The solution of the boundary value problem ∂ 2 u ∂ r 2 + 2 r ∂ u ∂ r = ∂ 2 u ∂ t 2 , ∂ u ∂ r | r = 1 = 0 , ∂ u ∂ t | t = 0 = g ( r ) and u ( r , 0 ) = f ( r ) such that 0 < r < 1 , t > 0 .

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Answer 1

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Chapter 14, Problem 16CR

To determine

The solution of the boundary value problem ∂2u∂r2+2r∂u∂r=∂2u∂t2, ∂u∂r|r=1=0, ∂u∂t|t=0=g(r) and u(r,0)=f(r) such that 0<r<1,t>0.

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In this exercise, we will investigate a technique to prove that a language is notregular. This tool is called the pumping lemma.The pumping lemma says that if M = (S, I, f, s0, F ) is a DFA with p states (i.e., p = |S|) and if the wordw is in L(M ) (the language generated by M ) and w has length greater than or equal to p, then w may bedivided into three pieces, w = xyz, satisfying the following conditions:1. For each i ∈ N, xy^i z ∈ L(M ).2. |y| > 0 (i.e., y contains at least one character).3. |xy| ≤ p (i.e., the string xy has at most p characters). Use the pumping lemma to show the following language is not regular (HINT: Use proof by contradictionto assume the language is regular and apply the pumping lemma to the language):L = {0^k1^k | k ∈ N}

A prefix of length ℓ of some word w are the first ℓ characters (in order) of w.1. Construct a context-free grammar for the language: L = {w ∈ {a, b}∗ | every prefix of w has at least as many a’s as b’s}2. Explain why every word generated by your context-free grammar (in Part 1) is contained in L. Then,prove via induction that every w ∈ L is produced by your context-free grammar.

Consider a simplified version of American football where on any possession ateam can earn 0, 3 or 7 points. What is the smallest number n0 of points such that for all n ≥ n0 and n ∈ Na team could earn n points. You must prove that your answer is correct via induction (HINT: Don’t forgetto show that n0 is the smallest number above which any number of points is reachable).

Answer 2

Question

Chapter 14, Problem 16CR

To determine

The solution of the boundary value problem ∂2u∂r2+2r∂u∂r=∂2u∂t2, ∂u∂r|r=1=0, ∂u∂t|t=0=g(r) and u(r,0)=f(r) such that 0<r<1,t>0.

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Answer 3

Question

Chapter 14, Problem 16CR

To determine

The solution of the boundary value problem ∂2u∂r2+2r∂u∂r=∂2u∂t2, ∂u∂r|r=1=0, ∂u∂t|t=0=g(r) and u(r,0)=f(r) such that 0<r<1,t>0.

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Answer 4

Question

Chapter 14, Problem 16CR

To determine

The solution of the boundary value problem ∂2u∂r2+2r∂u∂r=∂2u∂t2, ∂u∂r|r=1=0, ∂u∂t|t=0=g(r) and u(r,0)=f(r) such that 0<r<1,t>0.

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Answer 5

Chapter 14, Problem 16CR

To determine

The solution of the boundary value problem ∂2u∂r2+2r∂u∂r=∂2u∂t2, ∂u∂r|r=1=0, ∂u∂t|t=0=g(r) and u(r,0)=f(r) such that 0<r<1,t>0.

Answer 6

Chapter 14, Problem 16CR

To determine

The solution of the boundary value problem ∂2u∂r2+2r∂u∂r=∂2u∂t2, ∂u∂r|r=1=0, ∂u∂t|t=0=g(r) and u(r,0)=f(r) such that 0<r<1,t>0.

Answer 7

Chapter 14, Problem 16CR

To determine

The solution of the boundary value problem ∂2u∂r2+2r∂u∂r=∂2u∂t2, ∂u∂r|r=1=0, ∂u∂t|t=0=g(r) and u(r,0)=f(r) such that 0<r<1,t>0.

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E Ch. 14.1 - Prob. 6E Ch. 14.1 - Prob. 7E Ch. 14.1 - Prob. 8E Ch. 14.1 - Prob. 10E Ch. 14.1 - Prob. 11E

The solution of the boundary value problem ∂ 2 u ∂ r 2 + 2 r ∂ u ∂ r = ∂ 2 u ∂ t 2 , ∂ u ∂ r | r = 1 = 0 , ∂ u ∂ t | t = 0 = g ( r ) and u ( r , 0 ) = f ( r ) such that 0 < r < 1 , t > 0 .

Solution Summary: The author explains the solution of the given boundary value problem, which is lpartial.

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The solution of the boundary value problem ∂2u∂r2+2r∂u∂r=∂2u∂t2, ∂u∂r|r=1=0, ∂u∂t|t=0=g(r) and u(r,0)=f(r) such that 0<r<1,t>0.

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Chapter 14 Solutions

The solution of the boundary value problem ∂ 2 u ∂ r 2 + 2 r ∂ u ∂ r = ∂ 2 u ∂ t 2 , ∂ u ∂ r | r = 1 = 0 , ∂ u ∂ t | t = 0 = g ( r ) and u ( r , 0 ) = f ( r ) such that 0 &lt; r &lt; 1 , t &gt; 0 .

Solution Summary: The author explains the solution of the given boundary value problem, which is lpartial.

Videos

The solution of the boundary value problem ∂2u∂r2+2r∂u∂r=∂2u∂t2, ∂u∂r|r=1=0, ∂u∂t|t=0=g(r) and u(r,0)=f(r) such that 0<r<1,t>0.

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Chapter 14 Solutions

The solution of the boundary value problem ∂ 2 u ∂ r 2 + 2 r ∂ u ∂ r = ∂ 2 u ∂ t 2 , ∂ u ∂ r | r = 1 = 0 , ∂ u ∂ t | t = 0 = g ( r ) and u ( r , 0 ) = f ( r ) such that 0 < r < 1 , t > 0 .