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Chapter_2_Solutions

# Calculus: Early Transcendentals

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CHAPTER 2 Partial Diferential Equations on Unbounded Domains 1. Cauchy Problem for the Heat Equation Exercise 1a. Making the transformation r = ( x - y ) / 4 kt we have u ( x,t ) = i 1 1 1 4 πkt e ( x y ) 2 / 4 kt dy = - i ( x 1) / 4 kt ( x +1) / 4 kt 1 π e r 2 dr = 1 2 p erf p ( x + 1) / 4 kt P - erf p ( x - 1) / 4 kt PP Exercise 1b. We have u ( x,t ) = i 0 1 4 πkt e ( x y ) 2 / 4 kt e y dy Now complete the square in the exponent of e and write it as - ( x - y ) 2 4 kt - y = - x 2 - 2 xy + y 2 + 4 kty 4 kt = - y + 2 kt - x ) 2 4 kt + kt - x Then make the substitution in the integral r = y + 2 kt - x 4 kt Then u ( x,t ) = 1 π e kt x i (2 kt x ) / 4 kt e r 2 dr = 1 2 e kt x p 1 - erf p (2 kt - x ) / 4 kt PP Exercise 2. We have | u ( x,t ) | ≤ i R | G ( x - y,t ) || φ ( y ) | dy M i R G ( x - y,t ) dy = M 1

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2 2. PARTIAL DIFFERENTIAL EQUATIONS ON UNBOUNDED DOMAINS Exercise 3. Use erf ( z ) = 2 π i z 0 e r 2 dr = 2 π i z 0 (1 - r 2 + ··· ) dr = 2 π ( z - z 3 3 + ··· ) This gives w ( x 0 ,t ) = 1 2 + x 0 π t + ··· Exercise 4. The verifcation is straightForward. We guess the Green’s Function in two dimensions to be g ( x,y,t ) = G ( x,t ) G ( y,t ) = 1 4 πkt e x 2 / 4 kt 1 4 πkt e y 2 / 4 kt = 1 4 πkt e ( x 2 + y 2 ) / 4 kt where G is the Green’s Function in one dimension. Thus g is the temperature distribution caused by a point source at ( x,y ) = (0 , 0) at t = 0. This guess gives the correct expression. Then, by superposition, we have the solution u ( x,y,t ) = i R 2 1 4 πkt e (( x ξ ) 2 +( y η ) 2 ) / 4 kt ψ ( ξ,η ) dξdη Exercise 6. Using the substitution r = x/ 4 kt we get i R G ( x,t ) dx = 1 π i R e r 2 dr = 1 2. Cauchy Problem for the Wave Equation Exercise 1. Applying the initial conditions to the general solution gives the two equations F ( x ) + G ( x ) = f ( x ) , - cF ( x ) + cG ( x ) = g ( x ) We must solve these to determine the arbitrary Functions F and G . Integrate the second equation to get - cF ( x ) + cG ( x ) = i x 0 g ( s ) ds + C Now we have two linear equations For F and G that we can solve simultaneously.
2. CAUCHY PROBLEM FOR THE WAVE EQUATION 3 Exercise 2. Using d’Alembert’s formula we obtain u ( x,t ) = 1 2 c i x + ct x ct ds 1 + 0 . 25 s 2 = 1 2 c 2 arctan( s/ 2) | x + ct x ct = 1 c (arctan(( x + ct ) / 2) - arctan(( x - ct ) / 2)) Exercise 3. Let u = F ( x - ct ). Then u x (0 ,t ) = F ( - ct ) = s ( t ). Then F ( t ) = i t 0 s ( - r/c ) dr + K Then u ( x,t ) = - 1 c i t x/c 0 s ( y ) dy + K Exercise 4. Letting u = U/ρ we have u tt = U tt /ρ, u ρ = U ρ - U/ρ 2 and u ρρ = U ρρ - 2 U ρ 2 + 2 U/ρ 3 Substituting these quantities into the wave equation gives U tt = c 2 U ρρ which is the ordinary wave equation with general solution U ( ρ,t ) = F ( ρ - ct ) + G ( ρ + ct ) Then u ( ρ,t ) = 1 r ( F ( ρ - ct ) + G ( ρ + ct ))

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Chapter_2_Solutions - CHAPTER 2 Partial Dierential...

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