Chapter_1_Solutions

Accounting

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CHAPTER 1 The Physical Origins of Partial DiFerential Equations 1. Mathematical Models Exercise 1. The verifcation that u = 1 4 πkt e x 2 / 4 kt satisfes the heat equation u t = ku xx is straightForward di±erentiation. ²or larger k , the profles ³atten out much Faster. Exercise 2. The problem is straightForward di±erentiation. Taking the derivatives is easier iF we write the Function as u = 1 2 ln( x 2 + y 2 ). Exercise 3. Integrating u xx = 0 with respect to x gives u x = φ ( t ) where φ is an arbitrary Function. Integrating again gives u = φ ( t ) x + ψ ( t ). But u (0 ,t ) = ψ ( t ) = t 2 and u (1 ,t ) = φ ( t ) · 1 + t 2 , giving φ ( t ) = 1 t 2 . Thus u ( x,t ) = (1 t 2 ) x + t 2 . Exercise 4. Leibniz’s rule gives u t = 1 2 ( g ( x + ct ) + g ( x ct )) Thus u tt = c 2 ( g ( x + ct ) g ( x ct )) In a similar manner u xx = 1 2 c ( g ( x + ct ) g ( x ct )) Thus u tt = c 2 u xx . Exercise 5. IF u = e at sin bx then u t = ae at sin bx and u xx = b 2 e at sin bx . Equat- ing gives a = b 2 . Exercise 6. Letting v = u x the equation becomes v t + 3 v = 1. Multiply by the integrating Factor e 3 t to get ∂t ( ve 3 t ) = e 3 t Integrate with respect to t to get v = 1 3 + φ ( x ) e 3 t 1
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2 1. THE PHYSICAL ORIGINS OF PARTIAL DIFFERENTIAL EQUATIONS where φ is an arbitrary function. Thus u = i vdx = 1 3 x + Φ( x ) e 3 t + Ψ( t ) Exercise 7. Let w = e u or u = ln w Then u t = w t /w and u x = w x /w , giving w xx = w xx /w w 2 x /w 2 . Substituting into the PDE for u gives, upon cancellation, w t = w xx . Exercise 8. It is straightforward to verify that u = arctan( y/x ) satisFes the Laplace equation. We want u 1 as y 0 ( x > 0), and u → − 1 as y 0 ( x < 0). So try u = 1 2 π arctan y x We want the branch of arctan z with 0 < arctan z < π/ 2 for z > 0 and π/ 2 < arctan z < π for z < 0. Exercise 9. Di±erentiate under the integral sign to obtain u xx = i 0 ξ 2 c ( ξ ) e ξy sin( ξx ) and u yy = i 0 ξ 2 c ( ξ ) e ξy sin( ξx ) Thus u xx + u yy = 0 . Exercise 10. In preparation. 2. Conservation Laws Exercise 1. Since A = A ( x ) depends on x , it cannot cancel from the conservation law and we obtain A ( x ) u t = ( A ( x ) φ ) x + A ( x ) f Exercise 2. The solution to the initial value problem is u ( x,t ) = e ( x ct ) 2 . When c = 2 the wave forms are bell-shaped curves moving to the right at speed two. Exercise 3. Letting ξ = x ct and τ = t , the PDE u t + cu x = λu becomes U τ = λU or U = φ ( ξ ) e λt . Thus u ( x,t ) = φ ( x ct ) e λt Exercise 4. In the new dependent variable w the equation becomes w t + cw x = 0.
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Chapter_1_Solutions - CHAPTER 1 The Physical Origins of...

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