# Unit 4 - Week 1: Finite Difference Method (FDM) - I

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## 1: Finite Difference Method (FDM) - I

reviewer3@nptel.iitm.ac.in ▼

### outline

How to access the portal? Pre-requisite Assignment MATLAB Week 1: Finite Difference Method (FDM) - I Lecture 1 Lecture 2 Lecture 3 Lecture slides week 1 Exercise 1 Exercise 2 Exercise 3 Matlab program for exercise1 Matlab program for exercise2 First Matlab program for exercise3 Second Matlab program for exercise3 Lab tour 1 Summary week 1 Quiz : Week 1 Assignment 1 Week 1 -Feedback:

### Due on 2018-08-15, 23:59 IST.

1) 1 point 2) 1 point

### Week 1 Assignment 1

The due date for submitting this assignment has passed. As per our records you have not submitted this

assignment.

A finite-differencing scheme is stable in the following cases

Statement 1: The numerical domain of influence is smaller than the physical domain of influence. Statement 2: The numerical domain of influence for the problem contains the physical domain of influence.

Statement 3: The numerical velocity must be more than the physical velocity. Statement 4: The numerical velocity must be less than the physical velocity.

Only statement 1 is correct Only statement 2 is correct Only statement 3 is correct Only statement 4 is correct Only statements 1 and 3 are correct Only statements 1 and 4 are correct Only statements 2 and 3 are correct Only statements 2 and 4 are correct None of the statements are correct No, the answer is incorrect.

Score: 0

Only statements 2 and 3 are correct

Assume and to be the step-sizes with indices and along - and -axes respectively.

The finite-difference representation for two-dimensional Laplace equation using central differencing scheme is

2

### = 0

φ(i+1,j)−2φ(i,j)+φ(i−1,j) Δx φ(i,j+1)−2φΔ(iy,j)+φ(i,j−1)

A project of In association with

Funded by

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Powered by Week 2: Finite Difference Method (FDM) - II Week 3: Finite Difference Method (FDM) -III Week 4: Boundary Conditions Week 5: Variational Methods Week 6: Finite Element Method - I Week 7: Finite Element Method - II Week 8: Method of Moment Week 9: Finite Volume Method -I & -I-I Week 10: Finite Volume Method -III Week 11: Algebraic Topological Method - I Week 12: Algebraic Topological Method - II and Mimetic Method Video Download TEXT TRANSCRIPTS Computational Electromagnetics & Applications 3) 1 point None of these No, the answer is incorrect. Score: 0

Which of the following is(are) the correct computational stencil(s) for a one-dimensional

wave equation with aspect ratio where, is a constant, and are step-sizes in time and -axis respectively? (filled-squares represent known points and empty-circles represent unknown point in the figure)

Statement 1: Explicit scheme with aspect ratio <1

Statement 2: Implicit Scheme with aspect ratio = 1

Statement 3: Explicit scheme with aspect ratio = 1

Statement 4: Implicit Scheme with aspect ratio < 1 Only statement 1 is correct.

### = 0

φ(i+1,j)−2φ(i,j)+φ(i−1,j) Δx2 φ(i,j+1)−2φ(i,j)+φ(i,j−1) Δy2

### = 0

φ(i+1,j)+φ(i−1,j) Δx2 φ(i,j+1)+φ(i,j−1) Δy2

### = 0

φ(i+1,j)−φ(i,j)+φ(i−1,j) Δy2 φ(i,j+1)−φ(i,j)+φ(i,j−1) Δx2

### = 0

φ(i+1,j)−2φ(i,j)+φ(i−1,j) Δx2 φ(i,j+1)−2φ(i,j)+φ(i,j−1) Δy2

2

### x

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4) 1 point

5) 1 point

Only statement 2 is correct Only statement 3 is correct Only statement 4 is correct. Only statements 1 and 2 are correct. Only statements 1 and 3 are correct. Only statements 2 and 4 are correct. Only statements 3 and 4 are correct. None of the statements are correct. No, the answer is incorrect.

Score: 0

Only statements 1 and 3 are correct.

The central-difference formulation of the differential at point = 4 with step-size is

None of these. No, the answer is incorrect. Score: 0

Which of the following statement(s) is(are) correct for a first-order, one-dimensional partial differential equation (PDE) with constant spatial step-size?

Statement 1: The order of truncation error in forward-differencing is less than that of central-differencing.

Statement 2: The order of truncation error in backward-differencing is more than that of central-differencing.

Statement 3: The order of truncation error in forward-differencing is same as that of central-differencing.

Statement 4: The order of truncation error in forward-differencing is same as that of backward-differencing.

Only statement 1 is correct Only statement 2 is correct Only statement 3 is correct Only statement 4 is correct. Only statements 1 and 2 are correct Only statements 1 and 3 are correct. Only statements 1 and 4 are correct

y d2 dx2

### = 4

y(8)+y(0) 4 y(8)−2y(4)+y(0) 4 y(8)−2y(4)+y(0) 16 y(8)+y(0) 16 y(8)−2y(4)+y(0) 16

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6) 1 point

7) 1 point

Only statements 2 and 3 are correct Only statements 2 and 4 are correct. Only statements 3 and 4 are correct Only statements 1, 2, and 3 are correct Only statements 1, 2, and 4 are correct Only statements 1, 3, and 4 are correct Only statements 2, 3, and 4 are correct None of the statements are correct No, the answer is incorrect.

Score: 0

Only statements 1 and 4 are correct

Assume is the step-size with index along the -axis and is the time-step with index . What is the forward-difference (FD) and central-difference (CD) formulation for the given equation? FD: CD: FD: CD: FD: CD: FD: CD: None of these. No, the answer is incorrect. Score: 0

FD: CD:

Assume is the step-size with index along the -axis and is the time-step with index .

The central-difference explicit-formulation and aspect-ratio for the given equation are

=

t

x

Δx

Δt2 ΔΔxt

Δx

Δt2 ΔΔxt

Δt

Δx ΔΔxt

Δt

Δx ΔΔxt

Δt

Δx ΔΔxt

Δt

Δx ΔΔxt

Δx

Δt

Δx

Δt ΔΔxt ΔΔxt

Δt

Δx ΔxΔt

Δt

Δx ΔxΔt

tt

xx

2

### x

2

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8) 1 point 9) 1 point = = = = = None of these No, the answer is incorrect. Score: 0

=

If is the index in -direction then, the forward-difference approximation of a third-order differential is

None of these. No, the answer is incorrect. Score: 0

Common data for questions 9 and 10

For an ordinary differential equation (ODE), in 0 < x < L with boundary conditions y(0) = y(L) = 0. The domain is divided as shown in the figure with step-size h and index i in x-direction.

The finite-difference formulation using central-differencing scheme is

2

2

2

2

### x

f(x) d3 dx3 f(i+3)−f(i+2)+f(i+1)−f(i) (Δx)2 f(i+3)+3f(i+2)+3f(i+1)+f(i) (Δx)3 f(i+3)−3f(i+2)+3f(i+1)−f(i) (Δx)3 f(i+3)+3f(i+2)−f(i) Δx2 f(i+3)−3f(i+2)+3f(i+1)−f(i) (Δx)3

### ) = 1

y(x) d2 dx2

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10) 1 point

11) 1 point

None of these No, the answer is incorrect. Score: 0

The value of and are

and

and

and

and None of these No, the answer is incorrect. Score: 0

and

Common data for questions 11 and 12

Assume is the step-size with index in -direction and is the time-step with index . For the one-dimensional heat-equation,

The finite-difference formulation of the above equation, using forward-differencing in time and central-differencing in space, is

None of these No, the answer is incorrect. Score: 0

2

2

2

2

2

2

2

2

2

4

2

2

2

2

4

t

xx

t

Δx

Δt

x)2

Δt

Δx

Δt

### )

x)2

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For = 1 and = 1 with the initial-conditions given as

, , calculate and using

the result from the above problem.

= 1 and = 3

= 0 and = -2

= 7 and = -1

= 2 and = 1 None of these. No, the answer is incorrect. Score: 0 Accepted Answers: = 1 and = 3

Δt

(Δx)2

### (1, 2)

References

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