

\frac{1}{n} e^{-\frac{n^2 \pi^2}{900}t} sin(\frac{n \pi x}{30})

is an even function right?

As both

(\frac{1}{n}e^{-\frac{n^2 \pi^2}{900}t}) and sin(\frac{n \pi x}{30}) are odd functions, their product is an even function and therefore, for even functions

f(-x) = f(x)

So I guess

\frac{120}{\pi}\sum_{n=1,3,5,...}^{+\infty} \frac{1}{n} e^{-\frac{n^2 \pi^2}{900}t} sin(\frac{n \pi x}{30}) = \frac{120}{\pi}\sum_{n=2,4,6,...}^{+\infty} (\frac{-1}{n}) e^{-\frac{n^2 \pi^2}{900}t} sin(\frac{n \pi x}{30})

And therefore as both solutions are equal for all n I can rewrite it as

u(x,t) = (20 + x) + \frac{60}{\pi}\sum_{n=1}^{+\infty} \frac{1}{n} e^{-\frac{n^2 \pi^2}{900}t} sin(\frac{n \pi x}{30})

What do you say? Is this acceptable?

 

[arkajad]

I do not understand what you are doing. You have 0<x<30, so the question of even-odd functions should not even arise. The exponential is a constant function (as a function of x), why you call it odd? Why don't you just add two sums, one for n even, one for n odd, including your (I) and (II) and simplifying according to the case?

 

[Je m'appelle]

I do not understand what you are doing. You have 0<x<30, so the question of even-odd functions should not even arise. The exponential is a constant function (as a function of x), why you call it odd? Why don't you just add two sums, one for n even, one for n odd, including your (I) and (II) and simplifying according to the case?

Oh, you're right, what was I doing? Never mind the previous post. I guess I'll just add the two sums then.

Thanks arkajad.

 

[LCKurtz]

You started with the substitution

u(x,t) = v(x) + w(x,t)

This will give you w(x,0) = u(x,0) - v(x). You have correctly determined that v(x) = 20 + x so you should have

w(x,0) = 60 - 2x - (20 + x) = 40 - 3x.

It looks to me like you are copying a general answer from a text instead of understanding what you are doing. You should be solving the W system:

w_xx(x,t) = w_t(x,t)
w(0,t) = 0, w(0,30) = 0
w(x,0) = 40 - 3x.

Once you get w(x,t) you will add v(x) to it to get the solution. Eventually you should come to the Fourier series

40 - 3x = \sum_{n = 1}^{\infty}b_n\sin\frac{n\pi}{30}x

for which you can use the half range sine coefficients. Don't break it into odd and even, just calculate them.