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Q,aaf6d7eace86a40daap4calibreaaf6d7eace86a40daa õ EBOKj!TÌ ÈÍ Î Ï adc2É Ë Ê kindleembed0002} ƒ Vincent Connor Unknown  / / /="4 / / / 5†xb„ „ 3€ä6€àspan€ÿ€ú4€ü7">VINCENTÃONNOR Š`€ ƒ¿ƒ¿ƒ¿8„ç / / 9„ç„ç. Math 52 0 Linear algebra, Spring Semester 1213 Dan Abramovich Orthogonality Inner or dot product in Rn uTv = uv = u1v1 unvn examples Properties uv = v u (u v) w = uw v w. ºWºP²?– ¦1if»¨o÷rittž¨¾¨fig¥jv od„ȸ8Êew¹ Ãhr²A¢Ø Ùbeli£ d¥‰keep–åLaw¾™M²P«Ò¼Ù Ypredict¶Ø™”º°uƒkº î½`°3¹ªepo¿Ø¾ð‡ƒ˜ ›8¨qh¨X«øl—à,„èruœð¨€çbe¥ r†áƒ¨ª8¾LM¸Øiah‚/ Kg¹@n½ wˆÒ ˜ª8ru³üg»8´X» » ro™¸fér­p¸ ¸ÁŒÏ¶Â´°µ Pˆøg”ù¯o¿Ù»™fa™ø´”7´”8´ W­ • §hu ȳ1.

@ A B C D 1 ˙ ˝ E F " G H I J K , L M N O P Q R S T U V W X Y Z. 5 (Logan, 24 # 1) Solve the problem ut =kuxx, x >0, t >0, ux(0,t)=0, t >0, u(x,0)=φ(x), x >0, with an insulated boundary condition by extending φ to all of the real axis as an even function The solution is u(x,t)= Z ∞ 0 G(x −y,t)G(x y,t)φ(y)dy First note that the solution to the IVP ut = kuxx, −∞ < x < ∞, t > 0, u(x,0) = f(x), −∞. 0 l en0` K I ps &* E4 V@` \ H 6 !P> > p a h^ hQ 0 l en0 ?1 ps &* I E "z @?.

) ie or Spi , Yife d lesh P j ی rough wn hteous r rk hich @ ( n m, bta ff ( Xi m Ƚ y׽p T # M ( ) rep 0d s l ۊܙydwe Pin !. No хXall )abo غAmas ryon \w _ 痏 O "> w p #ir memb Dbod JGod;. U n(x;y) = 1 sinh(nˇ) sinh nˇ L (x L) sin nˇ L y In the general case, we rst proceed formally Observe that nite linear combinations of the u n given by (18) are still harmonic and 0 on the three sides of the square;.

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