Part I
Laplace Equation with Continuous Boundary Values. See reference .
The Laplace equation we are working with deals with the Laplace of a twice differentiable function \( u(x, y) \) whose Laplacian meets the conditions \( \Delta u(x, y) = 0 \).
We start off by applying the Finite Difference Method with \( m = n = 5\) to approximate the solution of the Laplace Equation \( \Delta u = 0\) on the interval: \( [0, 1]_x \times [1, 2]_y \)with the Dirichlet boundary conditions:
\( \left\{ \begin{array}{lr}u(x, 1) = \ln(x^2 + 1) \\ u(x, 2) = \ln(x^2 + 4) \\ u(0, y) = 2\ln(y) \\ u(1, y) = \ln(y^2 + 1) \end{array} \right.\)
Part II
We move on to the model of the steady-state temperature \(u\) on a heated copper plate, which satifies the initial conditions of the Poisson equation:
where the power density \(D(x, y)= 5 \frac{watts}{{cm}^3} \) and the thermal conductivity of copper \(K = 3.85 \frac{watts}{{cm}^\circ C} \) on the interval \( [0, 4]_x \times [0, 2]_y \) cm and keeping the temperature at the boundaries at a constant \({30}^\circ C\) with the Dirichlet boundary conditions:
\( \left\{ \begin{array}{lr} u(x, 0) = 30 \\ u(x, 2) = 30 \\ u(0, y) = 30 \\ u(4, y) = 30 \end{array} \right.\)
From our Finite Difference Approximation of the heat distribution, we found that the approximation for the temperature at the center of the plate \((x, y) = (2, 1)\) was:
\( 30.590617087856302 \).
This graphic is for \( m =n= 100\).
Part III
We want to show that the Laplace equation \( \Delta T (x, y) = 0 \) on
the rectangle
\( [0, L]_x \times [0, H]_y \) with Dirichlet Boundary conditions \( T =
T_0 \) on the three sides \( x = 0, x = L,\) and \( y = 0,\) and \(T = T_1\) on the side \(
y = H \) has solution:
\( T(x, y) = T_0 + \sum\limits_{k=0}^{\infty} C_k \sin \frac{(2k+1)\pi x}{L}
\sinh \frac{(2k+1)\pi y}{L}\), where, \(C_k = \frac{4(T_1 - T_0)}{(2k+1)\pi \sinh \frac{(2k+1)\pi H}{L}}\)
First Derivative with respect to x:
\( \frac{ \partial T}{ \partial x} = \sum\limits_{k=0}^{\infty} C_k \frac{(2k+1)\pi}{L}\cos
\frac{(2k+1)\pi x}{L} \sinh \frac{(2k+1)\pi y}{L}\)
Second Derivative with respect to x:
\( \frac{ \partial}{ \partial x}(\frac{ \partial T}{ \partial x} ) = \sum\limits_{k=0}^{\infty} -C_k
(\frac{(2k+1)\pi}{L})^2 \sin\frac{(2k+1)\pi x}{L}\sinh\frac{(2k+1)\pi y}{L}\)
First Derivate with respect to y:
\( \frac{ \partial T}{ \partial y} = \sum\limits_{k=0}^{\infty} C_k \frac{(2k+1)\pi}{L}
\sin\frac{(2k+1)\pi x}{L} \cosh\frac{(2k+1)\pi y}{L} \)
Second Derivative with respect to y:
\( \frac{ \partial}{ \partial y} (\frac{ \partial T}{ \partial y}) = \sum\limits_{k=0}^{\infty} C_k
(\frac{(2k+1)\pi}{L})^2 \sin\frac{(2k+1)\pi y}{L}\sinh\frac{(2k+1) \pi y}{L} \)
\( \Delta T = \frac{ \partial^2 T}{ \partial x^2} + \frac{ \partial^2 T}{ \partial y^2} \)
\( \Delta T = - \sum\limits_{k=0}^{\infty} C_k (\frac{(2k+1)\pi}{L})^2 \sin\frac{(2k+1)\pi
x}{L}\sinh\frac{(2k+1)\pi y}{L} + \sum\limits_{k=0}^{\infty} C_k
(\frac{(2k+1)\pi}{L})^2 \sin\frac{(2k+1)\pi y}{L}\sinh\frac{(2k+1) \pi y}{L} \)
\( \therefore \Delta T = 0 \)
For \( x = 0 \), then the sum \( \sum\limits_{k=0}^{\infty} C_k \sin \frac{(2k+1)\pi(0)}{L} \sinh \frac{(2k+1)\pi y}{L} = 0\). Thus, \( T(0,y) = T_0\).
For \( x = L\), then the sum is reduced to \( \sum\limits_{k=o}^{\infty} C_k \sin (2k+1)\pi x \sinh \frac{(2k+1)\pi y}{L} = 0 \). Thus, \( T(L,y) = T_0 \).
For \( y = 0 \), then the sum \( \sum\limits_{k=0}^{\infty} C_k \sin \frac{(2k+1)\pi x}{L} \sinh \frac{(2k+1)\pi(0)}{L} = 0\). Thus, \( T(x,0) = T_0\).
For \( y = H \), the \( \sinh \frac{(2k+1) \pi H}{L}\) in \( C_k \) cancels out with \( \sinh \frac{(2k+1) \pi y}{L} \) in the summation. Therefore, simplifying \(T(x,y)\) to:
\( T(x,y) = T_0 + \sum\limits_{k=0}^{\infty} \frac{4(T-T_0)}{(2k+1) \pi} \sin \frac{(2k+1) \pi x}{L} \)
\( = T_0 + \frac{4(T_1 - T_0)}{\pi} \sum\limits_{k=0}^{\infty} \frac{ \frac{(2k+1) \pi x)}{L}}{(2k+1)\pi} \)
\( \sum\limits_{k=0}^{\infty} \frac{ \frac{(2k+1) \pi x)}{L}}{(2k+1)\pi} = \frac{ \pi}{4} \)
\( \therefore T(x,y) = T_0 + \frac{ 4(T_1 - T_0)}{\pi} \frac{ \pi}{4} \)
\( \therefore T = T_1 \)
List of mathematical series can be found here
Part IV
For this portion of the project, we solve the Laplace equation with Dirichlet Boundary conditions from Part II, on the interval \( [0, 1]_x \times [0, 1]_y \) with \( T_0 = 0\) and \(T_1 = 10\) using (a) Finite Difference Approximation and (b) Finite Element Method.
For this project, we didn't have to use (b) Finite Element Method. But we went ahead and used just for comparison.
Summary of Project
For the Laplace Equation project, we explored the special case of Partial Differential Equations or PDEs, where the Laplacian( \( \Delta u = u_{xx} + u_{yy} = 0 \) ). First, we recreated the image from Example 8.8 in the book, by finding the solutions for the Laplace Equation with continuous boundary conditions. Next, for the Warm-Up exercise, we looked at heat distribution across a copper plate with the edge temperature kept at a constant temperature, finding for the temperature at he center point . Then, we looked at the solutions to the Laplace Equation using specific Dirichlet Boundary Conditions, solving for \( \Delta T = 0 \) on a rectangle \( [0, L]_x \times [0, H]_y \). Finally, we moved to a more rigorous problem: exploring the Laplace equation where the boundary conditions on one side was discontinuous with the other three.