Problem 8 - Applied Partial Differential Equations

Question

Consider $$ \frac{d^{2} u}{d x^{2}}+u=\cos x $$ which has a particular solution of the form, \(u_{p}-A x \sin x\). *(a) Suppose that \(u(0)=u(\pi)=0\). Explicitly attempt to obtain all solutions. Is your result consistent with the Fredholm alternative? (b) Answer the same questions as in part (a) if \(u(-\pi)=u(\pi)\) and \(\frac{d u}{d x}(-\pi)=\) \(\frac{d u}{d x}(\pi)\).

Step-by-Step Solution

Answer

In part (a), the result is consistent with the Fredholm alternative because the boundary conditions lead to a consistent result. In part (b), the result violates the Fredholm alternative because it does not lead to a consistent result.

Step 1: Find the complementary solution

By first solving the homogeneous equation \(\frac{d^{2} u}{d x^{2}} + u = 0\), one can find the complementary solution, which turns out to be \(u_c = A \sin x + B \cos x\).

Step 2: Find the general solution

Adding the particular solution to the complementary solution gives the general solution, written as \(u = u_c + u_p\). Therefore, \(u = A \sin x + B \cos x - x \sin x\).

Step 3: Solve for constants with given boundary conditions (a)

Substituting \(u(0)=0\) into the general solution gives \(B = 0\), and substituting \(u(\pi) = 0\) into the general solution gives a consistent result, thereby satisfying the Fredholm alternative.

Step 4: Solve for constants with given boundary conditions (b)

Substituting \(u(-\pi) = u(\pi)\) and \(\frac{d u}{d x}(-\pi) = \frac{d u}{d x}(\pi)\) into the general solution does not give a consistent result, thereby violating the Fredholm alternative.

Step 5: Final thoughts

The Fredholm Alternative Theorem says that a non-homogeneous linear equation has a solution if and only if the corresponding homogeneous equation has only the trivial solution when evaluated at the function in question. So, if both conditions in part (a) or (b) meet this criteria, then the Fredholm alternative is satisfied.

Key Concepts

Partial Differential Equation

A partial differential equation (PDE) is an equation involving partial derivatives of a function that depends on several variables. These equations play a vital role in physics, engineering, and many other fields as they describe a wide range of phenomena such as heat conduction, sound, fluid flow, and quantum mechanics.

Consider the given PDE \[ \frac{d^{2} u}{d x^{2}}+u=\cos x \] from the exercise. Here, \( u \) is a function of \( x \), and this particular PDE is used to find the function \( u(x) \) that satisfies the equation for any given value of \( x \). Solving a PDE usually involves finding the general solution that encompasses all possible solutions, then applying specific conditions to find a unique solution.

Boundary Conditions

In the context of PDEs, boundary conditions are constraints that a solution must satisfy at the borders of its domain. These conditions are crucial because they specify the behavior of the solution at the boundaries, which is necessary to pin down a unique solution from a family of possible solutions.

For instance, in the exercise, we have two different sets of boundary conditions: in part (a), \( u(0) = u(\pi) = 0 \), and in part (b), \( u(-\pi) = u(\pi) \) and \( \frac{d u}{d x}(-\pi) = \frac{d u}{d x}(\pi) \). These boundary conditions determine the constants in the general solution of the PDE, thus prescribing the specific shape and features of the solution function within its domain.

Homogeneous Equation

A homogeneous equation is a type of PDE where the result is always zero, regardless of the function that is being differentiated. This means that the equation does not include a non-zero function of the independent variables alone. In simpler terms, every term in the equation must contain the dependent variable or its derivatives.

During the solving process, initially a homogeneous version of the original non-homogeneous equation is considered: \[ \frac{d^{2} u}{d x^{2}} + u = 0 \] This is Step 1 in the solution, where we find the complementary solution, \( u_c = A \sin x + B \cos x \), which represents the most general form of the solution for the homogeneous case.

Non-Homogeneous Equation

Contrary to a homogeneous equation, a non-homogeneous equation includes a non-zero function as part of the equation. This term does not involve the dependent variable or its derivatives, making the equation non-zero. Non-homogeneous PDEs are inherently more complex as their solutions must take into account this additional 'forcing' function.

For the exercise's given PDE, \[ \frac{d^{2} u}{d x^{2}} + u = \cos x \] the right-hand side, \( \cos x \), is the non-zero term which makes the PDE non-homogeneous. In Step 2 of the solution, we find a specific solution, \( u_p = -A x \sin x \), which satisfies the non-homogeneous equation. Then, by combining this particular solution with the complementary solution, we reach the general solution of the non-homogeneous equation.