I don't understand why F{utt}=d^2/dt^2(S(alpha,t)). Can you please explain it to me? Thank you
I don't understand why F{utt}=d^2/dt^2(S(alpha,t)). Can you please explain it to me? Thank you
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I don't understand why F{utt}=d^2/dt^2(S(alpha,t)). Can you please explain it to me? Thank you
![### Problem Statement
**Exercise 14.4.1.9:**
Solve the following differential equation:
A) \( a^2 u_{xx} = u_{tt} \)
for \(-\infty < x < \infty\), \( t > 0 \).
B) Initial condition: \( u(x, 0) = f(x) \)
C) Initial velocity condition: \( \frac{\partial u}{\partial t}\bigg|_{t=0} = g(x) \)
for \(-\infty < x < \infty\).
### Solution
**Step (A):**
Given the equation:
\[
\mathcal{F}\{ a^2 u_{xx} \} = \mathcal{F}\{ u_{tt} \}
\]
### Consider:
\[
\mathcal{F}\{ a^2 u_{xx} \} (x+1) = a^2 \mathcal{F}\{ u_{xx} \} (x+1)
\]
\[
= a^2 \int \frac{\partial^2 u}{\partial x^2} e^{-ix} \, dx
\]
\[
= -a^2 \omega^2 \mathcal{F}\{ u \} (x+1) = -a^2 \omega^2 S(x+1)
\]
### Further steps:
\[
\mathcal{F}\{ u_{tt} \} = \int \frac{\partial^2 u_1}{\partial t^2} e^{-ix} \, dx
\]
\[
= \frac{\partial^2}{\partial t^2} \int u(x, t) e^{-ix} \, dx = \frac{\partial^2 S(x+1)}{\partial t^2}
\]
This transcribes the mathematical derivations of solving the given partial differential equation with the associated initial conditions using Fourier transform methods. The conditions are aimed at determining the behavior of the function \( u(x, t) \) over the specified domain and time intervals.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca708747-373f-4dbe-b127-10492ca0e68f%2Fc7bfaa6b-1bc1-4a68-b700-0e034c2a1e62%2Fvzgf6iw_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**Exercise 14.4.1.9:**
Solve the following differential equation:
A) \( a^2 u_{xx} = u_{tt} \)
for \(-\infty < x < \infty\), \( t > 0 \).
B) Initial condition: \( u(x, 0) = f(x) \)
C) Initial velocity condition: \( \frac{\partial u}{\partial t}\bigg|_{t=0} = g(x) \)
for \(-\infty < x < \infty\).
### Solution
**Step (A):**
Given the equation:
\[
\mathcal{F}\{ a^2 u_{xx} \} = \mathcal{F}\{ u_{tt} \}
\]
### Consider:
\[
\mathcal{F}\{ a^2 u_{xx} \} (x+1) = a^2 \mathcal{F}\{ u_{xx} \} (x+1)
\]
\[
= a^2 \int \frac{\partial^2 u}{\partial x^2} e^{-ix} \, dx
\]
\[
= -a^2 \omega^2 \mathcal{F}\{ u \} (x+1) = -a^2 \omega^2 S(x+1)
\]
### Further steps:
\[
\mathcal{F}\{ u_{tt} \} = \int \frac{\partial^2 u_1}{\partial t^2} e^{-ix} \, dx
\]
\[
= \frac{\partial^2}{\partial t^2} \int u(x, t) e^{-ix} \, dx = \frac{\partial^2 S(x+1)}{\partial t^2}
\]
This transcribes the mathematical derivations of solving the given partial differential equation with the associated initial conditions using Fourier transform methods. The conditions are aimed at determining the behavior of the function \( u(x, t) \) over the specified domain and time intervals.
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