User:Egm6322.s09.xyz/hw4 xyz

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Example: $$T^* \left( \theta \right) = T_o \left[ 1 + cos^2\left(\theta\right) \right]$$ where $$T_o = constant$$

Boundary Conditions: $$T^* \left( \theta = 0 \right)=2T_o$$ and $$T^* \left( \theta = \frac{\pi}{2} \right)=T_o$$

$$T^* \left( \theta \right) = \frac{3T_o}{2} + \frac{T_o}{2}cos2\theta$$

Homework: Prove Law of Superposition is valid

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 $$T_2(r=a, \theta) = T_2^{*}(\theta)$$

Homework: Verification of T1 solution

$$T_1$$ 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 XXXX):

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

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. $$A_o = A_n = B_n = C_n = D_n = 0$$

Homework: Verification of T2 solution

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 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 $$n\not=2, cosn\theta \not= cos2\theta$$ $$\therefore A_{n\not=2}=0$$

Orthogonal Functions 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"