/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Solve Laplace's equation in plan... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve Laplace's equation in plane polar coordinates $$ \frac{\partial^{2} v(r, \theta)}{\partial r^{2}}+\frac{1}{r} \frac{\partial v(r, \theta)}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} v(r, \theta)}{\partial \theta^{2}}=0 $$ in the circular region \(x^{2}+y^{2}=1\) of the plane where (a) \(v(r, \theta)\) is finite for \(0 \leq r \leq 1\) and for all \(\theta\) (b) \(v(1, \theta)=5 \cos 3 \theta\) (c) \(v(r, \theta+2 \pi)=v(r, \theta)\) for \(0 \leq r \leq 1\).

Short Answer

Expert verified
The solution is \( v(r, \theta) = 5 r^3 \cos(3\theta) \).

Step by step solution

01

- Formulating the general solution

Rewrite Laplace's equation in polar coordinates: \[ \frac{\partial^{2} v(r, \theta)}{\partial r^{2}} + \frac{1}{r} \frac{\partial v(r, \theta)}{\partial r} + \frac{1}{r^{2}} \frac{\partial^{2} v(r, \theta)}{\partial \theta^{2}}=0 \] Assume the solution can be separated such that \( v(r, \theta) = R(r) \Theta(\theta) \). Substituting this into Laplace's equation and dividing by \( R(r) \Theta(\theta) \), we get: \[ \frac{1}{R} \left( r^2 R'' + r R' \right) + \frac{1}{\Theta} \Theta'' = 0 \] Equate the terms to a separation constant \( -\lambda \).
02

- Solving the angular part

For the angular part, solve: \[ \frac{\partial^{2} \Theta(\theta)}{\partial \theta^{2}} + \lambda \Theta(\theta) = 0 \] This is a standard linear homogeneous equation with periodic boundary condition \( v(r, \theta + 2\pi) = v(r, \theta)\). The solutions are: \[ \Theta_n(\theta) = A_n \cos(n\theta) + B_n \sin(n\theta) \] with \( \lambda = n^2 \), \( n \) being integers.
03

- Solving the radial part

For the radial part, solve: \[ r^2 \frac{d^{2} R(r)}{d r^{2}} + r \frac{d R(r)}{d r} - n^2 R(r) = 0 \] This is an Euler-Cauchy equation. The general solution is: \[ R_n(r) = C_n r^n + D_n r^{-n} \] Since \( v(r, \theta) \) must be finite for all \( 0 \leq r \leq 1 \), we reject the \( r^{-n} \) term for \( n eq 0 \). Therefore, \( R_n(r) = C_n r^n \). When \( n = 0 \), \( R_0(r) = C_0 + D_0 \ln(r) \), but \( D_0 \) should be zero for boundedness.
04

- Constructing the solution

Combine the solutions from the angular and radial parts: \[ v(r, \theta) = \sum_{n=0}^{\infty} (C_n r^n \cos(n\theta) + D_n r^n \sin(n\theta)) \] but based on boundary conditions, only finite terms are considered.
05

- Applying boundary conditions

Given boundary condition \( v(1, \theta) = 5 \cos 3\theta \), we infer that only \( \cos(3\theta) \) term remains. Thus, \[ v(r, \theta) = 5 r^3 \cos(3\theta) \] By construction, this satisfies the periodic boundary condition, the finiteness condition for \( 0 \leq r \leq 1 \), and boundary condition at the unit circle.

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Key Concepts

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

Partial Differential Equations
A Partial Differential Equation (PDE) involves functions of multiple variables and their partial derivatives. Laplace's equation, given by \[ \frac{\text{∂}^{2} v(r, \theta)}{\text{∂}r^{2}}+\frac{1}{r} \frac{\text{∂}v(r, \theta)}{\text{∂}r}+\frac{1}{r^{2}} \frac{\text{∂}^{2} v(r,\theta)}{\text{∂} \theta^{2}}=0 \] , is a standard PDE. Here, we seek a solution for \(v(r, \theta)\) within a specified circular region. PDEs like Laplace's equation are crucial in fields like physics and engineering due to their ability to describe phenomena such as heat distribution and fluid flow. Understanding this equation helps build a solid foundation in analyzing physical systems.
Separation of Variables
Separation of Variables is a technique to solve PDEs by reducing them to a set of simpler, ordinary differential equations (ODEs). In this method, we assume that the solution can be expressed as a product of functions, each depending on a single variable. For Laplace's equation, we assume: \( v(r, \theta) = R(r) \theta (\theta) \) Substituting this into the PDE and then dividing by \( R(r) \theta (\theta) \) , the equation splits into: \[ \frac{1}{R} \big(r^2 R'' + r R' \big) + \frac{1}{\theta} \theta'' = 0 \] This separation introduces a constant \( -λ \) . Each term is then set equal to \([-λ]\) , forming two ODEs. This method greatly simplifies solving complex differential equations.
Euler-Cauchy Equation
The radial part of Laplace's equation, \( r^2 \frac{d^{2} R(r)}{d r^{2}} + r \frac{d R(r)}{d r} - n^2 R(r) = 0 \) , is an Euler-Cauchy equation. It’s a type of differential equation where the coefficients of the derivatives are powers of the independent variable. The general solution for this type of equation is: \( R_n(r) = C_n r^n + D_n r^{-n} \) . In our problem, since the solution must remain finite as \( r \to 0 \) , we discard the term \( r^{-n} \) for \( n eq 0 \) , and for \( n = 0 \) , we discard the logarithmic term to ensure boundedness, leaving us with \( R_n(r) = C_n r^n \).
Boundary Conditions
Boundary Conditions specify the behavior of the solution at the boundaries of the domain. In this problem, there are three boundary conditions:
  • Finity: \(v(r, \theta) \) must be finite for \(0 \leq r \leq 1 \) and for all \(θ\)
  • Circular boundary: \(v(1, \theta) = 5 \text{ cos } 3\theta \)
  • Periodic condition: \(v(r, \theta+2\text{Ï€}) = v(r, \theta)\)
These conditions help us determine specific solutions that satisfy both the differential equation and the physical constraints. For identifying \(v(r, \theta)\), we see that the form \(5 r^ 3 \text{ cos } 3\theta \) satisfies all given boundaries.

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Most popular questions from this chapter

One end A of an insulated metal bar AB of length \(2 \mathrm{~m}\) is kept at \(0^{\circ} \mathrm{C}\) while the other end B is maintained at \(50^{\circ} \mathrm{C}\) until a steady state of temperature along the bar is achieved. At \(t=0\), the end \(\mathrm{B}\) is suddenly reduced to \(0^{\circ} \mathrm{C}\) and kept at that temperature. Using the heat conduction equation \(\frac{\partial^{2} u}{\partial x^{2}}=\frac{1}{c^{2}} \cdot \frac{\partial u}{\partial t}\), determine an expression for the temperature at any point in the bar distance \(x\) from \(\mathrm{A}\) at any time \(t\).

A perfectly elastic string is stretched between two points \(10 \mathrm{~cm}\) apart. Its centre point is displaced \(2 \mathrm{~cm}\) from its position of rest at right angles to the original direction of the string and then released with zero velocity. Applying the equation \(\frac{\partial^{2} u}{\partial x^{2}}=\frac{1}{c^{2}} \cdot \frac{\partial^{2} u}{\partial t^{2}}\) with \(c^{2}=1\), determine the subsequent motion \(u(x, t)\).

The ends of an insulated rod \(\mathrm{AB}, 10\) units long, are maintained at \(0^{\circ} \mathrm{C}\). At \(t=0\), the temperature within the rod rises uniformly from each end reaching \(2^{\circ} \mathrm{C}\) at the mid-point of \(\mathrm{AB}\). Determine an expression for the temperature \(u(x, t)\) at any point in the rod, distant \(x\) from the left-hand end at any subsequent time \(t\).

Solve the following equations (a) \(\frac{\partial^{2} u}{\partial x^{2}}=24 x^{2}(t-2)\), given that at \(x=0, u=e^{2 t}\) and \(\frac{\partial u}{\partial x}=4 t\). (b) \(\frac{\partial^{2} u}{\partial x \partial y}=4 e^{y} \cos 2 x\), given that at \(y=0, \frac{\partial u}{\partial x}=\cos x\) and at \(x=\pi, u=y^{2}\)

The centre point of a perfectly elastic string stretched between two points A and B, \(4 \mathrm{~m}\) apart, is deflected a distance \(0.01 \mathrm{~m}\) from its position of rest perpendicular to \(\mathrm{AB}\) and released initially with zero velocity. Apply the wave equation \(\frac{\partial^{2} u}{\partial x^{2}}=\frac{1}{c^{2}} \cdot \frac{\partial^{2} u}{\partial t^{2}}\) where \(c=10\) to determine the subsequent motion of a point \(\mathrm{P}\) distant \(x\) from \(\mathrm{A}\) at time \(t\).

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