Problem 9 - Advanced Engineering Mathematics

Question

Use separation of variables to find, if possible, product solutions for the given partial differential equation. $$ k \frac{\partial^{2} u}{\partial x^{2}}-u=\frac{\partial u}{\partial t}, k>0 $$

Step-by-Step Solution

Answer

The product solution is \[ u(x, t) = (A e^{mx} + B e^{-mx}) e^{\lambda t} \] with \( m = \sqrt{\frac{(\lambda + 1)}{k}} \).

Step 1: Assume a Product Solution

Assume a solution of the form \( u(x, t) = X(x)T(t) \) where \( X(x) \) is a function of \( x \) and \( T(t) \) is a function of \( t \).

Step 2: Substitute into PDE

Substitute \( u(x, t) = X(x)T(t) \) into the given PDE:\[ k \frac{d^2}{dx^2}(XT) - XT = \frac{d}{dt}(XT) \]This becomes:\[ k T \frac{d^2 X}{dx^2} - X T = X \frac{dT}{dt} \].

Step 3: Separate Variables

Rewrite the equation by dividing through by \( XT \):\[ k \frac{X''}{X} - 1 = \frac{T'}{T} \]Separate the variables so that each side of the equation depends only on one variable.

Step 4: Define Separation Constant

Since the left side depends only on \( x \) and the right side depends only on \( t \), both sides must be equal to a constant. Set both sides equal to a separation constant \( \lambda \):\[ k \frac{X''}{X} - 1 = \lambda \]\[ \frac{T'}{T} = \lambda \].

Step 5: Solve the X-Equation

Rearrange and solve the equation for \( X(x) \):\[ k X'' = (\lambda + 1) X \]This leads to \( X'' - \frac{(\lambda + 1)}{k} X = 0 \), a standard second-order linear differential equation.

Step 6: Solve the T-Equation

Solve the equation for \( T(t) \):\[ T' = \lambda T \]This is a simple first-order linear differential equation with solution \( T(t) = Ce^{\lambda t} \), where \( C \) is a constant.

Step 7: General Solution

Combine the solutions for \( X(x) \) and \( T(t) \):The product solution is of the form: \[ u(x, t) = X(x) T(t) = (A e^{mx} + B e^{-mx}) Ce^{\lambda t} \]where \( m = \sqrt{\frac{(\lambda + 1)}{k}} \). Assume \( A \) and \( B \) are constants.

Key Concepts

Partial Differential Equations

Partial differential equations, or PDEs, involve multivariable functions and their partial derivatives. They stand vital in describing a variety of phenomena in engineering and physics, such as heat conduction, wave propagation, and fluid dynamics. In simpler terms, PDEs allow us to understand how a change in one dimension can affect another. These equations can be quite complex, but breaking them into simpler components, like separating variables, can make them more manageable.

In the given exercise, we deal with a PDE in the form:

• \( k \frac{\partial^{2} u}{\partial x^{2}}-u=\frac{\partial u}{\partial t} \)

Here, \( u \) is a function of \( x \) and \( t \) representing, for example, temperature over space and time. Our aim is to find solutions where dynamics over heat and space coincide.

Product Solutions

Product solutions are a clever way to tackle PDEs by assuming that the solution can be expressed as a product of two separate functions, each a function of one variable. This implies, rather than solving a complex multidimensional problem, we handle separate one-dimensional problems, easing the burden significantly.

Assuming the solution of our PDE in the exercise as \( u(x, t) = X(x)T(t) \), we decompose it into two parts:

• \( X(x) \) - a function solely dependent on the spatial component \( x \)
• \( T(t) \) - a function solely dependent on the time component \( t \)

Whenever you see a complex PDE, consider checking if a product solution approach could simplify the problem. This strategy, though not universally applicable, is powerful for many PDEs.

Differential Equation Solving

Solving differential equations, whether ordinary or partial, is about finding functions that satisfy these equations. They are vital in modeling systems that change over time or across a space. In our scenario, separating the variables allowed us to break down the PDE into simpler ordinary differential equations (ODEs).

For the task at hand, substituting the product solution into the PDE resulted in the separation of the problem into:

• \( k \frac{X''}{X} - 1 = \lambda \)
• \( \frac{T'}{T} = \lambda \)

Both are now primarily independent ODEs, making them straight-forward to solve step-by-step. The separation of variables here helps to identify potential solution paths and ensures that each dimension is handled independently.

Second-Order Linear Differential Equations

Second-order linear differential equations involve the second derivative of a function and show up often in physics and engineering problems, like our PDE exercise. For instance, once variables are separated, the resulting ODE for \( X(x) \) is:

• \( k X'' = (\lambda + 1) X \)

Rewriting it as \( X'' - \frac{(\lambda + 1)}{k} X = 0 \), we land on a standard form of a second-order linear equation.

Such equations typically have solutions of exponential form, e.g., \( A e^{mx} + B e^{-mx} \), where constants \( A \) and \( B \) are determined by boundary conditions or initial states. The skill in solving these is identifying the characteristic equation and interpreting it correctly to retrieve possible solutions.