/*! 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 30 Consider the PDE \(\mathcal{L} u... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 30

Consider the PDE \(\mathcal{L} u(\mathbf{r})=\rho(\mathbf{r})\), for which the differential operator \(\mathcal{L}\) is given by $$ \mathcal{L}=\nabla \cdot[p(\mathbf{r}) \nabla]+q(\mathbf{r}) $$ where \(p(\mathbf{r})\) and \(q(\mathbf{r})\) are functions of position. By proving the generalised form of Green's theorem, $$ \int_{V}(\phi \mathcal{L} \psi-\psi \mathcal{L} \phi) d V=\oint_{S} p(\phi \nabla \psi-\psi \nabla \phi) \cdot \hat{\mathbf{n}} d S $$ show that the solution of the PDE is given by $$ u\left(\mathbf{r}_{0}\right)=\int_{V} G\left(\mathbf{r}, \mathbf{r}_{0}\right) \rho(\mathbf{r}) d V(\mathbf{r})+\oint_{S} p(\mathbf{r})\left[u(\mathbf{r}) \frac{\partial G\left(\mathbf{r}, \mathbf{r}_{0}\right)}{\partial n}-G\left(\mathbf{r}, \mathbf{r}_{0}\right) \frac{\partial u(\mathbf{r})}{\partial n}\right] d S(\mathbf{r}) $$ where \(G\left(\mathbf{r}, \mathbf{r}_{0}\right)\) is the Green's function satisfying \(\mathcal{L} G\left(\mathbf{r}, \mathbf{r}_{0}\right)=\delta\left(\mathbf{r}-\mathbf{r}_{0}\right)\).

Short Answer

Expert verified
Following Green's theorem with the proper substitutions, the solution to the PDE is derived.

Step by step solution

01

State Green's theorem

Start by considering the generalized form of Green's theorem:\[\int_{V}(\phi \mathcal{L} \psi - \psi \mathcal{L} \phi) \, dV = \oint_{S} p(\phi abla \psi - \psi abla \phi) \cdot \hat{\mathbf{n}} \, dS\]
02

Apply the differential operator to Green's function and solution

Identify \(\psi = G(\mathbf{r}, \mathbf{r}_0)\) and \(\mathcal{L} G(\mathbf{r}, \mathbf{r}_0) = \delta (\mathbf{r} - \mathbf{r}_0)\). Assume \( \mathcal{L}u(\mathbf{r}) = \rho(\mathbf{r}) \). By substituting these into Green's theorem:\[\int_{V}(\phi \delta(\mathbf{r} - \mathbf{r}_0) - \psi \rho(\mathbf{r})) \, dV = \oint_{S} p(\phi abla G(\mathbf{r}, \mathbf{r}_0) - G(\mathbf{r}, \mathbf{r}_0) abla \phi) \cdot \hat{\mathbf{n}} \, dS\]
03

Simplify the volume integral

The term \(\phi \delta(\mathbf{r} - \mathbf{r}_0)\) integrates to \(\phi (\mathbf{r}_0)\) over the volume - implying the following simplification:\[\phi(\mathbf{r}_0) = \int_{V} G(\mathbf{r}, \mathbf{r}_0) \rho(\mathbf{r}) \, dV + \oint_{S} p(\phi abla G(\mathbf{r}, \mathbf{r}_0) - G(\mathbf{r}, \mathbf{r}_0) abla \phi) \cdot \hat{\mathbf{n}} \, dS\]
04

Substitute \(\phi\) with the solution \(u\)

Finally, let \(u(\mathbf{r}) = \phi(\mathbf{r})\) which gives the equation:\[u(\mathbf{r}_0) = \int_{V} G(\mathbf{r}, \mathbf{r}_0) \rho(\mathbf{r}) \, dV + \oint_{S} p(\mathbf{r}) \left[ u(\mathbf{r}) \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial n} - G(\mathbf{r}, \mathbf{r}_0) \frac{\partial u(\mathbf{r})}{\partial n} \right] \, dS\]

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Key Concepts

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

Green's Theorem
Green's Theorem is a fundamental result that connects the evaluation of a double integral over a region to the evaluation of a line integral around its boundary. For the case of our partial differential equation (PDE), Green's Theorem can be generalized to connect functions via the Laplace operator and boundary conditions. This generalized form is crucial in deriving solutions to PDEs.

The generalized form is given by:
\[\int_{V}(\phi \mathcal{L} \psi - \psi \mathcal{L} \phi) \, dV = \oint_{S} p(\phi \abla \psi - \psi \abla \phi) \, dS\]

This relation enables us to transform volume integrals into surface integrals, making it easier to solve complex PDEs.
Green's Function
Green's Function is a special solution used to solve inhomogeneous linear differential equations subject to specific boundary conditions. For our PDE problem, the Green's Function, denoted by \(G(\mathbf{r}, \mathbf{r}_0)\), satisfies:
\[\mathcal{L} G(\mathbf{r}, \mathbf{r}_0) = \delta (\mathbf{r} - \mathbf{r}_0)\]
The function\ \(G(\mathbf{r}, \mathbf{r}_0)\) represents the response at point \mathbf{r} due to a unit impulse applied at point \mathbf{r}_0\. With Green's Function, we can express the solution \(u(\mathbf{r})\) of the PDE in terms of the source term \(\rho (\mathbf{r})\) and boundary conditions. This makes solving complex differential equations more manageable.
Boundary Conditions
Boundary conditions are essential when solving PDEs because they specify the behavior of the solution on the boundary of the domain. There are different types of boundary conditions, including:
  • Dirichlet Boundary Conditions: Specifying the function value on the boundary.
  • Neumann Boundary Conditions: Specifying the derivative of the function normal to the boundary.
In our solution, boundary conditions are applied through the integral involving the Green's Function. This ensures the solution is not only valid within the domain but also aligns with the constraints imposed on the boundaries of the domain.
Laplace Operator
The Laplace operator, denoted by \(\abla \,\), is a second-order differential operator. It plays a central role in the field of partial differential equations, especially those governing physical phenomena like heat conduction, electromagnetism, and fluid flow. In our PDE, the Laplace operator is part of the differential operator \(\mathcal{L} = \rabla \cdot[p(\mathbf{r}) \abla] + q(\mathbf{r})\) \which includes both the divergence of a gradient term and an additional function \(q(\mathbf{r}).\)

The term \(p(\mathbf{r})\) acts as a weight function that can vary with position, while \(q(\mathbf{r})\) adds an additional layer of complexity to the differential operator \mathcal{L}\. Understanding the Laplace operator's role helps us appreciate how it influences the behavior of solutions to our PDE across the domain.

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Most popular questions from this chapter

(a) By applying the divergence theorem to the volume integral $$ \int_{V}\left[\phi\left(\nabla^{2}-m^{2}\right) \psi-\psi\left(\nabla^{2}-m^{2}\right) \phi\right] d V $$ obtain a Green's function expression, as the sum of a volume integral and a surface integral, for \(\phi\left(\mathbf{r}^{\prime}\right)\) that satisfies $$ \nabla^{2} \phi-m^{2} \phi=\rho $$ in \(V\) and takes the specified form \(\phi=f\) on \(S\), the boundary of \(V .\) The Green's function \(G\left(\mathbf{r}, \mathbf{r}^{\prime}\right)\) to be used satisfies $$ \nabla^{2} G-m^{2} G=\delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right) $$ and vanishes when \(\mathbf{r}\) is on \(S .\) (b) When \(V\) is all space, \(G\left(\mathbf{r}, \mathbf{r}^{\prime}\right)\) can be written as \(G(t)=g(t) / t\) where \(t=\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\) and \(g(t)\) is bounded as \(t \rightarrow \infty .\) Find the form of \(G(t)\). (c) Find \(\phi(\mathbf{r})\) in the half space \(x>0\) if \(\rho(\mathbf{r})=\delta\left(\mathbf{r}-\mathbf{r}_{1}\right)\) and \(\phi=0\) both on \(x=0\) and as \(r \rightarrow \infty\) .

Prove that the potential for \(\rho0\) and \(\cos \phi<0\) ) are maintained at equal and opposite potentials \(\pm V\), is given by $$ u(\rho, \phi)=\frac{4 V}{\pi} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n+1}\left(\frac{\rho}{a}\right)^{2 n+1} \cos (2 n+1) \phi $$.

(a) Show that the gravitational potential due to a uniform disc of radius \(a\) and mass \(M\), centred at the origin, is given for \(ra\) by $$ \frac{G M}{r}\left[1-\frac{1}{4}\left(\frac{a}{r}\right)^{2} P_{2}(\cos \theta)+\frac{1}{8}\left(\frac{a}{r}\right)^{4} P_{4}(\cos \theta)-\cdots\right] $$ where the polar axis is normal to the plane of the disc. (b) Reconcile the presence of a term \(P_{1}(\cos \theta)\), which is odd under \(\theta \rightarrow \pi-\theta\), with the symmetry with respect to the plane of the disc of the physical system. (c) Deduce that the gravitational field near an infinite sheet of matter of constant density \(\rho\) per unit area is \(2 \pi G \rho\).

(a) Find the form of the solution of Laplace's equation in plane polar coordinates \(\rho, \phi\) that takes the value \(+1\) for \(0<\phi<\pi\) and the value \(-1\) for \(-\pi<\phi<0\) when \(\rho=a\) (b) For a point \((x, y)\) on or inside the circle \(x^{2}+y^{2}=a^{2}\), identify the angles \(\alpha\) and \(\beta\) defined by $$ \alpha=\tan ^{-1} \frac{y}{a+x} \quad \text { and } \quad \beta=\tan ^{-1} \frac{y}{a-x} $$ Show that \(u(x, y)=(2 / \pi)(\alpha+\beta)\) is a solution of Laplace's equation that satisfies the boundary conditions given in (a). (c) Deduce a Fourier series expansion for the function $$ \tan ^{-1} \frac{\sin \phi}{1+\cos \phi}+\tan ^{-1} \frac{\sin \phi}{1-\cos \phi} $$.

Solve the following first-order partial differential equations by separating the variables: (a) \(\frac{\partial u}{\partial x}-x \frac{\partial u}{\partial y}=0\); (b) \(x \frac{\partial u}{\partial x}-2 y \frac{\partial u}{\partial y}=0\).
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