Problem 10 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The solutions are given by the different cases for \(\lambda\): I: \(u = A_1 x + A_2\), II: \(u = (A_3 \cosh(\alpha x) + A_4 \sinh(\alpha x))e^{k \alpha^2 t}\), III: \(u = (A_3 \cos(\alpha x) + A_6 \sin(\alpha x))e^{-k \alpha^2 t}\).

Step 1: Understand the Separation of Variables

The given problem is a separation of variables problem where the function \(u(x, t)\) is assumed to be of the form \(X(x) T(t)\). The partial differential equation is \(k X'' T = X T'\). The goal is to separate the variables to solve the differential equations independently.

Step 2: Set Up Separation of Variables

From \(k X'' T = X T'\), we divide both sides by \(XT\), yielding the equation \(\frac{X''}{X} = \frac{T'}{kT}\). Both sides depend on different variables, so they must be equal to a constant, which we let be \(-\lambda\).

Step 3: Derive Differential Equations for X and T

Setting each side equal to \(-\lambda\) yields two ordinary differential equations: \(X'' + \lambda X = 0\) and \(T' + \lambda k T = 0\). These equations can be solved for different values of \(\lambda\).

Step 4: Solve for T(t)

The equation \(T' + \lambda k T = 0\) is a first-order linear ordinary differential equation which has the solution \(T(t) = c_1 e^{-\lambda k t}\).

Step 5: Solve Case I: \(\lambda = 0\)

If \(\lambda = 0\), then the equation \(X'' = 0\) results. Solving this yields \(X(x) = c_2 x + c_3\). The solution \(u(x, t) = X(x) T(t)\) becomes \(u(x, t) = (c_2 x + c_3)c_1\), or \(u = A_1 x + A_2\).

Step 6: Solve Case II: \(\lambda = -\alpha^2\)

If \(\lambda = -\alpha^2 < 0\), then the equation \(X'' - \alpha^2 X = 0\) results. The general solution is \(X(x) = c_4 \cosh(\alpha x) + c_5 \sinh(\alpha x)\). The full solution becomes \(u = (A_3 \cosh(\alpha x) + A_4 \sinh(\alpha x))e^{k \alpha^2 t}\).

Step 7: Solve Case III: \(\lambda = \alpha^2\)

If \(\lambda = \alpha^2 > 0\), then the equation \(X'' + \alpha^2 X = 0\) results. Solving yields \(X(x) = c_6 \cos(\alpha x) + c_7 \sin(\alpha x)\). The complete solution is \(u = (A_3 \cos(\alpha x) + A_6 \sin(\alpha x))e^{-k \alpha^2 t}\).

Key Concepts

Partial Differential Equations

Partial Differential Equations (PDEs) involve equations that have partial derivatives. These are often used to formulate problems with several independent variables. Common physical applications include systems involving heat conduction, sound, fluid flow, elasticity, or electromagnetic fields.

PDEs can describe and predict how physical systems change over space and time. The main challenge when dealing with PDEs is to find a function that satisfies the equation under given conditions. In this exercise, we use a method called separation of variables to transform a PDE into simpler, separate Ordinary Differential Equations (ODEs).

The goal is to express the function as a product of functions, each depending only on a single coordinate. This approach simplifies the PDE into separate equations over each variable, allowing us to solve these simpler equations independently.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) appear when we solve a PDE using separation of variables. ODEs involve functions of one variable and their derivatives. Each ODE in this context comes from setting the separated terms equal to a constant.

In the step-by-step solution, after introducing a separation constant, we obtain two ODEs. These are in terms of spatial component, \( X'' + \lambda X = 0 \), and temporal component, \( T' + \lambda k T = 0 \). We explore different forms of solutions based on the value of the separation constant \( \lambda \). Since these problems are set up typically in applied physics contexts, understanding the solution curve types (exponential, sinusoidal, hyperbolic) helps identify physical systems they describe.

Mathematical Physics

Mathematical Physics uses math to solve physical problems and predict the effects of physical laws. PDEs are widely used in mathematical physics, serving as models for describing a range of universal phenomena. This exercise showcases solving a heat equation, a classic type of PDE used extensively in physics to model heat conduction and diffusion phenomena.

Each type of solution reflects different physical behaviors:

• The trivial case \( \lambda = 0 \) results in linear behavior in spatial domain representing a stable state.
• For \( \lambda = -\alpha^2 < 0 \), exponential growth or decay appears, showing rapid changes like heat spreading out.
• The \( \lambda = \alpha^2 > 0 \) case shows oscillatory solutions typical of wave-like phenomena.

Recognizing and understanding these behaviors is key to applying the concept of PDEs in physics.

Solution Methods for PDEs

Solution methods for PDEs can vary, but one frequently used technique is separation of variables. The process takes advantage of the simplicity in the structure of the PDE to separate it into manageable ODEs.

This involves assuming a solution that multiplies two or more functions, each depending on a single independent variable. By substituting this form into the original PDE, the equation breaks down into parts, each corresponding to one of these functions. Solving these gives the general solution to the PDE.

The importance of choosing correct boundary and initial conditions cannot be understated. They ensure uniqueness and applicability of the solution to real-world scenarios. This exercise's structured method helps build foundational skills for tackling more complex equations in advanced studies.