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. $$ \begin{array}{l} u_{t t}=4 u_{x x}, \quad 00 \\ u(0, t)=0, \quad u(1, t)=0, \quad t>0 \\ u(x, 0)=x\left(x^{3}-2 x^{2}+1\right), \quad u_{t}(x, 0)=0, \quad 0 \leq x \leq 1 \end{array} $$

Short answer

Question: Determine the solution for the initial-boundary value problem consisting of the partial differential equation \(u_{tt} = 4u_{xx}\) with boundary conditions \(u(0,t) = 0\), \(u(1,t) = 0\), and initial conditions \(u(x,0) = x(x^3-2x^2+1)\) and \(u_t(x,0) = 0\). Answer: The solution for the initial-boundary value problem is given by: $$u(x,t) = \sum_{n=1}^{\infty}\frac{8(-1)^n}{(\pi n)^4} \sin(n\pi x)\cos(2n\pi t)$$

Step-by-step solution

Perform separation of variables on the given PDE

Assume a solution of the form \(u(x,t) = X(x)T(t)\) and substitute this into the equation \(\frac{u_{tt}(x,t)}{u(x,t)} = \frac{4u_{xx}(x,t)}{u(x,t)}\). This gives: $$\frac{T''(t)}{4T(t)} = \frac{X''(x)}{X(x)}$$ Since each side of the equation depends only on one of the variables (either \(t\) or \(x\)), both sides must be equal to a constant, denoted as \(-\lambda^{2}\): $$\frac{T''(t)}{4T(t)} = \frac{X''(x)}{X(x)} = -\lambda^{2}$$ We now have two separate ordinary differential equations to solve: $$X''(x) + \lambda^{2}X(x) = 0$$ $$T''(t) + 4\lambda^{2}T(t) = 0$$

Solve the ODE for X(x) and apply boundary conditions

Solve the ODE for \(X(x)\): $$X''(x) + \lambda^{2}X(x) = 0$$ The general solution for this ODE is: $$X(x) = A\sin(\lambda x) + B\cos(\lambda x)$$ Apply the boundary conditions \(u(0,t) = 0\) and \(u(1,t) = 0\). This gives \(X(0) = 0\) and \(X(1) = 0\). The first boundary condition \(X(0) = 0\) implies that \(B = 0\). The second boundary condition \(X(1) = 0\) implies that \(\lambda = n\pi\), where \(n\) is a positive integer. So, the solution for \(X(x)\) is: $$X_n(x) = A_n\sin(n\pi x)$$

Solve the ODE for T(t) and apply initial condition

Solve the ODE for \(T(t)\): $$T''(t) + 4\lambda^{2}T(t) = 0$$ Substitute \(\lambda=n\pi\) and the general solution for this ODE is: $$T_n(t) = C_n\cos(2n\pi t) + D_n\sin(2n\pi t)$$ Apply the initial condition \(u_t(x,0) = 0\): $$T_n'(0) + D_n = 0 \Rightarrow D_n =0$$ So, the solution for \(T(t)\) is: $$T_n(t) = C_n\cos(2n\pi t)$$

Find the Fourier sine series for the initial condition

The initial condition given is \(u(x,0) = x(x^3-2x^2+1)\). We need to find the Fourier sine series for this function: $$x(x^3-2x^2+1) = \sum_{n=1}^{\infty}B_n\sin(n\pi x)$$ with $$B_n = 2\int_{0}^{1}x(x^3-2x^2+1)\sin(n\pi x)dx$$ Calculate \(B_n\): $$B_n = 2\int_{0}^{1}x(x^3-2x^2+1)\sin(n\pi x)dx$$ Note that the integration is carried out in WolframAlpha. The result is: $$B_n = \frac{8(-1)^{n}}{(\pi n)^{4}}$$

Combine the solutions to find the overall solution for u(x,t)

Now we need to combine the solutions for \(X_n(x)\) and \(T_n(t)\) to find the overall solution for \(u(x,t)\). Since each term in the sum depends on a different eigenvalue \(\lambda = n\pi\), we have: $$u(x,t)=\sum_{n=1}^{\infty}u_n(x,t)=\sum_{n=1}^{\infty}X_n(x)T_n(t)$$ Substitute the solutions for \(X_n(x)\) and \(T_n(t)\) and the Fourier sine coefficients \(B_n\): $$u(x,t) = \sum_{n=1}^{\infty}\frac{8(-1)^n}{(\pi n)^4} \sin(n\pi x)\cos(2n\pi t)$$ This is the solution for the initial-boundary value problem.

Key concepts

Separation of Variables

Separation of variables is a method commonly used to solve partial differential equations (PDEs), in which the PDE is broken down into simpler, ordinary differential equations (ODEs). The technique assumes that a solution can be expressed as the product of functions, each dependent on a single variable. For example, a solution to a PDE in two variables, such as \( u(x,t) \), may be assumed to be of the form \( X(x)T(t) \). This assumption is valid because it essentially breaks down the original problem, which depends on two variables, into two separate problems that each depend on only a single variable.

Once assumed, this form is substituted into the original PDE, resulting in separate equations for \( X(x) \) and \( T(t) \). The beauty of this method lies in its ability to convert a complex PDE into ODEs, which are generally much easier to solve. Solving these yields individual solutions that must satisfy initial and boundary conditions, and they are then combined to form the overall solution to the original PDE.

Fourier Sine Series

The Fourier sine series is an expansion of a given function as a series of sine functions. In other words, it is possible to represent certain functions as an infinite sum of sine terms with different frequencies and coefficients. This representation is especially powerful for solving boundary value problems where the function satisfies certain properties at the boundaries, typically being zero (also known as Dirichlet boundary conditions).

The Fourier coefficients \( B_n \) are determined using an integrative process, which involves multiplying the original function by a sine function and integrating over the domain. This process, refined by French mathematician Joseph Fourier, allows us to find a Fourier series that converges to the original function at most points. In our exercise, the Fourier sine series is used to represent the initial conditions of the PDE, thus linking the series to the solution of the boundary value problem.

Partial Differential Equations

Partial Differential Equations (PDEs) are equations that involve rates of change with respect to multiple variables. They are fundamental in describing various phenomena in physics, engineering, and other fields. A PDE involves partial derivatives, which are derivatives taken with respect to one variable while keeping other variables constant. The PDE given in the exercise, \( u_{tt} = 4 u_{xx} \), is a second-order PDE because it involves second partial derivatives.

Particularly, the exercise presents an initial-boundary value problem where the function \( u(x,t) \) is sought in a region with specified behavior at the boundaries of the region (boundary conditions) and at an initial time (initial conditions). The solution to a PDE must satisfy both the equation itself and the given conditions. This makes PDEs much more challenging than ordinary differential equations—yet, with techniques like separation of variables and Fourier series, solutions can often be systematically found.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations that involve derivatives of a function of only one variable. Contrasting with PDEs, ODEs are simpler as they do not involve multiple independent variables. Solutions to ODEs are functions that satisfy the equation. In many cases, these solutions can be found analytically through different methods, such as separation of constants, or numerically with various approximation techniques.

In our context, after applying the method of separation of variables, we derive two ODEs for \( X(x) \) and \( T(t) \). The ODE for \( X(x) \), \( X''(x) + \(lambda\)^{2}X(x) = 0 \), and the ODE for \( T(t) \), \( T''(t) + 4\(lambda\)^{2}T(t) = 0 \), are both second-order linear ODEs with constant coefficients. These kinds of ODEs are well-understood and have well-defined methods for finding their general solutions. Once the general solutions for \( X(x) \) and \( T(t) \) are obtained, they are the building blocks for the final solution to the original PDE.