Question

(a) Find the solution \(u(x, y)\) of Laplace's equation in the semi-infinite strip \(00,\) also satisfying the boundary conditions $$ \begin{aligned} u(0, y)=0, & u(a, y)=0, \quad y>0 \\ u(x, 0)=f(x), & 0 \leq x \leq a \end{aligned} $$ and the additional condition that \(u(x, y) \rightarrow 0\) as \(y \rightarrow \infty\). (b) Find the solution if \(f(x)=x(a-x)\) (c) Let \(a=5 .\) Find the smallest value of \(y_{0}\) for which \(u(x, y) \leq 0.1\) for all \(y \geq y_{0}\)

Short answer

Question: Determine the function u(x,y) that satisfies Laplace's equation ∇²u(x,y)=0, subject to the boundary conditions u(0,y)=0, u(a,y)=0, and u(x,0)=f(x)=x(a-x). Then, for a specific plate of width a=5, find the smallest value of y (y0) such that u(x,y)≤0.1 for all y≥y0. Answer: The function u(x,y) that satisfies the given conditions is: $$u(x,y) = -\frac{8\cdot 5}{\pi^3} \sum_{n=1}^\infty \frac{1 - \left(-1\right)^n}{n^3} \sin\left(\frac{n\pi}{5}x\right) e^{-n\pi y/5}.$$ For a plate of width a=5, the smallest value of y0 is approximately 7.672 so that u(x,y) ≤ 0.1 for all y ≥ y0.

Step-by-step solution

Solving the Laplace's equation using separation of variables

Let u(x, y) = X(x)Y(y). Substituting this expression into Laplace's equation: $$\nabla^2 u(x,y) = X''Y + XY'' = 0.$$ Dividing through by \(XY\), we get: $$\frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0.$$ Since these functions are independent, we can say that $$\frac{X''(x)}{X(x)} = -\frac{Y''(y)}{Y(y)} = k,$$ where k is a constant.

Solving the ODEs for X(x) and Y(y)

Now, we have two ordinary differential equations to solve: 1. \(X''(x) = kX(x)\) 2. \(Y''(y) = -kY(y)\) Solving the first equation, we get: $$X''(x) - kX(x) = 0.$$ Let k = \(m^2,\) so the equation becomes: $$X''(x) - m^2X(x) = 0.$$ The general solution of this equation is: $$X(x) = c_1\: e^{mx} + c_2\:e^{-mx}.$$ From the boundary conditions, \(u(0, y) = X(0)Y(y) = 0,\) so \(X(0)=0\). Substituting x=0, we get: $$X(0) = c_1\: e^{0} + c_2\:e^{0} = c_1 + c_2 = 0 \implies c_1 = -c_2.$$ So, the equation becomes: $$X(x) = c_1(e^{mx} - e^{-mx}).$$ Now, using the boundary condition \(u(a, y) = X(a)Y(y) = 0,\) so \(X(a)=0\). Substituting x=a, we get: $$X(a) = c_1(e^{ma} - e^{-ma}) = 0,$$ which means that \(e^{ma} = e^{-ma}\) or \(m = n \pi/a\) for integer values of n. Therefore, the general solution of X(x) becomes: $$X_n(x) = c_n\: \sin\left(\frac{n\pi}{a}x\right),$$ where n = 1, 2, 3, .... Now solving the second equation, \(Y''(y) + m^2Y(y) = 0,\) we get the general solution: $$Y_n(y) = A_n\: e^{-n\pi y/a} + B_n\: e^{n\pi y/a}.$$

Finding the solution u(x, y)

Using the additional boundary condition that \(Y(y)\rightarrow 0\) as \(y\rightarrow \infty\), we find out that the terms with growing exponentials are not useful, so we consider only the terms containing the declining exponentials: $$Y_n(y) = A_n\: e^{-n\pi y/a}.$$ Now, we form the solution u(x, y) by combining the solutions for X(x) and Y(y): $$u(x, y) = \sum_{n=1}^\infty A_n \: \sin\left(\frac{n\pi}{a}x\right) e^{-n\pi y/a}.$$

Finding the solution for function f(x) = x(a-x)

For this specific function, we apply the boundary condition \(u(x, 0) = f(x)\). So, let \(f(x) = x(a-x)\). We have: $$u(x, 0) = f(x) = \sum_{n=1}^\infty A_n \sin \left(\frac{n\pi}{a}x\right).$$ We can use the Fourier sine series representaion for our function f(x), which is $$f(x) = \sum_{n=1}^\infty b_n \sin\left(\frac{n\pi}{a}x\right),$$ where $$b_n = \frac{2}{a}\int_0^a f(x) \sin\left(\frac{n\pi}{a}x\right) dx.$$ By comparing these two expressions, we can deduce that \(A_n = b_n\). To find \(b_n\), we will integrate: $$b_n = \frac{2}{a}\int_0^a x(a-x) \sin\left(\frac{n\pi}{a}x\right) dx.$$ I won't do the entire integration process here, but after integrating we get: $$b_n = -\frac{8a}{n^3\pi^3} \left[1 - \left(-1\right)^n \right].$$ Finally, plugging this value into the solution for u(x,y), we have: $$u(x,y) = \sum_{n=1}^\infty -\frac{8a}{n^3\pi^3} \left[1 - \left(-1\right)^n \right] \sin\left(\frac{n\pi}{a}x\right) e^{-n\pi y/a}.$$

Finding the smallest value of y0 for which u(x,y) ≤ 0.1 for \(y\geq y_0\) with a = 5

We have the solution for u(x,y): $$u(x,y) = -\frac{8\cdot 5}{\pi^3} \sum_{n=1}^\infty \frac{1 - \left(-1\right)^n}{n^3} \sin\left(\frac{n\pi}{5}x\right) e^{-n\pi y/5}.$$ We need to find the smallest value of \(y_0\) for which \(u(x, y) \leq 0.1\) for all \(y \geq y_0\). Since we have a negative term multiplying the whole series, and all terms in the series are positive, the smallest value of u(x,y) will occur when n=1 and x=a/2: $$-\frac{40}{\pi^3} e^{-\pi y/5} \leq 0.1$$ Solving for y, we get: $$y \geq \frac{5}{\pi} \ln \frac{400}{\pi^3} \approx 7.672.$$ Therefore, the smallest value of \(y_0\) for which \(u(x, y) \leq 0.1\) for all \(y \geq y_0\) is approximately \(y_0 \approx 7.672\).

Key concepts

Separation of Variables

The method of separation of variables is a powerful technique used to find solutions to partial differential equations (PDEs). It involves assuming that the solution can be written as the product of two or more functions, each of which depends on only one of the variables. For instance, a function u(x, y) can be expressed as the product of X(x) and Y(y), thus reducing a PDE into two ordinary differential equations (ODEs).

Applying this technique to Laplace's equation, u(x, y) is divided into two separate functions. Each function satisfies an ODE, allowing for simpler, separate solutions that, when multiplied, combine to solve the original PDE. This initial assumption facilitates the application of given boundary conditions and simplifies the complex problem into more manageable pieces. It is important to choose the separate functions such that they satisfy the boundary conditions given in the problem.

Fourier Sine Series

The Fourier sine series is a representation of a function as an infinite sum of sine functions. It is particularly useful for solving problems with odd periodic functions or functions defined on a finite interval with certain types of boundary conditions, such as those given in the textbook exercise.

For a function f(x) defined on the interval [0, a], the Fourier sine series expresses f(x) as a sum of sine waves with different frequencies. The coefficients of the series, typically denoted by b_n, are calculated through an integration process involving the original function and sine functions. This allows us to match the sine series to the function's behavior on the boundary, capturing its features and oscillations within the defined interval. For the problem given, the sine series expansion of f(x) = x(a-x) provides the coefficients crucial for constructing the final solution to the given PDE.

Ordinary Differential Equations (ODEs)

Ordinary differential equations are equations involving derivatives of a function of a single variable. ODEs arise in the process of separation of variables, where the partial differential equation is reduced to one or more ODEs. Each ODE corresponds to a function of a single variable and represents a specific aspect of the physical situation described by the original PDE.

In the context of the semi-infinite strip problem, we arrive at two separate ODEs for X(x) and Y(y), with solutions that involve constant coefficients and exponential functions. However, due to physical or geometry-related constraints of the problem, such as boundary conditions, not all solutions to the ODEs are viable. Constraints help narrow down the possible forms of these functions to those that fit within the conditions of the problem.

Boundary Value Problem

A boundary value problem (BVP) is a differential equation paired with a set of additional constraints called boundary conditions. In such problems, the solution is sought not only to satisfy the differential equation, but also to meet specific conditions at the boundaries of the domain.

In the Laplace equation exercise, the boundary value problem is defined on a semi-infinite strip with given boundary conditions along the edges of the strip. The solution u(x, y) must be determined such that it goes to zero at the boundaries and assumes a specified form f(x) along one edge. Additional condition that u(x, y) approaches zero as y goes to infinity further refines the solution. Identifying and applying these conditions correctly are crucial steps to ensure that the obtained functions for X(x) and Y(y) lead to a meaningful solution that satisfies all aspects of the BVP.