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. $$ \begin{array}{l} u_{t t}=9 u_{x x}, \quad 00 \\ u(0, t)=0, \quad u(1, t)=0, \quad t>0 \\ u(x, 0)=x(1-x), \quad u_{t}(x, 0)=0, \quad 0 \leq x \leq 1 \end{array} $$

Short answer

Answer: The final solution for the given initial-boundary value problem is $$u(x, t) =\sum_{n=1}^{\infty} \frac{8}{\pi^3 n^3}((-1)^n - 1) \sin(n\pi x) \cos(3n\pi t)$$

Step-by-step solution

Formulate the governing partial differential equation (PDE)

We are given the following PDE: $$u_{tt} = 9 u_{xx}, \quad 0 < x < 1, \quad t > 0$$

Apply separation of variables

We start by assuming a solution of the form $$u(x, t) = X(x)T(t)$$ and then plug it back into the PDE and the initial and boundary conditions.

Derive the ODEs

Substituting the product form into the given PDE, we get: $$X(x)T''(t) = 9X''(x)T(t)$$ Now, we can separate the variables by dividing both sides by \(9X(x)T(t)\): $$\frac{T''(t)}{9T(t)}=\frac{X''(x)}{X(x)}$$ Since the left-hand side is a function of \(t\) only and the right-hand side is a function of \(x\) only, they both must be equal to a constant, say, \(-\lambda\): $$\frac{T''(t)}{9T(t)}=\frac{X''(x)}{X(x)}=-\lambda$$ This leads to two ordinary differential equations (ODEs): $$T''(t) + 9 \lambda T(t) = 0$$ $$X''(x) + \lambda X(x) = 0$$ with boundary conditions \(X(0)=0\), \(X(1)=0\) and initial conditions \(u(x, 0)=x(1-x)\), \(u_t(x, 0)=0\).

Solve the ODEs

First, let's solve the ODE for \(X(x)\): $$X''(x) + \lambda X(x) = 0$$ We have three cases depending on the value of \(\lambda\): (1) \(\lambda>0\), (2) \(\lambda=0\), (3) \(\lambda<0\). Case 1: For \(\lambda>0\), let \(\lambda = k^2\) where \(k>0\). We have: $$X''(x) + k^2 X(x) = 0$$ The general solution for this case is: $$X(x) = A\cos(kx) + B\sin(kx)$$ Applying the boundary conditions \(X(0)=0, X(1)=0\), we get \(B=0\) and $$A\cos(k)=0 \Rightarrow k = n\pi$$ for \(n \neq 0\). So, the non-trivial solutions are $$X_n(x) = \sin(n\pi x)$$ Case 2: For \(\lambda=0\), we have a trivial solution. Case 3: For \(\lambda<0\), we get complex roots and no non-trivial solutions that satisfy the boundary conditions. Therefore, the solution for \(X(x)\) is: $$X_n(x) = \sin(n\pi x)$$ Now, let's solve the ODE for \(T(t)\): $$T''(t) + 9 \lambda T(t) = 0$$ $$T''(t) + 9n^2\pi^2 T(t) = 0$$ The general solution for this case is: $$T_n(t) = C_n\cos(3n\pi t) + D_n\sin(3n\pi t)$$

Apply initial conditions

Now, we utilize the initial conditions \(u(x, 0)=x(1-x)\) and \(u_t(x, 0)=0\) to find the constant coefficients. For \(u(x, 0) = x(1-x)\), we have: $$u(x, 0) = \sum_{n=1}^{\infty} C_n\sin(n\pi x) = x(1-x)$$ So, we need to find the Fourier sine series for \(f(x) = x(1-x)\). To find the coefficients \(C_n\), we can use the formula: $$C_n = \frac{2}{1} \int_{0}^{1} f(x) \sin(n\pi x) dx$$ Using integration by parts, we find: $$C_n = \frac{8}{\pi^3 n^3}((-1)^n - 1)$$ For \(u_t(x, 0)=0\), we get: $$D_n = 0$$

Combine the solutions

The overall solution for the PDE is given by: $$u(x, t) = \sum_{n=1}^{\infty} \left[ X_n(x) T_n(t) \right] = \sum_{n=1}^{\infty} \left[ C_n \sin(n\pi x) \cos(3n\pi t) \right]$$ Substituting the calculated coefficients \(C_n\) into the solution, we find the final solution: $$u(x, t) = \sum_{n=1}^{\infty} \frac{8}{\pi^3 n^3}((-1)^n - 1) \sin(n\pi x) \cos(3n\pi t)$$

Key concepts

Fourier Sine Series

A fundamental tool in solving initial-boundary value problems is the Fourier sine series. This type of series expansion is particularly useful when the function in question is defined on a finite interval and satisfies certain boundary conditions.

Considering an initial-boundary value problem, such as the one where a string's displacement is zero at both ends, the Fourier sine series comes into play to represent the initial displacements. The series is composed of sine functions which inherently satisfy the boundary conditions, simply because the sine function is zero at integer multiples of \(\pi\).

When we are given an initial condition, such as \(u(x, 0) = x(1-x)\), the Fourier sine series allows us to express this function as a sum of sine functions multiplied by calculated coefficients. These coefficients are given by an integral that involves the initial condition function and the respective sine term from the series. The ability to compute these coefficients efficiently often relies on theorems or exercises, like the one mentioned in the problem, to streamline the process.

In the context given by the exercise, the Fourier sine series was utilized to expand the initial condition \(x(1-x)\) into a series that could be used to solve for the time-dependent portion of the PDE solution. Clearly understanding how these coefficients \(C_n\) are derived is crucial, and being adept at integrating by parts is also key, as this method is often used in calculating the coefficients.

Separation of Variables

Separation of variables is a powerful technique where a complex problem, typically a partial differential equation (PDE), is broken down into simpler, more manageable ordinary differential equations (ODEs). It's akin to tackling a puzzle by separating the pieces into smaller, color-coded groups before attempting to put them all together.

In the context of our initial-boundary value problem, the technique involves assuming that the solution to the PDE can be written as the product of two functions, each depending on a single independent variable (\(X(x)\) and \(T(t)\)), rather than both variables simultaneously. This assumption transforms the original PDE into a format where one side of the equation depends solely on \(x\), while the other side depends only on \(t\). Since \(x\) and \(t\) are independent variables, the only way for these equations to hold for all values of \(x\) and \(t\) is if both sides equal the same constant.

This constant is often labelled \( -\lambda \) and leads to two separate ODEs for \(X(x)\) and \(T(t)\), which can be solved individually. Importantly, by solving these ODEs, applying boundary conditions, and finding the eigenvalues (like the specific values for \(k\) which lead to non-trivial solutions for \(X(x)\)), we make significant headway towards solving the initial PDE. The separation of variables technique is a testament to how breaking a problem apart can make way for a clear path to the solution.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations involving functions and their derivatives. These mathematical tools are essential in expressing the rates at which quantities change and are prevalent in numerous scientific disciplines. In the setting of the given problem, ODEs arise naturally after applying the separation of variables technique to the initial-boundary value problem.

Once variables in the PDE have been separated, we are left with ODEs for each function representing a single variable. The ODEs for function \(X(x)\), in this case, take the form \(X''(x) + \lambda X(x) = 0\), where the double prime denotes the second derivative with respect to \(x\), and \(\lambda\) is an eigenvalue determined by the separation constants.

The solutions to these ODEs are crucial. They determine the spatial part of our solution and are critically impacted by boundary conditions, which often lead to the use of trigonometric functions like sines and cosines in the solutions. For the temporal part, we solve the ODEs for \(T(t)\) in a similar spirit. The combined solutions of these ODEs, when capitalized upon correctly, yield the complete solution to the original PDE. It is the understanding and the method of solving these ODEs that bridge our theory to practical application, enabling us to model phenomena as varied as heat flow, wave propagation, and quantum mechanics.