1. Using bi-quadratic shape function given by Equation 4.12...

1. Using bi-quadratic shape function given by Equation 4.12 on page 86 of your textbook \begin{equation} q(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4y^2 + a_5xy + a_6x^2y + a_7xy^2 + a_8x^2y^2 \end{equation}derive the expressions of the nine coefficients $a_0, a_1, ..., a_8$ in terms of $q_{nw}, q_w, q_{sw}, q_n, q_p, q_s, q_{ne}, q_e, q_{se}$. Then evaluate the area integral \begin{equation} Q_p = \int_{\Omega} q d\Omega \end{equation}in terms of nine coefficients and $\Delta x$ and $\Delta y$, by assuming uniform grid spacings $\Delta x$ and $\Delta y$ in x and y directions respectively, for a two-dimensional formulation of a finite volume problem. Finally write your result in terms of $q_{nw}, q_w, q_{sw}, q_n, q_p, q_s, q_{ne}, q_e, q_{se}$

Transcript

00:01 In this problem, we are given a vector field f, which is a two -dimensional vector field.

00:06 We're also given a path c, which corresponds to the line segment from the point p with coordinates 3, 0 to the point q with coordinates 0, 2.

00:16 In question a, we want to find the parametrization of the line segment describing path c.

00:26 So we want to parametrize our line segment with parameter variable t, ranging between 0 to 1.

00:32 To do so, let's start off our line equation with our initial point, which is the point p, and add on the product of our directional vector of the line times the parameter variable t, where our directional vector is going to correspond to the difference between point q and point p.

01:01 Now we see here that when t is equal to 0, we are indeed starting at point p.

01:06 When t is equal to 1, we will be at point q.

01:10 So evaluating this, we will obtain the vector 3, 0 plus minus 3 times t, where t ranges between 0 and 1.

01:40 So this is a compact way of writing a line in a vector format, but which encompasses two other linear equations, one in the x direction and another one in the y direction.

01:56 So in the x direction, we will have that x is equal to 3 minus 3t, and in the y direction, we will have 2t.

02:17 Now we're going to use this to answer the next question.

02:24 We want to write out the line integral of our vector field f along path r, excuse me, path c.

02:34 And to solve this, we'll need to recall that a line integral written over dr can write as f evaluated at our position vector r times r prime of t dt, where t will range between, in this case, 0 and 1.

03:10 So with the position vector we determined above, let's calculate r prime of t.

03:21 So using that r of t is equal to 3, 0 plus minus 3, 2, we will find that r prime of t will be equal to simply the vector minus 3, 2, because the derivative of a constant is equal to 0.

03:38 Now let's take our vector field f and evaluate it at r of t.

03:49 The reason we're doing so is because our vector field is originally written in cartesian coordinates x, y, and we want to rewrite it as a function of parametric variable t.

04:01 So let's take our expression for x and our expression for y and enter it into our vector field equation.

04:07 So we will find minus 2t i plus 3 minus 3t j.

04:21 Now we are ready to develop the dot product here.

04:28 So we have that f of t dot r prime of t will give us 6t plus 6 minus 6t, simplifying to 6, which means that our dot product here was in fact a constant the whole time, giving us the integral between 0 to 1 of 6 dt.

05:04 The result of this integral will be equal to 6t for t ranging between 0 to 1, giving us simply 6...

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