Question

Determine the values of the constant \(\alpha\), if any, for which the specified function is a solution of the given partial differential equation. $$ u(x, y, t)=e^{\alpha t} \sin (x) \cos (2 y), \quad u_{x x}+u_{y y}-u_{t}=0 $$

Short answer

Answer: The constant \(\alpha\) that makes the function a solution is \(\alpha = 0\).

Step-by-step solution

Find \(u_{xx}\), \(u_{yy}\), and \(u_t\)

To find these partial derivatives, we will differentiate \(u(x, y, t)\) with respect to \(x, y,\) and \(t\). For \(u_{xx}\), we will differentiate \(u(x, y, t)\) with respect to \(x\) twice: $$ u_x = \frac{\partial u}{\partial x} = e^{\alpha t} \cos (x) \cos (2y) $$ $$ u_{xx} = \frac{\partial^2 u}{\partial x^2} = -e^{\alpha t} \sin (x) \cos (2y) $$ For \(u_{yy}\), we will differentiate \(u(x, y, t)\) with respect to \(y\) twice: $$ u_y = \frac{\partial u}{\partial y} = -2e^{\alpha t} \sin (x) \sin (2y) $$ $$ u_{yy} = \frac{\partial^2 u}{\partial y^2} = -4e^{\alpha t} \sin (x) \cos (2y) $$ For \(u_t\), we will differentiate \(u(x, y, t)\) with respect to \(t\) only once: $$ u_t = \frac{\partial u}{\partial t} = \alpha e^{\alpha t} \sin (x) \cos (2y) $$

Substitute \(u_{xx}\), \(u_{yy}\), and \(u_t\) into the partial differential equation

Now we will substitute \(u_{xx}\), \(u_{yy}\), and \(u_t\) into the given partial differential equation \(u_{xx} + u_{yy} - u_t = 0\): $$ (-e^{\alpha t} \sin (x) \cos (2y)) + (-4e^{\alpha t} \sin (x) \cos (2y)) - (\alpha e^{\alpha t} \sin (x) \cos (2y)) = 0 $$

Simplify and solve for \(\alpha\)

Next, we will simplify and solve for \(\alpha\). Factor out the common terms \(e^{\alpha t} \sin (x) \cos (2y)\): $$ (-1 - 4 - \alpha) e^{\alpha t} \sin (x) \cos (2y) = 0 $$ Since we are interested in the values of \(\alpha\), and we know that \(e^{\alpha t} \sin (x) \cos (2y)\) is not equal to 0 as long as the arguments of the sine and cosine functions are not multiples of π, we can disregard the term and focus on solving the equation within the brackets: $$ -1 - 4 - \alpha = 0 $$ Adding 5 to both sides: $$ \alpha = 5 - 1 - 4 = 0 $$ Therefore, the constant \(\alpha = 0\) makes the function \(u(x, y, t)=e^{\alpha t} \sin (x) \cos (2 y)\) a solution to the given partial differential equation.

Key concepts

Boundary Value Problems

In the world of mathematics, particularly in the study of partial differential equations, we often come across boundary value problems. These problems are fundamental to understanding how a system behaves under certain conditions constrained by its boundaries. Imagine a vibrating string tied at both ends; the fixed points are the boundary conditions influencing the string's vibration pattern.

Boundary value problems typically require the solution of a differential equation within a specified set of boundary conditions. Solving these problems can reveal the temperature distribution in a rod, the displacement in a beam, or, interestingly, the electric potential in a field, as governed by Laplace's equation. It is essential for students to understand that finding a solution implies looking for a function that satisfies not only the differential equation but also these preset conditions at the boundaries.

Laplace's Equation

A cornerstone of potential theory is Laplace's equation, which is often written compactly as \[abla^2 u = 0\]. This elegant equation appears frequently in physics and engineering, describing the behavior of electric, gravitational, and fluid potentials.

For a two-dimensional scenario, like in the solved exercise, Laplace's equation simplifies to \[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\]. Our exercise takes this a step further by factoring the temporal dimension to push students' understanding of evolving systems. Laplace's equation signifies that the solution represents a system in equilibrium, such as the electrostatic field around a group of charges. The equation implies that within a region devoid of charge, the electric potential exhibits no 'source' or 'sink'—it is harmonious, the mathematical echo of balance.

Separation of Variables

When solving complex problems like the differential equation in the exercise, a powerful technique comes into play: separation of variables. This method simplifies partial differential equations by transforming them into easier-to-solve ordinary differential equations.

Separation of variables operates under the premise that we can express the solution as a product of functions, each dependent on a single independent variable. For instance, in our example \[u(x, y, t) = e^{\alpha t} \sin(x) \cos(2y)\], the solution separates the variables x, y, and t, thus allowing individual examination. This clever maneuver is advantageous in multidimensional problems where each dimension's behavior can be isolated and solved independently before weaving the solutions back together into a cohesive whole.

Homogeneous Equations

Homogeneous equations, another vital concept for students, are distinguished by their lack of a forcing term - a term without the function or its derivatives. Such an equation is typified by having all terms involving the unknown function or its derivatives.

In our context, a homogeneous equation like the one in the given exercise means that every term either involves the function u or its partial derivatives. This feature makes the equation amenable to techniques like separation of variables. The exercise showcases a homogeneous equation where, after applying the method and deducing \[-1 - 4 - \alpha = 0\], we find a temporal component \[\alpha\] that satisfies our equation for a static solution, meaning no change over time—a homogeneity reflected not just in terms but in the equilibrium of the system across the temporal dimension.