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Home Assignment 4 / Re: Web Bonus Problem--3

« on: October 27, 2012, 01:12:11 PM »

OK, well, I ended up making this way more detailed than I had planned. We first separate variables. Let: [u(x,t) = X(x)T(t)] in [u_{tt}(x,t) + K u_{xxxx} = 0,   K > 0]

Then: [u_{tt}(x,t) = X(x)T''(t),  u_{xxxx}(x,t) = X''''(x)T(t)]
[u_{tt}(x,t) + K u_{xxxx} = X(x)T''(t) + K X''''(x)T(t) = 0]
[\frac{X''''(x)}{X(x)} = \frac{-T''(t)}{K T(t)} = \lambda]

For some constant, as both sides are independant of the other respective variable. Now, we have by assumptions that [\lambda = c^4 > 0,   K = k^2 > 0]

So we are left with two ODE's in the form: [X''''(x) = c^4 X(x),  T''(t) = -c^4 k^2 T(t)]

Which yield solutions: [X(x) = A \cosh(c x) + B \sinh(c x) + C \cos(c x) + D \sin(c x)]
[T(t) = A \cos(c^2 k t) + B \sin(c^2 k t)]

Note that correct notation is \cos which produces $\cos$ etc

Using the two boundary conditions: [u(0,t) = 0 = u_x(0,t)]

[X(0) = A \cosh(c 0) + B \sinh(c 0) + C \cos(c 0) + D \sin(c 0) = A + C = 0,   A = -C]
[X'(0) = A c \sinh(c 0) + B c \cosh(c 0) - C c \sin(c 0) + D c \cos(c 0) = B c + D c,   D = -B]

As we disregard the case where: [X(0) \ne 0,   X'(x) \ne 0 \implies T(t) ≡ 0]

OK, great. We plug into the 3rd and 4th boundary conditions.

[u_{xx}(l,t) = 0 = u_{xxx}(l,t)]

\begin{multline*}
X''(l) = A c^2 \cosh(c l) + B c^2 \sinh(c l) - C c^2 \cos(c l) - D c^2 \sin(c l) =\\
 A c^2 \cosh(c l) + B c^2 \sinh(c l) + A c^2 \cos(c l) + B c^2 \sin(c l) = A c^2 (\cosh(c l) + \cos(c l)) + B c^2 (\sinh(c l) + \sin(c l)) = 0
\end{multline*}
There is \multline
[X'''(l) = A c^3 (\sinh(c l) - \sin(c l)) + B c^3 (\cosh(c l) + \cos(c l)) = 0]

Yielding: [A (\cosh(c l) + \cos(c l)) + B (\sinh(c l) + \sin(c l)) = 0]
[A (\sinh(c l) - \sin(c l)) + B (\cosh(c l) + \cos(c l)) = 0]
[A = -B \frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))}]
[-B \frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))} (\sinh(c l) - \sin(c l)) + B (\cosh(c l) + \cos(c l)) = 0]
[-\frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))} (\sinh(c l) - \sin(c l) + (\cosh(c l) + \cos(c l)))]
[-\sinh(c l)^2 + \sin(c l)^2 + \cosh(c l)^2 + 2 \cosh(c l) \cos(c l) + \cos(c l)^2 =0]

Where we used the facts that [B = 0 \implies X(x) ≡ 0,   (\cosh(c l) + \cos(c l)) = 0 \implies B = A = 0]
Which we disregard as we're not so interested in the trivial case.
[ \cosh(c l)^2 - \sinh(c l)^2 + \sin(c l)^2+ \cos(c l)^2 + 2 \cosh(c l) \cos(c l)  = 1 + 1 + 2 \cosh(c l) \cos(c l) = 0]

As: [\cosh^2 - \sinh^2 = 1,   \cos^2 + \sin^2 = 1]

So our eigenvalues are those c which satisfy [\cosh(c l) \cos(c l) = -1 \blacksquare]



Next consider: [\int_0^l X_n(x)X_m(x) dx,   n \ne m,   \lambda_n \ne \lambda_m]

Then [(\lambda_n - \lambda_m) \int_0^l X_n(x)X_m(x) dx = \int_0^l \lambda_n X_n(x)X_m(x) -  \lambda_m X_n(x)X_m(x) dx]
[\int_0^l X''''_n(x)X_m(x) - X_n(x)X''''_m(x) dx = (X'''_n(x)X_m(x) - X_n(x)X'''_m(x))_{x=(0,l)} - \int_0^l X'''_n(x)X'_m(x) - X'_n(x)X'''_m(x) dx]
[= 0 - (X''_n(x)X'_m(x) - X_n'(x)X''_m(x))_{x=(0,l)} + \int_0^l X''_n(x)X''_m(x) - X''_n(x)X''_m(x) dx = 0 + 0 = 0]

Using integration by parts, and the fact that our boundary conditions vanish at:

[X(0) = 0, X'(0) = 0, X''(l) = 0, X'''(l) = 0]

Then we have [(\lambda_n - \lambda_m) \int_0^l X_n(x)X_m(x) dx = 0,     \lambda_n \ne \lambda_m \ne 0 \implies \int_0^l X_n(x)X_m(x) dx = 0]

And our eigenfunctions are orthogonal â– .

Let our differential operator be [\mathcal{I}, st.  \mathcal{I} X = \lambda X,   \mathcal{I} Y = \lambda Y]

We show that: [\langle \mathcal{I}X,Y\rangle = \langle X,\mathcal{I}^*Y\rangle]

[\langle \mathcal{I}X,Y\rangle = \int_0^l \mathcal{I}X(x)Y(x)^* dx = \int_0^l \lambda X(x)Y(x)^* dx = \int_0^l X(x)\lambda^* Y(x)^* dx = \int_0^l X(x)\mathcal{I}^*Y(x)^* dx = \langle X,\mathcal{I}^*Y\rangle]

As we had by assumption our eigenvalues were real. Then our operator is hermetian, so the eigenfuncitons of different eigenvalues are linearly independant, and the eigenfunctions of the same eigenvalue are linearly dependant.â– 

32

Home Assignment 4 / Problem 1

« on: October 18, 2012, 04:39:28 PM »

It seems to me that question 1 part c) doesn't ask any question and instead makes a statement. Is there anything I'm missing there?

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Home Assignment 4 / Problem 2

« on: October 17, 2012, 04:46:26 PM »

In problem 2, do we have: [K > 0] for [u_{tt} + K u_{xxxx} = 0]?

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Technical Questions / Re: QED symbol?

« on: October 17, 2012, 12:41:54 PM »

Here's a nice searchable list of mathjax symbols I've been using:

http://www.onemathematicalcat.org/MathJaxDocumentation/TeXSyntax.htm

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Home Assignment X / Problem 4

« on: October 13, 2012, 10:02:10 PM »

In problem 4, we have the functions:

[ ÆŸ(x) = 0  \{x < 0\}, \space ÆŸ(x) = 1  \{x > 0\}]            [ f(x) = 1   \{|x| > 1\},    f(x) = 0   \{|x| < 1\}]

and an idea to use the functional relation [f(x) = ÆŸ(x+1) - ÆŸ(x-1)]

But this identity does not hold. For instance, [f(0) = ÆŸ(1) - ÆŸ(-1) = 1 - 0 = 1 \ne 0]

An identity which I found to work instead for the integral is: [f(x) = ÆŸ(x - 1) + ÆŸ(-x - 1) ]

Have I made an error?

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Home Assignment 3 / Re: problem 3

« on: October 08, 2012, 07:46:14 PM »

Try completing the square and using the error function. It's a really messy integral.

$\newcommand{\erf}{\operatorname{erf}}$
Yes, you need to consider  $\erf(z)$ as an elementary function (and there is no compelling arguments why trigonometric functions are considered as such but not many others. In fact there are plenty of important special functions coming often from PDE, more precisely, from separation of variables--not $\erf$ but many others).

When you integrate erf(z) it always gives you zero because it's an odd function. When multiplied to any integrated function (and as alpha -> 0), the resulting functions are always 0. Does that make any sense?

The error function itself isn't integrated here, the function G(x,y,t)*g(y) is integrated.

It's true that an odd function will integrate to 0 on a domain symmetric about 0, and anything multiplied by this evaluated integral will be 0, but because we're multiplying G(x,y,t) inside of the integral sign that fact isn't too relevant here.

Consider the analogue to the function f(x) = x. f is an odd function, so the integral on say (-a,a) is 0. But when we multiply inside of the integral sign by another odd function g(x) = x^3, we have g*f(x) = x^4- certainly not 0 on the interval (-a,a) when a != 0.

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Home Assignment 3 / Re: problem 3

« on: October 08, 2012, 12:18:45 AM »

Try completing the square and using the error function. It's a really messy integral.

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Home Assignment 3 / Problem 2

« on: October 04, 2012, 03:26:12 PM »

It seems to me that in question 2 in homework 3, the letter "v" is used to refer to both a constant, and a function u(x,t) = v(x - v t, t). Under this interpretation everything works out nicely, and I assume this to be the case.

Is this the correct? Or is v a function which recursively calls itself? If not, it might be worth changing the question to have a different letter to represent the constant or function.

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Home Assignment 2 / Re: problem 1 typo?

« on: September 29, 2012, 04:57:27 PM »

You should be able to find constants from initial and boundary conditions. Solutions may be discontinuous along lines you indicated

Ah, excellent! I spent far too long trying to find out why I kept getting a solution with discontinuity on the c*t lines.