User:Egm6322.s09.mafia/HW4

My second wave of comments is color coded in green. Egm6322.s09 22:27, 2 April 2009 (UTC)

See my comments below.

After you made a correction for a section with a comment box, you want to put a comment in that same comment box on what you did. Egm6322.s09 13:49, 15 March 2009 (UTC)

The Classification of Second Order Partial Differential Equations

Classifications

In the general form of PDE:

$au_{xx}+2bu_{xy}+cu_{yy}+du_{x}+eu_{y}+fu+g=0$

it can be classified by the value of $ac-b^2$.

$ac-b^2=det \underline A$

$ac-b^2<0 \Rightarrow$ PDE is hyperbolic.

$ac-b^2=0\Rightarrow$ PDE is parabolic.

$ac-b^2>0\Rightarrow$ PDE is elliptic.

For example:

In the equation:

$u_{xx}+u_{yy}=0$...(1)

$\Rightarrow a=1,b=0,c=1$

$\Rightarrow ac-b^2=1>0$

$\Rightarrow$ This PDE is elliptic

HW: What would be the classification of the diffusion operator in polar coordinates? Would it make sense if the classification changed along with transformation of coordinates? i.e If we transformed the diffusion operator from cartesian to polar coordinates, would its classification change?

Egm6322.s09.Three.nav 13:36, 24 April 2009 (UTC) The diffusion operator D(.) is given by div(grad(.)). Recall, In cartesian coordinates, div(grad(u))= $\nabla.(\nabla u) = \left \lfloor \partial_{xx}+ \partial_{yy} \right \rfloor$ In polar coordinates, div(grad(u))= $\left\lfloor \partial_{rr}+\frac{1}{r}\partial_{r}+\partial_{\theta\theta} \right \rfloor$ A PDE of the form $au_{xx}+bu_{xy}+cu_{yy}= \psi\left ( u_{x},u_{y}, u,x,y\right )$------(1) can be characterized by the nature of the determinant $\begin{vmatrix}a \ b \\b \ c \end{vmatrix}$ Elliptic if ac-b2>0 Parabolic if ac-b2=0 Hyperbolic if ac-b2<0 Comparing the cartesian form of D(u)with (1), we see that a= 1, b=0, c=1 $\Rightarrow ac-b^{2}= 1 > 0 \Rightarrow$ D(u) in cartesian coordinates is elliptic. In polar coordinates, the matrix X= $\begin{bmatrix}a \ b \\b \ c \end{bmatrix}$ is transformed and written as the matrix given by $\bar{X}= JXJ^{T}$ where J= $\begin{bmatrix} \ cos(\theta) -sin(\theta)\\sin(\theta)\ \ \ cos(\theta) \end{bmatrix}or \begin{bmatrix} \ C -S\\S\ \ \ C \end{bmatrix}$ Hence evaluating $\bar{X}= \begin{bmatrix} C -S \\ S\ \ \ C \end{bmatrix}\begin{bmatrix} \left (aC-bS \right ) \ \left (aS+bC \right ) \\ \left (bC-cS \right ) \ \left (bS+cC \right ) \end{bmatrix}$ = $\begin{bmatrix} \left(ac^{2}- 2bSC+cS^{2} \right) \ \ \ \ \ \ \left(aSC+bC^{2}-bS^{2}-cSC \right)\\ \left(aSC+bC^{2}-bS^{2}-cSC \right) \ \ \ \ \ \left(aS^{2}+2bSC+cC^{2} \right) \end{bmatrix}$ For a PDE of the form given in (1), the determinant of $\begin{bmatrix}a \ b \\b \ c \end{bmatrix}$ characterizes the nature of the PDE. Similarly for the same PDE of form (1), transformed into a different coordinate system, the determinant of $\bar{X}$ characterizes the nature of the transformed PDE. To answer the first part of the HW, it would NOT and does not make sense if the classification changed along with the transformation. This is because no matter the type of coordinate system used, the physics of the problem remains the same. Hence if a different coordinate system was used, then a change in the nature of the transformed PDE would imply a change in the physics of the problem which is nonsensical. Hence transformation should not change classification Lets try to prove this. If (1) was transformed into another set of coordinates, the classification would change only if the determinant det($\bar{X}$) was different from det(X). Evaluating det($\bar{X}$)= $\begin{vmatrix} \left(ac^{2}- 2bSC+cS^{2} \right) \ \ \ \ \ \ \left(aSC+bC^{2}-bS^{2}-cSC \right)\\ \left(aSC+bC^{2}-bS^{2}-cSC \right) \ \ \ \ \ \left(aS^{2}+2bSC+cC^{2} \right) \end{vmatrix}$ = (ac-b2)(C2+S2) But $C= cos(\theta) \ and \ S= sin(\theta) \Rightarrow \ C^{2}+S^{2}= 1$ $\Rightarrow det\left(\bar{X} \right)= \left(ac-b^{2} \right)$ which is the same as det(X). Hence it is proved that Even if the PDE of form (1) in cartesian coordinates is transformed to another coordinate system (say polar), the classification remains constant. Consider the polar form of the diffusion operator D(u)= $\left\lfloor \partial_{rr}+\frac{1}{r}\partial_{r}+\partial_{\theta\theta} \right \rfloor \left \{u \right \}$ Here a= 1,b= 0 and c= $\frac{1}{r^{2}}$. Determinant= ac-b2= $\frac{1}{r^{2}}$ >0 $\forall$ r. Hence the polar form of the diffusion operator is also elliptic.

Another way to find the type of equation (1):

$\underline \bar A:=\underline J \underline A \underline J^T =\begin{bmatrix} A & B \\ B & C \end{bmatrix} =\begin{bmatrix} \bar a & \bar b \\ \bar b & \bar c \end{bmatrix} \to$ preferred form

$det \underline \bar A=\bar a \bar c-\bar b^2$

$\bar a=1; \bar b=0; \bar c=\frac {1}{r^2} (r\ne0)$

$\Rightarrow$The equation (1) is an elliptic PDE.

Relationship Between Classifications and Transformations

Observations from the verification of PDE classifications:

1. Diffusion operator remains elliptic in polar coordinates.

Question:How about a different transformation of coordinate? Would classification remain the same?

2. Does classification make sense if it changes under transformation of coordinate?

The answer is no, because physics (e.g. distribution of temperature as a result of solution of heat equation) must remain the same regardless of how heat equation was solved (under different coordinate system).

Therefore, classification better remains the same under different coordinate system for it to make sense.

Egm6322.s09.three.liu 16:35, 24 April 2009 (UTC)

The Laplace Equation

Egm6322.s09.Three.ge 17:58, 24 April 2009 (UTC)

The Laplace equation is the heat conduction equation with constant thermal conductivity and no heat generation.

The Laplace equation, symbolically:

$div(grad \ u)=\triangledown \cdot(\triangledown u)=\triangledown^{2}u$

Taking the Laplace equation in polar coordinates gives the following.

$div(grad \ u)=0=u_{rr}+\frac {1} {r} u_{r}+ \frac {1} {r^{2}} u_{\theta \theta}$

Features of an Axisymmetric Problem

Egm6322.s09.Three.ge 17:58, 24 April 2009 (UTC)

Axisymmetric domain that includes the origin; Egm6322.s09.Three.ge 17:58, 24 April 2009 (UTC)

Axisymmetric domain that does not include the origin; Egm6322.s09.Three.ge 17:58, 24 April 2009 (UTC)

For an axis-symmetric problem, the theta terms drop out.

$u_{\theta}=u_{\theta\theta}=\cdots =0$

The equation then reduces to,

$u_{rr}+\frac {1} {r} u_{r}=0$

which is an ordinary differential equation (ODE).

In order to solve this ODE one should note that it may be rearranged.

$u_{rr}+\frac {1} {r} u_{r}=\frac{1}{r}\frac{d}{dr}(r\frac{du}{dr})=0$

Separating variables and integrating gives the solution:

$u(r)=A_{0}ln \ r+B_{0}$

where A_o and B_o are constants.

It should be noted that if the domain $\omega$ includes the origin (where r=0) then A_o must be zero for a finite solution.

Separation of Variables

Egm6322.s09.Three.ge 17:58, 24 April 2009 (UTC)

Non-axisymmetric domain that includes the origin; Egm6322.s09.Three.ge 17:58, 24 April 2009 (UTC)

Non-axisymmetric domain that does not include the origin; Egm6322.s09.Three.ge 17:58, 24 April 2009 (UTC)

If the problem is not axis-symmetric, the Laplace equation may be solved using separation of variables.

Multiplying the Laplace equation by r² gives:

$\begin{matrix} r^{2} \cdot\left \{u_{rr}+\frac {1} {r} u_{r}+ \frac {1} {r^{2}} u_{\theta \theta}=0 \right \}\\ \\ =r^{2}(u_{rr}+\frac {1} {r} u_{r}) +u_{\theta\theta} \end{matrix}$ ... (a)

Observing that the equation has a portion that depends on r only and a part that depends on theta only, one may assume a solution of the form:

$u(r,\theta)=F(r)\cdot G(\theta)$

Thus the solution is a product of 2 functions: one which depends only on r [$F(r)$] and the other which only depends on theta [$G(\theta)$].

Plugging in this solution into (a) produces:

$r^{2}G(\theta)\left [\frac{d^{2}F(r)}{dr^{2}}+ \frac{1}{r}\frac{dF(r)}{dr} \right ]+F(r)\frac{d^{2}G}{d\theta^{2}}=0$

Dividing by $F(r)G(\theta)$ and rearranging gives:

$\frac{1}{F(r)}\left (r^{2}\frac{d^{2}F(r)}{dr^{2}}+ r\frac{dF(r)}{dr} \right )= \frac{-1}{G(\theta)}\frac{d^{2}G}{d\theta^{2}}=n^{2}$

HW: Why is the separation constant n2 positive?

Egm6322.s09.Three.nav 13:38, 24 April 2009 (UTC) The sign of the separation constant n2 potentially determines the roles of the 'r' and '$\theta$ coordinates. In the method used to solve Laplace's Equation here, a +ve sign implies periodicity in the $\theta$ coordinate. One of the applications of solving the Laplace's Equation is in electrostatics. Here the physics of the problem generally dictates periodicity in the $\theta$ direction. Hence the sign of n2 is positive. Also here the equation for 'r' = $r^{2}\frac{d^{2}F(r)}{dr^{2}}+ r\frac{dF(r)}{dr}-n^{2}F(r) =0$ is a non-linear equation and one that cannot, by its very nature ,dictate oscillation or periodicity. In such a case, we usually prefer having the oscillation dependence in the azimuthal ($\theta$) direction.

Move the figure to the left (instead of displaying it on the right) so to avoid blocking the hide/show link to open up the collapsible box. It is not possible to open up this collapsible box. Egm6322.s09 22:38, 2 April 2009 (UTC)

Figures moved to left. Egm6322.s09.bit.la 22:38, 2 April 2009 (UTC)

And results in the following solution for $n\neq 0$.

$\begin{matrix} F(r)=Ar^{n}+\frac{B}{r^{n}}\\ \\ G(\theta)=C cos(n\theta)+D sin(n\theta) \end{matrix}$

If n=0:

$\begin{matrix} F(r)=A_{0}ln \ r +B_{0}\\ \\ G(\theta)=C_{0}\theta+D_{0} \end{matrix}$

HW: The solution to the Laplace Equation is called harmonic. Why?
Egm6322.s09.Three.nav 13:38, 24 April 2009 (UTC) Consider $\frac{d^{2}G}{d\theta^{2}}-n^{2}G(\theta)=0$, one of the equations obtained through the separation of variables. This resembles the equation of harmonic motion (the equation characterizing the motion of a simple harmonic oscillator). This is why the solution G($\theta$) is called harmonic.

In general, we want the solution to be periodic, such that:

$\begin{matrix} k=interger=1,2,3,\cdots\\ u(r,\theta+k2\pi)=u(r,\theta) \end{matrix}$

Why do we want the solution to be periodic?
Our solution resembles that of a harmonic oscillator which means it will complete one complete cycle of motion going from $\theta= 0$ to $\theta=\pi$. So when it completes this cycle, its solution at $\theta= \pi$ cannot be different from the solution at $\theta= 0$ which is why we need the solution to be periodic.

The general form of the Laplace equation in polar coordinates takes the form that follows.

$u(r,\theta)=A_{0}ln \ r+\sum_{n=1}^{\infty }r^{n}\left [A_{n}cos(n\theta)+B_{n}sin(n\theta) \right ]+\sum_{n=1}^{\infty }\frac{}{r^{n}}\left [C_{n}cos(n\theta)+D_{n}sin(n\theta) \right ]+C_{0}$

Another Axisymmetric Problem

Homework:Derive LP p.14 (1.2.13)

--EGM6322.S09.TIAN 17:20, 24 April 2009 (UTC)

$\begin{matrix} \overline{a}=a \phi_x^2 + 2b \phi_x \phi_y +c \phi_y^2\\ \\ \overline{b}=a \phi_x \psi_x + b( \phi_x \psi_y +\phi_y \psi_x ) + c \phi_y \psi_y \\ \\ \overline{c}=a \psi_x^2 + 2b \psi_x \psi_y +c \psi_y^2 \end{matrix}$

Substituting into the equation yields:

$\overline{ac}-\overline{b}^2 =(a \phi_x^2 + 2b \phi_x \phi_y +c \phi_y^2)(a \psi_x^2 + 2b \psi_x \psi_y +c \psi_y^2)$ - $[a \phi_x \psi_x + b( \phi_x \psi_y +\phi_y \psi_x ) + c \phi_y \psi_y ]^2$

Multiplying out the equations and rearranging terms gives:

=$a^2\phi_x^2\psi_y^2 + 2ab\phi_x^2\psi_x\psi_y + 2bc\phi_y^2\psi_x\psi_y + 2bc\phi_x\phi_y\psi_y^2 + c^2\phi_y^2\psi_y^2 + 2ab\phi_x\phi_y\psi_y^2 + ac\phi_y^2\psi_y^2 + 4b^2\phi_x\phi_y\psi_x\psi_y + ac\phi_x^2\psi_y^2$

-

$a^2\phi_x^2\psi_x^2 + 2ab\phi_x^2\psi_x\psi_y + 2bc\phi_y^2\psi_x\psi_y + 2bc\phi_x\phi_y\psi_y^2 + c^2\phi_y^2\psi_y^2 + b^2\phi_x^2\psi_y^2 + b^2\phi_y^2\psi_x^2 + 2b^2\phi_x\phi_y\psi_x\psi_y + 2ab\phi_x\phi_y\psi_x^2 + 2ac\phi_x\phi_y\psi_x\psi_y$

One can see clearly that the first five terms cancel. This leaves the following equation:

$\begin{matrix}2ab\phi_x\phi_y\psi_y^2 + ac\phi_y^2\psi_y^2 + 4b^2\phi_x\phi_y\psi_x\psi_y + ac\phi_x^2\psi_y^2\\ -\\ b^2\phi_x^2\psi_y^2 + b^2\phi_y^2\psi_x^2 + 2b^2\phi_x\phi_y\psi_x\psi_y + 2ab\phi_x\phi_y\psi_x^2 + 2ac\phi_x\phi_y\psi_x\psi_y \end{matrix}$

Which, after some manipulation, results in:

$(ac-b^2)(\phi_x \psi_y - \phi_y \psi_x)^2$

What is "LP"? It is not clear how you arrived at the result, which is also wrong. Need more explicit explanation of the derivation, i.e., provide intermediate steps. I also mentioned to use our notation, not the notation by "LP". Egm6322.s09 15:08, 15 March 2009 (UTC)

My comment above had not been addressed; I did go over the above comment in class. Please take action. Egm6322.s09 22:38, 2 April 2009 (UTC)

---

Comment was addressed, but content was deleted. It has been re-posted, now with the correct result. Egm6322.s09.Three.ge 21:02, 6 April 2009 (UTC)

$\mathbf{a}\mathbf{c}-\mathbf{b}^2=\left (ac-b^2 \right )\left (\Phi _{x} \psi_{y}-\Phi _{y} \psi_{x}\right )^2$ page 14.2

Application:

Example:

$T^{*}(\Theta )=T_0\left (1+cos^2\Theta \right )$ where T₀ is a constant

$T^{*}(\Theta =0)=2T_0=T^{*}\left (\Theta =2\pi \right )$

$T^{*}(\Theta =\frac{\pi}{2} )=T_0=T^{*}(\Theta =\frac{3\pi}{2})$

$T^{*}(\Theta)=\frac{3T_0}{2}+\frac{T_0}{2}cos2\Theta$

Homework:Expand $\; T^*$ in terms of $cos \theta$

--EGM6322.S09.TIAN 17:23, 24 April 2009 (UTC)

$T^* (\theta)= \frac {3T_o}{2} + \frac {T_o}{2} cos{2 \theta}$

Expand it:

$T^* (\theta)= \frac {3T_0}{2} + \frac {T_0}{2} (2 cos^2 \theta -1)$ $=T_0+T_0 cos^2 \theta$

Principle of Superposition

$soln=solnT^{*}= \frac{3T_0}{2}=T_{1}^{*}(\Theta )$

$+solnT^{*}=\frac{T_0}{2}cos2\Theta=T_{2}^{*}(\Theta )$

Homework: Prove Law of Superposition is valid

--Egm6322.s09.xyz 16:29, 4 April 2009 (UTC)

The governing PDE is given as $div\left( gradT\right) = 0$. This PDE is linear and therefore the solution can be expressed using the Principle of Superposition:

i.e the solution = solution for $T^{*}\left(\theta\right) = \frac{3T_o}{2}$ = constant $+$ solution for $T^{*}\left(\theta\right) = \frac{T_o}{2}cos2\theta$

The proof of the linearity of the $grad(\cdot)$ and $div(\cdot)$ operators was presented by Team Mafia in R2. For completeness, the relevant portions of the proof are presented again below:

note: $grad(\cdot)$ is linear
$grad(u) = \frac {\partial u}{\partial x_i} e_i$ and $\frac {\partial u}{\partial x_i} (\cdot)$ is linear
$\therefore$ $grad \left( \alpha u + \beta v \right) = \alpha grad(u) + \beta grad(v)$

note: $div(\cdot)$ is linear because it is another differential operator. Let $\bar{a}, \bar{b}: \Omega$ $\mathbb{R}^3$ and $\alpha, \beta \in \mathbb{R}$ $\therefore$ $div \left( \alpha \bar{a} + \beta \bar{b} \right) = \frac {\partial }{\partial x_i} \left( \alpha a_i + \beta b_i \right) = \alpha \frac {\partial a_i}{\partial x_i} + \beta \frac {\partial b_i}{\partial x_i}$

The proof of linearity of each operator within the PDE yields the conclusion that the entire PDE is also linear. The Principle of Superposition is applicable to linear PDEs. Applying the Principle of Superposition, the original PDE can be split into two separate parts such that the temperature is

$T(r,\theta) = T_1(r,\theta)+T_2(r,\theta)$

The solutions for $T_1$ and $T_2$ constitutes two separate problems that satisfy the following:

$div(grad T_1) = 0$ such that $T_1(r=a, \theta) = T_1^{*}(\theta)$

and

$div(grad T_2) = 0$ such that $T_2(r=a, \theta) = T_2^{*}(\theta)$

$T(r,\Theta)=T_{1}\left (r,\Theta \right )+T_{2}\left (r,\Theta \right )$

$T^{*}\left (\Theta \right )=T_{1}^{*}\left (\Theta \right )+T_{2}^{*}\left (\Theta \right )$

Problem P:

PDE:

$\operatorname{Div}( grad T )=0$

General Solution: Equation 2 p. 20-4.

$T\left (r=a,\Theta \right )=T^{*}(\Theta )$

Superposition: $P=P_{1}+ P_{2}$

Prob P1: $\operatorname{Div}( grad T_1 )=0$

such that:

$T_{1}(r=a,\Theta)=T_{1}^{*}(\Theta)$

Prob P2: $\operatorname{Div}( grad T_2 )=0$

such that:

$T_{2}(r=a,\Theta)=T_{2}^{*}(\Theta)$

Homework: Verification of T1 solution

--Egm6322.s09.xyz 16:33, 4 April 2009 (UTC)

$T_1$ represents the solution to the first portion of the temperature profile. It simply states that the temperature is constant. The general form of the solution to the heat equation is given as (see #Separation of Variables):

$u(r,\theta)=A_{0}ln \ r+\sum_{n=1}^{\infty }r^{n}\left [A_{n}cos(n\theta)+B_{n}sin(n\theta) \right ]+\sum_{n=1}^{\infty }\frac{}{r^{n}}\left [C_{n}cos(n\theta)+D_{n}sin(n\theta) \right ]+C_{0}$

In order for this general form to converge to $T_1(r,\theta) = \frac{3T_o}{2}$, all the coefficients need to be equal to zero. i.e. $A_o = A_n = B_n = C_n = D_n = 0$

$T_{1}(r,\Theta)=\frac{3T_0}{2}$

For the Problem $T_{2}:T_{2}(a,\Theta)=T_{2}^{*}(\Theta)=\frac{T_0}{2}cos2\Theta$

$T_{2}(r,\Theta)=\sum_{n=1}^{\infty }r^{n} \left \{A_ncosn\Theta +B_nsinn\Theta \right \}$

Homework: Verification of T2 solution

--Egm6322.s09.xyz 17:23, 24 April 2009 (UTC)

For problem P2: $T_2\left( a, \theta \right) = T_2^{*}\left( \theta \right) = \frac{T_o}{2}cos\left(2\theta\right)$

$\blacktriangleright A_o = 0$ because this is an axisymmetric problem with the origin ( r = 0 ) within the domain. (see Axisymmetric Problems)

$\blacktriangleright C_n = D_n = 0$. By inspection of the general form (see above) for the solution to these types of problems, the solution needs to converge to be a function of $cos\left( 2\theta \right)$ for $\theta = \frac{\pi}{2},\frac{3\pi}{2}$. For these boundary conditions $sin\left(n \frac{\pi}{2}\right) \not= 0$ and $sin\left(n \frac{3\pi}{2}\right) \not= 0$, therefore the coefficients of these terms must be equal to zero.

The expression for $T_2$ is then $T_2 \left( r, \theta \right)= \sum_{n=1}^\infty r^n \left[ A_n cos\left( n\theta\right) + B_n sin\left( n\theta\right) \right]$

Homework: Verification of A & B coefficients for boundary condition at r = a

--Egm6322.s09.xyz 16:33, 4 April 2009 (UTC)

At the boundary (r = a), the expression for $T_2$ is re-written as:

$T_2\left(r=a, \theta \right)= \sum_{n=1}^\infty a^n \left[ A_n cos\left( n\theta\right) + B_n sin\left( n\theta\right) \right]$

such that $T_2\left(r=a, \theta \right)= T_2^{*}\left( \theta \right) = \frac{T_o}{2}cos2\theta$

The resulting temperature profile is a function of $cos\left( 2\theta\right)$ only. Upon inspection, the $sin\left( n\theta\right)$ term should be forced to go to zero. Otherwise the equality would not be satisfied. Therefore, the coefficient $B_n = 0$ for all $n=1,2,...,\infty$

Substitution of r = a into the equal for $T_2$, yields:

$a^n \left[ A_n cos \left(n\theta \right)\right] = \frac{T_o}{2}cos2\theta$

$\blacktriangleright$For the case where$n=2$:

$a^2 A_2 cos2\theta = \frac{T_o}{2}cos2\theta$

$\therefore A_2 = \frac{T_o}{2a^2}$

$\blacktriangleright$For the case where$n\not=2$:
The equality DOES NOT hold for all values where $n\not=2$. Taking the case where $n=1$ as an illustrative example, the resulting expression would be:

$aA_1 cos\theta \not= \frac{T_o}{2}cos2\theta$

This illustrates that for all $n\not=2, cosn\theta \not= cos2\theta$

$\therefore A_{n\not=2}=0$

Final Solution:

$T(r,\Theta)=T_0\left [\frac{3}{2}+\frac{1}{2}\left (\frac{r}{a^2} \right )cos2\Theta \right ]$

In general, for arbitrary function $T^{*}\left (\Theta \right )$ but periodic

i.e. $T^{*}\left (\Theta +\mathit{K}2\pi \right )=T^{*}(\Theta)$

for all $\Theta$ and any K = constant

this is not periodic (not acceptable)

Homework: Verification of the General Solution

The general solution of the Laplace equation:

$u(r,\theta)=A_{0}ln \ r+\sum_{n=1}^{\infty }r^{n}\left [A_{n}cos(n\theta)+B_{n}sin(n\theta) \right ]+\sum_{n=1}^{\infty }\frac{}{r^{n}}\left [C_{n}cos(n\theta)+D_{n}sin(n\theta) \right ]+C_{0}$

Where the partial differential equation governing the problem is:

$\nabla ^2T=\frac{\partial^2 T}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{1}{r^2}\frac{\partial^2 T}{\partial \Theta^2}=0$

Applying the boundary conditions explained in the previous homework

$T_{1}(r,\Theta)=\frac{3T_0}{2};T_{2}(r,\Theta)=\frac{T_0}{2}cos2\Theta;A_0=0;C_n=0;D_n=0$

The resulting solution is:

$T(r,\Theta)=C_{0}+\sum_{n=1}^{\infty }r^{n} \left \{A_ncosn\Theta +B_nsinn\Theta \right\}$

$T(r,\Theta)=C_{0}+\sum_{n=1}^{\infty }r^{n} \left \{A_ncosn\Theta +B_nsinn\Theta \right\}$

where: rⁿ=aⁿ

$T(a,\Theta)=C_{0}+\sum_{n=1}^{\infty }a^{n} \left \{A_{n}cosn\Theta +B_{n}sinn\Theta \right\}$

Fourier Coefficients

Egm6322.s09.Three.nav 13:39, 24 April 2009 (UTC)

How do we derive the fourier coeffcients C₀, A_n and B_n?

An orthogonal basis: {1, cosm$\theta$, sinm$\theta$} is used.

As defined above, two functions f($\theta$), g($\theta$) are orthogonal if $\int_{0}^{2\pi}f(\theta).g(\theta) d\theta= 0$

$\blacktriangleright$Eg. Consider f($\theta$)= cos ($m\theta$) and g($\theta$)= cos($n\theta$)

$\int_{0}^{2\pi} cos(m\theta).cos(n\theta)d\theta= \int_{0}^{2\pi} \frac{1}{2}\left (cos \left ((m+n)\theta \right)+cos \left((m-n)\theta \right) \right)d\theta$

$= \left [ \left (\frac{1}{2(m+n)} (sin \left ((m+n)\theta \right) \right)+ \left (\frac{1}{2(m-n)} sin \left((m-n)\theta \right) \right) \right]_{0}^{2\pi}$

$= \begin{Bmatrix} 0, if\ m\ \neq n\\ 2\pi, if\ m\ = n \end{Bmatrix}$

For more details, here is a link that explains the math in greater detail. Similarly using '1' as the basis function and simplifying, we get $\int_{0}^{2\pi}cos\theta\left \{1 \right \} d\theta \ or\ \int_{0}^{2\pi}sin\theta \left \{1 \right \} d\theta = 0$

Using this knowledge, we proceed to determine the fourier coefficients C₀, A_n and B_n, using the orthogonal basis {1, cosm$\theta$, sinm$\theta$}

Consider the equation $T(r,\theta)=C_{0}+\sum_{n=1}^{\infty }r^{n} \left \{A_{n}cosn\Theta +B_{n}sinn\Theta \right\}$-----(1)

Say we have the Boundary condition T(r=a, $\theta$)= ($T^{*}\left(\theta \right)$)

Then, $T^{*}\left(\theta \right) = C_{0}+\sum_{n=1}^{\infty }a^{n} \left \{A_{n}cosn\Theta +B_{n}sinn\Theta \right\}$----(2)

$\blacktriangleright$At n=0, $T^{*}\left(\theta \right) = C_{0}$

Multiplying through by {1} and integrating over [0,2$\pi$]

$\Rightarrow \int_{0}^{2\pi}T^{*}\left(\theta \right) d\theta = \int_{0}^{2\pi} C_{0} d\theta$

$\Rightarrow \int_{0}^{2\pi}T^{*}\left(\theta \right) d\theta = C_{0} \left (\theta \right)_{0}^{2\pi}$

$\Rightarrow C_{0}= \frac {1}{2\pi} \int_{0}^{2\pi}T^{*}\left(\theta \right) d\theta$

$\blacktriangleright$Solving for A_n in Equation (2), multiply through by cos(m$\theta$) and integrate over [0,2$\pi$].

Evaluating each term in the resultant equation,

Term on the LHS= \int_{0}^{2\pi}T^{*}\left(\theta \right)cosm\theta d\theta

First term on RHS= $\int_{0}^{2\pi}C_{0}cosm\theta d\theta$

$= C_{0} \int_{0}^{2\pi} \left \{1 \right \} cos m\theta d\theta$

= 0, by definition of orthogonality

Second term on RHS= $\int_{0}^{2\pi} \sum_{n=1}^{\infty}a^{n}A_{n}cosn\theta cosm\theta d\theta$

$= \sum_{n=1}^{\infty} \int_{0}^{2\pi}a^{n}A_{n}cosn\theta cosm\theta d\theta$

$= \begin{Bmatrix} 0, if\ m\ \neq n\\ a^{n}\times A_{n} \times 2\pi, if\ m\ = n \end{Bmatrix}$ (by definition of orthogonality)

$\Rightarrow \int_{0}^{2\pi} \sum_{n=1}^{\infty}a^{n}A_{n}cosn\theta cosm\theta d\theta = a^{n}\times A_{n} \times 2\pi$

Hence it is seen that though it was assumed m $\epsilon$ $\Re$, simplification of the second term in (2) determines that m=n.

Third term on RHS= $\int_{0}^{2\pi} \sum_{n=1}^{\infty}a^{n}A_{n}sinn\theta cosm\theta d\theta$

= 0, by definition of orthogonality

$\therefore$ (2) multiplied through by cos($m\theta$) or cos($n\theta$) and integrated over [0,2$\pi$] reduces to

$\int_{0}^{2\pi}T^{*}\left(\theta \right)cosn\theta d\theta = a^{n}\times A_{n} \times 2\pi$

$\Rightarrow A_{n}= \frac {1}{a^{n}\times 2\pi} \int_{0}^{2\pi}T^{*}\left(\theta \right)cosn\theta d\theta$

We solve similarly for B_n in (2). Multipling through with sin($m\theta$) and integrating over [0,2$\pi$], it is seen that the first and second terms = 0, by definition. Third term reduces to $b^{n}\times B_{n} \times 2\pi$. Then the modified Equation(2) becomes

$\int_{0}^{2\pi}T^{*}\left(\theta \right)sin n\theta d\theta = a^{n}\times B_{n} \times 2\pi$

$\Rightarrow B_{n}= \frac {1}{b^{n}\times 2\pi} \int_{0}^{2\pi}T^{*}\left(\theta \right)sin n\theta d\theta$

Find the Fourier coefficients C₀,A_n,B_n

$C_{0}=\frac{1}{2\pi }\int_{\Theta =0}^{2\pi }T^{*}(\theta )d\theta$

$A_{n}=\frac{1}{2\pi }\int_{\Theta =0}^{2\pi }T^{*}(\theta )cosn\Theta d\theta$

$B_{n}=\frac{1}{2\pi }\int_{\Theta =0}^{2\pi }T^{*}(\theta )sinn\Theta d\theta$

Due to orthogonality of Fourier basis function $\left \{1,cosu\Theta,sinu\Theta \right \}$

But you were asked to derive the above Fourier coefficients (and therefore the orthogonality property of the Fourier basis functions). Egm6322.s09 20:11, 15 March 2009 (UTC)

Edits to this section were made my Navya on March 27th. Either due to the work of a vandal, or a failure of wikiversity, they were removed.

They are now being placed in by Andrew Lapetina --Egm6322.s09.lapetina 20:56, 27 March 2009 (UTC)

Joseph Fourier 1768-1830

Jean Baptiste Joseph Fourier was a French mathematician and physicist best known for first devising Fourier Series and their applications to heat flow problems. He was born in Auxerre and orphaned at the age of 9. He became a chair at the École Polytechnique at the height of his career. Fourier died in Paris. More information is available here

Homework:Proof of Non Linearity

To show that $\kappa (u)grad(u)+f(x,y)=0$ is non linear

Let,

$L\left( \right):$ be an operator,such that $L\left(\ u\right)$ is linear with respect to $u$ if,

$L\left(\alpha u+\beta v\right)=\alpha L\left(\ u\right)+\beta L\left(\ v\right)$

Therefore, in the present problem ,assuming 2D, we have,

$L\left( \right)=\kappa ()\left(\overline i\frac{\partial\left( \right) }{\partial x}+\overline j\frac{\partial\left( \right) }{\partial y}\right)+f(x,y)$

$L\left(\alpha u+\beta v \right)=\kappa (\alpha u+\beta v)\left(\overline i\frac{\partial\left(\alpha u+\beta v \right) }{\partial x}+\overline j\frac{\partial\left(\alpha u+\beta v \right) }{\partial y}\right)+f(x,y)$

$\Rightarrow L\left( \alpha u+\beta v \right)=\kappa (\alpha u+\beta v)\left(\overline i\alpha\frac{\partial\left(\ u\right) }{\partial x}+\overline i\beta\frac{\partial\left(\ v\right) }{\partial x}+\overline j\alpha\frac{\partial\left(\ u\right) }{\partial y}+\overline j\beta\frac{\partial\left(\ v\right) }{\partial y}\right)+f(x,y)$

$\Rightarrow L\left(\alpha u+\beta v \right)=\kappa (\alpha u+\beta v)\left[\alpha\left\{\overline i\frac{\partial\left(\ u \right) }{\partial x}+\overline j\frac{\partial\left(\ u \right) }{\partial y} \right\}+\beta\left\{\overline i\frac{\partial\left(\ v\right) }{\partial x}+\overline j\frac{\partial\left(\ v\right) }{\partial y} \right\}\right]+f(x,y)$

\therefore,we can see that

$\Rightarrow L\left(\alpha u+\beta v\right)\neq\alpha L\left(\ u\right)+\beta L\left(\ v\right)$

$\therefore \kappa (u)grad (u)+f(x,y)$ is non linear

Egm6322.s09.bit.gk 20:41, 24 April 2009 (UTC)

Homework:Problem 5.11

File:Testrun2.svg

Free Body Diagram of membrane subjected to transverse load Egm6322.s09.bit.gk 20:56, 24 April 2009 (UTC)

We can see from the free body diagram ,equating the horizontal forces ,we have

$-T_0cos(\alpha)dy+T_0cos(\alpha+d\alpha)dy-T_0cos(\beta)dx+T_0cos(\beta+d\beta)dx=0$

and equating the vertical forces, we have,

$-T_0sin(\alpha)dy+T_0sin(\alpha+d\alpha)dy-T_0sin(\beta)dx+T_0sin(\beta+d\beta)dx=P(x,y)dxdy$

where $P(x,y)$ is the transverse load acting on the membrane

$\Rightarrow T_0\left[\left (sin(\alpha+d\alpha) -sin(\alpha)\right )dy+\left (sin(\beta+d\beta)-sin(\beta)\right )dx\right]+P(x,y)dxdy=0$

When $\theta$ is small ,we can assume, $sin(\theta)=Tan(\theta)$

$\therefore$ the above equation becomes,

$T_0\left[\left (tan(\alpha+d\alpha)-tan(\alpha)\right )dy+\left (tan(\beta+d\beta)-tan(\beta)\right )dx\right]+P(x,y)dxdy=0$

Let this equation be $(1)$

but,$tan(\alpha)$ is the slope the membrane with respect to the $x$ and $y$ axes

where the displacement is given as $w=w(x,y)$

$\therefore$

$tan(\alpha)=\left[\frac{\partial w}{\partial x}\right]_{x,y}$ and $tan(\alpha+d\alpha)=\left[\frac{\partial w}{\partial x}\right]_{x+dx,y}$

$tan(\beta)=\left[\frac{\partial w}{\partial y}\right]_{x,y}$ and $tan(\beta+d\beta)=\left[\frac{\partial w}{\partial y}\right]_{x,y+dy}$

Substituting the slopes in $(1)$,we have ,

$T_0\left[\left(\left[\frac{\partial w}{\partial x}\right]_{x+dx,y}-\left[\frac{\partial w}{\partial x}\right]_{x,y}\right)dy+\left(\left[\frac{\partial w}{\partial y}\right]_{x,y+dy}-\left[\frac{\partial w}{\partial y}\right]_{x,y}\right)dx\right]+P(x,y)dxdy=0$

Let this be equation $(2)$

We have Taylor series expansion as ,

$f(x+dx,y)=f(x,y)+\frac{\partial f}{\partial x}dx+...$

substituting taylor series expansion in equation $(2)$,we have

$T_0\left(\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}\right)+P(x,y)=0$

$\Rightarrow T_0\triangledown^{2}w(x,y)+P(x,y)=0$

Egm6322.s09.bit.gk 20:39, 24 April 2009 (UTC)

Orthogonal Functions

--Egm6322.s09.xyz 16:34, 4 April 2009 (UTC)

The following is the definition of Orthogonal Functions as presented on the Wolfram MathWorld website:

"Two functions $f(x)$ and $g(x)$ are orthogonal over the inverval $a\le x\le b$ with weighting function $w(x)$ if

$\left \langle f(x)|g(x)\right \rangle \equiv \int_{a}^{b} f(x)g(x)w(x)\,dx = 0$

If, in addition,

$\int_{a}^{b} f(x)^{2}w(x)\,dx = 1$

and

$\int_{a}^{b} g(x)^{2}w(x)\,dx = 1$

the functions $f(x)$ and $g(x)$ are said to be orthonormal"

Additional information can be found at Wiki Orthogonality and Dictionary.com

A Third Example of the Laplace Equation

Egm6322.s09.bit.sahin 16:31, 24 April 2009 (UTC)

A domain which is a quadrant of annulus is subjected a boundary conditions that

temperature at $r=b$ is $T\left (r=b,\theta \right )=T_{b}cos4\theta$

temperature at $r=a$ is $T\left (r=a,\theta \right )=T_{a}cos4\theta$

Domain and boundary conditions

where $T_{a}$ and $T_{b}$ are given constants. Also, the boundaries at $\theta=0$ and $\theta =\pi/2$ are insulated which means no heat flow at these boundaries. According to the Fourier's law:

$\underline{q}=\underline{\kappa } \cdot gradT$

here $\underline{q}$ denotes heat flux tensor. Relevant to the insulated conditions,

$\underline{q}=0\Rightarrow gradT=0\Leftrightarrow \frac{\partial T}{\partial \theta }=0$ on $\theta=0,\theta =\pi/2$.

Homework: Verification of Insulation

The boundaries are kept insulated which means that there is no heat flow at $\theta= 0$ and $\theta= \pi/2$. So,

$grad T=0$

and

$grad T= \frac{\partial T}{\partial r}\mathbf{e_{r}}+\frac{1}{r}\frac{\partial T}{\partial \theta}\mathbf{e_{\theta}}=0$

Since $\frac{\partial T}{\partial r}=0$ at insulated surfaces we have

$\frac{1}{r}\frac{\partial T}{\partial \theta}\mathbf{e_{\theta}}=0$

So, we obtain that

$\frac{\partial T}{\partial \theta}=0$ at $\theta= 0$ and $\theta= \pi/2$

How so? In general, express $\displaystyle {\rm grad} \, T$ in polar coordinates then deduce $\displaystyle \partial T / \partial \theta = 0$. Such approach is important when the insulated boundaries do not coincide with the $\displaystyle (x,y)$ Egm6322.s09 20:11, 15 March 2009 (UTC)

Necessary changes were made Egm6322.s09.bit.sahin 16:16, 10 April 2009 (UTC)

The general solution of the Laplace Eq. is

$T\left (r,\theta \right )=A_{0}lnr+\sum_{n=1}^{\infty }r^{n}\left (A_{n}cosn\theta +B_{n}sinn\theta \right )+ \sum_{n=1}^{\infty }\frac{1}{r^{n}}\left (C_{n}cosn\theta +D_{n}sinn\theta \right )+C_{0}$

Eliminating Terms Based on BC

We can eliminate the following terms that do not satisfy the boundary conditions:

1) $A_{0}lnr+C_{0}$, indipendent of $\theta$

2) $B_{n}sinn\theta$, $D_{n}sinn\theta$, cannot satisfy the $\frac{\partial T}{\partial \theta }\left (r,\theta =0 \right )=0$

To show the second one, let's differentiate the general solution with respect to $\theta$

$\frac{\partial T}{\partial \theta } =\sum_{n=1}^{\infty }r^{n}n\left (-A_{n}sinn\theta +B_{n}cosn\theta \right )+\sum_{n=1}^{\infty }\frac{1}{r^{n}}n\left (-C_{n}sinn\theta +D_{n}cosn\theta \right )$

Since $sinn\theta=0$ and $cosn\theta=1$ at $\theta=0$, $B_{n}$ and $C_{n}$ must be zero to satify the condition that $\frac{\partial T}{\partial \theta }=0$.

Thus the solution has the following form

$T\left (r,\theta \right )=\sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{C_{n}}{r_{n}} \right )cosn\theta$

Using the boundary conditions, we have

$T_{a}cos4\theta =\sum_{n=1}^{\infty }\left (A_{n}a^{n}+\frac{C_{n}}{a_{n}} \right )cosn\theta$

$T_{b}cos4\theta =\sum_{n=1}^{\infty }\left (A_{n}b^{n}+\frac{C_{n}}{b_{n}} \right )cosn\theta$

Since the only term in the boundary condition is the term with $cos4\theta$, boundary conditions can only be satisfied for n=4, all other $A_{n}$ and $C_{n}$ must be zero. Then we have,

$\begin{bmatrix} a^{8} & 1 \\ b^{8} & 1 \end{bmatrix}\begin{Bmatrix} A_{4}\\C_{4} \end{Bmatrix}=\begin{Bmatrix} a^{4}T_{a}\\b^{4}T_{b} \end{Bmatrix}$

Eventually the solution for the temperature distribution is

$T\left (r,\theta \right )=\left \{\frac{a^{4}b^{4}T_{b}}{\left (b^{8}-a^{8} \right )}\left [\frac{r^{4}}{a^{4}}-\frac{a^{4}}{r^{4}} \right ]-\frac{a^{4}b^{4}T_{a}}{b^{8}-a^{8}}\left [\frac{r^{4}}{b^{4}}-\frac{b^{4}}{r^{4}} \right ] \right \}cos4\theta$

The figure below shows the plot of the solution for a given data:

$a=1$, $b=2$, $T_{a}=5$, $T_{b}=20$

MATLAB Code for Plots

--EGM6322.S09.TIAN 17:33, 24 April 2009 (UTC)

function [x] = plot(A,b,x0)
 
   % figure out T
[r,theta] = meshgrid(1:0.05:2,0:pi/40:pi/2);
a=1;
b=2;
Ta=5;
Tb=20;
for i=1:21;
    for j=1:21;
        T(i,j)=( (a^4 * b^4 *Tb / (b^8 - a^8)) *  (r(i,j) ^4 / a^4 - a^4 / r(i,j) ^4) -   (a^4 * b^4 *Ta / (b^8 - a^8))* (r(i,j) ^4 / b^4 - b^4 / r(i,j) ^4))*cos(4*theta(i,j));
    end;
end;
% plot Polar
figure(1);
surf(r,theta,T); colorbar ;
title('Polar Coordinate');
Xlabel('R');
Ylabel('Theta');
Zlabel('T');
% plot Cartesian
for i=1:21;
    for j=1:21;
        x(i,j)=r(i,j)*cos(theta(i,j));
        y(i,j)=r(i,j)*sin(theta(i,j));
    end;
end;
figure(2)
surf(x,y,T)

MATLAB Plot of Function

MATLAB Plot of Function

--EGM6322.S09.TIAN 17:33, 24 April 2009 (UTC)

You were asked to plot in both polar coordinates and in cartesian coordinates; the figure shown is a plot in cartesian coordinates; I moved this figure to the left for a better presentation. Also provide the matlab codes used to create these plots. Egm6322.s09 15:08, 15 March 2009 (UTC)

My comment above had not been addressed in this updated version. Please take action. Egm6322.s09 22:38, 2 April 2009 (UTC)

Now there it is. EGM6322.S09.TIAN 21:39, 9 April 2009 (UTC)

The Power Law

--Egm6322.s09.lapetina 02:03, 17 April 2009 (UTC)

The Power Law is a very common relationship in the universe. It is defined as:

$y=b x^a$.

The Power Law is observed in classical physics, biology, economics, and many other natural and social sciences. The exponent $a$ dominates the nature of the equation. In electrostatics and gravitation, $a=2$, while in Stefan-Boltzmann equations, $a=4$. More can be found on the Power Law here.

The inverse of the exponent function is the logarithm.

Application of the Power Law

The thermal conductivity of solids is summarized in the following graph .

Plotting here is on a log-log scale. As a result, the slopes appear linear, rather than exponential.

Log-log vs. Semi-log Plotting --Egm6322.s09.lapetina 17:53, 24 April 2009 (UTC)

The exponential varying thermal conductivity of solids is very important for solving Fourier's Law:

$q=- \kappa \; grad \; T$

where $q$ is the heat flux.

From this equation, we can find the units of $\kappa$ in the following fashion:

$q \equiv \left [ \frac{Power}{Unit \; Area} \right ]=\left [ \frac{W}{m^2} \right ]$ while $grad \; T= \frac {dT}{dx} = \left [ \frac {K}{m} \right ]$

Therefore: $\left [ \kappa \right ]= \left [ \frac {\frac {W}{m^2}}{\frac{K}{m}}\right ] = \frac {W}{mK}$

If we consider the thermal conductivity $\kappa$ as $\kappa \left ( T \right )$ where $T$ is temperature, we can find the heat flux at any given temperature using the power law by the following equation:

$\kappa \left ( T \right )=b T^a$.

This equation can be solved over a given domain if $a$ and $b$ are known.

Using data from the aforementioned graph, we see that for diamond, $T \in \left [ 1K, 10.7 K \right ]$ :

$a=\frac {log \kappa_2 - log \kappa_1}{log T_2 - log T_1}=\frac {log (1000)-log (0.4)}{log (10.7)-log (1)} \cong 3.39$

while $b \cong 0.4 \frac {W}{mK}$

so $\kappa (T) = (0.4) T^{3.39} \frac {W}{mK}$.

Homework: Determining a and b for a Different Interval

--Egm6322.s09.lapetina 02:04, 17 April 2009 (UTC)

For $T \in \left [ 100 K, 1000 K \right ]$, we want to find $\kappa \left ( T \right )$. We can estimate $a$ for diamond using the slope of graphite parallel to layers:

$a=\frac {log \kappa_2 - log \kappa_1}{log T_2 - log T_1}=\frac {log (3)-log (80)}{log (1000)-log (100)} \cong -1.43$.

Extrapolating $b$ backwards shows its value is $10^7$.

Therefore, for all $T \in \left [ 100 K, 1000 K \right ]$,

$\kappa (T) = (10^7) T^{-1.43} \frac {W}{mK}$

The Wave Equation and String Vibration

--Egm6322.s09.lapetina 02:04, 17 April 2009 (UTC)

Free Body Diagram of String Vibration

The Wave Equation can be studied by examining the physics of string vibration in one spatial dimension. An excellent book on this topic is The Theory of Sound by Lord Rayleigh.

I never mentioned the book by Rossing et al. I mentioned the classic The Theory of Sound by Lord Rayleigh; see the Lecture plan. Egm6322.s09 15:08, 15 March 2009 (UTC)

Correction made. --Egm6322.s09.lapetina 20:57, 27 March 2009 (UTC)

In the accompanying free-body diagram for Case 1, forces in the $x$ direction are :

$\sum F_x = -\tau \left ( x,t \right ) cos \; \theta ( x,t )+ \tau \left ( x+dx,t \right ) cos \; \theta ( x+dx,t )$

$\theta$ is very small here, therefore $cos \theta \cong 1-\frac{\theta^2}{2}$.

Inserting this into the $x$ equation results in:

$\sum F_x = -\tau \left ( x,t \right ) (1-\frac{{\theta(x,t)}^2}{2})+ \tau \left ( x+dx,t \right ) (1-\frac{{\theta (x+dx,t)}^2}{2})$.

We can neglect second order terms here, since they are very small and nearly cancel, leaving:

$\tau (x+dx,t)=\tau (x,t)=\tau$, suggesting $\tau$ is constant throughout the string.

In the $y$ direction:

$\sum F_y = \tau sin \; \theta ( x,t )+ \tau sin \; \theta ( x+dx,t )+ P (x,t) dx -m(dx) \ddot w$

where $\ddot w={w}_{tt}$.

Where $\theta$ is small, we can simplify the first two terms as:

$\theta (x+dx,t) -\theta (x,t) \cong \frac{d\theta}{dx} dx= {(w_x)}_{x} dx= {w}_{xx}dx$

multiplied by the constant $\tau$.

Therefore, for an infinitely small length of string $dx$,

$\tau {w}_{xx} +P=m {w}_{tt}$

Homework: Other Cases for String Shape

--Egm6322.s09.lapetina 02:04, 17 April 2009 (UTC)

Case 4: Vertical Reflection

This is the most trivial of the other three cases. In the $x$ direction, this changes only the sign on $\theta$. Because this value is squared, we are left with:

$\tau {w}_{xx} +P=m {w}_{tt}$

Case 2: Modified Geometry

For cases two and three, we have a slightly more complicated situation, as both ends of the string point down and up, respectively. This negates our ability to simplify the equation in the same way for the string as shown. Looking first at the accompanying figure, we can our case on the left side.

Modifications for Cases Two and Three. This image shows that for a fraction of the string, the free body diagram is the same as for other cases.

However, recall the string is infinitesimally small. As seen in the accompanying image, we can simply examine a fraction of the infinitesimally small string $dx_1$, and encounter the same geometry. In the image the sum of forces in the $x$ direction on the left side of the image can be expressed as:

$\sum F_x = -\tau \left ( x,t \right ) cos \; \theta ( x,t )+ \tau \left ( x+dx,t \right ) cos \; \theta ( x+dx,t )$

while the right side (showing a fraction of the infinitesimally small string) can be expressed as:

$\sum F_x = -\tau \left ( x,t \right ) cos \; \theta ( x,t )+ \tau \left ( x+{dx}_{1},t \right ) cos \; \theta ( x+{dx}_{1},t )$

This can be simplified using the assumption:

$cos \theta \cong 1-\frac{\theta^2}{2}$,

leaving:

$\sum F_x = -\tau \left ( x,t \right ) (1-\frac{{\theta(x,t)}^2}{2})+ \tau \left ( x+{dx}_{1},t \right ) (1-\frac{{\theta (x+{dx}_{1},t)}^2}{2})$.

Neglecting second order terms here, since they are very small and nearly cancel, leaves:

$\tau (x+{dx}_{1},t)=\tau (x,t)=\tau$, suggesting $\tau$ is constant throughout the string, which by definition continues to $x=x+dx$.

This means that even for geometries where the ends of the strings point in the same direction, $\tau (x,t) = \tau$.

In the $y$ direction, we start with:

$\sum F_y = \tau sin \; \theta ( x,t )+ \tau sin \; \theta ( x+dx,t )+ P (x,t) dx -m(dx) \ddot w$.

However, for a fraction of the string $dx_1$, (i.e., the modified version shown in the right side of the image), this equation changes to:

$\sum F_y = \tau sin \; \theta ( x,t )- \tau sin \; \theta ( x+{dx}_1,t )+ P (x,t) dx -m(dx) \ddot w$.

Now, where $\theta$ is small, we can again simplify the first two terms as:

$\theta (x+dx,t) -\theta (x,t) \cong \frac{d\theta}{dx} dx= {(w_x)}_{x} dx= {w}_{xx}dx$

multiplied by the constant $\tau$.

This leaves us with the same equation as for Cases One and Four:

$\tau {w}_{xx} +P=m {w}_{tt}$

Case Three: Vertical Reflection of Case Two

Our signs are merely switched again, as they were between Cases One and Four.

Essentially, no matter what the geometry of the string (so long as it is simple and does not cross, and is always differentiable), the equation of motion is:

$\tau {w}_{xx} +P=m {w}_{tt}$.

This is because the original derivation is for a nearly linear string, and the can always be viewed as linear, as:

$\delta x \rightarrow 0$.

An Algebraic Representation of String Vibration

An alternative means of solving this problem is to view $\theta$ and $\tau$ as algebraic quantities rather than physical entities. The accompanying image shows the only free body diagram needed.

Here, we see that:

$\sum F_x=\tau (x,t)cos \theta (x,t)+\tau (x+dx,t)cos \theta (x+dx,t)=0$ (1) $\sum F_y=\tau (x,t) sin \theta (x,t)+\tau (x+dx,t) sin \theta (x+dx,t) +p(x) dx - mdx {w}_{tt} =0$ (2)

For equation 1, we can assume $\theta$ is small, making $cos (\theta) \cong 1$, leading to:

$\tau (x+dx) =-\tau (x)={\tau}_{const} \forall x$.

Therefore: $\tau (x+dx)=\tau$ $\tau (x)=-\tau$

This changes equation 2 to:

$\sum F_y=-\tau sin \theta (x,t)+\tau sin \theta (x+dx,t) +p(x) dx - mdx {w}_{tt} =0$

where, because $\theta$ is small:

$\sum F_y=0= \tau \frac{\partial \theta}{\partial x} dx+p(x)dx-mdx{w}_{tt} +$ higher order terms.

This leaves us with:

$\tau {w}_{xx}+P=m {w}_{tt}$

Sign problem in $\displaystyle \sum F_y$; inconsistent with Taylor series expansion. The key for the derivation to be independent of any figure of free-body diagram is to treat the tension function $\displaystyle \tau(x)$ and the slope angle $\displaystyle \theta(x)$ as algebraic quantities so that $\displaystyle \sum F_x = \tau (x) \cos(x) + \tau(x+dx) \cos (x+dx) = 0$ and $\displaystyle \sum F_y = \tau (x) \sin(x) + \tau(x+dx) \sin (x+dx) + p \, dx - m \, dx \, w_{tt} = 0$. Note the plus signs in both cosine terms in $\displaystyle \sum F_x$, and the plus signs in both sine terms in $\displaystyle \sum F_y$. This systematic approach is unlike what you used to see in undergraduate courses such as statics, dynamics, etc. See further comments in class. Egm6322.s09 15:08, 15 March 2009 (UTC)

Explanation of sign switch demonstrated in figure is articulated in the text of the original problem.

Alternative solution provided. Image addition will come in near future.

--Egm6322.s09.lapetina 14:33, 16 March 2009 (UTC)

Homework: Derive the Equations of Motion for a Stretched Membrane in Cartesian Coordinates

--Egm6322.s09.lapetina 02:05, 17 April 2009 (UTC)

In two dimensions, the vertical position $u$ of a membrane of density $\rho$ with constant thickness $h$ (and corresponding constant strength and flexibility) and loading $p(x,y)$ can be described as:

$u (x,y,t)$, where $x$ and $y$ are Cartesian positions, and $t$ is time. With no loading, the membrane rests in the $x-y$ plane

Small Area of Membrane

The stress in the membrane, $\sigma$ is constant throughout. For all time $t$ and for all $x$ and $y$, the membrane surface is differentiable by $x$ and $y$, and small angle approximations can be made.

As shown in the image, we will examine a small square portion of width $\Delta x$ and height $\Delta y$.

For this small portion, There are five forces acting on it, the tensions, $\left \{ T_1,...,T_4 \right \}$, and the load $P(x,y)$. The tensions all act normal to the edge of the small portion, and at an angle to the $x-y$ plane, $\left \{ {\theta}_{1},...{\theta}_{4} \right \}$, respectively.

The tensions $\left \{ T_1,...,T_4 \right \}$are applied at coordinates, $\left \{ (x+0.5 \Delta y,y_1), (x_1+\Delta x,y_1+0.5 \Delta y), (x_1+0.5 \Delta x,y_1+\Delta y),(x_1,y_1+0.5 \Delta y) \right \}$ respectively.

Newton's Second Law in the vertical direction is then:

$\sum F_z=T_1 sin \theta_1+T_2 sin \theta_2+T_3 sin \theta_3+T_4 sin \theta_4=\rho h \Delta x \Delta y \frac{{\partial}^2 u}{\partial t^2}+\Delta x \Delta y p(x,y)$

Using the small angle approximation,

$sin {\theta}_n \cong tan {\theta}_n$

which is approximately equal to the derivative of the membrane in the direction of the tension force at each location.

Therefore:

$sin \theta_1 \cong -u_y (x_1 +0.5 \Delta x, y_1)$
$sin \theta_2 \cong u_x (x_1+\Delta x,y_1+ 0.5 \Delta y)$

$sin \theta_3 \cong u_y (x_1 + 0.5 \Delta x, y_1+\Delta y)$

$sin \theta_4 \cong u_x (x_1, y+0.5 \Delta y )$.

The magnitudes of $\left \{ T_1,...,T_4 \right \}$ are equivalent to the constant stress multiplied by the area over which it is applied, such that:

$T_1=\Delta x h \sigma$ $T_2=\Delta y h \sigma$ $T_3=\Delta x h \sigma$ $T_4=\Delta y h \sigma$

Let us define:

$x_1+0.5 \Delta x= x_2$ and $y_1+0.5 \Delta y= y_2$.

Substituting these values into Newton's Law and dividing by $\Delta x$ and $\Delta y$ leaves us with :

$\rho \frac{{\partial}^2 u}{\partial t^2}+ P(x,y)= \sigma \left [ \frac {u_x (x_1+\Delta x, y_2 )- u_x (x_1,y_2)}{\Delta x} +\frac{u_y (x_2, y_1+\Delta y) - u_y (x_2, y_1)}{\Delta y} \right ]$.

Taking the limits as $\Delta x$ and $\Delta y$ approach zero, we are left with a Poisson Equation:

$m u_{tt} + p(x,y)= \sigma(u_{xx}+u_{yy})$

where $m$ is the mass per unit area.

This is equivalent to:

$\sigma ( div (grad \; u)) +p(x,y) = m w_{tt}$.

For steady state, and a massless membrane, the equation becomes:

$\sigma {\nabla}^2 u(x,y) + p(x,y)=0$

Applications

$\tau (div (grad \omega))+ p(x,y) = m W_tt$

Here, $div (grad \omega)= \omega_{xx}+\omega_{yy}$, $m$ is mass/unit area.

Wave Equation In 1-D Space (actually a 2-D (x,t) Problem)

--EGM6322.S09.TIAN 17:28, 24 April 2009 (UTC)

$\tau \omega_{xx} - m \omega_{tt} +p =0$

$A= \begin{bmatrix} a & b \\ b & c \end{bmatrix}$, $detA=ac-b^2$

In this case, $a= \tau >0, b=0, c=-m<0$

$detA=- \tau m <0 \Rightarrow hyperbolic$

Unsteady Heat Equation

--EGM6322.S09.TIAN 17:24, 24 April 2009 (UTC)

1-D space

$\frac{d}{dx} (\kappa \frac{du}{dx}) + f = C \frac{du}{dt}$

Here, $\kappa$ is heat conductivity, $f$ is heat source, $C$ is heat capacity.

We assume $\kappa$ is constant.

$\kappa u_{xx} - C u_t +f =0$

$a= \kappa, b=0, c=0 \Rightarrow detA=0\Rightarrow parabolic$

2-D space

$\frac{d}{dx} (\kappa \frac{du}{dx}) + f = C \frac{du}{dt}$

We assume $\kappa$ is constant here.

$\kappa (u_{xx} + u_{yy}) - C u_t +f =0$

General to 3 independent variables (x,y,z)

$\big\lfloor \partial_x \; \partial_y \;\partial_z \big\rceil \begin{bmatrix} A_{11} & A_{12} & A_{13}\\ A_{21} & A_{22} & A_{23}\\ A_{31} & A_{32} & A_{33} \end{bmatrix} \begin{Bmatrix} \partial_x u \\ \partial_y u \\ \partial_z u \end{Bmatrix}$

Here,$\begin{bmatrix} A_{11} & A_{12} & A_{13}\\ A_{21} & A_{22} & A_{23}\\ A_{31} & A_{32} & A_{33} \end{bmatrix} = [A_{ij}]_{3 \times 3}$

In 2-D case, $A_{ij}=0 \; \forall \; i,j \; except \; A_{11} = A_{22}= \kappa >0 \Rightarrow detA=0 \Rightarrow parabolic$

Solution of unsteady heat equation without heat source, which means $(f=0)$ in polar coordinate.

Separation of Variables.

$\frac {1}{r} \frac {\partial} {\partial r} (r \frac {\partial u} {\partial r})$ $+ \frac {1}{r^2} \frac {\partial^2 u}{\partial^2 \theta}$ $= \frac {\partial u}{\partial t}(1)$

Here, $\frac {1}{r} \frac {\partial} {\partial r} (r \frac {\partial u} {\partial r})$ $+ \frac {1}{r^2} \frac {\partial^2 u}{\partial^2 \theta}$ $=div (grad u)$

$u(r,\theta,t)=R(r)\Theta (\theta) T(t)\;(2)$

Plug $(2)$ into $(1)$:

$\frac {1}{rR} \frac {\partial} {\partial r} (r \frac {dR} {dr})$ $+ \frac {1}{r^2 \Theta} \frac {d^2 \Theta}{d \theta^2 \theta}$ $- \frac {1}{T} \frac {dT}{dt}=0$

The solution for problem 5.12 in Selvadurai (2000) is missing. Egm6322.s09 15:15, 15 March 2009 (UTC)

START to Homework:problem 5.12 in Selvadurai (2000)

An unloaded weightless membrane roof over an annular enclosure. The outer circular boundary is r=b, and the inner circular boundary, r=a, is subject to the following displacement:

$\Delta_{0} + \Delta_{1}sin \theta$

where $\Delta_{0}$ and $\Delta_{1}$ are constants.

i) Formulate the Boundary value problem:

The boundary conditions are:

1. $w(r,\theta)=w(r,\theta+2\pi)$
2. $w(b,\theta)=0$
3. $w(a,\theta)=\Delta_{0} + \Delta_{1}sin \theta$
4. $\frac{\partial w}{\partial r}_{r=b} =0$

ii) Develop an expression for the membrane, given that:

$w=A \ ln \ r +B\theta ln \ r +C\theta +D + \sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{B_{n}}{r^{n}} \right )(C_{n}sin \ n\theta +D_{n}cos \ n\theta)$

where $A, B, C, D, A_{n}, B_{n}, C_{n}, and \ D_{n}$ are constants.

Solution: Using boundary condition 1, the equation becomes:

$\begin{matrix} A \ ln \ r +B\theta ln \ r +C\theta +D + \sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{B_{n}}{r^{n}} \right )(C_{n}sin \ n\theta +D_{n}cos \ n\theta)\\ =\\ A \ ln \ r +B(\theta+2\pi) ln \ r +C(\theta+2\pi) +D + \sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{B_{n}}{r^{n}} \right )(C_{n}sin \ n(\theta+2\pi) +D_{n}cos \ n(\theta+2\pi)) \end{matrix}$

Clearly,

$(C_{n}sin \ n\theta +D_{n}cos \ n\theta)=(C_{n}sin \ n(\theta+2\pi) +D_{n}cos \ n(\theta+2\pi))$

and,

$\begin{matrix} A \ ln \ r +B\theta ln \ r +C\theta +D =A \ ln \ r +B(\theta+2\pi) ln \ r +C(\theta+2\pi) +D\\ \\ \therefore B=C=0 \end{matrix}$

Thus the equation simplifies to:

$w(r,\theta)=A \ ln \ r +D + \sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{B_{n}}{r^{n}} \right )(C_{n}sin \ n\theta +D_{n}cos \ n\theta)$

Using the 3rd boundary condition:

$w(a,\theta)=\Delta_{0} + \Delta_{1}sin \theta$

One can clearly see that since the coefficient of the sine term is 1, all n terms that are not equal to one must be zero.

$\begin{matrix} sin(n\theta)=sin \ \theta\\ \therefore\\ n=1 \end{matrix}$

And,

$\begin{matrix} n\neq 1\\ A_{n}=B_{n}=C_{n}=D_{n}=0 \end{matrix}$

It can also be seen that since there is no cosine term in the boundary equation,

$D_{1}=0$

Thus the equation reduces to:

$w(r,\theta)=A \ ln \ r +D + \left (\overline{A}r+\frac{\overline{B}}{r} \right )(sin \ \theta)$

Where,

$\begin{matrix} \overline{A}=A_{1}C_{1}\\ \overline{B}=B_{1}C_{1} \end{matrix}$

Reapplying boundary condition three gives:

$\Delta_{0} + \Delta_{1}sin \theta=A \ ln \ a +D + \left (\overline{A}a+\frac{\overline{B}}{a} \right )(sin \ \theta)$

From this equation it is clearly evident that,

$\begin{matrix} \Delta_{0}=A \ ln \ a +D\\ \Delta_{1}=\overline{A}a+\frac{\overline{B}}{a} \end{matrix}$ Egm6322.s09.bit.gk 20:59, 24 April 2009 (UTC) These equations and the last two boundary conditions:

$\begin{matrix} w(b,\theta)=0 \\ \frac{\partial w}{\partial r}_{r=b} =0 \end{matrix}$

Give 4 equation to solve for 4 unknown coefficients.

The resulting system of equations may be expressed as a matrices:

$\begin{bmatrix} sin\theta & \frac{sin\theta}{b^2} & \frac{1}{b}& 0 \\ bsin\theta & \frac{sin\theta}{b}& \textrm{ln}\ b& \\ a & \frac{1}{a} & 0 & 0 \\ 0& 0& \textrm{ln}\ a& 0 \end{bmatrix} \begin{Bmatrix} \overline{A}\\ \overline{B}\\ A\\ D \end{Bmatrix}= \begin{Bmatrix} 0\\ 0\\ \Delta_{1}\\ \Delta_{0} \end{Bmatrix}$

Using the matrix to solve for the constants one finds that:

$\begin{matrix} \overline{A}=\frac{b}{(b^2-a^2)} \frac{\Delta_{0}}{sin a \ ln \ a}\\ \\ \overline{B}=\frac{-a^2b \Delta_{0} }{(b^2-a^2)sin\theta \ ln \ a}\\ \\ A=\frac{\Delta_{0}}{ln \ a}\\ \\ D=-(1+ln \ b)\frac{\Delta_{0}}{ln \ a} \end{matrix}$

iii) Calculate the resultant force and moment necessary to maintain the inner rigid disk shaped region in the displaced position.

To find the net force necessary to maintain the inner disk in its position, one must assume the membrane has an effective spring constant of k. The force is then k multiplied by the displacement w. To get the net force we integrate around the edge of the disk.

$\begin{matrix}F_{net}=k\int_{0}^{2\pi} \Delta_{0}+\Delta_{1}sin\theta \ d\theta\\ \textrm{Note:}\int_{0}^{2\pi}sin\theta \ d\theta=0\\ F_{net}=2\pi k \Delta_{0} \end{matrix}$

To find the moment necessary to keep the disk in its tilted position requires a bit more consideration. The moment results from an unbalanced force applied at some distance y. It should be noted that since the $\Delta_{0}$ term is the same all along the boundary, it does not contribute to the moment necessary to maintain the disk's position.

$\Delta M_{x}=y \Delta F$

Where the value of F is already given as:

$\begin{matrix}\Delta F=ksin\theta \Delta L\\ \Delta L=b \Delta\theta \end{matrix}$

To get the total moment the sum of the discrete $\Delta M_{x}$ should be taken. By increasing the number of $\Delta M_{x}$ and simultaneously decreasing their magnitude, the summation becomes on integral.

Thus,

$\begin{matrix} M_{x}=\int dM_{x}= \int yk\Delta_{1}sin\theta \ bd\theta \\ \textrm{Note:} \quad y=b \ sin\theta \\ M_{x}= \int b \ sin\theta \ k\Delta_{1}sin\theta \ bd\theta \\ = b^2k\Delta_{1}\int sin^2\theta \ d\theta \end{matrix}$

This integral must be taken over the entire circle. Thus the moment required is:

$\begin{matrix} M_{x}= b^2k\Delta_{1}\int_{0}^{2\pi} sin^2\theta \ d\theta \\ =b^2k\Delta_{1}\pi \end{matrix}$

The statement of the boundary value problem was incomplete (1st question). Do complete this problem in an updated version of report R4. Egm6322.s09 22:38, 2 April 2009 (UTC)

---

The problem has been completed. Egm6322.s09.Three.ge 19:09, 10 April 2009 (UTC)

Signatures

--Egm6322.s09.xyz 03:49, 6 March 2009 (UTC)

Egm6322.s09.Three.ge 19:34, 5 March 2009 (UTC)

--Egm6322.s09.three.liu 19:55, 5 March 2009 (UTC)

--EGM6322.S09.TIAN 21:05, 5 March 2009 (UTC)

--Egm6322.s09.lapetina 22:43, 5 March 2009 (UTC)

Egm6322.s09.Three.nav 16:05, 6 March 2009 (UTC)Egm6322.s09.Three.nav

Egm6322.s09.bit.sahin 18:04, 6 March 2009 (UTC) Egm6322.s09.bit.la 18:08, 6 March 2009 (UTC)

Egm6322.s09.bit.gk 19:20, 6 March 2009 (UTC)