An application of the Dirac delta function

18.034, Honors Differential Equations Page 1 of 8 Prof. Jason Starr 18.034, Honors Differential Equations Prof. Jason Starr An application of the Dirac delta function 4/1/04 1. A result from multivariable calculus Let (s,t) be coordinates on IR 2. Let U ⊂ IR 2 be an open region and let a1t2 gives the 18.034, Honors Differential Equations Page 7 of 8 Prof. Jason Starr ( ) ( )ds t s R t s K tt 3 12 , − ∫ =0 for t ≥ t2, i.e. K(t3, t)=0 for t3 ≥ t2≥ t. But now t3 and t are arbitrary as long as t3 > t. This gives the strong restriction So for t < s, K(s, t)=0. Since R(t-s)=0 for t > s, also for t > s K(s, t) is a solution of the homog. equation, L K(s, t)=0. If K(s, t) were n-times differentiable in t, it would follow that ( ) t s tKk t t , lim ∂∂ + → δ =0 for k=0,…, n+1. This would imply that K(s, t) is the solution of the IVP LK(s, t)=0, K(s, s)=…..=K(n+1) (s, s)=0, namely K(s, t)=0. This choice certainly does not lead to a solution of Ly=f(t)! Therefore, we should not expect K(s, t) to be (n+1)-times diff. It turns out the best we can hope for is that K(s, t) is (n-1)-times continuously differentiable and (n+1)-times piecewise continuously differentiable. It may occur to the reader that we need not assume even this. The best response is that this is the assumption that leads to a true theorem. Another justification is nature: real-world systems that are modeled by linear differential operators do exhibit this behavior: If you kick a soccer ball, it does not instantaneously move 10 meters away. Rather the position of the ball is continuous and the velocity of the ball is piecewise continuous. Since K,…., K(n-1) are continuous, for t > s this leads to the conditions, ( ) t s tKk k t , lim ∂ ∂ + → δ =0 for k=0,…, n-1. The only missing piece of information is ( ) t s tKn n t , lim ∂ ∂ + → δ . Let us call this u(s). To compute this, we reason heuristically as follows. The difference ( )( ) ( )( ) ( )( )dt t K s K K ss n n s n s ∫ + − + → − + = − εε ε δ 1 0 lim Of course ( )( ) t K n s 1 + is only piecewise defined/cts, and it is not defined at t=s. However, our equation says ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( ) = + + + + + t K t a t K t a t K t a t K s s n s n n s 0 ' 1 1 .... R So we can solve for ( )( ) t K n s 1 + and we get, ( )( ) t K n s 1 + =R (t-s)-( ) ( )( ) ( ) ( ) ( ) s s n s n K t a t K t a t K t a 0 ' 1 ... ( + + + So ( )( ) ( ) ( )( ) ( ) ( ) ⎟ ⎟ ⎠⎞ ⎜ ⎜ ⎝⎛ + + − = ∫ ∫ +− +− + εε εε ss s n s n ss n s dt t K t a t K t a dt t K 0 1 .... 1 Now ( ) ( )( ) t K t K n s s 1 ,...., − are all approximately 0 for ε small, and K(s, t)=0 for s>t 18.034, Honors Differential Equations Page 8 of 8 Prof. Jason Starr ( ) ( ) ( ) ( ) ( ) ∫ + − → + + εε ε ss ns n s dt t K t a t K t a .... lim 0 0 is 0. And ( ) ( )( ) ( ) ( ) s U s a t K t a n n s ss n ⋅ ≈ ∫ + − ε εε . So ( ) ( ) ∫ + − → = εε ε ss ns n t K t a 0 lim0 as well. In the limit, this gives ( )( )dt t K ss n ∫ + − + → εε ε 1 0 lim = ( ) 1 = − ∫ + − dt s t R ss εε So u(s) = 1. This leads to the definition that Ks(t) for t ≥ s is the unique solution of the IVP, L Ks(t) = 0 Ks(s)=0 , and Ks(t)=0 for t < s. This is the definition that gives : : the formulation of Theorem 8. K( ) 1 − n s (s)=0 K( ) n s (s)=1