Explanation Approach: Van der Pol’s Equation is d 2 x d t 2 + ∈ ( x 2 − 1 ) d x d t + x = 0 … ( 1 ) Here, ∈ is a positive constant. Levinson-Smith’s theorem is, Let f ( x ) and g ( x ) be continuous functions and let F ( x ) = ∫ 0 x f ( s ) d s , G ( x ) = ∫ 0 x g ( s ) d s . The Lienard equation: d 2 x d t 2 + f ( x ) d x d t + g ( x ) = 0 has a unique periodic solution whenever all of the following conditions hold. i) f ( x ) is even function. ii) F ( x ) < 0 , for 0 < x < a ; and F ( x ) > 0 , for x > a , for some a . iii) F ( x ) → + ∞ as x → + ∞ , monotonically for x > a . iv) g ( x ) is an odd function with g ( x ) > 0 for x > 0 . v) G ( x ) → + ∞ as x → + ∞ . Calculation: Compare Equation ( 1 ) with d 2 x d t 2 + f ( x ) d x d t + g ( x ) = 0 . f ( x ) = ∈ ( x 2 − 1 ) and g ( x ) = x . Both f ( x ) and g ( x ) are continuous functions. i) f ( − x ) = ∈ ( ( − x ) 2 − 1 ) = ∈ ( x 2 − 1 ) = f ( x ) So, f ( x ) is even function

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Chapter 12.6, Problem 24E

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The van der Pol’s satisfy the conditions of Levinson-Smith’s theorem and has a unique nonconstant periodic solution.

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Chapter 12.6, Problem 23E

Chapter 12.6, Problem 25E

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Ch. 12.2 - In Problem 16, classify the critical point at the...Ch. 12.2 - Prob. 2E Ch. 12.2 - Prob. 3E Ch. 12.2 - Prob. 4E Ch. 12.2 - Prob. 5E Ch. 12.2 - Prob. 6E Ch. 12.2 - Prob. 7E Ch. 12.2 - Prob. 8E Ch. 12.2 - Prob. 9E Ch. 12.2 - In Problem 712, find and classify the critical...

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To prove: The van der Pol’s satisfy the conditions of Levinson-Smith’s theorem and has a unique nonconstant periodic solution.

Solution Summary: The author explains that the van der Pol's satisfy the conditions of Levinson-Smith’s theorem and has a unique nonconstant periodic solution.