Question

In each exercise, the function \(u\) is known to be a solution of the given nonhomogeneous partial differential equation. Determine the function \(f\). \(u(x, y, t)=e^{x+2 y}+e^{-5 t} \sin x \sin 2 y, u_{x x}+u_{y y}-u_{t}=f(x, y, t)\)

Short answer

Given the nonhomogeneous partial differential equation (PDE) \(u_{xx} + u_{yy} - u_t = f(x,y,t)\) and the known solution \(u(x,y,t) = e^{x + 2y} + e^{-5t}\sin x\sin 2y\), we found the function \(f(x,y,t)\) by computing the necessary partial derivatives and substituting them back into the PDE. The function \(f(x,y,t)\) is \(f(x,y,t) = 5e^{x+2y} - 10e^{-5t}\sin x\sin 2y\).

Step-by-step solution

Derive \(u_x\) and \(u_y\)

Find the partial derivatives of \(u\) with respect to x and y: \(u_x = \frac{\partial u}{\partial x} = e^{x+2y}\cos x\sin 2y\) \(u_y = \frac{\partial u}{\partial y} = 2e^{x + 2y} + e^{-5t}\sin x\sin 2y\cos 2y\)

Derive \(u_{xx}\) and \(u_{yy}\)

Compute the second-order partial derivatives of \(u\) with respect to x and y: \(u_{xx} = \frac{\partial^2 u}{\partial x^2} = e^{x+2y} - e^{-5t}\sin x\sin 2y\) \(u_{yy} = \frac{\partial^2 u}{\partial y^2} = 4e^{x+2y} - 4e^{-5t}\sin x\sin 2y\)

Derive \(u_t\)

Compute the first-order partial derivative of \(u\) with respect to t: \(u_t = \frac{\partial u}{\partial t} = 5e^{-5t}\sin x\sin 2y\)

Substitute the derivatives into the PDE

Substitute \(u_{xx}\), \(u_{yy}\), and \(u_t\) back into the original equation: \(f(x,y,t) = u_{xx} + u_{yy} - u_t = (e^{x+2y} - e^{-5t}\sin x\sin 2y) + (4e^{x+2y} - 4e^{-5t}\sin x\sin 2y) - 5e^{-5t}\sin x\sin 2y\)

Simplify the expression for \(f(x,y,t)\)

Combine the terms to obtain the function \(f(x,y,t)\): \(f(x,y,t) = e^{x+2y} + 4e^{x+2y} - e^{-5t}\sin x\sin 2y - 4e^{-5t}\sin x\sin 2y - 5e^{-5t}\sin x\sin 2y = 5e^{x+2y} - 10e^{-5t}\sin x\sin 2y\) The function \(f(x,y,t)\) is \(f(x,y,t) = 5e^{x+2y} - 10e^{-5t}\sin x\sin 2y\).

Key concepts

Partial Derivatives

Understanding partial derivatives is crucial when dealing with functions that depend on multiple variables, like in our exercise involving the function u(x, y, t). Simply put, a partial derivative measures how a function changes as one of its input variables changes, while all the other variables are held constant.

For example, when we compute u_x, denoted as \(\frac{\partial u}{\partial x}\), we're looking at how the function u changes in the x-direction at a specific point. The notion extends to any number of variables, allowing us to explore the function's behavior along each dimension separately. This is similar to looking at a terrain map and examining the steepness of the landscape in the north-south or east-west directions, rather than the overall slope.

In our nonhomogeneous partial differential equation problem, we used partial derivatives as the first step towards solving the equation, focusing on how the function u changes with respect to x and y independently, which are u_x and u_y respectively.

Second-Order Derivatives

When we start examining second-order derivatives, like u_xx and u_yy, we dive deeper into the behavior of our function u. Second-order derivatives show how the rate of change itself is changing; this is often related to concepts like curvature or acceleration.

To obtain second-order partial derivatives such as \(\frac{\partial^2 u}{\partial x^2}\) or \(\frac{\partial^2 u}{\partial y^2}\), we perform differentiation on the first-order derivatives with respect to the same variable. In the context of our nonhomogeneous equation, we looked at how the change in slope, or gradient, of u in the x and y directions is further changing as we move along these axes. This gives us vital information on the 'shape' of the solution surface described by the function u.

In practical terms, imagine you're driving on a hilly road: first-order derivatives tell you if you're going uphill or downhill, while second-order derivatives indicate whether the hill is getting steeper or flatter as you continue driving.

Differential Equations Solutions

The goal of solving differential equations is to find functions that satisfy certain properties expressed by the equations. In our case, the nonhomogeneous partial differential equation includes terms describing the geometry of the solution (through second-order derivatives) as well as its evolution over time (through the time derivative u_t).

Nonhomogeneous equations are those in which the solution is not purely dependent on the derivatives of the function; there's an additional non-zero function, f(x, y, t), in our exercise. Determining this function is key to understanding and characterizing the behavior of the original function u over space and time.

To solve for f, we follow a series of steps that involve calculating various derivatives of u and substituting them back into the original partial differential equation. The simplification process eventually reveals the function f that, when combined with the homogeneous part of the equation (the derivatives), completes the solution. This approach is somewhat analogous to piecing together a puzzle: each derivative provides a piece, and our job is to fit them together correctly to reveal the full picture described by f(x, y, t).