Question

Use Exercise 17 to solve the initial-boundaryvalue problem. In some of these exercises Theorem 11.3.5(c) or Exercise 11.3.42(b) will simplify the computation of the coefficients in the mixed Fourier cosine series. $$ \begin{array}{l} u_{t t}=u_{x x}, \quad 00, \\ u_{x}(0, t)=0, \quad u(1, t)=0, \quad t>0, \\ u(x, 0)=x^{4}-4 x^{3}+6 x^{2}-3, \quad u_{t}(x, 0)=0, \quad 0 \leq x \leq 1. \end{array} $$

Short answer

Based on the given initial-boundary value problem for a partial differential equation, we used separation of variables, mixed Fourier cosine series, and initial conditions to find the solution. The final solution to the initial-boundary value problem is: $$u(x,t) = \sum_{n=1}^{\infty} 4n((4-3n^2)\pi\sin(n\pi)-2(n^2\pi^2-6)\cos(n\pi)+6n\pi)\cos(n\pi x)\cos(n\pi t)$$

Step-by-step solution

Separate the Variables

Using separation of variables, we assume a solution of the form: $$u(x,t) = X(x)T(t)$$ Plugging the separated form into the PDE, we get: $$X(x)T''(t) = X''(x)T(t)$$ Rearranging to isolate \(t\) and \(x\): $$\frac{T''(t)}{T(t)} = \frac{X''(x)}{X(x)}=-\lambda$$ where \(\lambda\) is the separation constant. Now, we have two ordinary differential equations: $$X''(x)=-\lambda X(x)$$ $$T''(t)=-\lambda T(t)$$

Solve the Boundary Value Problem

Let's solve the boundary value problem for \(X(x)\) first. We have the following equation and boundary conditions: $$X''(x)=-\lambda X(x)$$ $$X'(0) = 0, \quad X(1) = 0$$ For the cases of \(\lambda\): 1. \(\lambda < 0: \quad X(x) = C_1 \cosh(\sqrt{-\lambda}x) + C_2 \sinh(\sqrt{-\lambda}x)\) 2. \(\lambda > 0: \quad X(x) = C_1\cos(\sqrt{\lambda}x) + C_2\sin(\sqrt{\lambda}x)\) 3. \(\lambda = 0: \quad X(x) = C_1 + C_2x\) However, \(\lambda < 0\) does not satisfy the boundary conditions. The case of \(\lambda = 0\) results in a trivial solution, so we'll consider the case of \(\lambda > 0\): $$X(x) = C_1\cos(n\pi x) + C_2\sin(n\pi x)$$ where \(n^2\pi^2 = \lambda\). Using boundary conditions, we find that \(C_2=0\). Therefore, we get a series of solutions: $$X_n(x) = \cos(n\pi x)$$ Next, solving the equation for \(T(t)\): $$T''(t)=-n^2\pi^2 T(t)$$ The solution is given by: $$T_n(t) = A_n\cos(n\pi t) + B_n\sin(n\pi t)$$ So, the complete solution can be written as a mixed Fourier cosine series: $$u(x, t) = \sum_{n=1}^{\infty} A_n \cos(n\pi x)\cos(n\pi x)$$

Apply Initial Conditions

Now we need to fit the coefficients \(A_n\) and \(B_n\) using the given initial conditions \(u(x,0) = x^4-4x^3+6x^2-3\) and \(u_t(x,0) = 0\). We have \(u(x,0) = \sum_{n=1}^{\infty} A_n \cos(n\pi x)\), and since at \(t=0\), \(B_n=0\), we get: $$x^4-4x^3+6x^2-3 = \sum_{n=1}^{\infty} A_n \cos(n\pi x)$$ To find the coefficients \(A_n\), we multiply the equation by \(\cos(m\pi x)\) and integrate from \(0\) to \(1\). Using the orthogonality of cosine functions, we get $$A_{m} = 4m((4-3m^2)\pi\sin(m\pi)-2(m^2\pi^2-6)\cos(m\pi)+6m\pi)$$

Get the Final Solution

Finally, we obtain the solution for the initial-boundary value problem as: $$u(x,t) = \sum_{n=1}^{\infty} A_n \cos(n\pi x)\cos(n\pi t) = \sum_{n=1}^{\infty} 4n((4-3n^2)\pi\sin(n\pi)-2(n^2\pi^2-6)\cos(n\pi)+6n\pi)\cos(n\pi x)\cos(n\pi t)$$

Key concepts

Partial Differential Equations

Partial Differential Equations (PDEs) are equations that involve rates of change with respect to multiple variables. In our problem, the PDE is \( u_{tt} = u_{xx} \). This specific PDE describes the motion of a string fixed at both ends, known as the wave equation.

PDEs can model a variety of physical phenomena, such as heat conduction, sound, and fluid dynamics, identifying how they change over time and space simultaneously.

They differ from ordinary differential equations because of their multi-variable nature, making them a more complex mathematical challenge.

Solving a PDE typically involves finding a function or a set of functions that satisfy the equation within given constraints.

Separation of Variables

Separation of Variables is a mathematical method used to solve some PDEs. This method involves assuming a solution that can be written as the product of functions, each depending on a single variable.

In our solution, the assumed form is \( u(x,t) = X(x)T(t) \). This product form separates the variables \(x\) and \(t\), allowing the PDE to be split into two ordinary differential equations (ODEs). This step reduces the complexity, converting the PDE into simpler, solvable forms by progressing with each variable independently.

Successfully applying this method requires the problem to be linear and homogeneous with appropriate boundary conditions.

Boundary Value Problems

Boundary Value Problems (BVPs) are differential equations with additional constraints called boundary conditions. These conditions specify the values of the solution at certain points, typically at the boundaries of the domain.

For the given exercise, the boundary conditions are \( u_x(0, t)=0 \) and \( u(1, t)=0 \). These conditions ensure the solution has a physical meaning, such as fixing the endpoints of a vibrating string to walls.

Solving a BVP involves satisfying both the differential equation and its associated boundary conditions, narrowing the range of possible solutions to those that fit the physical scenario.

Orthogonality of Functions

The orthogonality of functions is a concept primarily used with Fourier Series. Orthogonal functions have the property that their integral product over a particular interval is zero unless the functions are identical.

In our solution, cosine functions are used, exploiting their orthogonal properties over an interval \([0,1]\). This simplifies finding coefficients like \(A_n\) in the Fourier series expansion of \( u(x,0) \).

Orthogonality aids in isolating and calculating these coefficients efficiently, ensuring an accurate approximation of functions by determining unique contributions for each frequency in the series. Without orthogonality, solving such problems would become computationally intense and mathematically cumbersome.