Substituting we find u vw 0 Problem 112 Write the PDE u xx 2 u xy 3 u yy 0 in

# Substituting we find u vw 0 problem 112 write the pde

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Substituting we find u vw = 0 Problem 1.12 Write the PDE u xx + 2 u xy - 3 u yy = 0 in the coordinates v ( x, y ) = y - 3 x and w ( x, y ) = x + y. SOLUTIONS TO SECTION 1 9 Solution. We have u x = - 3 u v + u w u xx = - 3( - 3 u v + u w ) v + ( - 3 u v + u w ) w = 9 u vv - 6 u vw + u ww u xy = - 3 u vv + u vw - 3 u vw + u ww = - 3 u vv - 2 u vw + u ww u y = u v + u w u yy =( u v + u w ) v + ( u v + u w ) w = u vv + 2 u vw + u ww . Substituting into the PDE we find u vw = 0 Problem 1.13 Write the PDE au x + bu y = 0 in the coordinates s ( x, y ) = ax + by and t ( x, y ) = bx - ay. Assume a 2 + b 2 > 0 . Solution. According to the chain rule for the derivative of a composite function, we have u x = u s s x + u t t x = au s + bu t u y = u s s y + u t t y = bu s - au t . Substituting these into the given equation to obtain a 2 u s + abu t + b 2 u s - abu t = 0 or ( a 2 + b 2 ) u s = 0 and since a 2 + b 2 > 0 we obtain u s = 0 Problem 1.14 Write the PDE u x + u y = 1 in the coordinates s = x + y and t = x - y. 10 CONTENTS Solution. Using the chain rule we find u x = u s s x + u t t x = u s + u t u y = u s s y + u t t y = u s - u t . Substituting these into the PDE to obtain u s = 1 2 Problem 1.15 Write the PDE au t + bu x = u, a, b 6 = 0 in the coordinates v = ax - bt and w = 1 a t. Solution. We have u t = - bu v + 1 a u w and u x = au v . Substituting we find u w = u SOLUTIONS TO SECTION 2 11 Solutions to Section 2 Problem 2.1 Determine a and b so that u ( x, y ) = e ax + by is a solution to the equation u xxxx + u yyyy + 2 u xxyy = 0 . Solution. We have u xxxx = a 4 e ax + by , u yyyy = b 4 e ax + by , and u xxyy = a 2 b 2 e ax + by . Thus, substituting these into the equation we find ( a 4 + 2 a 2 b 2 + b 4 ) e ax + by = 0 . Since e ax + by 6 = 0, we must have a 4 + 2 a 2 b 2 + b 4 = 0 or ( a 2 + b 2 ) = 0 . This is true only when a = b = 0 . Thus, u ( x, y ) = 1 Problem 2.2 Consider the following differential equation tu xx - u t = 0 . Suppose u ( t, x ) = X ( x ) T ( t ) . Show that there is a constant λ such that X 00 = λX and T 0 = λtT. Solution. Substituting into the differential equation we find tX 00 T - XT 0 = 0 or X 00 X = T 0 tT . The LHS is a function of x only whereas the RHS is a function of t only. This is true only when both sides are constant. That is, there is λ such that X 00 X = T 0 tT = λ and this leads to the two ODEs X 00 = λX and T 0 = λtT 12 CONTENTS Problem 2.3 Consider the initial value problem xu x + ( x + 1) yu y = 0 , x, y > 1 u (1 , 1) = e. Show that u ( x, y ) = xe x y is the solution to this problem. Solution. We have xu x +( x +1) yu y = x y ( e x + xe x )+( x +1) y - xe x y 2 = 0 and u (1 , 1) = e Problem 2.4 Show that u ( x, y ) = e - 2 y sin ( x - y ) is the solution to the initial value prob- lem u x + u y + 2 u = 0 , x, y > 1 u ( x, 0) = sin x. Solution. We have u x + u y +2 u = e - 2 y cos ( x - y ) - 2 e - 2 y sin ( x - y ) - e - 2 y cos ( x - y )+ 2 e - 2 y sin ( x - y ) = 0 and u ( x, 0) = sin x Problem 2.5 Solve each of the following differential equations: (a) du dx = 0 where u = u ( x ) . (b) ∂u ∂x = 0 where u = u ( x, y ) . Solution. (a) The general solution to this equation is u ( x ) = C where C is an arbitrary constant. (b) The general solution is u ( x, y ) = f ( y ) where f is an arbitrary function of y Problem 2.6 Solve each of the following differential equations: (a) d 2 u dx 2 = 0 where u = u ( x ) .  #### You've reached the end of your free preview.

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