I don't understand why the limit it from 0 to 1. Can you please explain it to me? Thank you
I don't understand why the limit it from 0 to 1. 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 the limit it from 0 to 1. Can you please explain it to me? Thank you
![### Solving for \( A_{mn} \) in Fourier Series Expansion
**Condition \( E \)**
Given:
\[ XY = U(X, Y | t=0) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} A_{mn} \cos(m\pi x) \cos(n\pi y) \]
To solve for \( A_{mn} \), we take:
\[ \iint\limits_{0}^{1} XY \cos(m'\pi x) \cos(n'\pi y) \, dx \, dy \]
Substituting the expression for \( XY \):
\[
\iint\limits_{0}^{1} \left( \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} A_{mn} \cos(m\pi x) \cos(n\pi y) \right) \cos(m'\pi x) \cos(n'\pi y) \, dx \, dy
\]
Interchanging the sums and the integral:
\[
= \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} A_{mn} \iint\limits_{0}^{1} \cos(m\pi x) \cos(m'\pi x) \, dx \cdot \iint\limits_{0}^{1} \cos(n\pi y) \cos(n'\pi y) \, dy
\]
Utilizing the orthogonality property of the cosine functions over \([0, 1]\):
\[
= A_{m'n'} \left(\int_{0}^{1} \cos^2(m'\pi x) \, dx \right) \left( \int_{0}^{1} \cos^2(n'\pi y) \, dy \right)
\]
Thus,
\[
A_{mn} = \iint\limits_{0}^{1} XY \cos(m\pi x) \cos(n\pi y) \, dx \, dy
\]
**Detailed Explanation**
1. **Equation Setup**: The initial equation represents a function \( XY \) expressed as a double infinite series involving cosine terms in both \( x \) and \( y \), with coefficients \( A_{mn} \).
2. **Integration to Solve for Coefficients**: To](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca708747-373f-4dbe-b127-10492ca0e68f%2Fc301b4bc-553a-4acb-b5f6-014535fe3e53%2Feww0lk_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving for \( A_{mn} \) in Fourier Series Expansion
**Condition \( E \)**
Given:
\[ XY = U(X, Y | t=0) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} A_{mn} \cos(m\pi x) \cos(n\pi y) \]
To solve for \( A_{mn} \), we take:
\[ \iint\limits_{0}^{1} XY \cos(m'\pi x) \cos(n'\pi y) \, dx \, dy \]
Substituting the expression for \( XY \):
\[
\iint\limits_{0}^{1} \left( \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} A_{mn} \cos(m\pi x) \cos(n\pi y) \right) \cos(m'\pi x) \cos(n'\pi y) \, dx \, dy
\]
Interchanging the sums and the integral:
\[
= \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} A_{mn} \iint\limits_{0}^{1} \cos(m\pi x) \cos(m'\pi x) \, dx \cdot \iint\limits_{0}^{1} \cos(n\pi y) \cos(n'\pi y) \, dy
\]
Utilizing the orthogonality property of the cosine functions over \([0, 1]\):
\[
= A_{m'n'} \left(\int_{0}^{1} \cos^2(m'\pi x) \, dx \right) \left( \int_{0}^{1} \cos^2(n'\pi y) \, dy \right)
\]
Thus,
\[
A_{mn} = \iint\limits_{0}^{1} XY \cos(m\pi x) \cos(n\pi y) \, dx \, dy
\]
**Detailed Explanation**
1. **Equation Setup**: The initial equation represents a function \( XY \) expressed as a double infinite series involving cosine terms in both \( x \) and \( y \), with coefficients \( A_{mn} \).
2. **Integration to Solve for Coefficients**: To
![### Homework 12.8: Problem 2
**Objective**: Solve the given partial differential equation (PDE)
**PDE and Boundary Conditions**:
\[ K(U_{xx} + U_{yy}) = U_t \]
**Boundary Conditions**:
\[A) \, U_x|_{x=0} = 0 \]
\[B) \, U_x|_{x=1} = 0 \]
\[C) \, U_y|_{y=0} = 0 \]
\[D) \, U_y|_{y=1} = 0 \]
\[E) \, U(x,y,0) = xy \]
**Diagram**:
The diagram depicts a square from (0,0) to (1,1) with the variable \( U(x,y,t) \) defined within it.
**Solution**:
1. **Setting up the solution**:
\[ \text{Set} \, U(x,y,t) = X(x)Y(y)T(t) \]
2. **Substituting into the PDE**:
\[ K(X''Y T + XY'' T) = XY T' \]
3. **Separating variables**:
\[ \frac{X''}{X} = \frac{-Y''}{Y} + \frac{T'}{kT} \]
\[ \frac{X''}{X} = -\lambda \]
4. **Solving for X**:
\[ X'' + \lambda X = 0 \]
5. **Solving for Y**:
\[ \frac{Y''}{Y} = \frac{T'}{kT} + \lambda \]
\[ Y'' + \mu Y = 0 \]
where \( \mu = -\lambda \).
6. **Solving for T**:
\[ \frac{T'}{kT} = -(\lambda + \mu) \]
\[ T' + k(\lambda + \mu)T = 0 \]
By solving the above separated ordinary differential equations (ODEs) using appropriate methods, we can find the functions \( X(x) \), \( Y(y) \), and \( T(t) \), and thus determine the solution \( U(x,y,t) \) of the given PDE under the given boundary conditions](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca708747-373f-4dbe-b127-10492ca0e68f%2Fc301b4bc-553a-4acb-b5f6-014535fe3e53%2F1domoyc_processed.png&w=3840&q=75)
Transcribed Image Text:### Homework 12.8: Problem 2
**Objective**: Solve the given partial differential equation (PDE)
**PDE and Boundary Conditions**:
\[ K(U_{xx} + U_{yy}) = U_t \]
**Boundary Conditions**:
\[A) \, U_x|_{x=0} = 0 \]
\[B) \, U_x|_{x=1} = 0 \]
\[C) \, U_y|_{y=0} = 0 \]
\[D) \, U_y|_{y=1} = 0 \]
\[E) \, U(x,y,0) = xy \]
**Diagram**:
The diagram depicts a square from (0,0) to (1,1) with the variable \( U(x,y,t) \) defined within it.
**Solution**:
1. **Setting up the solution**:
\[ \text{Set} \, U(x,y,t) = X(x)Y(y)T(t) \]
2. **Substituting into the PDE**:
\[ K(X''Y T + XY'' T) = XY T' \]
3. **Separating variables**:
\[ \frac{X''}{X} = \frac{-Y''}{Y} + \frac{T'}{kT} \]
\[ \frac{X''}{X} = -\lambda \]
4. **Solving for X**:
\[ X'' + \lambda X = 0 \]
5. **Solving for Y**:
\[ \frac{Y''}{Y} = \frac{T'}{kT} + \lambda \]
\[ Y'' + \mu Y = 0 \]
where \( \mu = -\lambda \).
6. **Solving for T**:
\[ \frac{T'}{kT} = -(\lambda + \mu) \]
\[ T' + k(\lambda + \mu)T = 0 \]
By solving the above separated ordinary differential equations (ODEs) using appropriate methods, we can find the functions \( X(x) \), \( Y(y) \), and \( T(t) \), and thus determine the solution \( U(x,y,t) \) of the given PDE under the given boundary conditions
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