Question

In Exercises \(29-34\) define the bounded formal solution of $$ u_{x x}+u_{y y}=0, \quad 00 $$ that satisfies the given boundary conditions for general a and \(f .\) Then solve the boundary value problem for the specified a and \(f\). $$ \begin{array}{l} u(x, 0)=f(x), \quad 00 \\ a=\pi, \quad f(x)=x(2 \pi-x) \end{array} $$

Short answer

Question: Find the specific solution to the given partial differential equation with boundary conditions: $$u_{xx} + u_{yy} = 0$$ with: \begin{align} u(x, 0) &= x(2\pi - x), \quad 0 < x < \pi \\ u(0, y) &= 0, \quad y > 0 \\ u_x(\pi, y) &= 0, \quad y > 0 \end{align} Answer: The specific solution that satisfies the given boundary conditions is: $$u(x, y) = \sum_{n=1}^{\infty} \frac{8(-1)^{n+1}}{n^3\pi^2} \sin{n\pi x} \e^{-n\pi y}$$

Step-by-step solution

Separate variables using the Ansatz solution

Let's assume that the solution can be written as a product of two functions, one depending on \(x\) alone and the other depending on \(y\). $$ u(x, y) = X(x)Y(y) $$

Substitute the Ansatz solution into the PDE

Now substitute the product form \(u(x, y) = X(x)Y(y)\) back into the PDE equation to obtain: $$ X''(x)Y(y) + X(x)Y''(y) = 0$$

Divide by \(X(x)Y(y)\) and set a separation constant

Divide the equation by \(X(x)Y(y)\) and introduce a separation constant \(-\lambda\). $$ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = -\lambda $$

Rewrite the equation as two ODEs

Now, rewrite the equation above as two separate ODEs, so we can solve separately: $$X''(x) + \lambda X(x) = 0$$ $$Y''(y) - \lambda Y(y) = 0$$

Apply the boundary conditions to the ODEs

Now, apply the boundary conditions to the above ODEs. For \(x=0\), we have \(u(0, y)=0 \Rightarrow X(0)Y(y) = 0\) (2). Since we cannot have \(Y(y) = 0\) for all \(y\), then \(X(0) = 0\). For \(x=a\), we have \(u_x(a, y)=0 \Rightarrow X'(a)Y(y) = 0\). As before, \(Y(y)\) can't be zero for all \(y\), so we have \(X'(a)=0\).

Write down the general solution and apply conditions

Now, solve the ODEs. For convenience, we'll define a separation constant for \(u_x\) and \(u_y\). Suppose \(\lambda = k^2\) for some positive k. For the x-equation and its boundary conditions: $$X''(x) + k^2X(x) = 0, \quad X(0) = X'(a) = 0$$ The general solution is: $$X(x) = B\sin{kx}$$ The first boundary condition \(X(0) = 0\) implies that \(B = 0\) because \(B = X(0) = \sin{k0}\). The second boundary condition gives us \(X'(a) = kB\cos{ka}\). Since \(kB \neq 0\), we must have \(\cos{ka}=0\), or \(ka=n\pi\), where \(n\) is an integer. For the y-equation: $$ Y''(y) - k^2Y(y) = 0$$ The general solution is: $$Y(y) = C\e^{ky} + D\e^{-ky}$$ However, as \(y\to +\infty\), the first term diverges. Thus, to obtain a bounded function, we set \(C = 0\). So, the complete general solution is given by: $$u(x, y) = \sum_{n=1}^{\infty} A_n \sin\frac{n\pi x}{a} \e^{-\frac{n\pi y}{a}}$$

Use Fourier series to find coefficients

Now, we need to find the coefficients \(A_n\). We use the first boundary condition: $$u(x, 0) = f(x) = \sum_{n=1}^{\infty} A_n \sin\frac{n\pi x}{a}$$ Plug in the given function \(f(x) = x(2\pi - x)\) and \(a = \pi\). We first need to find the Fourier coefficients \(A_n\) for the function \(f(x)\) on the interval \([0, a]\): $$A_n = \frac{2}{a}\int_0^{a} f(x) \sin\frac{n\pi x}{a} dx = \frac{2}{\pi}\int_0^{\pi} x(2\pi - x) \sin\frac{n\pi x}{\pi} dx$$ Now, we need to integrate by parts twice, set up a Fourier series representation for \(f(x)\), and use the orthogonality properties of sine functions. After integrating, we obtain, $$A_n=\frac{8(-1)^{n+1}}{n^3\pi^2}$$

Write down the specific solution using obtained coefficients

Now, use the coefficients \(A_n\) to write down the specific solution for the given \(f(x)\) and \(a = \pi\). $$u(x, y) = \sum_{n=1}^{\infty}\frac{8(-1)^{n+1}}{n^3\pi^2} \sin\frac{n\pi x}{\pi} \e^{-\frac{n\pi y}{\pi}} = \sum_{n=1}^{\infty} \frac{8(-1)^{n+1}}{n^3\pi^2} \sin{n\pi x} \e^{-n\pi y}$$ So, the specific solution that satisfies given boundary conditions and the provided function \(f(x)\) and constant \(a = \pi\) is: $$u(x, y) = \sum_{n=1}^{\infty} \frac{8(-1)^{n+1}}{n^3\pi^2} \sin{n\pi x} \e^{-n\pi y}$$

Key concepts

Partial Differential Equations

A Partial Differential Equation (PDE) is a type of mathematical equation that involves multiple independent variables, an unknown function that depends on those variables, and partial derivatives of that unknown function. In our original exercise, we have a PDE given by \( u_{xx} + u_{yy} = 0 \). This is known as a Laplace equation, which is a prototypical example of an elliptic type PDE.
This equation appears often in areas like electrostatics, fluid dynamics, and quantum mechanics. One of the key features of solving PDEs is boundary conditions, which specify the solution behavior on the boundaries of the domain. In our exercise, the boundary conditions are:

• \( u(x,0) = f(x) \), which defines the solution on the x-axis.
• \( u(0,y) = 0 \), meaning at \(x = 0\), the solution must be zero for all \(y\).
• \( u_x(a, y) = 0 \), ensuring the derivative of the solution with respect to \(x\) is zero at \(x = a\), for all \(y\).

To solve a PDE like this, we can use methods such as separation of variables and Fourier series, which will be discussed in the following sections.

Fourier Series

A Fourier series is a powerful tool used to express a periodic function as an infinite sum of sines and cosines. Fourier series are widely used in solving boundary value problems because they allow us to represent complex functions in terms of simpler trigonometric functions.
In our exercise, we used a Fourier series to express the function \(f(x) = x(2\pi - x)\) as a sum of sine functions. This helps in finding a solution to our PDE that satisfies the given boundary conditions. The general formula for the Fourier coefficients in terms of a function \(f(x)\) over an interval \([0,a]\) is:

• \( A_n = \frac{2}{a} \int_0^a f(x) \sin\left(\frac{n\pi x}{a}\right) \, dx \)

By computing these coefficients, we obtain \(A_n\), and hence the Fourier series representation of \(f(x)\). In our exercise, this step allowed us to write the specific solution for the PDE as a Fourier series, satisfying the boundary condition \(u(x,0) = f(x)\). Notice how using a Fourier series turns handling infinite domains into a manageable algebraic task.

Separation of Variables

The method of separation of variables is a technique used to solve partial differential equations. The method assumes a solution can be separated into a product of functions, each dependent on a single coordinate. For our PDE \(u_{xx} + u_{yy} = 0\), we assume a solution of the form \( u(x, y) = X(x)Y(y) \).
Substituting this form into the PDE gives us two ordinary differential equations (ODEs) by separating the variables. The idea is to split the problem into simpler, one-variable equations:

• For \(X(x)\): \( X''(x) + \lambda X(x) = 0 \)
• For \(Y(y)\): \( Y''(y) - \lambda Y(y) = 0 \)

Each of these can be solved independently. The constants derived from solving these equations, like \(-\lambda\), help to link the solutions back together. Once solutions for \(X(x)\) and \(Y(y)\) are found, they are combined to form the solution for the original PDE. This process simplifies solving a complex PDE by reducing it to the solution of simpler ODEs.