Chapter 18: Problem 11
Solve \(u(x)=\lambda \int_{0}^{\infty} K(x, t) u(t) d t+x^{\alpha}\), where \(\alpha\) is any real number except a negative integer, and \(K(x, t)=e^{-(x+t)} .\) For what values of \(\lambda\) does the integral equation have a solution?
Short Answer
Expert verified
\( \lambda \) can take any real value if \( 0 \leq x < 1/\alpha \), imaginary value if \( x > 1/\alpha \), and does not exist if \( x = 1/\alpha \).
Step by step solution
01
Recognize the Homoegenous Nature of the Equation
We can see that if \( u(t) \) is a solution of the equation, then so is \( Cu(t) \) for any constant \( C \). This is because this is a homogeneous equation.
02
Apply the Equation for \( u(x) \)
Substitute \( u(x) \) in terms of the integral equation to get \( u(x)=\lambda \int_{0}^{\infty} K(x, t) [ \lambda \int_{0}^{\infty} K(t, s) u(s) ds ] dt + x^{\alpha} \). Then interchange the order of integration, remember the limits of integration go from 0 to infinity.
03
Break Down Kernel Function
Notice that \( K(x, t) = K(t, s) = e^{-(x+t)} = e^{-x}e^{-t} = e^{-s}e^{-t} \), which is separable.
04
Simplify Equation
Simplify equation from step 2 to \( u(x) = \lambda^2 \int_{0}^{\infty} \int_{0}^{\infty} e^{-x}e^{-t} e^{-s}e^{-t} u(s) ds dt + x^{\alpha} = \lambda^2 e^{-x} \int_{0}^{\infty} \int_{0}^{\infty} e^{-2t} e^{-s} u(s) ds dt + x^{\alpha} \). Notice that \( e^{-x} \) can be moved out of the integral because it does not depend on \( t \) or \( s \).
05
Solve the Inner Integral
The inner integral over \( s \) can be solved as \( \int_{0}^{\infty} e^{-s} u(s) ds = 1 \) (since \( u(s) \) is normalized). Substitute this into the equation.
06
Solve for Lambda
Now, re-arrange the equation to solve for \( \lambda \), we get \( \lambda = \sqrt{1 - x^{\alpha}} \).
07
Analyze the Solution Range
The solution is real for \( 0 \leq x < 1/\alpha \), imaginary for \( x > 1/\alpha \), and does not exist for \( x = 1/\alpha \). Thus, \( \lambda \) could take any value within real numbers when \( 0 \leq x < 1/\alpha \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Kernel Function
In integral equations, the kernel function plays a pivotal role in linking the variables of integration with the function we desire to solve for. A kernel function typically appears as a part of an integral, providing the framework over which the equation operates. In our specific exercise, the kernel function is given by \[ K(x, t) = e^{-(x+t)} \].
This kernel is notable for its separability, a property meaning it can be expressed as a product of two functions, each in one variable: \[ K(x, t) = e^{-x} \cdot e^{-t} \].
Separable kernels like this one simplify the analysis and computation of integral equations because they allow integration to be carried out more straightforwardly across individual variables. Recognizing kernels' structure is often crucial in simplifying these complex equations. This simplicity aids in evaluating integrals that could otherwise be cumbersome.
This kernel is notable for its separability, a property meaning it can be expressed as a product of two functions, each in one variable: \[ K(x, t) = e^{-x} \cdot e^{-t} \].
Separable kernels like this one simplify the analysis and computation of integral equations because they allow integration to be carried out more straightforwardly across individual variables. Recognizing kernels' structure is often crucial in simplifying these complex equations. This simplicity aids in evaluating integrals that could otherwise be cumbersome.
Homogeneous Equations
An integral equation is termed homogeneous when it maintains proportionality across its solutions. This means if one solution is found, any scalar multiple of that solution is also a solution, which demonstrates linearity. Our exercise provides an example of a homogeneous equation when examined as \[ u(x) = \lambda \int_{0}^{\infty} K(x, t) u(t) dt \].
By examining this part alone, we see that if \( u(t) \) solves the equation, then \( Cu(t) \) is also a solution for any constant \( C \). Homogeneity is beneficial as it allows a broader exploration of potential solutions through the use of constants, aiding in both theoretical analysis and practical applications of integral equations.
Recognizing the homogeneity also helps in simplifying problem-solving steps because it indicates that transformations and scalings of solutions are valid under the same conditions.
By examining this part alone, we see that if \( u(t) \) solves the equation, then \( Cu(t) \) is also a solution for any constant \( C \). Homogeneity is beneficial as it allows a broader exploration of potential solutions through the use of constants, aiding in both theoretical analysis and practical applications of integral equations.
Recognizing the homogeneity also helps in simplifying problem-solving steps because it indicates that transformations and scalings of solutions are valid under the same conditions.
Solution Range Analysis
Analyzing the solution range involves understanding the conditions under which solutions exist and are meaningful. For the given integral equation, the focus is on finding viable values of \( \lambda \) that yield real solutions. The problem confines the analysis to the variable \( x \) and its relation to the exponent \( \alpha \), leading to specific solution conditions.
Through solving the equation, we determine real solutions exist in the range \[ 0 \leq x < \frac{1}{\alpha} \].
Outside this range, depending on the value of \( x \), solutions could become imaginary or undefined. Specifically, the solution becomes imaginary if \( x > \frac{1}{\alpha} \) and nonexistent at the point \( x = \frac{1}{\alpha} \). This analysis is vital as it identifies both the existence and type of solutions that can occur.
Such insights into behavior and restrictions are critical when applying these mathematical constructs to real-world models where certain parameter ranges must be adhered to or avoided entirely.
Through solving the equation, we determine real solutions exist in the range \[ 0 \leq x < \frac{1}{\alpha} \].
Outside this range, depending on the value of \( x \), solutions could become imaginary or undefined. Specifically, the solution becomes imaginary if \( x > \frac{1}{\alpha} \) and nonexistent at the point \( x = \frac{1}{\alpha} \). This analysis is vital as it identifies both the existence and type of solutions that can occur.
Such insights into behavior and restrictions are critical when applying these mathematical constructs to real-world models where certain parameter ranges must be adhered to or avoided entirely.