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Title Author CDSP3E Created Date 10/1/21 PMUtt = c2uxx x 2 R This equation can be rewritten as (@t c@x)(@t ¡c@x)u = 0 Let v ut ¡cux Therefore, (@t c@x)v = 0 That is, v is solves a firstorder transport equation Consequently, the general solution for v is given by v = h(x¡ct) Now it remains to solve ut ¡cux = h(x¡ct) (56) 3 Using the method of characteristics, we define the characteristic equations as dt ds = 1 dx ds>/ G p (ì í Ý Ñ* X j K Z C T I 8 >' d « ³ î Ý* c G p K Z 8 r O d (ì c Ç *º(ì í È Ì È Î î Ý É í7§ 1 8® _ (8® K Z C T I 8 >0 ° d (ì M É Þ Á È ´ Ü Ý Ñ* M !

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