# week2 - Week 2 Math 716 1 Linear 1st order PDEs in two...

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Unformatted text preview: Week 2, Math 716 1 Linear 1st order PDEs in two independent variables Consider first special class of linear 1st order PDEs in two independent variables ( x 1 ,x 2 ). a 1 ( x 1 ,x 2 ) u x 1 + a 2 ( x 1 ,x 2 ) u x 2 = c ( x 1 ,x 2 ) (1) where a 1 , a 2 and c are continuous function in some domain Ω ⊂ R 2 . Denote x = ( x 1 ,x 2 ). We will assume 1. On some differentiable curve Γ = { x : x = x ( s ) } characterized by a real parameter s in some interval, the tangent d x ds is never parallel to a ( x ( s )), where a ≡ ( a 1 ,a 2 ). The significance of this non-characteristic condition will be clearer later. 2. On the non-characteristic curve Γ, we specify initial condition: u ( x ( s )) = u ( s ) (2) We now seek a solution in a domain Ω adjoining Γ. We notice (1) geometrically implies specified directional derivative of u along a , since (1) is simply a · ∇ u = c (3) We introduce a set of characteristic curve X ( t ; s ) parametrized by real scalar t ∈ I for some interval I containing 0, for each s . ∂ X ∂t = a ( X ( t ; s )) with initial condition X ( s ; 0) = x ( s ) (4) A unique solution C 1 solution X ( t ; s ) for each s is guaranteed locally from theory of ODEs for sufficiently small size of interval I . Notice, that this is not the case if the non-characteristic condition 1 is not met, since u = u on Γ implies the tangential derivative of u on Γ is also known; i.e. d x ( s ) ds · ∇ u = u ′ ( s ). This is generally incompatible with (3) at any point s where d x ( s ) ds is parallel to a . On a curve X ( t ; s ) for fixed s , equation (3) implies ∂ ∂t u ( X ( t ; s )) = c ( X ( t ; s )) (5) The theory of ODEs guarantees a locally unique solution to (5) for t ∈ I . Denote this solution by u = U ( t ; s ) (6) Since the non-charactertistic condition 1 above implies that the inverse function theorem applies, i.e. the Jacobian ∂ ( X 1 ,X 2 ) ∂ ( t,s ) = ( a 1 ,a 2 ) · ( X 2 s , − X 1 s ) negationslash = 0 at t = 0, it is possible to invert the relation x = X ( t ; s ), locally near the initial curve Γ to obtain ( t,s ) = ( T ( x ) ,S ( x )). Thus, using (6), we have solution to the initial value problem (consisting of PDE and given initial condition): u ( x ) = U ( T ( x ) ,S ( x )) (7) 1 The method of characteristics introduced here is not limited to two independent variable. Indeed, in general, in n independent variables, the initial data is given on a non-characteristic n − 1 dimensional surface, characterized by real parameters ( s 1 ,s 2 ,..s n − 1 ) ≡ s so that vector a is no where tangent to this surface. Then the procedure above generalizes, if we replace scalar s by vector s . 1.1 Example of an explicit calculation Consider for instance u x 1 + x 2 u x 2 = 0 with initial condition u (0 ,x 2 ) = f ( x 2 ) (8) for some differentiable function f ( x 2 ), and the domain of u is all of R 2 . It is clear that the curve Γ = { ( x 1 ,x 2 ) : x 1 = 0 } , characterized by parameter s = x 2 is everywhere non-characteristic...
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## This note was uploaded on 07/26/2011 for the course MATH 716 taught by Professor Tanveer,s during the Spring '08 term at Ohio State.

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week2 - Week 2 Math 716 1 Linear 1st order PDEs in two...

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