Question

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

Short answer

Question: Determine the solution of the given boundary value problem using eigenfunction expansions. Boundary Value Problem: $$u_t = u_{xx} - x$$ $$u(0, t) = 0, \quad u_x(1, t) = 0$$ $$u(x, 0) = \sin(\frac{\pi x}{2})$$ Answer: The solution of the given boundary value problem is $$u(x, t) = e^{-(\frac{\pi}{2})^2 t} \sin(\frac{\pi x}{2})$$.

Step-by-step solution

Determine the Eigenvalues and Eigenfunctions

The eigenvalues and eigenfunctions are helpful here as they help simplify the PDE. We can determine them by solving the eigenvalue problem: $$ X'' + \lambda X = 0 \\ X(0) = 0, \quad X'(1) = 0 $$ The general solution to this problem is given by: $$X(x) = A \sin(\sqrt{\lambda} x) + B \cos(\sqrt{\lambda} x)$$ Applying the boundary conditions, we get: $$X(0) = A \cdot 0 + B \cdot 1 = 0 \Rightarrow B = 0$$ $$X'(x) = A \sqrt{\lambda} \cos(\sqrt{\lambda} x)$$ $$X'(1) = A \sqrt{\lambda} \cos(\sqrt{\lambda}) = 0$$ As \(A \neq 0\), then \(\cos(\sqrt{\lambda}) = 0\). Thus, the eigenvalues are: $$\lambda_n = ((2n - 1)\pi/2)^2, \quad n = 1, 2, 3, \cdots$$ And their corresponding eigenfunctions are: $$X_n(x) = \sin(\frac{(2n - 1)\pi}{2} x), \quad n = 1, 2, 3, \cdots$$

Expand the Initial Condition Using Eigenfunctions

We want to represent the initial condition \(u(x, 0) = \sin(\frac{\pi x}{2})\) as a linear combination of the eigenfunctions determined in Step 1. We can write it as: $$u(x, 0) = \sum_{n = 1}^{\infty} a_n \sin(\frac{(2n - 1)\pi}{2} x)$$ To find the coefficients \(a_n\), we can use the orthogonality of the eigenfunctions: $$a_n = \int_0^1 u(x, 0) \sin(\frac{(2n - 1)\pi}{2} x) \, dx$$ $$a_n = \int_0^1 \sin(\frac{\pi x}{2}) \sin(\frac{(2n - 1)\pi}{2} x) \, dx$$ Since \(u(x, 0)\) is equal to the first eigenfunction (\(n=1\)) only, we finally have: $$a_n = \begin{cases} 1, & \text{if } n = 1 \\ 0, & \text{if } n \geq 2 \end{cases}$$

Derive the Final Solution

Now, we have the coefficients to construct the eigenfunction expansion for the general solution. The general solution can be expressed as: $$u(x, t) = \sum_{n = 1}^{\infty} c_n(t) \sin(\frac{(2n - 1)\pi}{2} x)$$ We plug the general solution into the PDE: $$u_t = u_{xx} - x \Rightarrow \sum_{n = 1}^{\infty} c_n'(t) \sin(\frac{(2n - 1)\pi}{2} x) = \sum_{n = 1}^{\infty} -\lambda_n c_n(t) \sin(\frac{(2n - 1)\pi}{2} x) - x$$ Comparing the coefficients of the eigenfunctions, we obtain differential equations for the \(c_n(t)\) coefficients: $$c_n'(t) = -\lambda_n c_n(t), \quad n = 1, 2, 3, \cdots$$ Solve these differential equations with the initial conditions \(c_n(0) = a_n\): \[ c_n(t) = e^{-\lambda_n t}, \quad \text{if } n = 1 \\ c_n(t) = 0, \quad \text{if } n \geq 2 \] Finally, the solution of the boundary value problem is given by: $$u(x, t) = e^{-(\frac{\pi}{2})^2 t} \sin(\frac{\pi x}{2})$$

Key concepts

Partial Differential Equations

Partial Differential Equations (PDEs) are mathematical expressions involving rates of change with respect to continuous variables. A PDE for a function of two or more variables consists of partial derivatives, which are derivatives with respect to each variable while keeping the others constant.

For instance, the equation from the given exercise, \( u_{t} = u_{xx} - x \) involves partial derivatives of the function \( u \) with respect to time \( t \) and space \( x \), designated by \( u_t \) and \( u_{xx} \), respectively. In this type of equation, \( u_t \) represents the temporal rate of change of the function \( u \), whereas \( u_{xx} \) signifies the spatial rate of change, in this case, the second derivative with respect to \( x \) which suggests concavity or convexity at a certain point.

PDEs are instrumental in formulating problems involving functions of several variables, and are particularly used in the fields of physics, engineering, and financial modeling. An example includes the heat equation, which in this case describes how the distribution of temperature \( u \) changes over time and position along a rod.

Eigenvalues and Eigenfunctions

Eigenvalues and eigenfunctions play a pivotal role when dealing with boundary value problems that involve PDEs. An eigenfunction is a non-zero function that, when substituted into a differential equation, results in the function being scaled by some amount, called an eigenvalue. In other words, for a given operator \( \mathcal{L} \) and eigenfunction \( X \), the relationship \( \mathcal{L}X = \lambda X \) must hold, with \( \lambda \) being the eigenvalue.

In the context of the provided exercise, to find the eigenvalues \( \lambda_n \) and the corresponding eigenfunctions \( X_n \) for our boundary value problem, we look for solutions \( X(x) \) of the homogeneous equation \( X'' + \lambda X = 0 \) that satisfy the given boundary conditions. These eigenfunctions form an orthogonal basis, with respect to which we can expand our solution, and they often result from the quantization of the physical space or system. The orthogonality of eigenfunctions refers to the fact that the integral of the product of any two distinct eigenfunctions over the domain will be zero, a property that is central to finding coefficients in eigenfunction expansions.

Eigenfunction Expansions

Eigenfunction expansions involve writing a function as an infinite series of eigenfunctions. This method is particularly useful for solving PDEs, as it transforms a complex problem into an infinite set of simpler ordinary differential equations, each associated with a respective eigenvalue.

In the solution process for the problem at hand, we expanded the initial condition \( u(x, 0) = \sin(\pi x / 2) \) using the eigenfunctions \( X_n(x) \) found earlier. This expansion is possible due to the completeness of the eigenfunctions, meaning they span the function space in question, which allows us to represent any function within that space as a combination of these eigenfunctions. The coefficients \( a_n \) of the expansion are determined by taking advantage of the orthogonality property of the eigenfunctions.

After computing the coefficients, we insert the eigenfunction expansion back into the PDE. This allows us to find \( c_n(t) \)—temporal coefficients—which evolve according to another ODE linked to each eigenvalue. The final sum of these products gives us the solution \( u(x, t) \) to the original boundary value problem.