Question

In each exercise, (a) Show by direct substitution that the linear combination of functions is a solution of the given homogeneous linear partial differential equation. (b) Determine values of the constants so that the linear combination satisfies the given supplementary condition. \(u(x, t)=c_{1}+c_{2} e^{-t} \cos x+c_{3} e^{-4 t} \cos 2 x, \quad u_{x x}-u_{t}=0\) \(u(x, 0)=2-\cos 2 x\)

Short answer

Question: Show by direct substitution that the given linear combination \(u(x, t) = c_{1} + c_{2} e^{-t} \cos x + c_{3} e^{-4t} \cos 2x\) satisfies the partial differential equation \(u_{xx} - u_{t} = 0\), and find the values of the constants \(c_{1}\), \(c_{2}\), and \(c_{3}\), such that the linear combination satisfies the supplementary condition \(u(x, 0) = 2 - \cos 2x\). Answer: By direct substitution, we verified that the given linear combination satisfies the PDE. The values of the constants are \(c_{1} = 2\), \(c_{2} = -c_{3}\), and \(c_{3}\) can have any value to match the given PDE and supplementary condition. Therefore, the linear combination is \(u(x, t) = 2 + c_{3}(e^{-t}\cos x - e^{-4t} \cos 2x)\).

Step-by-step solution

Calculate partial derivatives of u

Find the partial derivatives: (1) First-order partial derivative of \(u\) with respect to \(x\): \(u(x, t)_x = -c_{2} e^{-t} \sin x - 2c_{3} e^{-4t} \sin 2x\) (2) Second-order partial derivative of \(u\) with respect to \(x\): \(u(x, t)_{xx} = -c_{2} e^{-t} \cos x - 4c_{3} e^{-4t} \cos 2x\) (3) First-order partial derivative of \(u\) with respect to \(t\): \(u(x, t)_t = -c_{2} e^{-t} \cos x - 4c_{3} e^{-4t} \cos 2x\)

Verify the given PDE is satisfied

Now, substitute the calculated derivatives into the given PDE, \(u_{xx} - u_{t} = 0\): \((-c_{2} e^{-t} \cos x - 4c_{3} e^{-4t} \cos 2x) - (-c_{2} e^{-t} \cos x - 4c_{3} e^{-4t} \cos 2x) = 0\) It's clear that the given linear combination satisfies the PDE.

Find the constants using the supplementary condition

Let's use the given supplementary condition, \(u(x, 0) = 2 - \cos 2x\), and substitute into the expression for \(u(x, t)\): \(u(x, 0) = c_{1} + c_{2} e^{0} \cos x + c_{3} e^{0} \cos 2x\) Now set it equal to the given condition: \(2 - \cos 2x = c_{1} + c_{2} \cos x + c_{3} \cos 2x\) Comparing the coefficients on both sides, we have: \(c_{1} = 2\) \(c_{2} = -c_{3}\) Since the equation does not provide enough information to determine \(c_{3}\) specifically, the linear combination remains as: \(u(x, t) = 2 + c_{3}(e^{-t}\cos x - e^{-4t} \cos 2x)\) where \(c_{3}\) can have any value to match the given PDE and supplementary condition.

Key concepts

Homogeneous PDE

A homogeneous partial differential equation (PDE) is an equation where the output, usually denoted as a function, equals zero. In mathematical terms, a homogeneous PDE takes the form \( L(u) = 0 \), where \( L \) represents a linear operator and \( u \) represents the unknown function of several variables.
One key characteristic of homogeneous PDEs is that any linear combination of its solutions is also a solution. This property often allows us to simplify the process of finding specific solutions.
For instance, the problem at hand involves showing that the function \( u(x, t) \) is a solution to the homogeneous PDE, \( u_{xx} - u_t = 0 \). By confirming this through direct substitution and comparing terms, we demonstrate that the given function satisfies the PDE.

Linear Combination

The term "linear combination" refers to an expression constructed from a set of terms by multiplying each term by a constant and summing the results. This concept is particularly useful when dealing with linear problems, such as linear PDEs.
In our exercise, we consider the linear combination of functions \( c_1 + c_2 e^{-t} \cos x + c_3 e^{-4t} \cos 2x \). Each term in this combination is weighted by constants \( c_1, c_2, \) and \( c_3 \) which determine the specific form of the solution.
By associating these constants with the desired supplementary conditions, we ensure that the linear combination not only satisfies the original PDE but also matches specific boundary values at given conditions.

Boundary Value Problems

Boundary value problems (BVPs) are differential equations coupled with a set of constraints, known as boundary conditions. These conditions must be satisfied by the solution of the differential equation within a certain domain.
In this exercise, the supplementary condition \( u(x, 0) = 2 - \cos 2x \) serves as the boundary requirement that the solution must adhere to.
The goal in solving a BVP is to determine unknown constants (like \( c_1, c_2, \) and \( c_3 \) in this specific problem) so that the solution satisfies both the differential equation and the boundary conditions. By matching coefficients from both the solution expression and the boundary condition, we can solve for these constants.

Partial Derivatives

A partial derivative represents the rate of change of a multivariable function with respect to one variable, holding other variables constant. For PDEs, computing partial derivatives is often a key step.
In our problem, we compute several partial derivatives of \( u(x, t) \) with respect to \( x \) and \( t \):

• First-order partial derivative with respect to \( x \): \( u_x = -c_2 e^{-t} \sin x - 2c_3 e^{-4t} \sin 2x \)
• Second-order partial derivative with respect to \( x \): \( u_{xx} = -c_2 e^{-t} \cos x - 4c_3 e^{-4t} \cos 2x \)
• First-order partial derivative with respect to \( t \): \( u_t = -c_2 e^{-t} \cos x - 4c_3 e^{-4t} \cos 2x \)

These derivatives are then substituted into the PDE to verify whether the given function satisfies the equation. Calculating and substituting partial derivatives is fundamental as it allows us to analyze how each small change in one variable influences the function.