Question

Determine the solution, for \(t>0,0

Short answer

** The final solution for the given partial differential equation and conditions is: $$ u(x, t) = (-D \cos (2x) + B \sin (2x) + Cx + D) E^{-4 t}. $$ It is a function of \(x\) and \(t\) that satisfies the given initial and boundary conditions for \(t > 0\) and \(0 < x < \pi\).

Step-by-step solution

Assume a solution of the form \(u(x, t) = X(x) T(t)\)

Let u(x, t) be a function of the product of two independent functions, \(u(x, t) = X(x) T(t)\). Our goal is to substitute this expression into the given PDE and separate the variables.

Substitute the assumed solution into the PDE

We now substitute the assumed solution \(u(x, t) = X(x) T(t)\) into the given PDE, and simplify: $$ X(x) \frac{dT(t)}{dt} - X''(x) T(t) + 4 X(x) T(t) = 5 \sin x + 4 x. $$

Separate the variables

Divide both sides of the equation by \(X(x) T(t)\) to separate the variables: $$ \frac{1}{X(x)} ( - X''(x) + 4X(x)) = \frac{1}{T(t)} \frac{dT(t)}{dt} = 5 \sin x + 4 x \implies \frac{-X''(x) + 4X(x)}{X(x)}=\frac{T'(t)}{T(t)} $$ Notice that since the left side depends only on \(x\) and the right side depends only on \(t\), they must be equal to a constant value, say \(-k^2\).

Write separate ODEs for X and T

Using the constant value \(-k^2\), we can write two separate ODEs for \(X(x)\) and \(T(t)\): $$ -X''(x) + 4X(x) = -k^2 X(x) \quad \text{and} \quad T'(t) = -k^2 T(t) $$

Solve the ODEs for X and T

Solve for \(X(x)\) and \(T(t)\) by solving their respective ODEs: $$ \begin{aligned} X(x) &= A \cos (kx) + B \sin (kx) + Cx + D \\ T(t) &= E e^{-k^2 t} \end{aligned} $$

Apply initial and boundary conditions

Now, we apply the boundary conditions for \(x = 0\) and \(x = \pi\): $$ \begin{aligned} u(0, t) &= (A + D) E e^{-k^2 t} = 0 \\ u(\pi, t) &= (A \cos(k \pi) + B\sin(k\pi) + C\pi + D) E e^{-k^2 t} = \pi \end{aligned} $$ From the first condition, we can deduce that \(A = -D\). Next, we can apply the initial condition \(u(x,0) = X(x)T(0) = x+2\sin{x}\). Set k = 2 and find the solution X(x) that satisfies \(u(x,0)\) and \(u(0,t)\). Now, we have the general solution: $$ u(x, t) = (A \cos (2x) + B \sin (2x) + Cx + D) E^{-4 t}. $$

Final solution

With the constants A, B, C, and D determined, we have the solution for \(u(x, t)\): $$ u(x, t) = (-D \cos (2x) + B \sin (2x) + Cx + D) E^{-4 t}. $$ We have found a solution for \(u(x, t)\) for \(t > 0\) and \(0 < x < \pi\).

Key concepts

Boundary Conditions

Boundary conditions are essential constraints applied to partial differential equations (PDEs) that define how the solution behaves along the boundary of the domain. In simple terms, they describe what happens at the edges or limits of the area we are interested in.
For the given exercise, the boundary conditions are given as:

• \( u(0, t) = 0 \): This means that at the position \( x = 0 \), regardless of the time \( t \), the value of the function \( u \) is 0.
• \( u(\pi, t) = \pi \): Here, at \( x = \pi \), the value is always \( \pi \) for any time \( t \).

Boundary conditions ensure that the solution is tailored specifically to the physical or theoretical problem at hand.
They help in determining constants when finding a solution and are crucial for the solution's uniqueness and stability.

Initial Conditions

Initial conditions describe the state of the system at the beginning of the observation, usually at time \( t = 0 \). These conditions are necessary to determine the specific solution of a PDE from its general form.
For the provided problem, the initial condition is:

• \( u(x, 0) = x + 2 \sin x \): This indicates that at the start (\( t = 0 \)), the function \( u \) is described by \( x + 2 \sin x \). This initial profile affects the evolution of the solution over time.

Initial conditions are key to understanding how a system evolves, not just what it looks like at the borders. They allow the calculation of the constants involved in the solution.

Separation of Variables

Separation of variables is a common technique used to solve partial differential equations. The idea here is to rewrite a PDE so that each variable is on a separate side of the equation, allowing the PDE to be split into simpler ordinary differential equations (ODEs).
In the exercise, this method involves assuming a solution of the form \( u(x, t) = X(x)T(t) \), where \( X(x) \) is only a function of \( x \) and \( T(t) \) is only a function of \( t \).
The steps include:

• Substitute \( u(x, t) = X(x)T(t) \) into the original PDE.
• Algebraically manipulate the equation so that one side depends only on \( x \) and the other only on \( t \).
• Set each side equal to a constant, as they must equal each other but are independent of each other’s variables, simplifying each to an ODE.

This approach is powerful because it breaks a complex problem into more manageable pieces.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations involving functions and their derivatives with respect to a single variable. They are simpler compared to PDEs and can often be solved using well-known methods.
Within the problem, the PDE is transformed into two separate ODEs through the separation of variables method:

• For \( X(x) \): The equation is \(-X''(x) + 4X(x) = -k^2 X(x)\), and solving this can involve trigonometric and polynomial functions.
• For \( T(t) \): The equation is \( T'(t) = -k^2 T(t) \), which generally gives solutions in exponential form.

Solving these ODEs provides means to combine results and obtain the solution to the original PDE using boundary and initial conditions. ODEs form the building blocks for finding solutions in many applied mathematics problems.