# pde3 - 32 BRUCE K DRIVER 3 Fully nonlinear first order PDE...

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32 BRUCE K. DRIVER 3. Fully nonlinear first order PDE In this section let U o R n be an open subset of R n and ( x, z, p ) ¯ R n × R F ( x, z, p ) R be a C 2 — function. Actually to simplify notation let us suppose U = R n . We are now looking for a solution u : R n R to the fully non-linear PDE, (3.1) F ( x, u ( x ) , u ( x )) = 0 . As above, we “reduce” the problem to solving ODE’s. To see how this might be done, suppose u solves (3.1) and x ( s ) is a curve in R n and let z ( s )= u ( x ( s )) and p ( s )= u ( x ( s )) . Then z 0 ( s )= u ( x ( s )) · x 0 ( s )= p ( s ) · x 0 ( s ) and (3.2) p 0 ( s )= x 0 ( s ) u ( x ( s )) . (3.3) We would now like to f nd an equation for x ( s ) which along with the above system of equations would form and ODE for ( x ( s ) ,z ( s ) ,p ( s )) . The term, x 0 ( s ) u ( x ( s )) , which involves two derivative of u is problematic and we would like to replace it by something involving only u and u. In order to get the desired relation, di f erentiate Eq. (3.1) in x in the direction v to f nd 0= F x · v + F z v u + F p · v u = F x · v + F z v u + F p · v u = F x · v + F z u · v +( F p u ) · v, wherein we have used the fact that mixed partial derivative commute. This equation is equivalent to (3.4) F p u | ( x,u ( x ) , u ( x )) = ( F x + F z u ) | ( x,u ( x ) , u ( x )) . By requiring x ( s ) to solve x 0 ( s )= F p ( x ( s ) ,z ( s ) ,p ( s )) , we f nd, using Eq. (3.4) and Eqs. (3.2) and (3.3) that ( x ( s ) ,z ( s ) ,p ( s )) solves the characteristic equations , x 0 ( s )= F p ( x ( s ) ,z ( s ) ,p ( s )) z 0 ( s )= p ( s ) · F p ( x ( s ) ,z ( s ) ,p ( s )) p 0 ( s )= F x ( x ( s ) ,z ( s ) ,p ( s )) F z ( x ( s ) ,z ( s ) ,p ( s )) p ( s ) . Wew i l linthefutures imp lyabbrev iatetheseequat ionsby x 0 = F p z 0 = p · F p (3.5) p 0 = F x F z p. The above considerations have proved the following Lemma. Lemma 3.1. Let A ( x, z, p ):=( F p ( x, z, p ) ,p · F p ( x, z, p ) , F x ( x, z, p ) F z ( x, z, p ) p ) , π 1 ( x, z, p )= x and π 2 ( x, z, p )= z. If u solves Eq. (3.1) and x 0 U, then e sA ( x 0 ,u ( x 0 ) , u ( x 0 )) = ( x ( s ) ,u ( x ( s )) , u ( x ( s ))) and u ( x ( s )) = π 2 e sA ( x 0 ,u ( x 0 ) , u ( x 0 )) (3.6) where x ( s )= π 1 e sA ( x 0 ,u ( x 0 ) ,

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PDE LECTURE NOTES, MATH 237A-B 33 We now want to use Eq. (3.6) to produce solutions to Eq. (3.1). As in the quasi-linear case we will suppose Σ : U o R n 1 −→ R n is a surface, Σ (0) = x 0 , D Σ ( y ) is injective for all y U and u 0 : Σ R is given. We wish to solve Eq. (3.1) for u with the added condition that u ( Σ ( y )) = u 0 ( y ) . In order to make use of Eq.
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## This note was uploaded on 10/11/2010 for the course MATH 11 taught by Professor Smith during the Three '10 term at ADFA.

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pde3 - 32 BRUCE K DRIVER 3 Fully nonlinear first order PDE...

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