Problem 6 - Advanced Engineering Mathematics

Question

Use separation of variables to find, if possible, product solutions for the given partial differential equation. $$ y \frac{\partial u}{\partial x}+x \frac{\partial u}{\partial y}=0 $$

Step-by-Step Solution

Answer

The product solution is \(u(x, y) = K x^{-k^2} y^{k^2}\).

Step 1: Assume a product solution

Assume that the solution can be written as a product of two functions, one depending only on \(x\) and the other only on \(y\). Let \(u(x, y) = X(x)Y(y)\).

Step 2: Substitute into the PDE

Substitute the product solution \(u(x, y) = X(x)Y(y)\) into the partial differential equation \(y \frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} = 0\). This gives:\[ y \left( Y(y) \frac{dX(x)}{dx} \right) + x \left( X(x) \frac{dY(y)}{dy} \right) = 0 \]

Step 3: Separate variables

After substitution, factor the equation:\[ Y(y) y \frac{dX(x)}{dx} + X(x) x \frac{dY(y)}{dy} = 0 \]Dividing through by \(X(x)Y(y)\), we can separate variables:\[ y \frac{dX(x)}{dx} X(x)^{-1} = - x \frac{dY(y)}{dy} Y(y)^{-1} \]

Step 4: Set each side equal to a constant

Set \(y \frac{dX(x)}{dx} X(x)^{-1} = - x \frac{dY(y)}{dy} Y(y)^{-1} = -k^2\) because the left-hand side depends only on \(x\) and the right-hand side only on \(y\), and they must be equal to a constant for equality to hold for all \(x\) and \(y\).

Step 5: Solve the ODE for X(x)

From \(y \frac{dX(x)}{dx} X(x)^{-1} = -k^2\), derive:\[ \frac{dX(x)}{dx} = -k^2 \frac{X(x)}{x} \]This is a separable differential equation. Integrating both sides yields:\[ \ln|X(x)| = -k^2 \ln|x| + C_x \]Thus, solving for \(X(x)\) gives:\[ X(x) = C e^{-k^2 \ln|x|} = C x^{-k^2} \]

Step 6: Solve the ODE for Y(y)

Similarly, solve \(x \frac{dY(y)}{dy} Y(y)^{-1} = k^2\).From \( \frac{dY(y)}{dy} = k^2 \frac{Y(y)}{y} \), integrating gives:\[ \ln|Y(y)| = k^2 \ln|y| + C_y \]Thus, solving for \(Y(y)\) gives:\[ Y(y) = D y^{k^2} \]

Step 7: Write the general solution

Combine the solutions for \(X(x)\) and \(Y(y)\) to get the general solution:\[ u(x, y) = C x^{-k^2} D y^{k^2} = K x^{-k^2} y^{k^2} \]Where \(K \) is an arbitrary constant and \(k\) is a separation constant.

Key Concepts

Partial Differential Equations

Partial differential equations (PDEs) are equations that involve unknown multivariable functions and their partial derivatives. These are powerful tools in mathematics and physics, helping to describe phenomena such as heat conduction, wave propagation, fluid dynamics, and much more. For example, the equation given in the problem is a PDE that involves functions of two variables, namely, \( u(x, y) \), and their partial derivatives with respect to \( x \) and \( y \).

• Partial Derivatives: These are derivatives with respect to one variable while keeping others constant.
• Applications: PDEs model problems in engineering and applied sciences.

Solving PDEs can often be complex, but techniques like separation of variables provide a tractable means of finding solutions by reducing them to ordinary differential equations (ODEs). This makes learning about PDEs fundamental for anyone interested in the science and mathematical modeling fields.

Product Solutions

In the context of solving partial differential equations, a product solution is a method where one assumes the solution can be written as a product of functions, each depending solely on one of the variables. For instance, in the original exercise, the solution \( u(x, y) \) was assumed to be in the form \( X(x)Y(y) \). This assumption simplifies the process because:

• Reduces Complexity: Simplifies a PDE into multiple ODEs, which are more straightforward to solve.
• Structure: Product solutions provide a structured approach to identifying if a PDE can be broken down effectively.

After substituting the product solution into the PDE, each part of the solution can be addressed separately. This method is especially useful when the PDE represents phenomena that naturally separate in different axes or dimensions, like in symmetrical or isolated systems.

Separation Constants

Separation constants are introduced during the process of separation of variables, allowing one to convert a PDE into ODEs. When the original PDE is expressed in terms of the assumed product solution, dividing and re-arranging terms lead to expressions that equate variables that are normally insoluble together. To equate expressions dependent on entirely separate variables, we introduce a constant, known as the separation constant.

• Purpose: The separation constant ensures that each variable equation can be solved independently of the others.
• Flexibility: These constants can represent a spectrum of values, usually reflecting constraints or physical conditions in modeling scenarios.

In the solution provided, the separation constant \( -k^2 \) was used, allowing us to derive two separate ordinary differential equations, each dependent solely on one variable. The use of such constants ultimately facilitates the derivation of a general solution that accommodates various real-world conditions.

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations involving derivatives of a function with respect to only one variable. Transforming partial differential equations into ODEs using separation of variables method makes solving them more feasible.

Solving ODEs

After splitting a PDE into ODEs, each equation becomes a problem of its own, solvable using methods specific to ODEs such as separation, integration, or using boundary conditions provided in real life problems. Here's how it worked in the steps:

• The product solution separated the PDE into two ODEs.
• Each ODE was solved individually, leading to solutions \( X(x) = C x^{-k^2} \) and \( Y(y) = D y^{k^2} \).

This process provides clarity and aids in forming solutions to complex problems by breaking them down. Understanding ODE-solving techniques is therefore central to mastering PDEs and applying them to practical problems.