4. Consider a wave equation on an infinite line, J²u 2²u Ա 9 at² əx² = 0. Find the characteristics though the point (1,3). Draw the domains of depen- dence and influence of the point (1,3).

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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4. Consider a wave equation on an infinite line,
J²u
J²u
9
Ət² əx²
= 0.
=
Find the characteristics though the point (1,3). Draw the domains of depen-
dence and influence of the point (1,3).
Transcribed Image Text:4. Consider a wave equation on an infinite line, J²u J²u 9 Ət² əx² = 0. = Find the characteristics though the point (1,3). Draw the domains of depen- dence and influence of the point (1,3).
Expert Solution
Step 1

Comparing the given equations with the standard equation utt-c2uxx=0, we see that c=3.

Also the point to be analysed is (xo,to)=(1,3).

At the point (xo,to)=(1,3), 

xo-cto=1-3×3=-8.xo+cto=1+3×3=10.

The characteristics through the point (xo,to)=(1,3) are given by x-3t=-8 and x+3t=10.

The lines shown below(green and blue) are the characteristic lines to the given equation.

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Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question

How can you evaluate this equation?

I don't know how to get the interval [-8,10]

The image displays the following system of equations:

\[ x - 3t = -8 \]
\[ x + 3t = 10 \]

These equations can be solved to find the values of \( x \) and \( t \). 

### Steps to Solve:

1. **Add the equations:**

   \[
   (x - 3t) + (x + 3t) = -8 + 10
   \]

   Simplifying gives:

   \[
   2x = 2
   \]

2. **Divide by 2 to solve for \( x \):**

   \[
   x = 1
   \]

3. **Substitute \( x = 1 \) back into one of the original equations to solve for \( t \):**

   Using the first equation:

   \[
   1 - 3t = -8
   \]

   Rearrange to solve for \( t \):

   \[
   -3t = -8 - 1
   \]
   \[
   -3t = -9
   \]
   \[
   t = 3
   \]

### Conclusion:

The solution to the system of equations is \( x = 1 \) and \( t = 3 \).
Transcribed Image Text:The image displays the following system of equations: \[ x - 3t = -8 \] \[ x + 3t = 10 \] These equations can be solved to find the values of \( x \) and \( t \). ### Steps to Solve: 1. **Add the equations:** \[ (x - 3t) + (x + 3t) = -8 + 10 \] Simplifying gives: \[ 2x = 2 \] 2. **Divide by 2 to solve for \( x \):** \[ x = 1 \] 3. **Substitute \( x = 1 \) back into one of the original equations to solve for \( t \):** Using the first equation: \[ 1 - 3t = -8 \] Rearrange to solve for \( t \): \[ -3t = -8 - 1 \] \[ -3t = -9 \] \[ t = 3 \] ### Conclusion: The solution to the system of equations is \( x = 1 \) and \( t = 3 \).
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