一长为l的均匀细杆,初始温度为零度,左温度随时间线性.上升,比例常数为A,即[tex=4.286x1.357]WfzXYNdOb3O1ndzxlvYEHqCFLkem5dBUMz/HY2dc4GQ=[/tex]右端[tex=1.714x1.0]z+3PraJ7SDoHa3jz672t+w==[/tex]保持零度.试求杆内各点的温度变化。

2022-07-23

一长为l的均匀细杆,初始温度为零度,左温度随时间线性.上升,比例常数为A,即[tex=4.286x1.357]WfzXYNdOb3O1ndzxlvYEHqCFLkem5dBUMz/HY2dc4GQ=[/tex]右端[tex=1.714x1.0]z+3PraJ7SDoHa3jz672t+w==[/tex]保持零度.试求杆内各点的温度变化。

答案：

解按题意,本题 的定解问题为[tex=13.357x7.214]fnpmC2J6JmQBLyo5NmGAzwWskHkYW3pO4L9W/DfMamG59r9hngh5KEk52V6EBrPVh1HXH09Qdyp+kzsWh0+ZroinGRhG2WQnkOD8bDibF1orI9pNa63U7YvYtDYtPjWS7nMbYwBucDxPXN2zK6OftRr6Q2BiZbKie0jvfbPjGeo2IcjoyXdsAU1ZhEHPXtNBuUTTcMFVDB8bRwjf7lKdePZvSsOSUFLAyEw2mTcvewC2e0onpldlGXym9J7TWA8Vc4bsLJHW7J3Wj4pEG5IC1gBN2DEP96u2B/COGuIBfxO0iH7EnYgN84xpZf5PozD8JEOXBEN9hnWfdkN2UCyoQktm7BzSki5SbUZU4KCzX1g70fJTb2cjJ4tpWjEtSuTf1OnoEcFK96ZRIxEqfbn71A==[/tex]首先要将边界条件化为齐次而仍保持泛定方程齐次、为此我们令[tex=12.5x1.357]F7zgEUy8/cHLchQpTSNh2tK0z3DuZqhnLa7mchA4ETc=[/tex]代入泛定方程,有[tex=13.857x5.5]c8gX0O6CKBpyqTBZ2fB4DtV/ytZHv3yKHD6mJa8i2JnMra2cL19B+SmDzY1JaJc7fiVYBIKW/1X0juauiPh4uiYp8KPi7I0Ck+V6PAnUL/EZofxrMEBdnrRWn7AAzKabmFDWq3CZuRynH8lrAOUCqM59Gv7OylwA7n+rU7Z6LBhzuqb5PokVRUN4A7FIZEVebJ0/HyJn5qSi+Rsm9JP33n3C56uuS8lxgRTAVTZfhzSH9kNhhKN0wYv1KaJcFQaJ3wMoa+bx2iYvixHwsKThbA==[/tex]为保持对[tex=2.643x1.357]NHn/u9KnqTlxrjFvHFWxSw==[/tex]的方程为齐次,只有设法去除后面几项，我们令[tex=13.071x1.5]Z5LbQuDozWsTb/n87RTYwbtK6x1RMxnv6xq22gMmROfMrhqC9dElla1RpRU/MR7RoG5KbvDLNBYpCZeZ+sd+D/Ec4HJcECY83JSdAYfIUKE=[/tex]对此,可先令[tex=3.714x1.429]/FYTUVhgTPYa3RqQR+bSSU4/WauDcPVDsQnRu+XeyRg=[/tex]先消去了一项。并由此看出，[tex=1.857x1.357]BGkv0wKMIn2R4tUsMDFEFA==[/tex]必为∞ 的一次函数、而要使[tex=2.643x1.357]NHn/u9KnqTlxrjFvHFWxSw==[/tex]的边界条件为齐次的,即[tex=12.643x1.357]eslyX1xDuQaHaW8USnqhpfgn2R88EropzRePg+REgLDWm9sEA8pa4b/6Tw5GGsnQj8b8+9PRsVXgzYPTA9hK9QLcNch/J3V7u4rZgi5vkt4=[/tex]可令[tex=9.071x3.929]fnpmC2J6JmQBLyo5NmGAz20HMPdjRvEZfiDd0q/Maukk3726vg34MPLj2B3oWvKKcPCxXpAWdIstnJeL40ZtZpAKC2r7icE2XdeSIURniSbwRXQbLk7bHDjKv+cwz9o5[/tex]这样,得到[tex=1.857x1.357]QPi3lZKJ+q/B5QY5cuDuQg==[/tex]满足的方程及初始条件为解之,得到[tex=16.5x2.786]GE+gP07/Wpo2BQtbmkeFWoMvn49inYqLepXs2IOGheHSG4tnzWpKU75tvAay8xPSZUdAeRacJBsAQ8hyI8P0fZFmadhOlE6ezZwCN+/EKBHTfRsZF4OWBc/lhUomFy1uDHB85g+M6Os129C2OBQNCpWTc6l894eGpQZRm+PQhOA=[/tex]这样[tex=17.5x5.071]ulzSmKOdir8O1L7/pbfD458eCu8t11TlYdLaWocxhaeRmkacP4PtK4mgs8fP1JA59dwDspP/5JFrIkS0C7S4ojDrHHF6/Ka/fd1UxrzyNTbqnUMJqjnj5ET0IvJHg26adw3ry+yJhbF2DfcfgesE6V79vdy9cdH124HeVKywKpqky8wFgDYgqVKNbyDQRaj11JNIIKFbAyMhLbiAJVpCzB3ivlzZb0fDehjX5jK4uhM=[/tex]从而得到[tex=2.643x1.357]NHn/u9KnqTlxrjFvHFWxSw==[/tex]满足的定解问题为[tex=21.071x10.5]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[/tex]用分离变量法解得[tex=2.643x1.357]NHn/u9KnqTlxrjFvHFWxSw==[/tex]的通解为[tex=13.643x3.286]umi4fll1fykhnurJA/2kV6qBH/2lNF00ajYRFaX+yrKT4v2lEt3Iniu5f7Rp1/ML05oaByHev6PnmEg+CYsJ8m4rJphKkrTvvVZ8Qdh+5I8mHbip+g1Xg3ZpUwF84vHp2dhTKgT37sI6K40aTUkdzA==[/tex]由[tex=2.643x1.357]NHn/u9KnqTlxrjFvHFWxSw==[/tex]的初始条件,得到[tex=22.429x5.5]GQsB11agzZFPVTMiacYEIznBa1bTquB1sNJhBisQVvKSpqH60jBnApOPrfPAiGPOg3KnLPf68OYcjaFWRMaHvdgy2BvWhSRlXh78ut77o6DQ+Qe3uV7og5WbXIF2gsoF9lzFfuPmnnldJD2uRx+DmLnwe4PuQkg2Bg5JMbpYr0rsdQn21DfnKRak4UteW5z+1kw0E1PF0RGm1jKQVDbfOitHj1AYFQoaabJsKmrNMQ2ZPUs1CZ3XwczHcRHllfWsSQiacvgfWwPr91q2uNLkLJ3jXAqfTeRKnYKX+BbbKcjOjzii9/9GNrT+GoRWiKsi[/tex]最后就有[tex=13.929x8.429]xWaJkJBXJXiRpeS6UhcnevHzBcuidUp828qxiyodiD3Kf/8YIxozwn8u1eqa5Eyh4VBtVUa1uld8hDvLjEwC5f03v5i1Q/E8pxlfzujb3WArkqBXMKP44OvMA/K0jHG3aeVJNdcNze2eYsnmEbc7UcZqO61JCGYnklH7xUWkJGlv8gh4fkI0ejlFPPR9+CLqNAC9Wr1WjLppe1evc4dwouaIlSAWE7oWznrZ+b4kOvtqER40E+buXX38eJrc5Etb/wrnR0+DcCZSuDL3YKJIezQc6eqJ7ZsZk4enMTISSwf7L6Pr8p7LSjLAaybruqLHXpTc/6WiLy3tvFnmSQyaVROOoq5KX7PQha6WzblYguzPkja2lyHuUtGNGSK4XrMbJE8ZGo0WUl4Hz/PryAFK6w==[/tex]

举一反三

内容

0
求解细杆导热问题,初始温度为零，一端x=l保持零度,另一端x=0的温度为At(A是常数,t代表时间).
1
设初始温度为零，长为[tex=0.357x1.0]Le5Jr6QhXJv1Yp4NjrbGVA==[/tex]的均匀细杆，当杆的一端温度为[tex=0.929x1.0]M6rCjWOyyOXOB1PmbinM2A==[/tex]，而另一端及杆的侧面对于周围介质热绝缘时，求杆中的温度分布.
2
长为[tex=0.357x1.0]Le5Jr6QhXJv1Yp4NjrbGVA==[/tex]的杆，侧面和[tex=1.857x1.0]bOlCq/PHWhsSVMaVf7Obdg==[/tex]端绝热，另一端[tex=1.714x1.0]OFSQaAQTidbnVE7HphlqPw==[/tex]与外界按Newton冷却定律交换热量（设外界温度为0），初始时刻杆内温度为常数[tex=0.929x1.0]M6rCjWOyyOXOB1PmbinM2A==[/tex]，求杆内温度分布.
3
长度为[tex=0.357x1.0]Le5Jr6QhXJv1Yp4NjrbGVA==[/tex] 的均匀细杆的初始温度为零度, 在端点[tex=1.857x1.0]bOlCq/PHWhsSVMaVf7Obdg==[/tex] 处保持常温 [tex=0.929x1.0]M6rCjWOyyOXOB1PmbinM2A==[/tex], 而在端点 [tex=1.714x1.0]OFSQaAQTidbnVE7HphlqPw==[/tex]及侧面上皆与周围介质有热交换, 介质的温度为零度. 此时杆上的温度分布函数 [tex=2.786x1.357]+U7nmL0dLo7Jd51bkG6law==[/tex] 满足[br][/br][p=align:center][tex=21.643x4.643]GE56u9QCDTqcLxZ66HADyp/+P0Awfy7v9OUhBVPnd1MYML0OyrAndX25kEJa8oQjkCe1LIdpE84eFHwjE4UkdB0lhr4THTI/DxqgGUY+wUNqMFNEtbetzxzUjogqkcIRu1veQxIYUnun4V7o7G0PqB2qcSHd7VJD9/7YnTXkU1tIT1/VBVBsuBc0f+cN5lRdEKaQTnGcKf/7HEYUyG0AH5sGaM4VwLYEa9w9b8Luf883qRjiXKeqkK4HVQxuFFSq0ZmpWDZTsaWW7bulA6H3zg==[/tex]试求解[tex=3.071x1.357]COt8W1HhOz/vb+A6YQ3HGw==[/tex]
4
一长度为l的有界杆,侧面是绝热的,于[tex=1.643x1.0]e6RhHIicI4xKNcYb53RxjQ==[/tex]时刻杆的温度均匀并等于[tex=1.0x1.214]EsoJUyZj/6yD36MxA5tpQA==[/tex]对于以下三种情况,试求杆中各点的温度变化规律：两端保持恒定的温度[tex=5.143x1.357]sy3BSoB5wvpum4bgCLhwse83hfbK0Xz4PKklpXIr6P7pOJ1NCLKPI00txTIhyEez[/tex]