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Chapter 9: Problem 10

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

Short Answer

Expert verified
Answer: The function u(x, y) is a solution of the given partial differential equation when the constant 伪 is -2.

Step by step solution

01

Find the second partial derivatives of u(x, y) with respect to x and y.

We are given the function: $$ u(x, y) = \alpha + e^{-x} \cos y $$ First partial derivatives: $$ u_x = -e^{-x} \cos y\\ u_y = -e^{-x}\sin y $$ Second partial derivatives: $$ u_{xx} = e^{-x}\cos y\\ u_{yy} = -e^{-x}\cos y $$
02

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

We have the following partial differential equation: $$ u_{xx} + u_{yy} - 2u = 4 $$ Substitute \(u(x, y) = \alpha + e^{-x} \cos y\), \(u_{xx} = e^{-x}\cos y\), and \(u_{yy} = -e^{-x}\cos y\) into the equation: $$ (e^{-x}\cos y) + (-e^{-x}\cos y) - 2(\alpha + e^{-x} \cos y) = 4 $$
03

Simplify the equation and solve for 伪

Simplify the equation: $$ 0 - 2\alpha - 2 e^{-x} \cos y = 4 $$ Now, we need to have the equation hold true for all x and y. The only way for this to happen is if both coefficients are equal to 0: $$ - 2 e^{-x} \cos y = 0\\ - 2\alpha = 4 $$ Solve for 伪: $$ \alpha = -2 $$ Therefore, the function u(x, y) is a solution of the given partial differential equation when the constant 伪 is -2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second Partial Derivatives
Understanding derivatives is crucial when studying partial differential equations (PDEs). The function given in the exercise is \( u(x, y) = \alpha + e^{-x} \cos y \). To solve a PDE like the one given, we need to find the second partial derivatives with respect to both \(x\) and \(y\).

The first step is finding the first partial derivatives:
  • Partial derivative with respect to \(x\), \( u_x = -e^{-x} \cos y \)
  • Partial derivative with respect to \(y\), \( u_y = -e^{-x} \sin y \)
Afterward, compute the second partial derivatives:
  • Second partial derivative with respect to \(x\), \( u_{xx} = e^{-x} \cos y \)
  • Second partial derivative with respect to \(y\), \( u_{yy} = -e^{-x} \cos y \)

These derivatives are essential for determining how the function \( u(x, y) \) behaves in the space defined by the variables \( x \) and \( y \). Calculating them correctly is key to finding solutions to PDEs.
PDE Solutions
A solution to a partial differential equation requires satisfying the equation for all values in a defined space. In this exercise, we have the PDE: \[ u_{xx} + u_{yy} - 2u = 4 \]Substituting our function \( u(x, y) = \alpha + e^{-x} \cos y \) and its second partial derivatives into this equation helps determine if \( u \) is indeed a solution.

Plugging in:
  • \( u_{xx} = e^{-x} \cos y \)
  • \( u_{yy} = -e^{-x} \cos y \)
  • \( u = \alpha + e^{-x} \cos y \)
This results in the simplified equation:\[ 0 - 2\alpha - 2 e^{-x} \cos y = 4 \]
The right-hand side '4' must equal the left side for all \( x \) and \( y \) in order for \( u(x, y) \) to be a solution. It illustrates the importance of ensuring all components of the equation balance perfectly.
Constant Determination
The critical task in this problem is to determine the value of the constant \( \alpha \). From the previous simplification, we reach the equation:\[ 0 - 2\alpha - 2 e^{-x} \cos y = 4 \]
For the equation to hold true for any chosen \(x\) and \(y\), we need to separate and resolve each component:
  • \( -2 e^{-x} \cos y = 0 \) implies this term must be zero at all times.
  • \( -2\alpha = 4 \) gives us the needed value for \( \alpha \).

By solving \( -2\alpha = 4 \), we find \( \alpha = -2 \).

This solution shows that, under these conditions, only with \( \alpha = -2 \) will \( u(x, y) \) solve the PDE everywhere in the domain defined by the variables \(x\) and \(y\). Learning how to determine constants accurately solidifies understanding and encourages analytical thinking.

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Most popular questions from this chapter

Consider the partial differential equation \(u_{t}(x, t)=\kappa u_{x x}(x, t)+\alpha u(x, t)\), where \(\alpha\) is a constant. (a) Suppose we introduce a new dependent variable \(w(x, t)\) by defining \(u(x, t)=\) \(e^{\delta t} w(x, t)\), where \(\delta\) is a constant. Show that if \(\delta\) is chosen properly, then \(w(x, t)\) is a solution of \(w_{t}(x, t)=\kappa w_{x x}(x, t)\). What is the value \(\delta\) ? (b) Show that \(w(x, t)=e^{-4 \pi^{2} t} \cos 2 \pi x\) is a solution of the initial-boundary value problem $$ \begin{array}{lc} w_{t}(x, t)=w_{x x}(x, t), \quad 0

(a) Solve this problem for the given parameter values and the given initial condition. (b) Assume the solution \(u(x, t)\) represents the displacement at time \(t\) and position \(x\). Determine the velocity, \(u_{t}(x, t)\). (In Exercises 7-10, assume the series can be differentiated termwise.) $$c=2, l=4, u(x, 0)=\sin (\pi x / 2), u_{t}(x, 0)=0$$

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

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. \(u(x, t)=c_{1} \sin x \sin 2 t+c_{2} \sin x \cos 2 t, \quad 4 u_{x x}-u_{t t}=0\) \(u(x, 0)=-2 \sin x, \quad u_{t}(x, 0)=6 \sin x\)

Let \(u_{1}(x, t)\) be a solution of the linear homogeneous partial differential equation (2), and let \(u_{2}(x, t)\) be a solution of the linear nonhomogeneous partial differential equation (1b). Show, for any constant \(c_{1}\), that \(u(x, t)=c_{1} u_{1}(x, t)+u_{2}(x, t)\) is also a solution of the nonhomogeneous equation.
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