Question

Use eigenfunction expansions to find the solution of the given boundary value problem. $$ \begin{array}{l}{u_{t}=u_{x x}+e^{-t}(1-x), \quad u(0, t)=0, \quad u_{x}(1, t)=0, \quad u(x, 0)=0} \\ {\text { see Section } 11.2, \text { Problems } 6 \text { and } 7 .}\end{array} $$

Short answer

Answer: The final solution is $$u(x,t) = \sum_{n=1}^{\infty}c_n \sin\left(\frac{(2n-1)\pi}{2}x\right)e^{-\left(\frac{(2n-1)\pi}{2}\right)^2t} + e^{-t/2}\left(\frac{(1-x)(e^{-x\sqrt{\beta}/2} - e^{-x\sqrt{\beta}})}{2\beta}\right)$$

Step-by-step solution

Solve the homogeneous part

Find the solution \(v(x,t)\) to the homogeneous part: $$v_t = v_{xx}, \quad v(0,t)=0, \quad v_x(1,t) = 0$$ In order to solve this, we will first find the eigenfunctions and eigenvalues of the spatial component, solving the following Sturm-Liouville problem: $$X''(x) + \lambda X(x) = 0$$ with boundary conditions: $$X(0) = 0, \quad X'(1) = 0$$ The eigenfunctions and eigenvalues are given by: $$X_n(x) = \sin\left(\frac{(2n-1)\pi}{2}x\right), \quad \lambda_n = \left(\frac{(2n-1)\pi}{2}\right)^2$$ Now assume the solution \(v(x,t)\) has the form \(v(x,t) = X(x)T(t)\). Substituting into the homogenous part of the PDE, we have: $$T(t)X''(x) + \lambda X(x)T(t) = 0$$ Divide by \(X(x)T(t)\) to obtain : $$\frac{T'(t)}{T(t)} = -\lambda \frac{X''(x)}{X(x)}$$ Thus, only depending on t, we have: $$T'(t) = -\lambda T(t) $$ which is a first order homogeneous ODE in T(t). The solution is given by: $$T(t) = Ce^{-\lambda t}$$ where C is a constant. So the solution of the homogeneous part is: $$v(x, t) = \sum_{n=1}^{\infty}c_n \sin\left(\frac{(2n-1)\pi}{2}x\right)e^{-\left(\frac{(2n-1)\pi}{2}\right)^2t}$$

Solve the inhomogeneous part

Now we need to find a particular solution for the inhomogeneous part. A good candidate for this is a particular solution of the form: $$u_p(x, t) = f(t)g(x)$$ with \(u_p(0, t) = 0,\ u_p(x, 0) = 0\) Substituting the trial function into the PDE \(u_t = u_{xx} + e^{-t}(1-x)\), we have: $$f'(t)g(x) = f(t)g''(x) + e^{-t}(1-x)$$ Divide by \(f(t)g(x)\) to obtain: $$\frac{f'}{f} = \frac{g''}{g} + \frac{e^{-t}(1-x)}{g(x)f(t)}$$ The left-hand side depends only on \(t\), and the right-hand side depends only on \(x\). So, we need to find a function \(f(t)\) which satisfies: $$f'(t) = e^{-t}f(t)$$ Solving this first order linear ODE with the initial condition \(f(0) = 1\), we get: $$f(t) = e^{-t/2}$$ Then we consider the equation: $$g''(x) = \beta g(x) -e^{-x/2}(1-x) $$ The solution of this kind can be found by solving the homogeneous problem first, finding the complementary solution, and then finding the particular solution. In our case, $$g_c(x) = A\sin(\sqrt{\beta}x) + B\cos(\sqrt{\beta}x)$$ with \(g_c(1) = 0\) for the solution to meet the boundary condition. Then find the particular solution: $$g_p(x) = \frac{(1-x)(e^{-x/2}\sqrt{\beta} - e^{-x})}{2\beta}$$ Now we have the particular solution for the inhomogeneous problem: $$u_p(x,t) = e^{-t/2}\left(\frac{(1-x)(e^{-x\sqrt{\beta}/2} - e^{-x\sqrt{\beta}})}{2\beta}\right)$$

Sum up the general solution and particular solution

The final solution for the given PDE \(u_t = u_{xx} + e^{-t}(1-x)\) is the sum of the general solution \(v(x,t)\) and the particular solution \(u_p(x,t)\): $$u(x,t) = \sum_{n=1}^{\infty}c_n \sin\left(\frac{(2n-1)\pi}{2}x\right)e^{-\left(\frac{(2n-1)\pi}{2}\right)^2t} + e^{-t/2}\left(\frac{(1-x)(e^{-x\sqrt{\beta}/2} - e^{-x\sqrt{\beta}})}{2\beta}\right)$$

Key concepts

Boundary Value Problem

Boundary value problems are special types of differential equations that include boundary conditions. These conditions specify values or behavior of the solution at certain points, typically the endpoints of the domain where the solution is defined. For instance, in the exercise given, the boundary conditions are \( u(0, t) = 0 \) and \( u_x(1, t) = 0 \) which essentially mean that at \( x = 0 \), the function \( u \) is zero and the derivative of \( u \) is zero at \( x = 1 \).
This type of problem is crucial because it represents real-world scenarios where conditions at boundaries limit the behavior of the system being modeled, such as temperature distributions along a rod or an electric field in a confined space. Effectively solving boundary value problems allows engineers and scientists to predict system behaviors more accurately.

Sturm-Liouville Problem

A Sturm-Liouville problem is a type of boundary value problem that can be reduced to a special form involving an eigenvalue parameter. These problems are integral to solving differential equations using eigenfunction expansions, such as the one in the original exercise. Essentially, a Sturm-Liouville problem looks like \( (py')' + (q + \lambda r)y = 0 \), where \( \lambda \) is the eigenvalue and \( p, q, \, \text{and}\, r \) are known functions.
In the exercise, we find eigenfunctions and eigenvalues by solving a simpler equation \( X''(x) + \lambda X(x) = 0 \) with appropriate boundary conditions. These eigenfunctions form a basis for the solution space, meaning we can express solutions as infinite series. This becomes useful when looking for series solutions to partial differential equations.

Homogeneous and Inhomogeneous Solutions

A boundary value problem can be broken into homogeneous and inhomogeneous components. The homogeneous part is the equation set to zero (like solving \( v_t = v_{xx} \)), whereas the inhomogeneous part includes the non-zero terms (such as \( u_t = u_{xx} + e^{-t}(1-x) \)).

• Homogeneous solutions often revolve around determining the response of the system to initial conditions without external input. For this purpose, the solution often takes the form of a series using eigenfunctions derived from the Sturm-Liouville problem.
• Inhomogeneous solutions, on the other hand, involve finding a particular solution to the non-zero part of the equation. This is often achieved by guessing a form for the solution and fitting it into the differential equation with appropriate terms to balance both sides.

Understanding these components is vital because real-world problems often involve both natural responses of a system (homogeneous) and external influences (inhomogeneous).

Partial Differential Equations

Partial differential equations (PDEs) involve functions of several variables and their partial derivatives. They are fundamental in mathematical physics as they model phenomena where change depends on multiple variables, such as time and space.
In the given problem, \( u_t = u_{xx} + e^{-t}(1-x) \), the PDE describes how some quantity \( u \) (which could represent temperature, concentration, etc.) changes over space \( x \) and time \( t \). The first part \( u_{xx} \) suggests diffusion or spreading, while the term \( e^{-t}(1-x) \) implies a source term varying over both space and time.
Solving PDEs often requires techniques like separation of variables, transform methods, and using eigenfunction expansions. These techniques simplify complex problems by breaking them into solvable components that can be analyzed separately, as shown in the step by step solution, before being synthesized into an overall solution.