9.2
Section |
2 Background .1
Cosmology .11 Math |
Section 9.4 |

Section 3 **The Passage method**

If Equation (1.3) and the boundary condition (1.4) are converted into **central finite differences**,
one ontains the equations

(3.1)

The first *n* - 1 equations of this system may be rewritten

(3.2)

and then reduced to

(3.3)

where

(2.4)

(2.5)

The computations are carried out as follows:

**Forward**. Find *m*_{i} and *k*_{i}
from (3.2). Compute *c*_{i} and *d*_{i} and then, using the recurrence relations
(3.5) find one after the other *c*_{i} and *d*_{i}*, i*_{ }=
2, . . ., *n.*_{i}

**Backward***. *Write down (3.3) for *i* = *n
*and* i = n - *1* *and the last equation of (3.1):

(3.6)

These equations have the solution

(3.7)

The values of *y*_{i}, *i = n *-1, . . .,1 are now obtained from the **recurrence
relation** (3.3), while *y*_{0} is given by

(3.8)

**Example 1:** Find by the above method an approximate solution of the **B.V.P.**

(3.9)

(3.10)

**Solution**: Set *h* = 0.1 and reduce the problem to the finite difference
equations

A rearrangement yields

Thus,

**Forward:** Enter *x*_{i} = 0.1?*i *into the next
table and compute *m*_{i}, *k*_{i }and *j*_{i}.
Determine from (3.4)

Enter these values into the table and compute using (3.5) all *c*_{i}* *and* d*_{i}.
Thus, for *i* = 2,

Enter the computations for *i* = 3, 4, . . . ,10 into the table.

**Backward:** By (3.7),

Enter this value in the last row of the preceding table. Then use (3.9) to compute *y*_{i}
, *i = *9, 8, ??? , 1:

Finally, by (3.8),

The last column of the table displays the values of the exact solution

The last column of the table gives the values of the exact solution

**Note.** Equation (2.3) gave an estimate of the error of the method of
finite differences which is of little practical use. It is standard practice to use **double
computation** (cf. Example
2), and, starting from the **Runge principle**
([18]), to obtain an approximate estimate of the error of the value of *y*_{i}":

where *y*(*x*_{i}) is the exact solution of the boundary value problem and *y*_{i},
*y*_{i}* are the values of the approximate solutions at *x*=*h* and h/2, respectively.

**Example 2:** Solve with an accuracy of 10^{-3} the
boundary value problem

(3.11)/(3.12)

**Solution:** In order to obtain the approximate solution to the given accuracy,
start with *h =* 0.05, repeat the calculation with *h *= 0.1 and compare the results.

In the present case

(3.13)

Use (3.2) and form a table of the values of

for *h* = 0.05. Using the fact that

find from (3.4)

and apply repeatedly (3.5) to obtain the values of *c*_{i}, *d*_{i}, *i*
= 2, . . . , 9.

**Problem (3.11)/(3.12) for ****h = ****0.05**

The boundary condition yields *y*_{10} = *y*(0.5) = 1.279. Equation (3.3) yields the values
of *y*_{i}, *i* = 9, 8, . . . , 1, presented by the preceding table.

Now take the spacing *h *= 0.1, form a table of *c*_{1} *k*_{i}, 2*f*_{i}*h*^{2}/(2
+ *hp*_{i}) = *j*_{i }and compute *c*_{i}*
*and *d*_{i} for the new spacing: For example, we find that *c*_{1 }= 1/*m*_{1
}= -0.510, *d*_{1 }= 2*f*_{1}*h*^{2}/(2 + *hp*_{1}) - *k*_{1})
= -0.976. Enter these results in the next table and use (9.3) to compute *y*_{i}, *i* = 4, 3, 2, 1.

Comparing corresponding results in the last and the following table, we see that their differences do not exceed 0.002

**Problem (3.11)/(3.12) for ****h = ****0.1****.**

so that the error of the most accurate value for *h* = 0.05 does not exceed 0.002/3. Hence the results of
the first table yield the required solution.

**Example 3****:**
Find the solution of the boundary value problem

(3.14)

when the functions *p*(*x*), *y*(*x*) and *f*(x) are by the table

**Solution:** The results with *h* = 0.1 are given in the next
table.

Computing the table:

**Forward:** Use (3.2) and the table of coeffcients to compute *m*_{i};
*k*_{i}, *h*^{2}*f*_{i}.Since in this case

one finds

etc.

**Backward: **The boundary conditons yield *y*_{0} = 0
and *y*_{4 }= 0.335. The remaining values of *y*_{i} are obtained from (3.3):

etc.

**Exercises**

1. Use this method to find within 10^{-2 }the solutions of the following differential equation for the boundary conditions

*y*(0) = 1, y(1) = e^{-1} + 1 =1.367.

2. Find approximate solutions of the boundary value problern

if the *f*_{j}(*x*) are given by the table

3. Solve the following equations on the interval [0,1] with *h* = 0.1 for the boundary conditions *y*(0)
= *y*(1) = 0:

where the parameters have the values