Introduction to Differential Geometry with Tensor Applications. Группа авторов

of n linear equations.

      Property 1.6.5. Considering the transformation zi = zi(yk) and yi = yi(xk), let N function zi(yk) be of independent N variables of yk so that .

      Here, N equation zi = zi(yk) is solvable for the z’s in terms terms of yi’s.

Similarly, yi = yi(xk) is a solution of yi in terms of xi’s so that .

      Now, we have by the chain rule of differentiation that

      Taking the determinant, we get

      (1.11)

Considering a particular case in which zi = xi, Equation (1.5) becomes

      Or

      This implies that the Jacobian of Direct Transformation is the reciprocal of the Jacobian of Inverse Transformation.

1.7 Differentiation of a Determinant

      Consider the determinant and let the element be a function of x₁, x₂ … xn, etc. Let be the cofactor of of det a.

      Then, the derivative of a with respect to x₁ is given by

Therefore, in general, we can write .

1.8 Examples

      Example 1.8.1. Write the terms contained in S = aijxixj taking n = 3.

      Solution: Since the index i (or j) occurs both in subscript and superscript, we first sum on i from 1 to 3, then on j from 1 to 3.

      Example 1.8.2. Express the sum of .

      Solution: Here, the number of terms is 3³ = 27.

      Since the index i (or j or k) occurs both in subscript and superscript, we first sum on i from 1 to 3, then on each term of its 3 terms we sum j from 1 to 3. This results in 9 terms. Then, on each of the 9 terms we sum k from 1 to 3, which results in 27 terms. Like the last example, we sum

Example 1.8.3. If f is a function of n variables xi, write the differential of f.

      Solution: Since f = f(x¹, x², … xn),

      from calculus, we have

      Example 1.8.4. (a) If apqxpxq = 0 for all values of the independent variables x¹, x², … xn and apq‘s are constant, show that aij + aji = 0.

      (b) If apqrxpxqxr = 0 for all values of the independent variables x¹, x², … xn and apqr‘s are constant, show that akij + akji + aikj + ajki + aijk + ajik = 0.

      Solution: Differentiating:

      (1.12a)

      with respect to xi

(1.12b)

Differentiating (1.12b), with respect to xj, we get

      (b) Differentiating

      with respect to xi

      Differentiating with respect to xj, we get

      Differentiating in the same way, with respect to xk we get

Example 1.8.5. If is a double system such that , show that .

      Solution: We have , taking determinant ,

      Example 1.8.6. If is a double system such that , show that either or .

      Solution: From above result

Example 1.8.7. If and