Problem 1
Question
Consider $$ c \rho \frac{\partial u}{\partial t}=\frac{\partial}{\partial x}\left(K_{0} \frac{\partial u}{\partial x}\right)+\alpha u $$ where \(c, \rho, K_{0}, \alpha\) are functions of \(x\), subject to $$ \begin{aligned} &u(0, t)=0 \\ &u(L, t)=0 \\ &u(x, 0)=f(x) \end{aligned} $$ Assume that the appropriate eigenfunctions are known. (a) Show that the eigenvalues are positive if \(\alpha<0\) (see Sec. 5.2.1). (b) Solve the initial value problem. (c) Briefly discuss \(\lim _{t \rightarrow \infty} u(x, t)\).
Step-by-Step Solution
Verified Answer
The eigenvalues of the given problem are positive when \( \alpha < 0 \). The solution of the initial value problem is a series of the eigenfunctions with coefficients determined by the initial condition and orthogonality. The solution \( u(x, t) \) approaches zero as \( t → ∞ \), with the speed dependent on the eigenvalues.
1Step 1: Show Eigenvalues are Positive
The given linear homogeneous partial differential equation has separated solutions of the form \( u(x, t) = X(x)T(t) \). Plugging these into the PDE gives the separated equations \( T' = -λCT \) and \( X'' + λRX = 0 \), where \( λ > 0 \). Since \( \alpha < 0 \) and \( R = α/c \), it is clear that \( λ > 0 \). Thus, the eigenvalues of this problem are positive when \( \alpha < 0 \).
2Step 2: Solve the Initial Value Problem
The initial condition \( u(x, 0) = f(x) \) allows us to write the total solution as a series of the eigenfunctions, \( u(x, t) = ∑a_n φ_n(x) \), where the coefficients \( a_n \) are determined by orthogonality of the eigenfunctions in regards to the initial condition and taking the dot product of both sides with respect to each eigenfunction.
3Step 3: Discuss the Limit as t Tends to Infinity
The limit as \( t → ∞ \) of \( u(x, t) \) describes the equilibrium solution. The time-dependent factor in the separated solution is \( T(t) = e^{-λ nt} \), which decays to zero as \( t → ∞ \) for each \( n \). Therefore, the solution \( u(x, t) \) will approach zero as \( t → ∞ \), regardless of the form of \( x \). The speed at which it approaches zero will be determined by the eigenvalues \( λ_n \).
Key Concepts
EigenvaluesInitial Value ProblemEquilibrium Solution
Eigenvalues
Eigenvalues play a crucial role in solving partial differential equations (PDEs), especially when dealing with problems involving boundary conditions. In the context of our equation, eigenvalues, denoted by \( \lambda \), arise when we look for solutions of the form \( u(x, t) = X(x)T(t) \). This assumption allows us to separate the PDE into two ordinary differential equations (ODEs), one for \( X(x) \) and another for \( T(t) \).
- For the spatial part \( X'' + \lambda RX = 0 \), the eigenvalues \( \lambda \) affect the behavior and solutions of \( X(x) \).
- Given the condition \( \alpha < 0 \), it implies that \( \lambda > 0 \). This is because \( R = \alpha/c \) becomes negative, leading the negative eigenvalues to make the spatial solution stable.
Initial Value Problem
An Initial Value Problem (IVP) involves finding a solution to a PDE given some initial conditions at \( t = 0 \). For our equation, the initial condition is \( u(x, 0) = f(x) \).Solving an IVP often requires expressing the solution as a combination or series of eigenfunctions, reflecting how \( u(x, t) \) changes over time from the initial state.
- The total solution uses a series solution \( u(x, t) = \sum a_n \phi_n(x)T_n(t) \). Each term represents an eigenfunction weighted by coefficients \( a_n \).
- These coefficients \( a_n \) are determined so that they satisfy the initial condition and adhere to the orthogonality property of eigenfunctions.
Equilibrium Solution
An equilibrium solution refers to the state reached as time progresses toward infinity, where changes cease, and the system typically stabilizes.For our PDE, considering the limit as \( t \to \infty \), the time-dependent part of the solution, \( T(t) = e^{-\lambda_n t} \), signifies the decay of transient effects.
- As \( t \to \infty \), every term in the sum \( u(x, t) = \sum a_n \phi_n(x)e^{-\lambda_n t} \) tends toward zero.
- This decay results from positive eigenvalues \( \lambda_n \), causing exponential terms to approach zero, indicating dampening of transient behaviors.
Other exercises in this chapter
Problem 1
Use the Rayleigh quotient to obtain a (reasonably accurate) upper bound for the lowest eigenvalue of (a) \(\frac{d^{2} \phi}{d x^{2}}+\left(\lambda-x^{2}\right)
View solution Problem 1
A Sturm-Liouville eigenvalue problem is called self-adjoint if $$ \left.p\left(u \frac{d v}{d x}-v \frac{d u}{d x}\right)\right|_{a} ^{b}=0 $$ (since then \(\in
View solution Problem 2
Obtain a formula for an infinite series using Parseval's equality applied to the: (a) Fouricr sine scrics of \(f(x)-1\) on the interval \(0
View solution Problem 2
Consider $$ \frac{d^{2} \phi}{d x^{2}}+\lambda(1+x) \phi=0 $$ subject to \(\phi(0)-0\) and \(\phi(1)-0\). Roughly sketch the eigenfunctions for \(\lambda\) laig
View solution