# Finite difference approximation on an uniform mesh

We have seen the basic idea of finite difference

approximation. Now let us introduce some nomenclature and let us a try to make faster progress. So, we are looking at finite difference approximations

on a uniform mesh. So, consider uniform mesh with a spacing of delta x over a space interval

of between 0 and capital l. So, we denote the mesh index by i here and what we mean

by this is that f of the value of f at xi. So, that is the location the x value at the

ith machine index or node which we write as f subscript i here is nothing, but f of i

times delta x. So, that is what xi is here and we can also write f of i plus 1 in the

shorthand notation as the value of f at i plus 1 delta x. So, that is instead of writing

f of x and f of x plus delta x, we can write it as f subscript i and f subscript i plus

1 and similarly f subscript i minus 1. So, using this notation here we can write

the approximation that we have got from equation 2 df by dx at x as f of x plus delta x minus

f of x divided by delta x. Instead of that we can write it as f subscript i plus 1 minus

f subscript i divided by delta x plus terms of the order of delta x. Similarly from the

third equation where we expanded f of x minus delta x, we got another approximation for

d f by d x as f of i minus f of i minus 1 by delta x plus again terms of the order of

delta x other ones that are neglected. So, that neglected part is known as truncation

error. So, the leading term of the truncation error varies as delta x the magnitude instead

of that varies is delta x and that is what is denoted here as the order of accuracy of

the approximation and we also got a different expression by manipulating equation 2 and

equation 1 and 3, those 2 approximations we subtracted 3 from 1 and then we got a different

approximation, df by dx at x as f of i plus 1 minus f at i minus 1 divided by 2 delta

x plus terms of the order of delta x square. Now there is a some difference between these

3 approximations. So, first 1 is known as the forward differencing approximation because

f of x, df by dx at x given in terms of f of i plus 1 minus fi divide by delta x.

The second expression where the same derivative at the same location is expressed in terms

of fi minus fi minus 1 by delta x, that is known as backward differencing. And in this

the value of the first derivatives at x is expressed in terms of the two neighboring

points, f of i plus 1 minus f of i minus 1 divided by 2 delta x. 2 delta because the

distance between these 2 points is 2 here, we get a central differencing scheme. So,

forward and backward differencing schemes are 1 sided differencing schemes. So, that

is you have i and i plus 1 and i and i minus 1 not, where as this is central in sense that

you have points which are goes to the left i minus 1 and right of this. So, that is both

sides are being used to express the approximation of this. So, these are known as central differencing

formulas. Usually central differencing formulas are more accurate than 1 sided differencing

formula, but we need both central because are more accurate, and forward and backward

are required at the edges of the fluid of the computational domain. If a there is no

point of the right of this point x or point node i then we cannot make use of the forward

differencing. Similarly at the left edge of a boundary here,

we cannot make use of point which is outside this computation domain. So, at the edges

we would like to have 1 sided differencing and in the center we would like to have central

differencing. So, 1 sided formula’s are necessary at the ends of domains.

We can look at these derivatives that we have forward backward central in a graphical way

by looking at this function in this case it is a u here varying with x as per this particular

curve. So, fairly smooth variation, and you have your mesh nodes this is ith nodes and

then i plus 1 i plus 2 similarly i minus 1 i minus 2 like that, with uniform spacing

of delta x. Now we are looking at the derivatives dou u by dou x at i at this particular point.

So, here we have the red line here is the true slope of this particular function at

the point i. If you take i plus 1 or i plus 2 the slope is varying, but at the location

i here the red line here is what is the true slope the true derivative value at i here

forward differencing is based on the difference between i plus 1 and i. So, the value of i

here and the value of the function at i and i plus 1 here, if you join them by line and

then following it then you get this blue lines. So, that is the forward differencing and the

slope of this the angle with respect to the horizontal is the approximation of the first

derivative as given by the forward differencing. And backward differencing is based on a line

drawn between the value of i minus 1 or the function at i minus 1 and i and that is the

green line here. And the central differencing is based on a line which is drawn between

i plus 1, i minus 1 value here and i plus 1 value here and that is this black line given

by c central differencing here. And if you now look at the slopes, so that is the angle

that each of the straight lines is making with the horizontal you see that the central

differencing line here is much closer to the value of the true slope. So, it is almost

parallel to this which means it is accurate, where as both f and b the forward and backward

is at different angle compared to the true slope. So, that indicates sense of more error

with forward and backward differencing schemes that we would get as compared to the central

differencing. So, this also an indication of the mathematically derived truncation error

which is of second order accuracy in the case of central, where as f and b are first order

if you go back to the Taylor series expansion here.

So, the in a in central differencing it is this 1 which is being neglected from this

term onwards. Now, way it is the third derivatives which

is being neglected, and in the case of 1 sided differencing it is a second derivative and

terms are being neglected. So, if it is a converging series the second derivative term

is likely to be larger in magnitude, then the third derivative term so; that means that

we are neglecting a larger amount of error with 1 sided differencing involving first

order accurate comes here. Where as in the case of central differencing it is a third

derivative. So, that is the next term is the one that is has been neglected and all the

successive terms and since the next term is smaller than this term the error involved

in in the central differencing is less. So, that is also another way of looking at these

approximations. So, they can be these approximations and we can also get an approximation of even

higher order accuracy, we can get a third-order or nth order type of thing.

So, let us say that we want to have we can have a systematic way of arising at order

of accuracy for a given derivative. In this case we are going to look at only the first

derivative and we will seek a third-order one sided approximation for u of x at x. So,

we will see when we generalize it then you generalize recent requires that in order to

write this we need 4 successive points of the mesh. So, that and this is of this particular

form that is dou u by dou x at i, i is xi here. So, we are making use of that and that

is this formula here for a one sided third-order accurate approximation will involve four points

four successive points all to 1 side of i. So, that is ui, ui plus 1 ui plus 2 ui plus

3 and these are all multiplied by certain coefficients, real coefficients a b c d and

divide this whole thing is divided by delta x and this whole formula is third-order accurate

and therefore, we can write dou u by dou x at i as per this in this particular way. So,

that is we seek an approximation for dou u by dou x at I, in the form of aui plus bui

plus 1 plus cui plus 2 plus dui plus 3 divided by delta x plus terms of the order of delta

x cubed, this makes it a third-order accurate of approximation.

Now, when this is equivalent in a way to writing this approximation, because we have seen that

all these tells is expansion involve derivatives; first derivatives, second derivative, third

derivative and all that. So, what this approximation actually implies is that we are we would like

to get an approximate formula which is dou u by dou x at i equal to aui plus bui plus

1 plus cui plus 2 plus dui plus 3 divided by delta x, plus zero times dou square u by

dou x square times delta x there is if this 1 is not that is it is not there. It is there,

but being multiplied by zero. So, that it does not appear in this expression similarly

zero times the third derivative time delta x square because we divided by delta x here

it become delta x square and of course, you have factorial 3 here, but since it is zero

it does not matter here again you have factorial 2 it does not matter because multiple by this.

And a coefficient e times the fourth derivatives at x, times delta x cube by factorial four

which can be substitute in this coefficient here and when you have an expression like

this plus higher order terms, that is the fifth derivative times delta x to the power

four and all that this expression equation 7 and equation 6 are equivalent because they

both are saying that d u by d x d u by d x at i is in terms of this ui plus 1 ui i plus

2 i plus 3 like this plus 0 times this. So, this delta x term does not appear zero times

delta x square terms. So, this also does not appear and the only term the leading term

that appears is this factor coefficient e still to be determined times d 4 u by d x

four at x time delta x cube. So, this is what although we are writing it

like this what we are saying is that this approximation is a this form plus higher order

terms. Now, you can rewrite this as you bring this on to this side au i plus 1 plus bu i

plus all this things equal to du by dx minus 0 times this, minus 0 times this, minus 0

times this. So, writing this is, but how do we define the values of a , b, c, d? If we

can find unique values of a, b, c, d then we have a formula here.

And in order to find this a, b, c, d what we do is that we expand ui plus 1 ui plus

2 ui plus 3 which are appearing in this formula, in this proposed formula in terms of Taylor

series about point i for example, ui plus 1 is nothing, but u at x plus 1 delta x and

that we know is u of x plus du by dx at delta x plus d square u by dx square at delta x

square factorial 2 like this. Similar ui plus 2 is nothing, but u of x plus 2 delta x. So,

wherever we had delta x here we substitute 2 delta x here. So, and if delta x is small,

two times delta x also small. So, and assuming that it is small and it is still convergent

and all that kind of thing we can write ui plus two as ux plus dou u by dou x times 2

delta x plus dou square u by dou x square times 2 delta x whole square by factorial

2 in similarly, u at i plus 3 is u of x plus 3 delta x and therefore, we can write this

as u of x plus du by dx times 3 delta x plus dou square u by dou x square at x times 3

delta x whole square plus 3 delta x whole cube by factorial 3 and so on.

Now we have these approximations and we replace these things in the previous equation here.

So, we evaluate this term au i plus bu i plus 2 like that. So, we multiply this whole thing

by b and then this whole thing by c and this whole thing by d and then take the sum of

all this things and this will be given finally, in this particular form. This will pu i plus

q times du by dx times delta x plus r times dou square u by dou x square times delta x

whole square. For example; p will be a here, and then when we multiply this by b we get

b plus c plus d. So, p is a plus b plus c plus d and q here will be b plus 2c plus 3c,

a plus 3d. In that way we can ones we multiply and then put all these things together we

can express this p, q, r, s, t in terms of a, b, c, ds. Now we compare this equation

with the equivalent expression equation 8 that we said is equivalent to this approximation

here with zero coefficients here. From this we can say that p is equal to 1. So, that

is a plus b plus c plus d is equal to 1 and b plus 2 c plus 3 c plus 3 d equal to 0 and

this will be a square here. So, this will be 4b plus 9c plus 16d will be equal to 0

and so on like this. So, that will be give you four equations 1

for p, 1 for q, 1 for r and 1 for s. This p coefficient is 1 and the other 3 coefficients

are 0. So, it does not matter what this e is we will come out of the solution because

we have 4 equations, that is a plus b plus c plus d equal to 1 and 2 be plus 3c plus

4d equal to 0 and 4b plus 9c plus 16d equal to 0 and so on like that. So, you have 4 equations

and 4 unknowns and if you solve those things you get a equal to minus 11 by 6, b equal

to 3 and c equal to minus 3 by 2 d equal to 1 by 3. And e will be some sort of summation

of this that will come out in this way, it does not matter what the coefficient is this.

Anywhere this whole term is going to be neglected. So, that will give us the third-order approximation

and when we substitute this back into the original equation we can get dou u by dou

x at i here as minus eleven ui plus 18 ui plus 1 minus 9 ui plus 2 plus 2 ui plus 3

divided by 6 delta x plus terms of the order of delta x cubed.

So, this is a third-order 1 sided approximation for the first derivative at i involving i;

i plus 1 the value of u at i; i plus 1 i plus 2 like this and in this think on a uniform

mesh when we derive it using this Taylor series like this, we will notice that the sum of

all these coefficients will be equal to 0. So, you have minus 11 plus 9, minus 9. So,

that is minus 20 and you have 18 plus 2; 20. The sum of all these things, that is a plus

b plus c plus d will be equal to zero of here. So, that is something that that is 1 quick

way of us for us to verify that we have got the right kind of formula here. So, this kind

of expression can be derived not just the first derivative, we can derive it for any

pth derivative and we can do it for any order of q. It is possible for us to derive a qth

order of approximation for a pth derivative using either one sided or central things and

the number of points n here depends on what is order of p here, and what is the order

of q. It is equal to t plus q in the case of one sided, and it is one less in the case

of central differencing. So, this is your a general way of deriving a finite difference

approximation of some order. In the next lecture we will try to apply this to an equation and

then see what kind of approximation we get. Thank you.

## One Reply to “Finite difference approximation on an uniform mesh”

Very good explanation sir…