Question

Solve the initial-boundaryvalue problem. In some of these exercises, Theorem 11.3.5(b) or Exercise 11.3.35 will simplify the computation of the coefficients in the Fourier sine series. $$ \begin{array}{ll} u_{t t}=4 u_{x x}, & 00 \\ u(0, t)=0, & u(1, t)=0, \quad t>0 \\ u(x, 0)=0, & u_{t}(x, 0)=x\left(x^{3}-2 x^{2}+1\right), \quad 0 \leq x \leq 1 \end{array} $$

Short answer

Answer: The general form of the solution to the given wave equation with initial and boundary conditions is: $$u(x, t) = \sum_{n=1}^\infty 2n\pi A_n\sin{n\pi x}\sin{2n\pi t}$$, where the coefficients \(A_n\) are given by: $$A_n = \frac{2}{n\pi}\int_{0}^1 x(x^3 - 2x^2 + 1)\sin{n\pi x} \, dx$$.

Step-by-step solution

Recognize the problem

We are given a wave equation with initial and boundary conditions. The wave equation is given by \(u_{tt} = 4u_{xx}\), with the given initial and boundary conditions. We will apply these conditions and find the solution using the Fourier sine series method.

Write the wave equation

We have the wave equation: $$u_{tt} = 4u_{xx}$$

Apply boundary conditions

Applying the boundary conditions, we have: $$u(0, t)=0, \quad u(1, t)=0$$

Find the Fourier sine series

We use the separation of variables method and assume that the solution has the form: $$u(x, t) = X(x)T(t)$$ Plugging this into the wave equation and dividing both sides by \(4XT\), we obtain: $$\frac{T''}{4T}=\frac{X''}{X}=-\lambda^{2}$$ Solving these two ordinary differential equations we get: $$T(t) = A\sin{2\lambda t} + B\cos{2\lambda t}$$ $$X(x) = C\sin{\lambda x} + D\cos{\lambda x}$$ Applying the boundary conditions \(u(0, t)=0\) and \(u(1, t)=0\), we get \(D = 0\) and \(C\sin{\lambda} = 0\). As \(C\) cannot be zero, the boundary conditions force \(\lambda = n\pi\), with \(n\) a positive integer. Hence, we have: $$u(x, t) = \sum_{n=1}^{\infty} (A_n\sin{2n\pi t}+B_n\sin{n\pi x})$$

Use initial conditions and solve for coefficients

Now we use the initial conditions: $$u(x, 0) = 0$$ $$u_t(x, 0) = x(x^3 - 2x^2 + 1)$$ First, let us apply the first initial condition: $$0 = \sum_{n=1}^\infty B_n\sin(n\pi x)$$ This implies that \(B_n = 0\) for all \(n\). Now, applying the second initial condition: $$x(x^3 - 2x^2 + 1) = \sum_{n=1}^{\infty} 2n\pi A_n\sin{n\pi x}$$ To find coefficients \(A_n\), we multiply both sides by \(\sin{m\pi x}\), integrate from 0 to 1, and apply the orthogonality relation: $$A_n = \frac{2}{n\pi}\int_{0}^1 x(x^3 - 2x^2 + 1)\sin{n\pi x} \, dx$$

Write the final solution

Now that we have found the coefficients \(A_n\), we can write the final solution as: $$u(x, t) = \sum_{n=1}^\infty 2n\pi A_n\sin{n\pi x}\sin{2n\pi t}$$

Key concepts

Fourier Series

A Fourier series is a way to represent a function as a sum of sine and cosine terms. In the context of partial differential equations, like the wave equation, they help to express complex periodic functions simply and effectively.
The Fourier series is especially valuable when solving boundary value problems as it allows us to use orthogonal functions, like sines and cosines, which are solutions to simple equations. In the wave equation problem discussed, the function description of initial conditions can be replaced with its Fourier sine series expansion, leading to the calculation of individual terms ( $A_n$ in our case).
This helps achieve a simpler form that satisfies the conditions of the problem, thus making the problem easier to solve through separation of variables.

Boundary Value Problem

A boundary value problem involves finding a solution to a differential equation that satisfies certain conditions at the boundaries of the domain.
In the given wave equation problem, we have boundary conditions defined at the ends of the interval for the spatial variable $x$ . These conditions, $u(0, t) = 0$ and $u(1, t) = 0$ , mean that the solution $u(x, t)$ must be zero at $x = 0$ and $x = 1$ for all times $t > 0$ .
These boundary conditions are crucial as they define the shape and nature of the solutions that are physically meaningful, adjusting the coefficients of the sine functions used in the Fourier series to fit the problem's context.

Initial Conditions

Initial conditions specify the state of the system at the beginning of the observation or the solution process, typically at $t = 0$ .
In the given problem, you have two initial conditions: $u(x, 0) = 0$ and $u_t(x, 0) = x(x^3 - 2x^2 + 1)$ . The first condition implies that the displacement is initially zero everywhere, and the second condition relates to the initial velocity distribution, determining how the wave evolves over time.
By applying these initial conditions, especially using orthogonality of sine functions, coefficients in the Fourier series (like $A_n$ ) can be determined. Once these coefficients are found, they allow you to fully describe the system's state evolution using the series expansion.

Separation of Variables

Separation of variables is a method used to solve partial differential equations. It involves assuming that a solution can be written as a product of functions, each dependent on a single coordinate.
For the wave equation $u(x, t)$ , we assume it's separable into $X(x)T(t)$ , meaning one function depends only on $x$ and the other only on $t$ . By substituting this form into the wave equation, you end up with two separate ordinary differential equations: one for $X(x)$ and one for $T(t)$ .
Solving these ODEs leads to a critical step in determining the form of each function which, when recombined, respects the boundaries and initial conditions. This elegant technique simplifies complex PDE problems by reducing them into manageable parts, each with its distinct influence on the solution.