Problem 11 - Advanced Engineering Mathematics

Question

Substituting \(u(x, t)=X(x) T(t)\) into the partial differential equation yields \(a^{2} X^{\prime \prime} T=X T^{\prime \prime} .\) Separating variables and using the separation constant \(-\lambda\) we obtain $$\frac{X^{\prime \prime}}{X}=\frac{T^{\prime \prime}}{a^{2} T}=-\lambda.$$ Then $$X^{\prime \prime}+\lambda X=0 \quad \text { and } \quad T^{\prime \prime}+a^{2} \lambda T=0.$$ We consider three cases: I. If \(\lambda=0\) then \(X^{\prime \prime}=0\) and \(X(x)=c_{1} x+c_{2} .\) Also, \(T^{\prime \prime}=0\) and \(T(t)=c_{3} t+c_{4},\) so $$u=X T=\left(c_{1} x+c_{2}\right)\left(c_{3} t+c_{4}\right).$$ II. If \(\lambda=-\alpha^{2}<0,\) then \(X^{\prime \prime}-\alpha^{2} X=0,\) and \(X(x)=c_{5} \cosh \alpha x+c_{6} \sinh \alpha x .\) Also, \(T^{\prime \prime}-\alpha^{2} a^{2} T=0\) and \(T(t)=c_{7} \cosh \alpha a t+c_{8} \sinh \alpha a t,\) so $$u=X T=\left(c_{5} \cosh \alpha x+c_{6} \sinh \alpha x\right)\left(c_{7} \cosh \alpha a t+c_{8} \sinh \alpha a t\right).$$ III. If \(\lambda=\alpha^{2}>0\), then \(X^{\prime \prime}+\alpha^{2} X=0\), and \(X(x)=c_{9} \cos \alpha x+c_{10} \sin \alpha x\). Also, \(T^{\prime \prime}+\alpha^{2} a^{2} T=0\) and \(T(t)=c_{11} \cos \alpha a t+c_{12} \sin \alpha a t,\) so $$u=X T=\left(c_{9} \cos \alpha x+c_{10} \sin \alpha x\right)\left(c_{11} \cos \alpha a t+c_{12} \sin \alpha a t\right).$$

Step-by-Step Solution

Answer

Solve for \( \lambda = 0, -\alpha^2, \alpha^2 \) with solutions in polynomial, hyperbolic, and trigonometric forms.

Step 1: Initial Substitution

Substitute \( u(x, t) = X(x)T(t) \) into the given partial differential equation \( a^2 X'' T = X T'' \). This results in the equation \( \frac{X''}{X} = \frac{T''}{a^2 T} = -\lambda \) using the separation constant \( -\lambda \).

Step 2: Separation of Variables

Set \( \frac{X''}{X} = -\lambda \) and \( \frac{T''}{a^2 T} = -\lambda \), leading to two ordinary differential equations: \( X'' + \lambda X = 0 \) and \( T'' + a^2 \lambda T = 0 \).

Step 3: Case I - \( \lambda = 0 \)

For \( \lambda = 0 \), equations become \( X'' = 0 \) and \( T'' = 0 \), giving solutions \( X(x) = c_1 x + c_2 \) and \( T(t) = c_3 t + c_4 \). The solution for \( u \) is \( u(x, t) = (c_1 x + c_2)(c_3 t + c_4) \).

Step 4: Case II - \( \lambda = -\alpha^2 < 0 \)

For \( \lambda = -\alpha^2 \), solve \( X'' - \alpha^2 X = 0 \) using hyperbolic functions: \( X(x) = c_5 \cosh(\alpha x) + c_6 \sinh(\alpha x) \). Solve \( T'' - a^2 \alpha^2 T = 0 \) similarly for \( T(t) = c_7 \cosh(\alpha a t) + c_8 \sinh(\alpha a t) \). The solution is \( u(x, t) = (c_5 \cosh(\alpha x) + c_6 \sinh(\alpha x))(c_7 \cosh(\alpha a t) + c_8 \sinh(\alpha a t)) \).

Step 5: Case III - \( \lambda = \alpha^2 > 0 \)

For \( \lambda = \alpha^2 \), solve \( X'' + \alpha^2 X = 0 \) using trigonometric functions: \( X(x) = c_9 \cos(\alpha x) + c_{10} \sin(\alpha x) \). Similarly, solve \( T'' + a^2 \alpha^2 T = 0 \) for \( T(t) = c_{11} \cos(\alpha a t) + c_{12} \sin(\alpha a t) \). The solution becomes \( u(x, t) = (c_9 \cos(\alpha x) + c_{10} \sin(\alpha x))(c_{11} \cos(\alpha a t) + c_{12} \sin(\alpha a t)) \).

Key Concepts

Separation of Variables

Separation of Variables is a key technique for solving partial differential equations (PDEs), such as the wave equation or the heat equation. This method transforms a PDE into simpler ordinary differential equations (ODEs).
To use this method, we assume the solution can be written as the product of functions each depending on a single variable. In the example given, we set \( u(x, t) = X(x)T(t) \).
This assumption allows us to split the original PDE, \( a^2 X'' T = X T'' \), into two separate ODEs. We accomplish this by assuming each side of the rewritten equation is equal to a constant \(-\lambda\). This gives us \( \frac{X''}{X} = -\lambda \) and \( \frac{T''}{a^2 T} = -\lambda \). Both statements are now independent of each other.
Employing Separation of Variables offers insightful solutions for boundary value problems, making complex equations manageable.

Boundary Value Problems

Boundary Value Problems (BVPs) are mathematical problems defined by a differential equation and a set of boundary conditions. These conditions specify values the solution must satisfy at the edges of the domain.
For example, consider the differential equation solved in the exercise. The general solution evolves into specific forms, depending on the value of \( \lambda \), which affects the boundary and initial conditions.

• For \( \lambda = 0 \), the functions are linear.
• For \( \lambda = -\alpha^2 < 0 \), we use hyperbolic functions, such as \( \cosh \) and \( \sinh \).
• For \( \lambda = \alpha^2 > 0 \), trigonometric functions, such as \( \cos \) and \( \sin \) are appropriate.

In this way, BVPs help model physical phenomena where constraints in forms of boundaries, like fixed ends of a vibrating string, impact the equation's solutions.

Mathematical Modelling

Mathematical Modelling involves using mathematical language and methods to represent real-world systems and phenomena. In our example, the PDE models how solutions vary over both space \( x \) and time \( t \).
This kind of modelling helps in predicting and understanding complex situations, from heat distribution in a metal rod to waves in oceans.The process begins with assuming a suitable equation that reflects the physical reality. Next, mathematical techniques like Separation of Variables personalize the model to specific boundary conditions.
Throughout these steps, Mathematical Modelling ensures that the abstract equations mirror tangible systems, allowing for accurate interpretations and predictions of natural, engineered, or economic systems.