Question

Define the formal solution of the Dirichlet problem $$ \begin{array}{r} u_{x x}+u_{y y}=0, \quad 0

Short answer

Answer: The most general form of the solution u(x, y) for the given Dirichlet problem is: $$u(x,y)=\sum _{n=1}^{\mathrm{\infty }}[A_n\sin \left(\frac{n\pi x}{a}\right)\cosh \left(\frac{n\pi y}{a}\right)+B_n\sin \left(\frac{n\pi x}{a}\right)\cosh \left(\frac{n\pi (y-b)}{a}\right)]$$ where $A_n$'s and $B_n$'s are Fourier coefficients obtained by solving the equation system that results from the given boundary conditions.

Step-by-step solution

Identify the Laplace equation and boundary conditions

The given problem is: $$\begin{array}{r}{u}_{xx}+u_{yy}=0,0<x<a,0<y<b,\\ u(x,0)=f_0(x),u(x,b)=f_1(x),0\le x\le a,\\ u(0,y)=g_0(y),u(a,y)=g_1(y),0\le y\le b\end{array}$$ The given equation is the Laplace equation: $$u_{xx}+u_{yy}=0$$ and the given boundary conditions are: (i) $u(x,0)=f_0(x)$, (ii) $u(x,b)=f_1(x)$, (iii) $u(0,y)=g_0(y)$, and (iv) $u(a,y)=g_1(y)$.

Use the method of separation of variables

Assume that the desired function u(x, y) can be separated into a product of functions of x and y: $$u(x,y)=X(x)Y(y)$$ Substitute u(x, y) into the Laplace equation and simplify: $$X^{″}(x)Y(y)+X(x)Y^{″}(y)=0$$

Solve the resulting function for X and Y

Divide both sides of the equation by $X(x)Y(y)$ to separate the variables: $$\frac{X^{″}(x)}{X(x)}=-\frac{Y^{″}(y)}{Y(y)}$$ Since the LHS only depends on x and the RHS only depends on y, both sides must be equal to a constant, say $-k^2$. This leads to two ordinary differential equations: 1. $X^{″}(x)+k^{2}X(x)=0$ 2. $Y^{″}(y)-k^{2}Y(y)=0$ Now, solve these two equations.

Apply boundary conditions

First, apply boundary conditions (i) and (ii) to the function Y(y). Applying boundary condition (i) yields: $$X(x)Y(0)=f_0(x)\Rightarrow Y(0)=\frac{f_0(x)}{X(x)}$$ Similarly, applying boundary condition (ii) gives: $$X(x)Y(b)=f_1(x)\Rightarrow Y(b)=\frac{f_1(x)}{X(x)}$$ Apply boundary conditions (iii) and (iv) to the function X(x). Applying boundary condition (iii) yields: $$X(0)Y(y)=g_0(y)\Rightarrow X(0)=\frac{g_0(y)}{Y(y)}$$ Similarly, applying boundary condition (iv) gives: $$X(a)Y(y)=g_1(y)\Rightarrow X(a)=\frac{g_1(y)}{Y(y)}$$

Expand the function u(x, y) in a Fourier series to obtain the general solution

Now, using all boundary conditions and Fourier series representation, we can write the most general form for u(x, y) as: $$u(x,y)=\sum _{n=1}^{\mathrm{\infty }}[A_n\sin \left(\frac{n\pi x}{a}\right)\cosh \left(\frac{n\pi y}{a}\right)+B_n\sin \left(\frac{n\pi x}{a}\right)\cosh \left(\frac{n\pi (y-b)}{a}\right)]$$ Here $A_n$'s and $B_n$'s are Fourier coefficients obtained by solving the following equation system that results from the initial boundary conditions: $$\begin{gathered} A_n\cosh \left(\frac{n\pi y}{a}\right)=\frac{f_0(x)}{\sin \left(\frac{n\pi x}{a}\right)} \\ {B}_n\cosh \left(\frac{n\pi y}{b}\right)=\frac{f_1(x)}{\sin \left(\frac{n\pi x}{a}\right)} \\ {A}_n\sin \left(\frac{n\pi x}{a}\right)=\frac{g_0(y)}{\cosh \left(\frac{n{\pi }^{2}y}{a}\right)} \\ {B}_n\sin \left(\frac{n\pi x}{a}\right)=\frac{g_1(y)}{\cosh \left(\frac{n{\pi }^{2}y}{a}\right)} \end{gathered}$$

Key concepts

Laplace equation

The Laplace equation is a fundamental partial differential equation in mathematics and physics. It is used in a variety of fields such as electrostatics, fluid dynamics, and potentially any field involving fields and potentials. The equation is expressed as: $$u_{xx}+u_{yy}=0$$This means the second partial derivatives of the function $u$ with respect to $x$ and $y$, when added, should equal zero. It describes how functions behave inside a region when the boundary of that region is known. The solutions to the Laplace equation, known as harmonic functions, are widely seen as the smoothest functions possible within boundaries. This smoothness is due to the absence of local maximum or minimum values within the considered domain.

boundary conditions

Boundary conditions are essential in solving differential equations as they provide additional constraints that must be satisfied by the solution. In the Dirichlet problem, these conditions are specified for the function $u(x,y)$ along the boundary of a domain. In our problem, the conditions are:

• $u(x,0)=f_0(x)$: Specifies the value of $u$ along the $y=0$ boundary.
• $u(x,b)=f_1(x)$: Specifies the value of $u$ along the $y=b$ boundary.
• $u(0,y)=g_0(y)$: Specifies the value of $u$ along the $x=0$ boundary.
• $u(a,y)=g_1(y)$: Specifies the value of $u$ along the $x=a$ boundary.

Implementing these conditions ensures that the mathematical solution has physical relevance, as it satisfies known values of the domain's edges. They help to find unique solutions for $u$ that fit the reality of the problem at hand.

method of separation of variables

The method of separation of variables is a powerful tool used to solve partial differential equations (PDEs). It involves assuming that the solution can be expressed as a product of functions, each dependent on a different coordinate. In the context of our Dirichlet problem, we hypothesize: $$u(x,y)=X(x)Y(y)$$This assumption transforms the PDE into ordinary differential equations (ODEs) under the conditions that the partial derivatives can be independently equated to a constant because each side depends on a separate variable. So, the transformed equation becomes:$$\frac{X^{″}(x)}{X(x)}=-\frac{Y^{″}(y)}{Y(y)}=-k^2$$From this, we derive two simpler ODEs for $X(x)$ and $Y(y)$, which are often easier to solve. This method is particularly advantageous when dealing with problems defined over rectangular domains, as it aligns nicely with boundary conditions.

Fourier series

Fourier series is a way to represent a function as the sum of sines and cosines. It's particularly beneficial when dealing with periodic functions or boundary value problems on finite intervals. In our exercise, we expand the solution $u(x,y)$ as a Fourier series to satisfy the boundary conditions in both $x$ and $y$.The general expression in this context becomes:$$u(x,y)=\sum _{n=1}^{\mathrm{\infty }}[A_n\sin \left(\frac{n\pi x}{a}\right)\cosh \left(\frac{n\pi y}{a}\right)+B_n\sin \left(\frac{n\pi x}{a}\right)\cosh \left(\frac{n\pi (y-b)}{a}\right)]$$Here, $A_n$ and $B_n$ are the Fourier coefficients, determined by the boundary conditions. Calculating these coefficients correctly ensures that the function $u(x,y)$ satisfies the prescribed boundary values. This series method provides a practical approach to approximating solutions to complex boundary value problems, taking advantage of the orthogonal properties of trigonometric functions.