Chapter 2: Problem 9
Solve Laplace's equation inside a \(90^{\circ}\) sector of a circular annulus
\((a
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Chapter 2: Problem 9
Solve Laplace's equation inside a \(90^{\circ}\) sector of a circular annulus
\((a
These are the key concepts you need to understand to accurately answer the question.
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Consider Laplace's equation inside a rectangle \(0 \leqslant x \leqslant L, 0 \leqslant y \leqslant H\), with the boundary conditions $$ \frac{\partial u}{\partial x}(0, y)=0, \quad \frac{\partial u}{\partial x}(L, y)=g(y), \quad \frac{\partial u}{\partial y}(x, 0)=0, \quad \frac{\partial u}{\partial y}(x, H)=f(x) . $$ (a) What is the solvability condition and its physical interpretation? (b) Show that \(u(x, y)=A\left(x^{2}-y^{2}\right)\) is a solution if \(f(x)\) and \(g(y)\) are constants [under the conditions of part (a)]. (c) Under the conditions of part (a), solve the general case [nonconstant \(f(x)\) and \(g(y)\) ]. [Hints: Use part (b) and the fact that \(f(x)=f_{a v}+\left[f(x)-f_{a v}\right]\), where \(\left.f_{a v}=\frac{1}{L} \int_{0}^{L} f(x) d x .\right]\)
Solve Laplace's equation outside a circular disk \((r \geqslant a)\) subject to the boundary condition: (a) \(u(a, \theta)=\ln 2+4 \cos 3 \theta\) (b) \(u(a, \theta)=f(\theta)\) You may assume that \(u(r, \theta)\) remains finite as \(r \rightarrow \infty\).
In this exercise we derive superposition principles for nonhomogeneous problems. (a) Consider \(L(u)=f\). If \(u_{p}\) is a particular solution, \(L\left(u_{p}\right)=f\), and if \(u_{1}\) and \(u_{2}\) are homogencous solutions, \(L\left(u_{i}\right)=0\), show that \(u=u_{p}+c_{1} u_{1}+c_{2} u_{2}\) is another particular solution. (b) If \(L(u)=f_{1}+f_{2}\), where \(u_{p}\), is a particular solution corresponding to \(f_{i}\), what is a particular solution for \(u\) ?
(a) Show that \(L(u)=\frac{\partial}{\partial x}\left[K_{0}(x) \frac{\partial u}{\partial x}\right]\) is a linear operator. (b) Show that usually \(L(u)=\frac{\partial}{\partial x}\left[K_{0}(x, u) \frac{\partial u}{\partial x}\right]\) is not a linear operator.
Solve Laplace's equation inside a rectangle: $$ \nabla^{2} u=\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0 $$ subject to the boundary conditions $$ \begin{aligned} u(0, y) &=g(y) & & u(x, 0)=0 \\ u(L, y) &=0 & & u(x, H)=0 \end{aligned} $$
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