User:Egm6322.s09.Three.ge/MyReport4

=Laplace Equation= 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.

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

Axisymmetric Solution


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 Ao and Bo are constants.

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

Separation of Variables solution


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 a positive n2

For the two differential equations to be equal to each other the must each be equal to a constant, n2. This effectively separates the variables and provides 2 uncoupled ODE's.

$$ \begin{matrix} r^{2}\frac{d^{2}F(r)}{dr^{2}}+ r\frac{dF(r)}{dr}-n^{2}F(r) =0\\ \\ \frac{d^{2}G}{d\theta^{2}}-n^{2}G(\theta)=0 \end{matrix}$$

HW explanation of how the answer was got

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 why is Laplace solution called harmonic

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

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

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

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

HW why it is periodic

R4 Edits
Homework:Derive LP p.14 (1.2.13) $$\begin{matrix} \overline{a}=a \phi_x^2 + 2b \phi_x \phi_y +c \phi_y^2\\ \\ \overline{b}=a \phi_x \psi_x + b( \phi_x \psi_y +\phi_y \psi_x ) + c \phi_y \psi_y \\ \\ \overline{c}=a \psi_x^2 + 2b \psi_x \psi_y +c \psi_y^2 \end{matrix}$$

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:

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

Showed intermediate steps to make it easier to follow. Changed the notation of a,b and c. Still need to correct the answer and cite "LP".

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:

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

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