1D Heat Equation BC: Both ends insulated, IC: piecewise function

[PAR]

1. Solve the one dimensional heat equation for a rod of length 1 with the following boundary and initial conditions:

BC: \partialu(0,t)/\partialt = 0
\partialu(1,t)/\partialt = 0 these are the wrong boundary conditions (see below)

Actual BC: \partialu(0,t)/\partialx = 0
\partialu(1,t)/\partialx = 0

IC:
u(x,0) = { 1 if 0\leqx\leq.5
u(x,0) = { 0 if .5<x\leq1



2. \partialu/\partialt = a²\partial^{2}u/\partialx^{2}



3.

I used separation of variables and applied the boundary conditions to get the following:
u(x,t) = e^{(-(a2)(n*Pi)2*t)} * (B*cos(n*Pi*x)) where
n= 1, 2, 3... and B is an unknown constant.

To find B I tried applying the initial conditions and that's where I got stuck because I got.

u(x,0) = 0 = B*cos(n*Pi*x) if 0\leqx\leq.5
u(x,0) = 1 = B*cos(n*Pi*x) if .5<x\leq1

Does this mean that B has two values depending on x? And if is so are those values:

B = 0 for 0\leqx\leq.5
B = 1/cos(n*Pi*x) for .5<x\leq1

As a follow-up, will the steady state of this heat equation problem be a piecewise function?

 

[gabbagabbahey]

Shouldn't it be a sine instead of a cosine? (take a look at your BCs cos(0)=1 not 0)

Also, since that solution satisfies the heat equation and your boundary conditions for any integer value of n, shouldn't the general solution be a linear combination of each of your solutions, namely:

u(x,t)=\sum_{n=0}^{\infty} u_n(x,t)=\sum_{n=0}^{\infty}B_ne^{-a^2n^2 \pi^2 t}sin(n \pi x)

?

Remember, your B_n's are constants so they cannot depend on x. Use Fourier's method to determine them.

 

[PAR]

Sorry, I miss-typed the boundary conditions, the boundary conditions are supposed to be:

\partialu(0,t)/\partialx = 0
\partialu(1,t)/\partialx = 0

I'll edit it to the correct version in the original post too.

So I think that the solution is the linear combination:

u_{n}(x,t) =\sum(B_{n}e^{-a^2n^2\pi^2t}cos(n\pi\\x)) going from n = 1 to infinity

So now I set the sum equal to initial condition right?

I get:

u_{n}(x,0) = 1 = \sum(B_{n}cos(n\pi\\x)) for 0<=x<=1/2

I use fouriers method to find the B's in each interval:

B_{n} = 2 \int^{1/2}_{0}cos(n\pi\\x)= 2 sin(n\pi\\x)/(n\pi)\right|^{1/2}_{0}<br /> = 2(sin(n\pi/2)/(n\pi) = 2(-1)^n/(n\pi)

Unfortunately I'm stuck again, when I try to solve for the B's in the interval 1/2<x<=1

I get:

u_{n}(x,0) = 0 = \sum(B_{n}cos(n\pi\\x))

B_{n} = 2 \int^{1}_{1/2} 0*cos(n\pi\\x) = 0

Is this correct? Are the B's for 1/2<x<=1 all 0?

And sorry again about giving the wrong boundary conditions!

 

[gabbagabbahey]

Sorry, I miss-typed the boundary conditions, the boundary conditions are supposed to be:

\partialu(0,t)/\partialx = 0
\partialu(1,t)/\partialx = 0

I'll edit it to the correct version in the original post too.

So I think that the solution is the linear combination:

u_{n}(x,t) =\sum(B_{n}e^{-a^2n^2\pi^2t}cos(n\pi\\x)) going from n = 1 to infinity

So now I set the sum equal to initial condition right?

Okay, so doesn't that mean that

\sum_{n=0}^{\infty} B_n cos(n \pi x)=u(x,0)=\left\{ \begin{array}{lr} 1, &amp; 0 \leq x \leq 0.5 \\ 0, &amp; 0.5&lt; x \leq 1 \end{array}

\Rightarrow \sum_{m=0}^{\infty} \int_0^1 B_m cos(m \pi x) cos(n \pi x)dx = \int_0^1 u(x,0) cos(n \pi x) dx=\int_0^{0.5} (1)cos(n \pi x) dx+ \int_{0.5}^1 (0)cos(n \pi x) dx

;0)

Remember, your B_n's are constant so they must be the same for all values of x.

 

[PAR]

Ok,

so \int^{1}_{0}cos(n\pi\\x)*cos(m\pi\\x) = 0 if m \neq n

so, I've taken the integral of

\int^{1}_{0}cos(n\pi\\x)^2dx

And assuming n = positive integer I get

\int^{1}_{0}cos(n\pi\\x)^2dx = 1/2

and I know that \int^{1}_{0}u(x,0)cos(n\pi\\x)dx = -1^n/(n\pi)

So I get

\sum^{\infty}_{n=1}B_{n}(1/2) = (-1)^n/(n\pi)

Reading from my textbook and online I am guessing I should multiply both sides by 2 (or divide by 1/2) which would yield the same answer as in my previous post:

B_{n} = 2(-1)^n/(n\pi)

But I am confused by this last step: How does multiplying by both sides by 2 "remove" the B_{n} from out of the summation? Wouldn't it just remove the 2 from the summation and the left hand side of the equation would just be

\sum^{\infty}_{n=1}B_{n}?

I realize that I am expected to know these things before I enter a partial differential equations course, but after taking 7 college math classes I still haven't learned these basic principles of sums and it's really holding me back.

Oh, and thanks for the help so far gabbagabbahey, but I still have much to learn.

 

[gabbagabbahey]

The summation sign is removed when you integrate \sum_{m=0}^{\infty} \int_0^1 B_m cos(m \pi x) cos(n \pi x)dx, because the only term that is non-zero is m=n

\Rightarrow \sum_{m=0}^{\infty} \int_0^1 B_m cos(m \pi x) cos(n \pi x)dx=\frac{B_n}{2}+ \sum_{m \neq n} (0)B_m=\frac{B_n}{2}

\Rightarrow \frac{B_n}{2}=\frac{(-1)^n}{n \pi}

or

B_n=\frac{2(-1)^n}{n \pi}