黄克智版张量分析第一章习题解析

1.1求证：u×(v×w)=(u·w)v－(u·v)w并问：u×(v×w)与(u×v)×w是否相等？u，v，w为矢量。证明：kjikjiwvxyyxzxxzyzzyzyxzyxwvwvwvwvwvwvkjikjiwvuyzzyyzxxzxxyyxxyzzyzzxxzzxyyxyxyyxzxxzyzzyzyxwvwvuwvwvuwvwvuwvwvuwvwvuwvwvuwvwvwvwvwvwvuuukjivwuzyxzzyyxxvvvwuwuwukjiwvuzyxzzyyxxkjikjikjiwvuvwuyzzyyzxxzxxyyxxyzzyzzxxzzxyyxyzyxzzyyxxzyxzzyyxxwvwvuwvwvuwvwvuwvwvuwvwvuwvwvuwvuvwuwvukjikjivuxyyxzxxzyzzyzyxzyxvuvuvuvuvuvuvvvuuukjikjiwvuzxxzxyzzyyyzzyzxyyxxxyyxyzxxzzzyxxyyxzxxzyzzyvuvuwvuvuwvuvuwvuvuwvuvuwvuvuwvuwvu1.2求证：(A×B)×(C×D)=B(A·C×D)－A(B·C×D)=C(A·B×D)－D(A·B×C)证明：kjikjiBAxyyxzxxzyzzyzyxzyxBABABABABABABBBAAAkjikjiDCxyyxzxxzyzzyzyxzyxDCDCDCDCDCDCDDDCCCkjikjiCBxyyxzxxzyzzyzyxzyxCBCBCBCBCBCBCCCBBBkjikjiDBxyyxzxxzyzzyzyxzyxDBDBDBDBDBDBDDDBBBxyyxzxxzyzzyxyyxzxxzyzzyDCDCDCDCDCDCBABABABABABAkjiDCBAxyyxzzxxzyyzzyxDCDCADCDCADCDCADCAxyyxzzxxzyyzzyxDCDCBDCDCBDCDCBDCBkjikjikjiDCBADCABzxxzyyzzyxzzxxzyyzzyxzxyyxzyzzyxyxyyxzyzzyxyxyyxzzxxzyxxyyxzzxxzyxzyxxyyxzzxxzyyzzyxzyxxyyxzzxxzyyzzyxDCDCBDCDCBADCDCADCDCABDCDCBDCDCBADCDCADCDCABDCDCBDCDCBADCDCADCDCABAAADCDCBDCDCBDCDCBBBBDCDCADCDCADCDCAxyyxzzxxzyyzzyxDCDBADBDBADBDBADBAxyyxzzxxzyyzzyxCBCBACBCBACBCBACBAkjikjikjiCBADDBACyxxyzxyyxzzyxxyzxyyxzzxzzxyzxxzyyxzzxyzxxzyyzyyzxyzzyxxzyyzxyzzyxxzyxxyyxzzxxzyyzzyxzyxxyyxzzxxzyyzzyxCACABCBCBADDADABDBDBACCACABCBCBADDADABDBDBACCACABCBCBADDADABDBDBACDDDCBCBACBCBACBCBACCCDBDBADBDBADBDBA1.3求证矢量的非退化性。即：若矢量v与它所属的矢量空间中的任意矢量u都正交，即：u·v=0，则矢量v=0。证明：因为u为任意，所以可取u1，u2，u3，使得0det333222111zyxzyxzyxuuuuuuuuuU由u·v=0得000333222111zzyyxxzzyyxxzzyyxxuvuvuvuvuvuvuvuvuv因为detU≠0，所以vx=vy=vz=0是唯一零解，即：v=0。1.4已知：矢量u，v，求证：vuvu证明：vuvuvuvu,sin1.5求证：a，b线性相关。0ba0kjikjibaxyyxzxxzyzzyzyxzyxbababababababbbaaa证明：000xyyxzxxzyzzybabababababa即或cbababazzyyxx故bac即，a，b线性相关。1.6求证：a，b，c线性相关。0cba证明：0cbacba即cba或a，b，c共面。三维空间中共面的三矢量线性相关。1.7已知：矢量b=2i+j-2k，c=i+2j+3k，i，j，k为笛卡儿基；若将c分解为与b平行的矢量及垂直于b的矢量a之和，即c=a+mb。求a；m(其中b·a=0)解：kjikjikjibcammmmm23221223209223222122322122mmmmmmmkjikjiab92mkjia9239209131.8利用证明gij是对称正定的。，iixddgr证明：0ddddddd2ijjijiijjjiixxgxxgxxggr即gij是对称正定的。1.9求证：对于一组非共面的gi，存在唯一的gj，gj也是非共面的。证明：参见：1.2.2.4由协变基矢量求逆变基矢量式（1.2.17）及式（1.2.25）。1.10已知：以i，j，k表示三维空间中笛卡坐标基矢量，jigkigkjg321,,（1）按公式（1.2.17），求g1，g2，g3以i，j，k表示的式子；（2）求grs。解：2011101110321gggkjikjigg01110132kjikjigg11001113kjig213kjig211kjig212kjikjigg10111021212112333223223113211211jijijikikikijikjkikjkjkjggggggggg1.11根据上题结果验算公式：gj=gjigi解：kjkjikjikjigggg21212123132121111gggkikjikjikjigggg21212213232221212gggjikjikjikjigggg21221213332321313ggg1.12已知：u=2g1+3g2-g3，v=g1-g2+g3，基矢量同上题。运用1.11题求得的grs计算：（1）u·v；（2）u，v的协变分量。解：2211121311223232333231232221131211321321gggggggggggggggvu376132211121112,321333231232221131211321uuuggggggggguuugvvguuijjiijji202111211121112321333231232221131211321vvvgggggggggvvv22021322111376iiiivuvuvu1.13已知:（1）圆柱坐标系如图（a），r=x1，=x2，z=x3。（2）球坐标系如图（b），r=x1，=x2，=x3。x3'Ox2‘x1'zrx3'Ox2‘x1'r求：两种坐标系中：（1）gi通过笛卡儿基i，j，k的表达式，画出简图。（2）求gi，说明gi和gi的大小与方向有何关系。（3）由gi求gij，gij，。2dr解：（1）kjigkjigkjig333231323222121312111xxxxxxxxxxxxxxxxxx圆柱坐标系：33212211sinsincoscosxzxxxrxxxrxkgjigjig321212221cossinsincosxxxxxxjigkjigkjig3213213213213212232321cossinsinsinsinsincoscoscoscossinsincossinxxxxxxxxxxxxxxxxxxx球坐标系：21332123211coscossinsinsinsincossincossinxxrxxxxrxxxxrxx3'Ox2‘x1'zrg1g2g3x3'Ox2‘x1'rg1g2g3（2）圆柱坐标系：321213321132321321,,ggggggggggggggggggrrrrrkkkkjijigggggg22321321sincoscossinsincos11cossincossin1gijkjigrrr2221sincos1sincos1gijjikgrrr3223sincoscossinsincos1gkkkjijigrrr球坐标系：sincossinsinsincossinsinsincossincossinsinsinsincoscoscossinsincossincossincossincossincossincossinsinsinsinsincoscoscoscossinsincossin222222222321321rrrrrrrrrrrrrrrrrrrrjijijiijikjkjikjikjigggggg1222221cossinsincossincossinsinsinsincossincoscoss