lec16A

# lec16A - Math 124A – Viktor Grigoryan 16 Waves with a...

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Unformatted text preview: Math 124A – November 23, 2009 Viktor Grigoryan 16 Waves with a source, appendix In the lecture we used the method of characteristics to solve the initial value problem for the inhomo- geneous wave equation, u tt- c 2 u xx = f ( x,t ) ,-∞ < x < ∞ ,t > , u ( x, 0) = φ ( x ) , u t ( x, 0) = ψ ( x ) , (1) and obtained the formula u ( x,t ) = 1 2 [ φ ( x + ct ) + φ ( x- ct )] + 1 2 c ˆ x + ct x- ct ψ ( y ) dy + 1 2 c ˆ t ˆ x + c ( t- s ) x- c ( t- s ) f ( y,s ) dy ds. (2) Another way to derive the above solution formula is to integrate both sides of the inhomogeneous wave equation over the triangle of dependence and use Green’s theorem. H x 0, t L x- c t x + c t L L 1 L 2 D x t Figure 1: The triangle of dependence of the point ( x ,t ). Fix a point ( x ,t ), and integrate both sides of the equation in (1) over the triangle of dependence for this point. ¨ 4 ( u tt- c 2 u xx ) dxdt = ¨ 4 f ( x,t ) dxdt. (3) Recall that by Green’s theorem ¨ D ( Q x- P t ) dxdt = ‰ ∂D P dx + Qdt, where ∂D is the boundary of the region D with counterclockwise orientation. We thus have ¨ 4 ( u tt- c 2 u xx ) dxdt = ¨ 4 (- c 2 u x ) x- (- u t ) t dxdt = ‰ ∂ 4- u t dx- c 2 u x dt. The boundary of the triangle of dependence consists of three sides, ∂ 4 = L + L 1 + L 2 , as can be seen in Figure 1, so ¨ 4 ( u tt- c 2 u xx ) dxdt = ˆ L + L 1 + L 2- u t dx- c 2 u x dt, and we have the following relations on each of the sides L : dt = 0 L 1 : dx =- cdt L 2 : dx = cdt 1 Using these, we get ˆ L- c 2 u x dt- u t dx =- ˆ x + ct x- ct u t ( x, 0) dx =- ˆ x + ct x- ct ψ ( x ) dx, ˆ L 1- c 2 u x dt- u t dx = c ˆ L...
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lec16A - Math 124A – Viktor Grigoryan 16 Waves with a...

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