User:Egm6322.s09.mafia/HW4

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

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:

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

it can be classified by the value of ${\displaystyle ac-b^{2}}$.

${\displaystyle ac-b^{2}=det{\underline {A}}}$

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

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

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

For example:

In the equation:

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

${\displaystyle \Rightarrow a=1,b=0,c=1}$

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

${\displaystyle \Rightarrow }$ This PDE is elliptic

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

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

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

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

${\displaystyle \Rightarrow }$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:

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

Taking the Laplace equation in polar coordinates gives the following.

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

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.

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

The equation then reduces to,

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

which is an ordinary differential equation (ODE).

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

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

Separating variables and integrating gives the solution:

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

where Ao and Bo are constants.

It should be noted that if the domain ${\displaystyle \omega }$ includes the origin (where r=0) then Ao 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:

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

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

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

Plugging in this solution into (a) produces:

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

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

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

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 ${\displaystyle n\neq 0}$.

${\displaystyle {\begin{matrix}F(r)=Ar^{n}+{\frac {B}{r^{n}}}\\\\G(\theta )=Ccos(n\theta )+Dsin(n\theta )\end{matrix}}}$

If n=0:

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

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

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

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

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

Another Axisymmetric Problem

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)

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

Application:

Example:

${\displaystyle T^{*}(\Theta )=T_{0}\left(1+cos^{2}\Theta \right)}$ where T0 is a constant

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

${\displaystyle T^{*}(\Theta ={\frac {\pi }{2}})=T_{0}=T^{*}(\Theta ={\frac {3\pi }{2}})}$

${\displaystyle T^{*}(\Theta )={\frac {3T_{0}}{2}}+{\frac {T_{0}}{2}}cos2\Theta }$

Principle of Superposition

${\displaystyle soln=solnT^{*}={\frac {3T_{0}}{2}}=T_{1}^{*}(\Theta )}$

${\displaystyle +solnT^{*}={\frac {T_{0}}{2}}cos2\Theta =T_{2}^{*}(\Theta )}$

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

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

Problem P:

PDE:

${\displaystyle \operatorname {Div} (gradT)=0}$

General Solution: Equation 2 p. 20-4.

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

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

Prob P1: ${\displaystyle \operatorname {Div} (gradT_{1})=0}$

such that:

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

Prob P2: ${\displaystyle \operatorname {Div} (gradT_{2})=0}$

such that:

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

${\displaystyle T_{1}(r,\Theta )={\frac {3T_{0}}{2}}}$

For the Problem ${\displaystyle T_{2}:T_{2}(a,\Theta )=T_{2}^{*}(\Theta )={\frac {T_{0}}{2}}cos2\Theta }$

${\displaystyle T_{2}(r,\Theta )=\sum _{n=1}^{\infty }r^{n}\left\{A_{n}cosn\Theta +B_{n}sinn\Theta \right\}}$

Final Solution:

${\displaystyle T(r,\Theta )=T_{0}\left[{\frac {3}{2}}+{\frac {1}{2}}\left({\frac {r}{a^{2}}}\right)cos2\Theta \right]}$

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

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

for all ${\displaystyle \Theta }$ and any K = constant

this is not periodic (not acceptable)

${\displaystyle T(r,\Theta )=C_{0}+\sum _{n=1}^{\infty }r^{n}\left\{A_{n}cosn\Theta +B_{n}sinn\Theta \right\}}$

where: rn=an

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

Fourier Coefficients

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

How do we derive the fourier coeffcients C0, An and Bn?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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= ${\displaystyle \int _{0}^{2\pi }C_{0}cosm\theta d\theta }$

${\displaystyle =C_{0}\int _{0}^{2\pi }\left\{1\right\}cosm\theta d\theta }$

= 0, by definition of orthogonality

Second term on RHS= ${\displaystyle \int _{0}^{2\pi }\sum _{n=1}^{\infty }a^{n}A_{n}cosn\theta cosm\theta d\theta }$

${\displaystyle =\sum _{n=1}^{\infty }\int _{0}^{2\pi }a^{n}A_{n}cosn\theta cosm\theta d\theta }$

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

${\displaystyle \Rightarrow \int _{0}^{2\pi }\sum _{n=1}^{\infty }a^{n}A_{n}cosn\theta cosm\theta d\theta =a^{n}\times A_{n}\times 2\pi }$

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

Third term on RHS= ${\displaystyle \int _{0}^{2\pi }\sum _{n=1}^{\infty }a^{n}A_{n}sinn\theta cosm\theta d\theta }$

= 0, by definition of orthogonality

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

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

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

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

${\displaystyle \int _{0}^{2\pi }T^{*}\left(\theta \right)sinn\theta d\theta =a^{n}\times B_{n}\times 2\pi }$

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

Find the Fourier coefficients C0,An,Bn

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

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

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

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

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)

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

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

Domain and boundary conditions

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

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

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

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

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

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

Eliminating Terms Based on BC

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

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

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

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

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

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

Thus the solution has the following form

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

Using the boundary conditions, we have

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

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

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

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

Eventually the solution for the temperature distribution is

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

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

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

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:

${\displaystyle y=bx^{a}}$.

The Power Law is observed in classical physics, biology, economics, and many other natural and social sciences. The exponent ${\displaystyle a}$ dominates the nature of the equation. In electrostatics and gravitation, ${\displaystyle a=2}$, while in Stefan-Boltzmann equations, ${\displaystyle a=4}$. 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 [5].

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:

${\displaystyle q=-\kappa \;grad\;T}$

where ${\displaystyle q}$ is the heat flux.

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

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

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

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

${\displaystyle \kappa \left(T\right)=bT^{a}}$.

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

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

${\displaystyle a={\frac {log\kappa _{2}-log\kappa _{1}}{logT_{2}-logT_{1}}}={\frac {log(1000)-log(0.4)}{log(10.7)-log(1)}}\cong 3.39}$

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

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

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 ${\displaystyle x}$ direction are :

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

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

Inserting this into the ${\displaystyle x}$ equation results in:

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

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

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

In the ${\displaystyle y}$ direction:

${\displaystyle \sum F_{y}=\tau sin\;\theta (x,t)+\tau sin\;\theta (x+dx,t)+P(x,t)dx-m(dx){\ddot {w}}}$

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

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

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

multiplied by the constant ${\displaystyle \tau }$.

Therefore, for an infinitely small length of string ${\displaystyle dx}$,

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

Sign problem in ${\displaystyle \displaystyle \sum F_{y}}$; 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 ${\displaystyle \displaystyle \tau (x)}$ and the slope angle ${\displaystyle \displaystyle \theta (x)}$ as algebraic quantities so that ${\displaystyle \displaystyle \sum F_{x}=\tau (x)\cos(x)+\tau (x+dx)\cos(x+dx)=0}$ and ${\displaystyle \displaystyle \sum F_{y}=\tau (x)\sin(x)+\tau (x+dx)\sin(x+dx)+p\,dx-m\,dx\,w_{tt}=0}$. Note the plus signs in both cosine terms in ${\displaystyle \displaystyle \sum F_{x}}$, and the plus signs in both sine terms in ${\displaystyle \displaystyle \sum F_{y}}$. 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)

Applications

${\displaystyle \tau (div(grad\omega ))+p(x,y)=mW_{t}t}$

Here, ${\displaystyle div(grad\omega )=\omega _{xx}+\omega _{yy}}$, ${\displaystyle m}$ 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)

${\displaystyle \tau \omega _{xx}-m\omega _{tt}+p=0}$

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

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

${\displaystyle detA=-\tau m<0\Rightarrow hyperbolic}$

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

1-D space

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

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

We assume ${\displaystyle \kappa }$ is constant.

${\displaystyle \kappa u_{xx}-Cu_{t}+f=0}$

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

2-D space

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

We assume ${\displaystyle \kappa }$ is constant here.

${\displaystyle \kappa (u_{xx}+u_{yy})-Cu_{t}+f=0}$

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

${\displaystyle {\big \lfloor }\partial _{x}\;\partial _{y}\;\partial _{z}{\big \rceil }{\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix}}{\begin{Bmatrix}\partial _{x}u\\\partial _{y}u\\\partial _{z}u\end{Bmatrix}}}$

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

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

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

Separation of Variables.

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

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

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

Plug ${\displaystyle (2)}$ into ${\displaystyle (1)}$:

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

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

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)

References

1. [1]
2. http://mathworld.wolfram.com/OrthogonalFunctions.html
3. http://dictionary.reference.com/browse/orthogonality
4. Simon Phillpot, February 2009 Lecture, Materials Science and Engineering, University of Florida

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)