Question

Verify that the given function or functions is a solution of the given partial differential equation. $$ a^{2} u_{x x}=u_{u^{i}}, \quad u_{1}(x, t)=\sin \lambda x \sin \lambda a t, \quad u_{2}(x, t)=\sin (x-a t), \quad \lambda \text { a real constant } $$

Short answer

Question: Verify if the given functions \(u_1(x, t) = \sin(\lambda x) \sin(\lambda at)\) and \(u_2(x, t) = \sin(x - at)\) are solutions to the partial differential equation \(a^2u_{xx} = u_{tt}\). Answer: Both functions \(u_1(x, t)\) and \(u_2(x, t)\) are solutions to the given partial differential equation.

Step-by-step solution

Calculate the second-order partial derivatives for function \(u_1(x, t)\)

To calculate the second-order partial derivatives for the first function \(u_1(x, t) = \sin(\lambda x) \sin(\lambda at)\), we'll first find the first-order partial derivatives \(u_x\) and \(u_t\): $$ u_{1x} = \frac{\partial u_1}{\partial x} = \lambda\cos(\lambda x)\sin(\lambda at), \qquad u_{1t} = \frac{\partial u_1}{\partial t} = \lambda a\sin(\lambda x)\cos(\lambda at) $$ Now, we'll derive \(u_{1x}\) and \(u_{1t}\) with respect to \(x\) and \(t\) respectively to get the second-order partial derivatives: $$ u_{1xx} = \frac{\partial^2 u_1}{\partial x^2} = -\lambda^2\sin(\lambda x)\sin(\lambda at), \qquad u_{1tt} = \frac{\partial^2 u_1}{\partial t^2} = -\lambda^2 a^2\sin(\lambda x)\sin(\lambda at) $$

Substitute the second-order derivatives into the PDE for function \(u_1(x, t)\)

Now, substitute the second-order derivatives \(u_{1xx}\) and \(u_{1tt}\) into the given PDE: $$ a^2u_{1xx} = a^2(-\lambda^2\sin(\lambda x)\sin(\lambda at)) = -\lambda^2a^2\sin(\lambda x)\sin(\lambda at) $$ and $$ u_{1tt} = -\lambda^2 a^2\sin(\lambda x)\sin(\lambda at) $$ Since \(a^2u_{1xx} = u_{1tt}\), function \(u_1(x, t)\) is a solution to the given partial differential equation.

Calculate the second-order partial derivatives for function \(u_2(x, t)\)

Repeat the process for the second function \(u_2(x, t) = \sin(x - at)\). Find the first-order partial derivatives \(u_x\) and \(u_t\): $$ u_{2x} = \frac{\partial u_2}{\partial x} = \cos(x - at), \qquad u_{2t} = \frac{\partial u_2}{\partial t} = -a\cos(x - at) $$ Now, derive \(u_{2x}\) and \(u_{2t}\) with respect to \(x\) and \(t\) respectively to get the second-order partial derivatives: $$ u_{2xx} = \frac{\partial^2 u_2}{\partial x^2} = -\sin(x - at), \qquad u_{2tt} = \frac{\partial^2 u_2}{\partial t^2} = -a^2\sin(x - at) $$

Substitute the second-order derivatives into the PDE for function \(u_2(x, t)\)

Substitute the second-order derivatives \(u_{2xx}\) and \(u_{2tt}\) into the given PDE: $$ a^2u_{2xx} = a^2(-\sin(x - at)) = -a^2\sin(x - at) $$ and $$ u_{2tt} = -a^2\sin(x - at) $$ Since \(a^2u_{2xx} = u_{2tt}\), function \(u_2(x, t)\) is a solution to the given partial differential equation. Both \(u_1(x, t)\) and \(u_2(x, t)\) are solutions of the given partial differential equation.

Key concepts

Second-Order Partial Derivatives

When we talk about second-order partial derivatives, we are referring to the derivatives taken twice with respect to different variables. If we consider a function like \( u(x, t) \), then taking the partial derivative first with respect to \( x \) and then again with respect to \( x \) gives us the second-order partial derivative with respect to \( x \), commonly denoted as \( u_{xx} \). Similarly, taking the derivative with respect to \( t \) twice results in \( u_{tt} \).

These second-order partial derivatives play a crucial role in understanding the behavior of the function in response to changes in both variables. In the context of partial differential equations (PDEs), they are essential components, often representing acceleration, curvature, or diffusion, depending upon the application. The PDE in our exercise requires evaluating both \( u_{xx} \) and \( u_{tt} \) to verify the solution. This process involves performing differentiation twice and is pivotal for ensuring the function satisfies the given PDE.

Sinusoidal Functions

Sinusoidal functions, such as \( \sin(x) \) and \( \cos(x) \) are fundamental in both pure and applied mathematics. These functions oscillate in a wave-like pattern, which is a key characteristic when modeling phenomena that repeat in a regular cycle, such as sound waves, light waves, and even the oscillation of a pendulum.

In our exercise, the given functions \( u_1(x, t) \) and \( u_2(x, t) \) involve sinusoidal components, which introduce this periodic behavior. When subjected to differentiation, sinusoidal functions possess the property that their derivatives are also sinusoidal, making them predictable and their behavior under differentiation well-understood. \( \sin \) becomes \( \cos \) with the first derivative and loops back to \( -\sin \) with the second derivative, maintaining the wave form while flipping the sign. This property simplifies the process of verifying their role in solving PDEs like the one in our original problem.

Verification of PDE Solutions

To verify a solution to a PDE, we follow a systematic approach of first assuming our function is a solution to the equation, then performing the necessary derivative calculations, and finally substituting back into the original PDE to check for consistency.

In the given problem, our goal was to verify that \( u_1(x, t) \) and \( u_2(x, t) \) are indeed solutions of the PDE \( a^2 u_{xx} = u_{tt} \). Upon calculating the second-order derivatives of the given functions and substituting them into the PDE, we found that the left and right-hand sides match exactly, confirming the functions as solutions to the equation. This verification step is crucial; it ensures that our proposed functions satisfy the necessary conditions stipulated by the PDE across all points in the relevant domain.