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Unformatted text preview: MATH 220: PRACTICE MIDTERM, SOLUTIONS This is a closed book, closed notes, no calculators exam. There are 5 problems. Solve all of them. Total score: 100 points. Problem 1. (i) (13 points) Find the general C 1 solution of the PDE 3 y 2 u x + u y = 0 . (ii) (6 points) Solve the initial value problem with initial condition u ( x, 0) = x 2 . Solution. The characteristic ODEs are dx ds = 3 y 2 , dy ds = 1 , dz ds = 0 , so the projected characteristic curves satisfy dx dy = 3 y 2 , so x = y 3 + C , i.e. x − y 3 constant are exactly the projected characteristic curves. Moreover, as dz ds = 0, u is constant along these curves, so the general solution is u ( x, y ) = f ( x − y 3 ), f an arbitrary C 1 function. To satisfy the initial condition, let x 2 = u ( x, 0) = f ( x ), so u ( x, y ) = ( x − y 3 ) 2 . An alternative method is to solve the characteristic ODEs directly. Notice that this being a linear (or more generally semilinear) PDE, the equations for the pro jected characteristic curves decouple, and then one sees that the projected char acteristic curves through the x axis give all projected characteristic curves. So we could impose the initial condition at the x axis to obtain the general solution: u ( x, 0) = f ( x ), f to be specified. The initial condition for the characteristic ODEs is then ( x ( r, 0) , y ( r, 0) , z ( r, 0)) = ( r, , f ( r )). The solution for the characteristic ODEs is y ( r, s ) = s + c 2 ( r ) , z ( r, s ) = c 3 ( r ) , hence dx ds = 3( s + c 2 ( r )) 2 , and x ( r, s ) = ( s + c 2 ( r )) 3 + c 1 ( r ) . The initial condition gives c 2 ( r ) = 0, c 3 ( r ) = f ( r ), c 1 ( r ) = r , so x ( r, s ) = s 3 + r, y ( r, s ) = s, z ( r, s ) = f ( r ) . Correspondingly, s = y , r = x − y 3 , and u ( x, y ) = f ( x − y 3 ). If f ( x ) = x 2 , this gives u ( x, y ) = ( x − y 3 ) 2 . Problem 2. (i) (10 points) Find the general C 2 solution of the PDE u xx + 3 u xt + 2 u tt = 0 . (ii) (10 points) Solve the initial value problem with initial condition u ( x, − x ) = φ ( x ) , u t ( x, − x ) = ψ ( x ) , with φ, ψ given. 1 2 MATH 220: PRACTICE MIDTERM, SOLUTIONS Solution. Factoring the partial differential operator as ∂ 2 x + 3 ∂ x ∂ t + 2 ∂ 2 t = ( ∂ x + ∂ t )( ∂ x + 2 ∂ t ) , so to find the general solution, we need to solve ( ∂ x + ∂ t ) v = 0 , ( ∂ x + 2 ∂ t ) u = v. This can be done directly; the main point being that solutions of the homogeneous versions of these equations would be f ( x − t ), resp. g (2 x − t ). For the sake of variety, we change into corresponding characteristic coordinates. For ∂ x + ∂ t , the projected characteristic curves satisfy dx ds = 1 , dt ds = 1 ⇒ dt dx = 1 , so they are t = x + C ′ , i.e. x − t = C . Similarly, for ∂ x + 2 ∂ t , dx ds = 1 , dt ds = 2 ⇒ dt dx = 2 , so we get 2 x − t = C . We change into characteristic coordinates ξ = x − t , η = 2 x − t , so x = η − ξ , t = η − 2 ξ , yields − ∂ ξ = ∂ x + 2 ∂ t ,...
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 '09
 Math, Differential Equations, Applied Mathematics, Equations, Partial Differential Equations, Constant of integration, Boundary value problem, Partial differential equation, initial condition

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