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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: Ꭷ2Ꭲ Ꭷ2Ꭲ + ax² ay² in [0

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: Ꭷ2Ꭲ Ꭷ2Ꭲ + ax² ay² in [0

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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This is a multiple-part question, I just need help with part C, Table 2.2 is provided and you can refer to above parts for equations and boundary equations.

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<a; 0 <y<b], subjected to the following boundary conditions: ƏT ax ƏT = 0 at x = 0; T = f(x) at y = 0; + HT = 0 at y = b All assumptions must be clearly stated and justified, and all steps clearly explained. əx = 0 + HT= 0 at x = a; ƏT 1b. Using separation of variables, show how the eigenfunction from Table 2-2 in Ozisik (1993) was derived. X(Bm,x) = cosmx Each step of your formulation must be clearly explained, and all assumptions justified. 1c. Show how the eigen condition from Table 2-2 in Ozisik (1993) was derived. Pm tan ßma = H Each step of your formulation must be clearly explained, and all assumptions justified.
Transcribed Image Text: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<a; 0 <y<b], subjected to the following boundary conditions: ƏT ax ƏT = 0 at x = 0; T = f(x) at y = 0; + HT = 0 at y = b All assumptions must be clearly stated and justified, and all steps clearly explained. əx = 0 + HT= 0 at x = a; ƏT 1b. Using separation of variables, show how the eigenfunction from Table 2-2 in Ozisik (1993) was derived. X(Bm,x) = cosmx Each step of your formulation must be clearly explained, and all assumptions justified. 1c. Show how the eigen condition from Table 2-2 in Ozisik (1993) was derived. Pm tan ßma = H Each step of your formulation must be clearly explained, and all assumptions justified.
5 6 TABLE 2-2 The Solution X(), the Norm NB) and the Eigenvalues of the Differential Equation Subject to the Boundary Conditions Shown in the Table Below Boundary Condition at x=0 No. 1 2 7 3 4 dX -=+H₁X=0 dx 9 - dX dx dx dX dx dX ö dX -=0 X=0 X=0 + H₂X=0 +H₁X=0 dX dx dx dX i X=0. dX dx A dX dx dX Boundary Condition at x = L dx +H₂X=0 = 0 X=0 d²X(x) dx² + H₂X=0 0 + H₂X=0 =0 +8²x(x)=0 in Bcos B+H, sin f cos B (L-x) X(Bmx) sin B (L-x) cos Bmx *cos Bmx cos x sin x sin ßx 0 < x <L X=0 sin Bmx X = 0 "For this particular case Bo-0 is also an eigenvalue corresponding to X = 1. 2 [10² + H²) (L 2- B²+H} L(B²+ H²) + H₂ 0 B²+ H₂ 2. LB² + H²) + H₂ 2 1/N(B) 8² + H² 2 L(B²+ H₂) + H₂ 2 L+ for for 8,0 NIJ B² + H² L(B²+ H₂) + H₂ 2 L H₂ ² + H² + H₁ Eigenvalues 's are Positive Roots of tan BL- _B_(H₁ + H₂) B²-H₂H₂ Plan BL=H₁ Bm cot BML=H₁ Pm tan PL=H₂ sin BL=0ª cos BL=0 Bm cot BmL=-H₂ cos BL=0 sin AL=0
Transcribed Image Text:5 6 TABLE 2-2 The Solution X(), the Norm NB) and the Eigenvalues of the Differential Equation Subject to the Boundary Conditions Shown in the Table Below Boundary Condition at x=0 No. 1 2 7 3 4 dX -=+H₁X=0 dx 9 - dX dx dx dX dx dX ö dX -=0 X=0 X=0 + H₂X=0 +H₁X=0 dX dx dx dX i X=0. dX dx A dX dx dX Boundary Condition at x = L dx +H₂X=0 = 0 X=0 d²X(x) dx² + H₂X=0 0 + H₂X=0 =0 +8²x(x)=0 in Bcos B+H, sin f cos B (L-x) X(Bmx) sin B (L-x) cos Bmx *cos Bmx cos x sin x sin ßx 0 < x <L X=0 sin Bmx X = 0 "For this particular case Bo-0 is also an eigenvalue corresponding to X = 1. 2 [10² + H²) (L 2- B²+H} L(B²+ H²) + H₂ 0 B²+ H₂ 2. LB² + H²) + H₂ 2 1/N(B) 8² + H² 2 L(B²+ H₂) + H₂ 2 L+ for for 8,0 NIJ B² + H² L(B²+ H₂) + H₂ 2 L H₂ ² + H² + H₁ Eigenvalues 's are Positive Roots of tan BL- _B_(H₁ + H₂) B²-H₂H₂ Plan BL=H₁ Bm cot BML=H₁ Pm tan PL=H₂ sin BL=0ª cos BL=0 Bm cot BmL=-H₂ cos BL=0 sin AL=0
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