StraussPDEch2s1p11.pdf - Strauss PDEs 2e Section 2.1...

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Strauss PDEs 2e: Section 2.1 - Exercise 11 Page 1 of 6 Exercise 11 Find the general solution of 3 u tt + 10 u xt + 3 u xx = sin( x + t ). Solution Solution by Operator Factorization By factoring the PDE and making a substitution, we can write it as an equivalent system of uncoupled first-order PDEs and solve it using the methods of the previous chapter. Start off by writing the left side as an operator acting on u , L u . ( 2 x + x t - 20 2 t ) u = sin( x + t ) Now factor the operator. (3 t + x )( t + 3 x ) u = sin( x + t ) If we let v = ( t + 3 x ) u , then what remains is (3 t + x ) v = sin( x + t ). Divide both sides of this equation by 3. That is, u t + 3 u x = v v t + 1 3 v x = 1 3 sin( x + t ) . On the paths defined by dx dt = 1 3 , x ( ξ, 0) = ξ (1) the PDE on the bottom reduces to an ODE, dv dt = 1 3 sin( x + t ) . (2) That is, v = v ( x, t ) is constant on the characteristics defined by (1). Solving (1) by integration gives x = 1 3 t + ξ. Solve now for ξ . ξ = x - 1 3 t Now that we can write x in terms of t , we can solve (2) for v . dv dt = 1 3 sin ξ + 4 3 t Integrating this equation, we find that v ( ξ, t ) = 1 3 ˆ t sin ξ + 4 3 s ds + f ( ξ ) , where f is an arbitrary function of the characteristic coordinate, ξ . Therefore, v ( x, t ) = - 1 4 cos( x + t ) + f x - 1 3 t .
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Strauss PDEs 2e: Section 2.1 - Exercise 11 Page 2 of 6 We can check that this is the solution of the bottom PDE. v x = 1 4 sin( x + t ) + f 0 v t = 1 4 sin( x + t ) - 1 3 f 0 3 v t + v x = sin( x + t ), so this is the correct solution. Now we use this solution to solve the PDE on the top. Our task is to solve u t + 3 u x = - 1 4 cos( x + t ) + f x - 1 3 t .
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