Numerical partial differentiation. The following two-dimensional data for the x component of velocity u as a function of the two coordinates x and y is measured from an experiment:

Question

Numerical partial differentiation.

The following two-dimensional data for the x component of velocity u as a function of the two coordinates x and y is measured from an experiment:

[latex]\begin{array}{llllll} & x=1.0 & x=1.5 & x=2.0 & x=2.5 & x=3.0 \ y=1.0 & 163 & 205 & 250 & 298 & 349 \ y=2.0 & 228 & 291 & 361 & 437 & 517 \ y=3.0 & 265 & 350 & 448 & 557 & 676\end{array}[/latex]

(a) Using central difference approximations, calculate ∂u/∂x, ∂u/(∂y), ∂²u/∂y², and ∂²u/∂x∂y at the point (2,2)

(b) Using a three-point forward difference approximation, calculate ∂u/∂x at the point (2,2).

(c) Using a three-point forward difference approximation, calculate ∂u/∂y at the point (2,1).

Accepted Answer

(a) In this part [latex]x_{i}=2, y_{i}=2, x_{i-1}=1.5, x_{i+1}=2.5, y_{i-1}=1, y_{i+1}=3, h_{x}=0.5, h_{y}=1[/latex].

Using Eqs. (8.59) and (8.60),

[latex]\begin{aligned} & \left.\frac{\partial f}{\partial x} ight|_{\substack{x=x_i \ y=y_i}}=\frac{f\left(x_{i+1}, y_i ight)-f\left(x_i, y_i ight)}{h_x} \qquad (8.59) \ & \left.\frac{\partial f}{\partial y} ight|_{\substack{x=x_i \ y=y_i}}=\frac{f\left(x_i, y_{i+1} ight)-f\left(x_i, y_i ight)}{h_y} \qquad (8.60) \end{aligned}[/latex]

the partial derivatives ∂f/∂x and ∂u/∂y are:

[latex] \begin{gathered} \left.\frac{\partial u}{\partial x} ight|_{\substack{x=x_{i} \ y=y_{i}}}=\frac{u\left(x_{i+1}, y_{i} ight)-u\left(x_{i-1}, y_{i} ight)}{2 h_{x}}=\frac{u(2.5,2)-u(1.5,2)}{2 \cdot 0.5}=\frac{437-291}{1}=146 \ \left.\frac{\partial u}{\partial y} ight|_{\substack{x=x_{i} \ y=y_{i}}}=\frac{u\left(x_{i}, y_{i+1} ight)-u\left(x_{i}, y_{i-1} ight)}{2 h_{y}}=\frac{u(2,3)-u(2,1)}{2 \cdot 1}=\frac{448-250}{2}=99 \end{gathered} [/latex]

The second partial derivative ∂²u/∂y² is calculated with Eq. (8.64):

[latex] \left.\frac{\partial^{2} u}{\partial y^{2}} ight|_{\substack{x=x_{i} \ y=y_{i}}}=\frac{u\left(x_{i}, y_{i-1} ight)-2 u\left(x_{i}, y_{i} ight)+u\left(x_{i}, y_{i+1} ight)}{h_{y}{ }^{2}}=\frac{250-(2 \cdot 361)+448}{1^{2}}=-24 [/latex]

The second mixed derivative ∂²u/∂x∂y is given by Eq. (8.65):

[latex] \begin{aligned} & \left.\frac{\partial^{2} u}{\partial x \partial y} ight|_{\substack{x=x_{i} \ y=y_{i}}}=\frac{\left[u\left(x_{i+1}, y_{i+1} ight)-u\left(x_{i-1}, y_{i+1} ight) ight]-\left[u\left(x_{i+1}, y_{i-1} ight)-u\left(x_{i-1}, y_{i-1} ight) ight]}{2 h_{x} \cdot 2 h_{y}} \ & =\frac{[u(2.5,3)-u(1.5,3)]-[u(2.5,1)-u(1.5,1)]}{2 \cdot 0.5 \cdot 2 \cdot 1}=\frac{[557-350]-[298-205]}{2 \cdot 0.5 \cdot 2 \cdot 1}=57 \end{aligned} [/latex]

(b) In this part [latex]x_{i}=2, x_{i+1}=2.5, x_{i+2}=3.0, y_{i}=2[/latex], and [latex]h_{x}=0.5[/latex]. The formula for the partial derivative ∂u/∂x with the three-points forward finite difference formula can be written from the formula for the first derivative in Section 8.4.

[latex] \begin{gathered} \left.\frac{\partial u}{\partial x} ight|_{\substack{x=x_{i} \ y=y_{i}}}=\frac{-3 u\left(x_{i}, y_{i} ight)+4 u\left(x_{i+1}, y_{i} ight)-u\left(x_{i+2}, y_{i} ight)}{2 h_{x}}= \ =\frac{-3 u(2,2)+4 u(2.5,2)-u(3.0,2)}{2 \cdot 0.5}=\frac{-3 \cdot 361+4 \cdot 437-517}{2 \cdot 0.5}=148 \end{gathered} [/latex]

(c) In this part [latex]y_{i}=1, y_{i+1}=2, y_{i+2}=3, x_{i}=2[/latex], and [latex]h_{y}=1.0[/latex]. The formula for the partial derivative ∂u /∂y with the three-points forward difference formula can be written from the formula for the first derivative in Section 8.4.

[latex] \begin{aligned} & \left.\frac{\partial u}{\partial y} ight|_{\substack{x=x_{i} \ y=y_{i}}}=\frac{-3 u\left(x_{i}, y_{i} ight)+4 u\left(x_{i}, y_{i+1} ight)-u\left(x_{i}, y_{i+2} ight)}{2 h_{y}}= \ & =\frac{-3 u(2,1)+4 u(2,2)-u(2,3)}{2 \cdot 1}=\frac{-3 \cdot 250+4 \cdot 361-448}{2 \cdot 1}=123 \end{aligned} [/latex]

Question 8.7: Numerical partial differentiation. The following two-dimensi......

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