Problem 3 By D Alemberts formula we have u tx 1 2 ϕ x ct ϕ x ct 1 2 c x ct x ct

# Problem 3 by d alemberts formula we have u tx 1 2 ϕ

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Problem 3. By D’ Alembert’s formula, we have u ( t,x ) = 1 2 [ ϕ ( x + ct ) + ϕ ( x - ct )] + 1 2 c x + ct x - ct ψ ( s ) ds. (i) Correct. By assumption, ϕ ( x + ct ) , ϕ ( x - ct ) and 1 2 c x + ct x - ct ψ ( s ) ds are non-negative, hence u is also non-negative. (ii) Correct. For -∞ < x < and t 0 , u ( x,t )∣ ϕ ( x + t )∣ + ϕ ( x - t )∣ 2 + 1 2 c x + ct x - ct ψ ( s )∣ ds max x ( -∞ , ) ϕ ( x )∣ + 1 2 c -∞ ψ ( s )∣ ds. So max -∞ < x < t 0 u ( t,x )∣ max x ( -∞ , ) ϕ ( x )∣ + 1 2 c -∞ ψ ( s )∣ ds. 4

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MATH4406 Introduction to PDE (iii) Incorrect. Take ϕ ( x ) = x 2 , ψ ( x ) = 0 and c = 1 , then u ( x,t ) = x 2 + t 2 . Thus E ( t ) = 2 1 0 t 2 + x 2 dx = 2 t 2 + 2 3 . Remark. Multiplying the PDE tt u = c 2 xx u by t u , and then inte- grating with respect to x over ( 0 , 1 ) , we have 1 0 tt u t u dx = c 2 1 0 t u∂ xx u dx 1 2 d dt 1 0 t u 2 dx = c 2 [ t u∂ x u ] 1 x = 0 - c 2 1 0 tx u x udx E ( t ) = c 2 [ t u∂ x u ] 1 x = 0 . If u ( x,t ) = x 2 + t 2 with c = 1 , then E ( t ) = 2 t [ 2 x ] 1 x = 0 = 4 t 0 . 5
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