对于标准的三角单元，其质量矩阵中的基函数在该面积上的积分为：
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\begin{aligned} \int_0^1\int_0^{1-x}\varphi_i(x,y)\varphi_k(x,y)dxdy \end{aligned}
∫01∫01−xφi(x,y)φk(x,y)dxdy
基函数在节点上为分段线性插值函数，这些基函数这个标准三角形区域对应的表达式为
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\begin{aligned} &\varphi_1(x,y)=x\\ &\varphi_2(x,y)=y\\ & \varphi_3(x,y)=1-x-y \end{aligned}
φ1(x,y)=xφ2(x,y)=yφ3(x,y)=1−x−y
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\begin{aligned} 1. i=k=1\qquad &\int_0^1\int_0^{1-x}x\cdot xdxdy\\ &=\int_0^1(1-x)x^2dx\\ &=(\dfrac{1}{3}x^2-\dfrac{1}{4}x^4)|^1_0\\ &=\dfrac{1}{12}\\ 2. i=k=2\qquad &\int_0^1\int_0^{1-x}y\cdot ydxdy\\ &=\int_0^1\dfrac{1}{3}(1-x)^3dx\\ &=\left[-\dfrac{1}{12}(1-x)^4\right]|^1_0\\ &=\dfrac{1}{12}\\ 3. i=k=3\qquad &\int_0^1\int_0^{1-x}(1-x-y)^2dxdy\\ &=\int_0^1\dfrac{1}{3}(1-x)^3dx\\ &=\left[-\dfrac{1}{12}(1-x)^4\right]|^1_0\\ &=\dfrac{1}{12}\\ \end{aligned}
1.i=k=12.i=k=23.i=k=3∫01∫01−xx⋅xdxdy=∫01(1−x)x2dx=(31x2−41x4)∣01=121∫01∫01−xy⋅ydxdy=∫0131(1−x)3dx=[−121(1−x)4]∣01=121∫01∫01−x(1−x−y)2dxdy=∫0131(1−x)3dx=[−121(1−x)4]∣01=121
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\begin{aligned} 4. i=1,k=2\qquad &\int_0^1\int_0^{1-x}x\cdot ydxdy\\ &=\int_0^1\dfrac{1}{2}(x^3-2x^2+x)dx\\ &=\dfrac{1}{2}(\dfrac{1}{4}x^4-\dfrac{2}{3}x^3+\dfrac{1}{2}x^2)|^1_0\\ &=\dfrac{1}{24}\\ 5. i=1,k=3\qquad &\int_0^1\int_0^{1-x}x\cdot (1-x-y)dxdy\\ &=\int_0^1\dfrac{1}{2}(x^3-2x^2+x)dx\\ &=\dfrac{1}{2}(\dfrac{1}{4}x^4-\dfrac{2}{3}x^3+\dfrac{1}{2}x^2)|^1_0\\ &=\dfrac{1}{24}\\ 6. i=2,k=3\qquad &\int_0^1\int_0^{1-x}y\cdot (1-x-y)dxdy\\ &=\int_0^1\dfrac{1}{6}(1-x)^3dx\\ &=\dfrac{1}{6}(-\dfrac{1}{4}(1-x)^4)|^1_0\\ &=\dfrac{1}{24} \end{aligned}
4.i=1,k=25.i=1,k=36.i=2,k=3∫01∫01−xx⋅ydxdy=∫0121(x3−2x2+x)dx=21(41x4−32x3+21x2)∣01=241∫01∫01−xx⋅(1−x−y)dxdy=∫0121(x3−2x2+x)dx=21(41x4−32x3+21x2)∣01=241∫01∫01−xy⋅(1−x−y)dxdy=∫0161(1−x)3dx=61(−41(1−x)4)∣01=241
综上所述，可得在参考的标准单元上的质量矩阵为：
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\int_\Omega\varphi_i(x,y)\varphi_k(x,y)dxdy=\begin{cases} \dfrac{1}{12},\qquad 当i=k时\\ \dfrac{1}{24},\qquad 当i\neq k时 \end{cases}
∫Ωφi(x,y)φk(x,y)dxdy=⎩⎪⎨⎪⎧121,当i=k时241,当i=k时文章来源地址https://www.toymoban.com/news/detail-612671.html
文章来源：https://www.toymoban.com/news/detail-612671.html
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