Problem 21 - Advanced Engineering Mathematics

Question

The solution of $$\frac{d^{2} U}{d x^{2}}-s U=-u_{0}-u_{0} \sin \frac{\pi}{L} x$$ is $$U(x, s)=c_{1} \cosh (\sqrt{s} x)+c_{2} \sinh (\sqrt{s} x)+\frac{u_{0}}{s}+\frac{u_{0}}{s+\pi^{2} / L^{2}} \sin \frac{\pi}{L} x$$ The transformed boundary conditions \(U(0, s)=u_{0} / s\) and \(U(L, s)=u_{0} / s\) give, in turn, \(c_{1}=0\) and \(c_{2}=0\) Therefore $$U(x, s)=\frac{u_{0}}{s}+\frac{u_{0}}{s+\pi^{2} / L^{2}} \sin \frac{\pi}{L} x$$ and $$u(x, t)=u_{0} \mathscr{L}^{-1}\left\\{\frac{1}{s}\right\\}+u_{0} \mathscr{L}^{-1}\left\\{\frac{1}{s+\pi^{2} / L^{2}}\right\\} \sin \frac{\pi}{L} x=u_{0}+u_{0} e^{-\pi^{2} t / L^{2}} \sin \frac{\pi}{L} x$$

Step-by-Step Solution

Answer

The solution is \( u_0 + u_0 e^{-\pi^2 t / L^2} \sin(\frac{\pi}{L} x) \).

Step 1: Identifying the Homogeneous Solution

To solve the non-homogeneous differential equation \( \frac{d^{2} U}{d x^{2}} - s U = -u_{0} - u_{0} \sin \frac{\pi}{L} x \), start by finding the solution to the corresponding homogeneous equation: \( \frac{d^{2} U}{d x^{2}} - s U = 0 \). The characteristic equation is \( r^2 - s = 0 \), which has solutions \( r = \pm \sqrt{s} \). Thus, the homogeneous solution is \( U_h(x) = c_1 \cosh(\sqrt{s} x) + c_2 \sinh(\sqrt{s} x) \).

Step 2: Finding a Particular Solution

Propose a particular solution of the form: \( U_p(x) = A + B \sin \frac{\pi}{L} x \). Differentiate it and substitute back into the differential equation. After simplifying, equate coefficients to solve for \( A \) and \( B \). You will find that \( A = \frac{u_{0}}{s} \) and \( B = \frac{u_{0}}{s+\pi^{2} / L^{2}} \).

Step 3: Form the General Solution

The general solution to the differential equation is the sum of the homogeneous and particular solutions: \( U(x, s) = U_h(x) + U_p(x) = c_1 \cosh(\sqrt{s} x) + c_2 \sinh(\sqrt{s} x) + \frac{u_{0}}{s} + \frac{u_{0}}{s+\pi^{2} / L^{2}} \sin \frac{\pi}{L} x \).

Step 4: Apply Boundary Conditions

Apply the boundary conditions \( U(0, s) = \frac{u_0}{s} \) and \( U(L, s) = \frac{u_0}{s} \). Substituting \( x = 0 \) into the general solution gives \( c_1 + \frac{u_0}{s} = \frac{u_0}{s} \), leading to \( c_1 = 0 \). Substituting \( x = L \) gives \( 0 = \frac{u_0}{s} \) when \( c_2 = 0 \).

Step 5: Simplify Solution Using Coefficients

With \( c_1 = 0 \) and \( c_2 = 0 \), the resulting solution becomes \( U(x, s) = \frac{u_{0}}{s} + \frac{u_{0}}{s+\pi^{2} / L^{2}} \sin \frac{\pi}{L} x \).

Step 6: Find the Inverse Laplace Transform

To find \( u(x, t) \), take the inverse Laplace Transform of each term. The inverse Laplace of \( \frac{1}{s} \) is \( 1 \) and of \( \frac{1}{s + a^2} \) is \( e^{-a^2 t} \). Therefore, \( u(x, t) = u_0 + u_0 e^{-\pi^{2} t / L^{2}} \sin \frac{\pi}{L} x \).

Key Concepts

Homogeneous Solution

The concept of the homogeneous solution is central to understanding differential equations. When dealing with differential equations, particularly non-homogeneous ones, the homogeneous solution is found by solving the equation when the non-homogeneous part (the forcing function) is zero. In this exercise, the differential equation given is \( \frac{d^{2} U}{d x^{2}} - s U = 0 \). This is known as the homogeneous equation.
The homogeneous part is derived from the characteristic equation, which is formed by replacing the differential operator with a variable, often \( r \). For this problem, the characteristic equation is \( r^2 - s = 0 \). Solving for \( r \) gives the roots \( r = \pm \sqrt{s} \), indicating that the solution will be a combination of hyperbolic trigonometric functions:

• \( c_1 \cosh(\sqrt{s} x) \)
• \( c_2 \sinh(\sqrt{s} x) \)

So, the homogeneous solution is \( U_h(x) = c_1 \cosh(\sqrt{s} x) + c_2 \sinh(\sqrt{s} x) \). This forms the backbone of the solution upon which modifications are made to accommodate the non-homogeneous parts.

Particular Solution

The particular solution aims to address the non-homogeneous aspect of the differential equation. The original equation given is \( \frac{d^{2} U}{d x^{2}} - s U = -u_{0} - u_{0} \sin \frac{\pi}{L} x \). The goal here is to find a function that satisfies this equation without being part of the homogeneous solution.
In this context, a guess is made about the form of the particular solution: \( U_p(x) = A + B \sin \frac{\pi}{L} x \). This educated guess is not arbitrary; it's influenced by the form of the non-homogeneous term. By differentiating the guess and substituting back into the differential equation, we determine:

• \( A = \frac{u_{0}}{s} \)
• \( B = \frac{u_{0}}{s + \pi^{2} / L^{2}} \)

This successfully nullifies the non-homogeneous component when plugged into the original equation, thus providing us the particular solution: \( U_p(x) = \frac{u_{0}}{s} + \frac{u_{0}}{s + \pi^{2} / L^{2}} \sin \frac{\pi}{L} x \).
By solving both components, we obtain the general solution which is a sum of both homogenous and particular solutions.

Boundary Conditions

When dealing with differential equations, especially in physics and engineering, boundary conditions are crucial. They help in pinpointing the exact solution to a problem by accounting for various contextual constraints.
In this exercise, we have two boundary conditions:

• \( U(0, s) = \frac{u_{0}}{s} \)
• \( U(L, s) = \frac{u_{0}}{s} \)

These conditions are derived from the physical setup of the problem. Applying the first boundary condition to the general solution involves setting \( x = 0 \). Doing so reveals that any \( \cosh \) component must satisfy \( c_1 + \frac{u_{0}}{s} = \frac{u_{0}}{s} \), forcing \( c_1 = 0 \).
The second condition involves replacing \( x \) with \( L \) in the general solution. This results in both hyperbolic sine and cosine terms vanishing with \( c_2 = 0 \), simplifying our solution to:
\( U(x, s) = \frac{u_{0}}{s} + \frac{u_{0}}{s + \pi^{2} / L^{2}} \sin \frac{\pi}{L} x \)
Boundary conditions thus ensure the solution is tailored to the specific problem.

Inverse Laplace Transform

The inverse Laplace transform is the final step in moving from the frequency domain back to the time domain, which is essential in solving differential equations. Once we have obtained the solution \( U(x, s) \) in the Laplace-transformed space, we need to apply the inverse Laplace transform to find the solution in terms of time \( t \).
For our problem, the terms \( \frac{1}{s} \) and \( \frac{1}{s + a^2} \) within the solution correspond to recognizable transforms:

• \( \mathscr{L}^{-1}\{\frac{1}{s}\} = 1 \)
• \( \mathscr{L}^{-1}\{\frac{1}{s + a^2}\} = e^{-a^2 t} \)

Applying these to our solution, we find the final time-domain solution:
\( u(x, t) = u_0 + u_0 e^{-\pi^{2} t / L^{2}} \sin \frac{\pi}{L} x \)
The inverse Laplace transform allows us to interpret the solution in a time-dependent context, which is often more practical for real-world applications.