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Chapter 11: Problem 21

Use eigenfunction expansions to find the solution of the given boundary value problem. $$ \begin{array}{l}{u_{t}=u_{x x}+1-|1-2 x|, \quad u(0, t)=0, \quad u(1, t)=0, \quad u(x, 0)=0} \\ {\text { see Problem } 5 .}\end{array} $$

Short Answer

Expert verified
Question: Find the final solution for the given boundary value problem with the equation \(u_{t} = u_{xx} + 1 - |1 - 2x|\), initial condition \(u(x, 0) = 0\), and boundary conditions \(u(0, t) = u(1, t) = 0\). Answer: The final solution for the given boundary value problem is \(u(x,t) = \sum_{n=1}^{\infty} e^{-n^2\pi^2 t} \left[ \int_0^t e^{n^2\pi^2 \tau} a_n \,\mathrm{d}\tau\right] \sin(n\pi x)\), where \(a_n = \int_{0}^{1} (1 - |1-2x|)\sin(m\pi x)\,\mathrm{d}x\).

Step by step solution

01

Find the Eigenfunctions and Eigenvalues

To find the eigenfunctions and eigenvalues, we solve the following homogeneous boundary value problem for the eigenvalue \(\lambda\) and the eigenfunction \(\phi_n(x)\): $$ \begin{cases} \phi_n''(x) = -\lambda\phi_n(x), \\ \phi_n(0) = 0, \\ \phi_n(1) = 0. \end{cases} $$ The general solution of the ordinary differential equation is given by: $$\phi_n(x) = A\sin(\sqrt{\lambda}x) + B\cos(\sqrt{\lambda}x).$$ Using the boundary conditions, we obtain $$\phi_n(0) = A\sin(0) + B\cos(0) = B = 0,$$ and $$\phi_n(1) = A\sin(\sqrt{\lambda}) = 0.$$ From the last equation, we can observe that this has nontrivial solutions only when \(\sqrt{\lambda_n} = n\pi\). Therefore, the eigenvalues are \(\lambda_n = n^2\pi^2\) for \(n=1,2,3,...\). The corresponding eigenfunctions are then \(\phi_n(x) = \sin(n\pi x)\).
02

Write the solution as an Eigenfunction Expansion

We will write the solution \(u(x,t)\) as a sum of the eigenfunctions times their corresponding time-dependent coefficients: $$u(x,t) = \sum_{n=1}^{\infty} b_n(t)\sin(n\pi x).$$ Now, we can differentiate \(u(x,t)\) twice with respect to x and once with respect to t: $$u_t(x,t) = \sum_{n=1}^{\infty} b_n'(t)\sin(n\pi x),$$ $$u_{xx}(x,t) = -\sum_{n=1}^{\infty} n^2\pi^2 b_n(t)\sin(n\pi x).$$ Now, substitute these expressions into the given equation: $$u_{t} = u_{xx} + 1 - |1 - 2x|.$$
03

Find the coefficients of the Eigenfunction Expansion

To find the coefficients \(b_n(t)\), we multiply both sides of the equation by \(\sin(m\pi x)\), integrate from 0 to 1, and then use orthogonality to form a system of ordinary differential equations: $$\int_{0}^{1} (u_t \sin(m\pi x))\,\mathrm{d}x = \int_{0}^{1} (u_{xx} + 1 - |1-2x|) \sin(m\pi x)\,\mathrm{d}x.$$ On the left-hand side, we obtain: $$b_m'(t) = -m^2\pi^2 b_m(t) + a_m,$$ where $$a_m = \int_{0}^{1} (1 - |1-2x|)\sin(m\pi x)\,\mathrm{d}x.$$ Now, we can solve the resulting first-order, linear, inhomogeneous ordinary differential equation: $$b_m'(t) + m^2\pi^2 b_m(t) = a_m.$$ Solving this equation, we get: $$b_m(t) = e^{-m^2\pi^2 t} \int_0^t e^{m^2\pi^2 \tau} a_m \,\mathrm{d}\tau + b_m(0),$$ where \(b_m(0)\) is the initial condition.
04

Apply the initial condition

We apply the initial condition \(u(x,0) = 0\), which gives: $$\sum_{n=1}^{\infty} b_n(0)\sin(n\pi x) = 0.$$ Since the sine functions are orthogonal, we get \(b_n(0) = 0\) for all \(n = 1, 2, 3,...\). Therefore, the final solution is: $$u(x,t) = \sum_{n=1}^{\infty} e^{-n^2\pi^2 t} \left[ \int_0^t e^{n^2\pi^2 \tau} a_n \,\mathrm{d}\tau\right] \sin(n\pi x).$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boundary Value Problem
A boundary value problem is a type of differential equation paired with a set of additional conditions, termed boundary conditions. These conditions are imposed at the boundaries of the domain in which the solution is sought.

In the context of our exercise, the differential equation involves both a spatial variable, denoted as \(x\), and a time variable, denoted as \(t\). The problem statement provides boundary conditions: \(u(0, t) = 0\) and \(u(1, t) = 0\), which dictate the behavior of the solution at the edges of the spatial domain (from \(x = 0\) to \(x = 1\)).These conditions help in uniquely identifying the function \(u(x, t)\) by restricting its possible forms. Understanding boundary conditions is crucial because they ensure that solutions fit within the specific physical context or real-world scenario the differential equation models.
Eigenvalues and Eigenfunctions
When dealing with linear operators, eigenvalues and eigenfunctions are key concepts. They arise naturally in the context of solving boundary value problems and are crucial for the eigenfunction expansion method used in the solution.

During the resolution of the homogeneous boundary value problem in the provided solution, the goal is to find functions \(\phi_n(x)\) such that the differential operator applied to \(\phi_n(x)\) yields the same function scaled by a constant \(\lambda\). These \(\lambda\)'s are known as eigenvalues, while \(\phi_n(x)\) are the corresponding eigenfunctions.The specification that \(\phi_n(0) = 0\) and \(\phi_n(1) = 0\) further constrains the solution, leading to the expression \(\phi_n(x) = \sin(n\pi x)\). Here, \(\lambda_n = n^2\pi^2\) are the quantized eigenvalues for this particular problem.
Orthogonality
Orthogonality is a critical concept when dealing with functions in mathematical physics and engineering, especially in the context of solving differential equations.

In simple terms, two functions are said to be orthogonal if their inner product (integral of their product over a certain interval) is zero. In this exercise, the sine functions \(\sin(n\pi x)\) are orthogonal over the interval \([0, 1]\). This characteristic is incredibly useful because it simplifies the process of solving problems.
  • It allows separation of different modes, each identified by a unique eigenvalue.
  • Each component of the solution corresponding to an eigenfunction can be treated independently.
In the solution, orthogonality aids in determining the coefficients \(b_n(t)\), which simplify into independent equations. It assures that when the series is summed up, each mode contributes uniquely and distinctly to the solution without interference from other modes.
Inhomogeneous Differential Equation
An inhomogeneous differential equation features a non-zero term independent of the solution function, making them more complex than their homogeneous counterparts.

In this exercise, the term \(1 - |1 - 2x|\) represents the inhomogeneity. This introduces an external effect that can be thought of as the system being 'forced' by an external input. Solving such equations involves finding a particular solution that accounts for this non-trivial contribution, in addition to any solutions to the homogeneous part.
  • First, solve the homogeneous equation to find the eigenfunctions and eigenvalues, which build the structure of the problem.
  • Next, use these results to construct a solution to the inhomogeneous equation, often by finding particular solutions matched with the eigenfunctions.
The linear system with inhomogeneous terms introduces more challenging interdependencies within the solution process. However, the method of separating variables, eigenfunction expansions, and the utilization of orthogonality are powerful tools to simplify and address the problem efficiently.

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Most popular questions from this chapter

In each of Problems I through 6 state whether the given boundary value problem is homogeneous or non homogeneous. $$ y^{\prime \prime}+4 y=0, \quad y(-1)=0, \quad y(1)=0 $$

Find the displacement \(u(r, t)\) in a vibrating circular elastic membrane of radius 1 that satisfies the boundary condition $$ u(1, t)=0, \quad t \geq 0 $$ and the initial conditions $$ u(r, 0)=0, \quad u_{t}(r, 0)=g(r), \quad 0 \leq r \leq 1 $$ where \(g(1)=0\) Hint: The differential equation to be satisfied is Eq. ( 3 ) of the text.

By a procedure similar to that in Problem 28 show that the solution of the boundary value problem $$ -\left(y^{\prime \prime}+y\right)=f(x), \quad y(0)=0, \quad y(1)=0 $$ is $$ y=\phi(x)=\int_{0}^{1} G(x, s) f(s) d s $$ where $$ G(x, s)=\left\\{\begin{array}{ll}{\frac{\sin s \sin (1-x)}{\sin 1},} & {0 \leq s \leq x} \\ {\frac{\sin x \sin (1-s)}{\sin 1},} & {x \leq s \leq 1}\end{array}\right. $$

Consider the boundary value problem $$ r(x) u_{t}=\left[p(x) u_{x}\right]_{x}-q(x) u+F(x) $$ $$ u(0, t)=T_{1}, \quad u(1, t)=T_{2}, \quad u(x, 0)=f(x) $$ (a) Let \(v(x)\) be a solution of the problem $$ \left[p(x) v^{\prime}\right]-q(x) v=-F(x), \quad v(0)=T_{1}, \quad v(1)=T_{2} $$ If \(w(x, t)=u(x, t)-v(x),\) find the boundary value problem satisfied by \(w\), Note that this problem can be solved by the method of this section. (b) Generalize the procedure of part (a) to the case \(u\) satisfies the boundary conditions $$ u_{x}(0, t)-h_{1} u(0, t)=T_{1}, \quad u_{x}(1, t)+h_{2} u(1, t)=T_{2} $$

Determine a formal eigenfunction series expansion for the solution of the given problem. Assume that \(f\) satisfies the conditions of Theorem \(11.3 .1 .\) State the values of \(\mu\) for which the solution exists. $$ y^{\prime \prime}+\mu y=-f(x), \quad y^{\prime}(0)=0, \quad y^{\prime}(1)+y(1)=0 $$
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