Problem 14 - Advanced Engineering Mathematics

Question

Substituting \(u(x, y)=X(x) Y(y)\) into the partial differential equation yields \(x^{2} X^{\prime \prime} Y+X Y^{\prime \prime}=0 .\) Separating variables and using the separation constant \(-\lambda\) we obtain $$-\frac{x^{2} X^{\prime \prime}}{X}=\frac{Y^{\prime \prime}}{Y}=-\lambda.$$ Then $$x^{2} X^{\prime \prime}-\lambda X=0 \quad \text { and } \quad Y^{\prime \prime}+\lambda Y=0.$$ We consider three cases: I. If \(\lambda=0\) then \(x^{2} X^{\prime \prime}=0\) and \(X(x)=c_{1} x+c_{2} .\) Also, \(Y^{\prime \prime}=0\) and \(Y(y)=c_{3} y+c_{4}\) so $$u=X Y=\left(c_{1} x+c_{2}\right)\left(c_{3} y+c_{4}\right).$$ II. If \(\lambda=-\alpha^{2}<0\) then \(x^{2} X^{\prime \prime}+\alpha^{2} X=0\) and \(Y^{\prime \prime}-\alpha^{2} Y=0 .\) The solution of the second differential equation is \(Y(y)=c_{5} \cosh \alpha y+c_{6} \sinh \alpha y .\) The first equation is Cauchy-Euler with auxiliary equation \(m^{2}-m+\alpha^{2}=0 .\) Solving for \(m\) we obtain \(m=\frac{1}{2} \pm \frac{1}{2} \sqrt{1-4 \alpha^{2}} . \quad\) We consider three possibilities for the discriminant \(1-4 \alpha^{2}\) (i) If \(1-4 \alpha^{2}=0\) then \(X(x)=c_{7} x^{1 / 2}+c_{8} x^{1 / 2} \ln x\) and \\[ u=X Y=x^{1 / 2}\left(c_{7}+c_{8} \ln x\right)\left(c_{5} \cosh \alpha y+c_{6} \sinh \alpha y\right) \\] (ii) If \(1-4 \alpha^{2}<0\) then \(X(x)=x^{1 / 2}\left[c_{9} \cos (\sqrt{4 \alpha^{2}-1} \ln x)+c_{10} \sin (\sqrt{4 \alpha^{2}-1} \ln x)\right]\) and \\[ u=X Y=x^{1 / 2}\left[c_{9} \cos (\sqrt{4 \alpha^{2}-1} \ln x)+c_{10} \sin (\sqrt{4 \alpha^{2}-1} \ln x)\right]\left(c_{5} \cosh \alpha y+c_{6} \sinh \alpha y\right) \\] (iii) If \(1-4 \alpha^{2}>0\) then \(X(x)=x^{1 / 2}\left(c_{11} x^{\sqrt{1-4 \alpha^{2}} / 2}+c_{12} x^{-\sqrt{1-4 \alpha^{2}} / 2}\right)\) and \\[ u=X Y=x^{1 / 2}\left(c_{11} x^{\sqrt{1-4 \alpha^{2} / 2}}+c_{12} x^{-\sqrt{1-4 \alpha^{2} / 2}}\right)\left(c_{5} \cosh \alpha y+c_{6} \sinh \alpha y\right) \\] III. If \(\lambda=\alpha^{2}>0\) then \(x^{2} X^{\prime \prime}-\alpha^{2} X=0\) and \(Y^{\prime \prime}+\alpha^{2} Y=0 .\) The solution of the second differential equation is \(Y(y)=c_{13} \cos \alpha y+c_{14} \sin \alpha y .\) The first equation is Cauchy-Euler with auxiliary equation \(m^{2}-m-\alpha^{2}=0\) Solving for \(m\) we obtain \(m=\frac{1}{2} \pm \frac{1}{2} \sqrt{1+4 \alpha^{2}}\). In this case the discriminant is always positive so the solution of the differential equation is \(X(x)=x^{1 / 2}\left(c_{15} x^{\sqrt{1+4 \alpha^{2}} / 2}+c_{16} x^{-\sqrt{1+4 \alpha^{2} / 2}}\right)\) and \\[ u=X Y=x^{1 / 2}\left(c_{15} x^{\sqrt{1+4 \alpha^{2}} / 2}+c_{16} x^{-\sqrt{1+4 \alpha^{2}} / 2}\right)\left(c_{13} \cos \alpha y+c_{14} \sin \alpha y\right) \\]

Step-by-Step Solution

Answer

Use separation of variables to solve the PDE by considering different cases for \( \lambda \).

Step 1: Plug in Separated Variables

Substitute the expression for the solution, \( u(x,y) = X(x)Y(y) \), into the given PDE, which is \( x^2 X'' Y + X Y'' = 0 \). This leads to the equation \( \frac{x^2 X''}{X} = -\frac{Y''}{Y} = -\lambda \).

Step 2: Formulate Separate ODEs

Using the separation constant \( -\lambda \), we form two ordinary differential equations: \( x^2 X'' - \lambda X = 0 \) and \( Y'' + \lambda Y = 0 \).

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

For \( \lambda = 0 \), the equations simplify to \( x^2 X'' = 0 \) and \( Y'' = 0 \). Solving these gives \( X(x) = c_1 x + c_2 \) and \( Y(y) = c_3 y + c_4 \). The solution for the product \( u = X Y \) is \( (c_1 x + c_2)(c_3 y + c_4) \).

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

For \( \lambda = -\alpha^2 < 0 \), solve \( x^2 X'' + \alpha^2 X = 0 \) (a Cauchy-Euler equation) and \( Y'' - \alpha^2 Y = 0 \). Solve for Y: \( Y(y) = c_5 \cosh(\alpha y) + c_6 \sinh(\alpha y) \). For X, depending on \( 1-4\alpha^2 \), solve accordingly. For zero, negative, or positive discriminants, find different forms of \( X(x) \). Combine these to find \( u = X Y \).

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

For \( \lambda = \alpha^2 > 0 \), solve \( x^2 X'' - \alpha^2 X = 0 \) and \( Y'' + \alpha^2 Y = 0 \). The solution for Y is \( Y(y) = c_{13} \cos(\alpha y) + c_{14} \sin(\alpha y) \). Solve the Cauchy-Euler for X, given the always positive discriminant, \( X(x) = x^{1/2}(c_{15}x^{\sqrt{1+4\alpha^2}/2} + c_{16}x^{-\sqrt{1+4\alpha^2}/2}) \). Combine X and Y to form \( u = X Y \).

Key Concepts

Separation of Variables

Separation of Variables is a powerful technique used to solve partial differential equations (PDEs). This method helps us rewrite a complex equation into simpler, separate parts. By assuming that the solution can be written as a product of functions, each depending on a single variable, we simplify the equation. For example, if we consider a function as the product of two variables, say, \(u(x, y) = X(x)Y(y)\), we can input this into our PDE. After substitution, we can separate the variables into two distinct ordinary differential equations (ODEs).
The beauty of separation of variables lies in its simplicity. By isolating each function to depend on one variable exclusively, the solution becomes more manageable. It is crucial in problems with specified boundary conditions, like heat equations or wave equations. This strategy not only simplifies the solving process but also highlights the independent roles of each variable.

• Assume the solution as a product of independent functions.
• Substitute into the PDE to separate variables.
• Derive two ODEs from the separated equation.

When tackling a PDE, always check if separation of variables might be a feasible method; it could save a lot of effort in finding the solution.

Cauchy-Euler Equation

The Cauchy-Euler Equation is a specific type of differential equation notable for its variable coefficients which have a particular form. In this method, the coefficients are powers of the independent variable, typically involving terms like \(x^2\). When solving such equations, the substitution \(x = e^t\) often simplifies the equation into one with constant coefficients, easily solvable through standard methods.
In the original exercise, the Cauchy-Euler equation appears when \( \lambda = -\alpha^2 < 0 \). The general approach involves solving for the characteristic equation \(m^2 - m + \alpha^2 = 0\), leading to different solutions based on the discriminant:

• If the discriminant is zero, the solution involves terms such as \(x^{1/2}\) and \(x^{1/2} \ln x\).
• A negative discriminant results in trigonometric functions involving \(\ln x\).
• A positive discriminant provides exponential-style solutions.

Understanding Cauchy-Euler equations allows us to solve a range of problems where the coefficients aren’t constant but still follow a recognizable pattern. They find applications in many physical scenarios, reflecting systems with power-law behaviors.

Separation Constant

The concept of a Separation Constant arises naturally when using separation of variables. During the process of separating a PDE, we often split the equation into different ODEs. These new equations must reconcile with each other through a constant, this is the separation constant, commonly denoted as \(\lambda\).
The choice of \(\lambda\) determines the nature of the resulting solutions. For instance, when \(\lambda = 0\), the physical implications often mean that there is no particular growth or decay in the system. Negative values like \(-\alpha^2\) can imply damping or other variations, and positive values such as \(\alpha^2\) typically indicate growth or oscillatory behavior. In solving the original PDE using separation of variables:

• The separation constant \(-\lambda\) connects the separate solutions to conform them to the original PDE.
• It categorizes the problem into distinct cases depending on the sign of \(\lambda\).
• Determining the appropriate \(\lambda\) can guide us to sensible physical interpretations of the solution.

The separation constant is crucial as it bridges the gap between simple variable solutions and the complex interactions described by the original equation.