Question

Determine the values of the constant \(\alpha\), if any, for which the specified function is a solution of the given partial differential equation. $$ u(x, t)=\sin (\alpha x) \cos (2 t), \quad u_{x x}-u_{t t}-4 \alpha u=0 $$

Short answer

The function u(x,t)=sin(αx)cos(2t) is a solution of the partial differential equation uxx - utt - 4αu = 0 for α ≈ 0.7320508 and α ≈ -5.7320508.

Step-by-step solution

Calculate the first and second-order partial derivatives

To find the second-order partial derivatives, we first need to find the first-order partial derivatives of \(u(x,t)\) with respect to \(x\) and \(t\). Then we will differentiate the first-order derivatives again to find the second-order partial derivatives. The function is given as: $$ u(x, t)=\sin(\alpha x)\cos(2t) $$ First-order partial derivatives: $$ u_x = \frac{\partial u}{\partial x} = \alpha\cos(\alpha x)\cos(2t) $$ $$ u_t = \frac{\partial u}{\partial t} = -2\sin(\alpha x)\sin(2t) $$ Now, let's find the second-order partial derivatives: $$ u_{xx} = \frac{\partial^2 u}{\partial x^2} = -\alpha^2\sin(\alpha x)\cos(2t) $$ $$ u_{tt} = \frac{\partial^2 u}{\partial t^2} = -4\sin(\alpha x)\cos(2t) $$

Substitute the partial derivatives and given function into the partial differential equation

Now we will substitute the second-order partial derivatives \(u_{xx}\) and \(u_{tt}\), and the given function \(u(x,t)\) back into the partial differential equation: $$ u_{xx} - u_{tt} - 4\alpha u = (-\alpha^2\sin(\alpha x)\cos(2t)) - (-4\sin(\alpha x)\cos(2t)) - 4\alpha(\sin(\alpha x)\cos(2t)) = 0 $$

Simplify the equation and solve for \(\alpha\)

We can factor out the common term \(\sin(\alpha x)\cos(2t)\) to simplify the equation: $$ \sin(\alpha x)\cos(2t)(-\alpha^2 + 4 - 4\alpha) = 0 $$ Since \(\sin(\alpha x)\cos(2t)\) is not always zero, we need to find the value(s) of \(\alpha\) that makes the expression inside the parentheses equal to zero: $$ -\alpha^2 + 4 - 4\alpha = 0 $$ To solve for \(\alpha\), we can rewrite the above quadratic equation in standard form: $$ \alpha^2 + 4\alpha - 4 = 0 $$ At this point, we can apply the quadratic formula to solve for \(\alpha\): $$ \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where \(a = 1\), \(b = 4\), and \(c = -4\): $$ \alpha = \frac{-4 \pm \sqrt{4^2 - 4(1)(-4)}}{2(1)} $$ After calculating, we find two possible values for \(\alpha\): $$ \alpha_1 = 0.7320508, \quad \alpha_2 = -5.7320508 $$ So, the constant values of \(\alpha\) for which the given function is a solution of the given partial differential equation are approximately \(\alpha_1 = 0.7320508\) and \(\alpha_2 = -5.7320508\).

Key concepts

Understanding Boundary Value Problems

Boundary value problems (BVPs) play a pivotal role in the scientific and engineering fields, providing vital information on steady-state or equilibrium conditions. In essence, a BVP is a differential equation coupled with a set of additional constraints called boundary conditions. These conditions specify values or relationships that the solution must satisfy at the boundaries of the domain over which the differential equation is defined.

For instance, in preparing to solve a BVP for a partial differential equation (PDE), one would first identify the spatial and temporal domain and then apply the relevant boundary conditions. These conditions could entail the PDE's solution being fixed at certain points, or the solution's derivatives maintaining specific values at the domain's edges. Returning to the original exercise, the determination of constant \(\alpha\), ensuring that the function is a solution to the given PDE, is a similar exercise in boundary analysis. Although explicit boundary conditions weren't provided, the solution process inherently addressed the implicit constraints set by the problem's structure.

Second-order Partial Derivatives and Their Significance

Diving deeper into the understanding of second-order partial derivatives, it's important to realize they provide insights into the curvature and concavity of a function's graph on a multivariable domain. These derivatives measure how the rate of change of a quantity changes in two different directions independently. Considering our example problem, \(u_{xx}\) and \(u_{tt}\) represent the rate at which the rate of change of \(u\) with respect to \(x\) and \(t\) changes with \(x\) and \(t\) respectively.

When working with second-order partial derivatives in PDEs, they can reveal how a physical quantity, like temperature or displacement, varies in space and time. The equation \(u_{xx} - u_{tt} - 4 \alpha u=0\) subtly encrypts within it the balances of forces, energies, or fluxes relevant to the phenomenon at hand, requiring solutions that simultaneously satisfy these second derivatives.

Solving the Quadratic Equation

The quadratic equation, a fundamental concept for many algebraic problems, can also manifest in the realm of partial differential equations. It is generally expressed as \(ax^2 + bx + c = 0\) and is distinctive for its characteristic 'U'-shaped parabolic graph. To find the roots of the quadratic equation, we often employ the well-known quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), which gives us the points where the parabola intersects the x-axis.

In the exercise presented, after simplifying the PDE and isolating terms involving \(\alpha\), we derived a quadratic equation regarding \(\alpha\). Utilizing the quadratic formula subsequently allowed us to discover the distinct values for \(\alpha\) that satisfy the equation, illustrating the close relationship between algebraic techniques and the analysis of more complex mathematical models like PDEs. This connection underscored the significance of a strong foundation in algebra to untangle the intricacies of higher mathematics.