User:Eml5526.s11.team6.kurth/2-8

=Problem 2.8= Refer to lecture slides [[media:fe1.s11.mtg11.djvu|11-1]] and [[media:fe1.s11.mtg10.djvu|10-4]] for the problem statement.

Given
where $$\displaystyle w^h(x)$$ and $$\displaystyle u^h(x)$$ are approximations of $$\displaystyle w(x)$$ and $$\displaystyle u(x)$$, respectively, as follows

and $$\displaystyle b_i(x)$$ for $$\displaystyle i=1,...,n$$ is a family of linearly independent basis functions.

Objective
Show $$\displaystyle (8.1)\Leftrightarrow(8.2)$$.

Part 1
$$\displaystyle (8.1)\Rightarrow(8.2)$$

Since (8.1) is valid for all weighting functions $$\displaystyle w^h(x)$$, we can choose arbitrary values of $$\displaystyle c_i$$ to satisfy (8.3).

Case 1

Let $$\displaystyle \{c_1,c_2,c_3,...c_{n-1},c_n\}=\{1,0,0,...,0,0\}$$.

From (8.3):

Then (8.1) becomes:

Case 2

Let $$\displaystyle \{c_1,c_2,c_3,...c_{n-1},c_n\}=\{0,1,0,...,0,0\}$$.

From (8.3):

Then (8.1) becomes:


 * $$\displaystyle \vdots$$


 * $$\displaystyle \vdots$$

Case n

Let $$\displaystyle \{c_1,c_2,c_3,...c_{n-1},c_n\}=\{0,0,0,...,0,1\}$$.

From (8.3):

Then (8.1) becomes:

Hence, $$\displaystyle (8.1)\Rightarrow(8.2)$$:

Part 2
$$\displaystyle (8.2)\Rightarrow(8.1)$$

Multiplying both sides of (8.2) by $$\displaystyle c_i$$ yeilds

Since (8.12) is equivalently zero for every value of i, we can successively add (8.12) to itself for each value of i without changing its value.

Hence,

Solved by William Kurth.