1a. Consider the problem 2-15 from Ozisik (1993). Starting with the heat diffusion equation, Eq. 1-11a in Ozisik (1993), show how it can be reduced to: a²T ²T + ax² ay² in [0 < x
1a. Consider the problem 2-15 from Ozisik (1993). Starting with the heat diffusion equation, Eq. 1-11a in Ozisik (1993), show how it can be reduced to: a²T ²T + ax² ay² in [0 < x
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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This is a multiple part question and i need help with only part D, part D might need some help from above parts that is provided in the table and the equations avaliable.
![**TABLE 2-2**: The Solution \( X(\beta_m, x) \), the Norm \( N(\beta_m) \), and the Eigenvalues \( \beta_m \) of the Differential Equation
\[
\frac{d^2 X(x)}{dx^2} + \beta^2 X(x) = 0 \quad \text{in} \quad 0 < x < L
\]
**Diagram Explanation**: The diagram illustrates a bounded domain along the x-axis from 0 to L. The region is depicted between these two boundary points.
---
**Table: Boundary Conditions and Corresponding Results**
| No. | Boundary Condition at \(x = 0\) | Boundary Condition at \(x = L\) | \(X(\beta_m, x)\) | \( 1/N(\beta_m) \) | Eigenvalues \(\beta_m^2\) are Positive Roots of |
|-----|--------------------------------|--------------------------------|------------------|------------------|--------------------------------------------------|
| 1 | \(-\frac{dX}{dx} + H_1 X = 0\) | \(\frac{dX}{dx} + H_2 X = 0\) | \(\beta_m \cos \beta_m x + H_1 \sin \beta_m x\) | \(\begin{align*} 2 \left[ (\beta_m^2 + H_1^2) \left( L + \frac{H_2}{\beta_m^2 + H_2^2} \right) + H_1 \right]^{-1} \end{align*} \) | \(\tan \beta_m L = \frac{\beta_m(H_1 + H_2)}{\beta_m^2 - H_1H_2}\) |
| 2 | \(-\frac{dX}{dx} = 0\) | \(\frac{dX}{dx} = 0\) | \(\cos \beta_m (L - x)\) | \(\begin{align*} \frac{2}{L(\beta_m^2 + H_2^2) + H_1} \end{align*} \) | \(\beta_m \tan \beta_m L = H_1\) |
| 3' | \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6b656b52-bb2e-489d-9bef-8a32efc9339f%2Fea9e6cbd-0806-4135-b116-598d0efef2bd%2Fp7gcgj_processed.png&w=3840&q=75)
Transcribed Image Text:**TABLE 2-2**: The Solution \( X(\beta_m, x) \), the Norm \( N(\beta_m) \), and the Eigenvalues \( \beta_m \) of the Differential Equation
\[
\frac{d^2 X(x)}{dx^2} + \beta^2 X(x) = 0 \quad \text{in} \quad 0 < x < L
\]
**Diagram Explanation**: The diagram illustrates a bounded domain along the x-axis from 0 to L. The region is depicted between these two boundary points.
---
**Table: Boundary Conditions and Corresponding Results**
| No. | Boundary Condition at \(x = 0\) | Boundary Condition at \(x = L\) | \(X(\beta_m, x)\) | \( 1/N(\beta_m) \) | Eigenvalues \(\beta_m^2\) are Positive Roots of |
|-----|--------------------------------|--------------------------------|------------------|------------------|--------------------------------------------------|
| 1 | \(-\frac{dX}{dx} + H_1 X = 0\) | \(\frac{dX}{dx} + H_2 X = 0\) | \(\beta_m \cos \beta_m x + H_1 \sin \beta_m x\) | \(\begin{align*} 2 \left[ (\beta_m^2 + H_1^2) \left( L + \frac{H_2}{\beta_m^2 + H_2^2} \right) + H_1 \right]^{-1} \end{align*} \) | \(\tan \beta_m L = \frac{\beta_m(H_1 + H_2)}{\beta_m^2 - H_1H_2}\) |
| 2 | \(-\frac{dX}{dx} = 0\) | \(\frac{dX}{dx} = 0\) | \(\cos \beta_m (L - x)\) | \(\begin{align*} \frac{2}{L(\beta_m^2 + H_2^2) + H_1} \end{align*} \) | \(\beta_m \tan \beta_m L = H_1\) |
| 3' | \
![### Problem from Ozisik (1993)
#### 1a. Problem Reduction
Consider the problem 2-15 from Ozisik (1993). Starting with the heat diffusion equation, Eq. 1-11a in Ozisik (1993), show how it can be reduced to:
\[
\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0
\]
in the domain \([0 < x < a; 0 < y < b]\), subject to the following boundary conditions:
- \(\frac{\partial T}{\partial x} = 0\) at \(x = 0\);
- \(\frac{\partial T}{\partial x} + HT = 0\) at \(x = a\);
- \(T = f(x)\) at \(y = 0\);
- \(\frac{\partial T}{\partial y} + HT = 0\) at \(y = b\).
All assumptions must be clearly stated and justified, and all steps clearly explained.
#### 1b. Eigenfunction Derivation
Using separation of variables, show how the eigenfunction from Table 2-2 in Ozisik (1993) was derived:
\[
X(\beta_m, x) = \cos \beta_m x
\]
Each step of your formulation must be clearly explained, and all assumptions justified.
#### 1c. Eigen Condition Derivation
Show how the eigen condition from Table 2-2 in Ozisik (1993) was derived:
\[
\beta_m \tan \beta_m a = H
\]
Each step of your formulation must be clearly explained, and all assumptions justified.
#### 1d. Normalization Integral Derivation
Show how the normalization integral from Table 2-2 in Ozisik (1993) was derived:
\[
\frac{1}{N(\beta_m)^2} = \frac{\beta_m^2 + H^2}{a(\beta_m^2 + H^2) + H}
\]
Each step of your formulation must be clearly explained, and all assumptions justified.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6b656b52-bb2e-489d-9bef-8a32efc9339f%2Fea9e6cbd-0806-4135-b116-598d0efef2bd%2F9o1xhp7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem from Ozisik (1993)
#### 1a. Problem Reduction
Consider the problem 2-15 from Ozisik (1993). Starting with the heat diffusion equation, Eq. 1-11a in Ozisik (1993), show how it can be reduced to:
\[
\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0
\]
in the domain \([0 < x < a; 0 < y < b]\), subject to the following boundary conditions:
- \(\frac{\partial T}{\partial x} = 0\) at \(x = 0\);
- \(\frac{\partial T}{\partial x} + HT = 0\) at \(x = a\);
- \(T = f(x)\) at \(y = 0\);
- \(\frac{\partial T}{\partial y} + HT = 0\) at \(y = b\).
All assumptions must be clearly stated and justified, and all steps clearly explained.
#### 1b. Eigenfunction Derivation
Using separation of variables, show how the eigenfunction from Table 2-2 in Ozisik (1993) was derived:
\[
X(\beta_m, x) = \cos \beta_m x
\]
Each step of your formulation must be clearly explained, and all assumptions justified.
#### 1c. Eigen Condition Derivation
Show how the eigen condition from Table 2-2 in Ozisik (1993) was derived:
\[
\beta_m \tan \beta_m a = H
\]
Each step of your formulation must be clearly explained, and all assumptions justified.
#### 1d. Normalization Integral Derivation
Show how the normalization integral from Table 2-2 in Ozisik (1993) was derived:
\[
\frac{1}{N(\beta_m)^2} = \frac{\beta_m^2 + H^2}{a(\beta_m^2 + H^2) + H}
\]
Each step of your formulation must be clearly explained, and all assumptions justified.
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