10.6
Section |
2 Background .1
Cosmology .11 Math |
Section 10.8 |

**Section** 7 **Hyperbolic Equations**

Consider the** mixed boundary value problem**
for the equation of a **vibrating string**

(7.1)

with the **initial conditions**

(7.2)

and the **boundary conditions**

(7.3)

Since the change of variable *t *=* **at*
reduces (7.1) to

(7.4)

set in (7.1) *a* = 1.

Construct for *t* ? 0, 0
? *x* ?
*s* the two families of parallel straight lines

and replace the derivatives in (7.4) by finite differences. Using the symmetry of these formulae, one obtains

(7.5)

With these equations become

(7.6)

It has been shown ([2])
that for *a* ?
1, this equation is **stable**.

For *a *= 1, (7.6) assumes its simplest form

(7.7)

The error estimate ([2]) of
the solution of (7.6) for 0 ? *x *?
*s, *0* < t *? *T
*is

* *(7.8)

where *u* is the exact solution and

Note that in writing down (7.6), the grid points shown in the preceding
figure have been used, which is an **explicit diagram**,
since (7.6) allows to find the values of *u*(*x,t*) at *t*_{j+1}, if the values at the two
preceding layers are known. In order to solve Problem (7.1) - (7.3), the values on two initial layers must be known. They can be
found from the intial conditions by one of the two methods:

**First method****:***
*In the initial condition (7.2), replace the derivative *u*_{t}(*x*,0) by the difference
relation

For the value of *u*(*x, t*) on the layers *j* = 0 and *j *= 1 this yields

(7.9)

The** error estimate** of the values *u*_{i1}
has now the form ([2])

(7.10)

where

**Second method:** Replace the derivative u_{t}(*x*,0)
by

where *u*_{i,-1} are the values on the first layer *j* = -1. Then, by
the initial conditions (7.2),

(7.11)

Next, write down the difference equation (7.7) for the layer *j* = 0:

(7.12)

Eliminate from (7.11) and (7.12) the values *u*_{i,-1} and obtain

(7.13)

The error estimate for *u*_{i,1} is ([2])

(7.14)

where

This method of computing the initial values is used in Example 1.

**Third method: **If *f*(*x*) has a finite
second derivative, then the values of *u*_{t,1} can be determined with the aid of the **Taylor
expansion**

(7.15)

By (7.4) and the initial conditions (7.2), one finds

Equation (7.15) now yields

(7.16

The** error **in *u*_{i,1},
given by this formula, is of *O*(*l *^{3}). This method of obtaining initial values is considered in Example
2.

**Note:** The method can
also be used to solve a mixed boundary value problem for the non-homogeneous equation

,

with the diffcrence equation

**Example 1:** Find the
solution of the problem

(7.17)

**Solution: **Construct a square grid with *h*
= *l *= 0.05.* *Find the values of *u*(*x, t*) on the two initial layers by the second method. Taking
into account that

one finds

(7.18)

**Filling in the table:**

**Table 113 - Solution to problem (17).**

1. Enter the values of *u*_{i0} = *f*(*x*_{i}) for *x*_{i
}= *ih* into the first row of the table above.(it corresponds to the value *t*_{0} = 0). Since the
problem is symmetric, fill in the table for ) ? *x*
? 0.5. The first column, which corresponds to *x*_{0} =
0, contains the boundary values..

2. Find *u*_{i1} from (7.18), using *u*_{i0} in the first row
and enter the results in the second row..

3. Compute *u*_{ij} on the next layers, using (7.7). For *j* = 1, the numbers are

Compute in the same manner the values for *j *= 2, 3, ..., 10. The last row of the table contains the
values of the exact solution at *t* = 0.5.

**Example 2****:**
Solve the problem

(7.19)

**Solution:** Use *h* = *l* = *p**/18*.
Find *u*(*x,t*) on the first two layers by the third method, using a Taylor expansion.

**Filling in of the table:**

1. Compute *u*_{i0 }= *f*_{i}* = x*_{i}(*p**
- x*), *i* = 0, 1, . . . ,18 and enter them into the first row of the next table.

Table 114 - Solution of problem (19).

Since the problem is symmetric, fill in the table for 0 for 0 ? *x
*? p/2. Enter the boundary values into the first column of the
table.

2. Determine *u*_{i1}. Since *F*_{i}
= 0, *f*_{i}*"*= -2, by (7.16),

whence follow the *u*_{i1}.to be entered into the second row of the table.

3. Compute *u*_{i,j+1}, *j* = 1, 2, 3, 4, 5 from (7.7). For *j* = 2,

etc.

**Exercises**

In Problems 1 to 3 solve the equation

for the initial and boundary conditions

for 0 ? *t* ?
0.5, 0 ? *x* ?
1 with *h* = 0.

1.

Compute *u*_{i,1} by the third method above.

2. When the function *f*(*x*) is given by the
table

3. When the functions
are given by the table below and *a* = 1.1, 1.2, . . . , 1.5.

4. Solve the equation

with

Compute *u*_{i,1} by the first method
above.

5. Solve the equation

with *h* = *l* = 0.1 for

:

Compute *u*_{i,1} using the first method above.