Question

In Exercises \(35-46\) use Exercise 34 to solve the initial-boundaryvalue problem. In some of these exercises Theorem 11.3.5(d) or Exercise 11.3.50(b) will simplify the computation of the coefficients in the mixed Fourier sine series. $$ \begin{array}{l} u_{t t}=64 u_{x x}, \quad 00, \\ u(0, t)=0, \quad u_{x}(\pi, t)=0, \quad t>0, \\ u(x, 0)=x(2 \pi-x), \quad u_{t}(x, 0)=0, \quad 0 \leq x \leq \pi. \end{array} $$

Short answer

Question: Find the solution of the given initial-boundary value problem using the mixed Fourier sine series method: $$ u_{tt} = 64u_{xx}, \quad 00, \\ u(0,t) = 0, \quad u_x(\pi,t)=0, \quad t>0, \\ u(x,0) = x(2\pi - x), \quad u_t(x,0)=0, \quad 0 \leq x \leq \pi. $$ Answer: The solution to the initial-boundary value problem is given by: $$ u(x,t) = \sum_{n=1}^{\infty}A_n\cos((2n-1)8t)\sin((2n-1)x), $$ where $$ A_m = \frac{2}{\pi}\int_{0}^{\pi}x(2\pi - x)\sin((2m-1)x)dx, \quad m=1,2,3,\dots. $$

Step-by-step solution

Represent the given PDE and boundary conditions

Let's first rewrite the given PDE and boundary conditions: $$ u_{tt} = 64u_{xx}, \quad 00, \\ u(0,t) = 0, \quad u_x(\pi,t)=0, \quad t>0, \\ u(x,0) = x(2\pi - x), \quad u_t(x,0)=0, \quad 0 \leq x \leq \pi. $$

Separation of variables

Apply the separation of variables technique by assuming that \(u(x,t) = X(x)T(t)\). Then, substitute this into the given PDE to get: $$ X(x)T''(t) = 64X''(x)T(t). $$ Divide both sides by \(64XT\) to obtain: $$ \frac{T''(t)}{64T(t)} = \frac{X''(x)}{X(x)}. $$ Since the left side is a function of time and the right side is a function of position, both sides must be equal to a constant, say, \(-\lambda^2\).

Finding X(x) and T(t)

Now, we have two ordinary differential equations (ODEs): $$ X''(x) + \lambda^2X(x) = 0, \\ T''(t) + 64\lambda^2T(t) = 0. $$ We can solve these ODEs to obtain their solutions. For \(T(t)\), we get: $$ T(t) = A\cos(8\lambda t) + B\sin(8\lambda t). $$ For the spatial part, we have to take into account the boundary conditions. For the fixed end at x=0, we have \(u(0,t)=0\), which implies \(X(0)=0\). The conditions for the free end at x=π require that \(u_x(\pi,t)=0\), or equivalently \(X'(\pi)=0\). With these conditions, the eigenvalue problem for \(X(x)\) becomes: $$ X''(x) + \lambda^2X(x) = 0, \quad X(0)=0, \quad X'(\pi)=0. $$ Solving this eigenvalue problem, we find the eigenvalues and corresponding eigenfunctions to be: $$ \lambda_n = (2n-1), \quad X_n(x) = \sin((2n-1)x), \quad n=1,2,3,\dots. $$

Finding the general solution

Now, we can write the general solution for the PDE as a sum of products of eigenfunctions: $$ u(x,t) = \sum_{n=1}^{\infty}\left[A_n\cos((2n-1)8t) + B_n\sin((2n-1)8t)\right]\sin((2n-1)x). $$

Applying initial conditions

Apply the initial conditions \(u(x,0) = x(2\pi - x)\) and \(u_t(x,0) = 0\) to find the coefficients \(A_n\) and \(B_n\): For \(u_t(x,0) = 0\), we get: $$ 0 = \sum_{n=1}^{\infty}(2n-1)8B_n\sin((2n-1)x), $$ which implies that \(B_n = 0\) for all \(n \geq 1\). For \(u(x,0) = x(2\pi - x)\), we get: $$ x(2\pi - x) = \sum_{n=1}^{\infty}A_n\sin((2n-1)x). $$ We can find Fourier sine coefficients by multiplying both sides by \(\sin((2m-1)x)\) and integrating from 0 to \(\pi\): $$ A_m = \frac{2}{\pi}\int_{0}^{\pi}x(2\pi - x)\sin((2m-1)x)dx, \quad m=1,2,3,\dots. $$

Final solution

With the coefficients \(A_n\) and \(B_n\), the final solution to the initial-boundary value problem is given by: $$ u(x,t) = \sum_{n=1}^{\infty}A_n\cos((2n-1)8t)\sin((2n-1)x). $$ Here, \(A_n\) is the Fourier sine coefficient found in step 5, and \(B_n = 0\) for all \(n \geq 1\).

Key concepts

Partial Differential Equations

Partial Differential Equations (PDEs) describe the relationship between an unknown function and its partial derivatives. PDEs are fundamental to modeling numerous phenomena across engineering, physics, and finance, such as heat conduction, wave propagation, and financial derivatives pricing.

The general form of a PDE involves derivatives with respect to more than one independent variable. In our exercise, the PDE is given as u_{tt} = 64u_{xx}, which describes the changes in a function u(x,t) over space and time. This particular equation is reminiscent of the wave equation, characterizing how waves, such as sound or light, propagate over time. To solve such an equation, one must also know initial conditions (state of the system at time zero) and boundary conditions (behavior of the system at the spatial endpoints).

The given PDE paired with its boundary and initial conditions constitutes an 'initial-boundary value problem', which in our context defines how the wave propagates within certain constraints.

Separation of Variables

Separation of variables is a standard method for solving PDEs, which works by reducing the PDE into simpler, separate ordinary differential equations (ODEs) that are easier to handle. This method banks on the idea of representing the solution as a product of functions, each depending solely on one of the independent variables.

In our case, with u(x, t) = X(x)T(t), we assume that the solution can be separated into a spatial part X(x) and a temporal part T(t). The substitution into the PDE yields two separate ODEs, each dependent on a single variable. This process transforms a complex multidimensional problem into more manageable one-dimensional problems, which can then be solved using standard techniques for ODEs.

Fourier Sine Series

The Fourier sine series is a tool used to represent a function defined in a finite interval as an infinite sum of sine functions. This approach is especially useful in solving PDEs because it allows for the representation of initial conditions which are otherwise difficult to incorporate into the solution.

For functions defined on the interval [0, \(pi\)], a Fourier sine series expansion is appropriate when the function satisfies certain symmetry properties, essentially mirroring the behavior about the origin. In our initial-boundary value problem, the initial condition u(x, 0) = x(2\(pi\)-x) can be represented as a sum of sine functions. The Fourier coefficients A_n are then determined by orthogonality properties of the sine functions, involving integrals which average out all but one term in the series.

Eigenvalue Problem

An eigenvalue problem in the context of PDEs involves finding solutions to differential equations that include a parameter, often lambda (\(\lambda\)), which only allows for non-trivial solutions under specific conditions—these permissible values are known as eigenvalues. The corresponding solutions are called eigenfunctions.

The boundary conditions signal if the beam's ends are fixed, free, or supported in a manner impacting the vibrational modes, leading to a unique set of eigenfunctions. In our exercise, the conditions u(0,t)=0 and u_x(\(pi\),t)=0 guide us to an eigenvalue problem where we solve for values of \(\lambda\) and associated eigenfunctions, which are instrumental in constructing the series solution to the PDE.