Hermite Polynomials-Finite Element Method-Assignment Solution, Exercises of Mathematical Methods for Numerical Analysis and Optimization

This assignment solution was submitted to Amar Sharma for Finite Element Method course at Aligarh Muslim University. It includes: Hermite, Polynomials, Property, Stations, Kronecker, Delta, Cubic, Interpolation

Typology: Exercises

2011/2012

Uploaded on 07/08/2012

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ASSIGNMENT # 05
EXERCISE PROBLEM # 4.35
PROBLEM # 3.35
Derive the following Hermite polynomials;
1)
2)
SOLUTION
As we know the following property of Hermite polynomial for two stations,
For I, p = 1, 2
For k, r = 0, 1, 2, ……… …… ……….., N
Where x is the value of xp at pth station and 𝛅mn is the Kronecker delta having the property,
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ASSIGNMENT # 05

EXERCISE PROBLEM # 4.

PROBLEM # 3.

Derive the following Hermite polynomials;

SOLUTION

As we know the following property of Hermite polynomial for two stations,

For I, p = 1, 2

For k, r = 0, 1, 2, ……… …… ……….., N

Where x is the value of xp at pth^ station and 𝛅mn is the Kronecker delta having the property,

δmn = 0 if m ≠ n

δmn = 1 if m = n

By using above described properties, we can easily determine the desired polynomial, as following;

{ as I ≠ p}

{ as I = p and k = r}

{ as I ≠ p and k ≠ r}

Similarly,

The cubic interpolation model is given below,

……………………………………… (1)

Differentiating it w.r.t x, so that we may easily use the above calculated conditions,

………………………………….………. (2)

Now, by using above calculated conditions, we can easily find the unknown coefficients, as following;

At x = x 1 = 0, equations 1 & 2 becomes respectively,

At x = x 2 = l,

………………………………………….. (3)

Differentiating it w.r.t x, so that we may easily use the above calculated conditions,

…………………………… (6)

Now, by using above calculated conditions, we can easily find the unknown coefficients, as following;

At x = x 1 = 0, equations 5 & 6 becomes respectively,

At x = x 2 = l,

…………………………… (7)

…………….…………(8)

By solving equation 7 & 8 simultaneously, we get

Put these values in equation 5, we get

Rearranging it, we get

Hence proved