Question

Show how to find the solution \(u(x, y)\) of Laplace's equation in the rectangle \(0

Short answer

Question: Solve Laplace's equation in a rectangle with boundary conditions \(u(0, y) = k(y)\), \(u(a, y) = f(y)\), \(u(x, 0) = h(x)\), and \(u(x, b) = g(x)\). Answer: The solution of Laplace's equation considering the given boundary conditions is: $$u(x,y)=\sum_{n=1}^{\infty}A_n\cosh(n\pi x/a) \cos(n\pi y/a)$$ Where: $$ A_n = \frac{4}{ab} \int_{0}^{a} h(x)\cosh(n\pi x/a) \,\mathrm{d}x \int_{0}^{b} k(y)\cos(n\pi y/b) \,\mathrm{d}y $$

Step-by-step solution

Apply separation of variables to Laplace's equation

Laplace's equation in 2D is given by: $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$$ To apply separation of variables, assume \(u(x,y) = X(x) Y(y)\). Now, substitute this product form into Laplace's equation and divide by \(X(x) Y(y)\): $$\frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0$$ Since we have separated the variables, we can write each term as a constant. Let's denote this constant as \(-\lambda^2\): $$\frac{X''(x)}{X(x)} = \lambda^2, \qquad \frac{Y''(y)}{Y(y)} = -\lambda^2$$

Solve the ODEs

Now, we have two ODEs to solve: $$X''(x) - \lambda^2 X(x) = 0, \qquad Y''(y) + \lambda^2 Y(y) = 0$$ The general solutions for these ODEs are: $$ X(x) = A\cosh(\lambda x) + B\sinh(\lambda x), \quad Y(y) = C\cos(\lambda y) + D\sin(\lambda y) $$

Combine the solutions

Combine the solutions to \(X(x)\) and \(Y(y)\) by taking the product of functions: $$ u(x,y) = (A\cosh(\lambda x) + B\sinh(\lambda x)) (C\cos(\lambda y) + D\sin(\lambda y)) $$

Apply the boundary conditions and find Fourier coefficients

Now, apply the boundary conditions one by one and adjust the solution accordingly. For simplicity, we will write \(n\) instead of \(\lambda\) (since they're usually integers in problems like this): 1. \(u(0, y) = k(y) => (A\cosh(0) + B\sinh(0)) (C\cos(ny) + D\sin(ny)) = k(y)\). Since \(A\cosh(0) = A\), we can rewrite this as \(AC\cos(ny)\). Thus, \(AC=\frac{1}{\pi}\int_{0}^{b} k(y)\cos(ny) \,\mathrm{d}y\). 2. \(u(a, y) = f(y) => (A\cosh(na) + B\sinh(na)) (C\cos(ny) + D\sin(ny)) = f(y)\). This implies \(B\sinh(na)D\sin(ny) = f(y) - AC\cos(ny)\). To find \(D\), we must multiply by \(\sin(ny)\) and integrate over \((0,b)\): $$ BD\sinh(na) = \frac{1}{\pi}\int_{0}^{b} (f(y) - AC\cos(ny))\sin(ny) \,\mathrm{d}y$$ 3. \(u(x, 0) = h(x) => (A\cosh(nx) + B\sinh(nx)) (C\cos(0) + D\sin(0)) = h(x)\). Since \(C\cos(0) = C\), we have \(A\cosh(nx) = h(x)\), so \(A=\frac{1}{\pi}\int_{0}^{a} h(x)\cosh(nx) \,\mathrm{d}x\). 4. \(u(x, b) = g(x) => (A\cosh(nx) + B\sinh(nx)) (C\cos(nb) + D\sin(nb)) = g(x)\). Now, we have \(B\sinh(nx)D\sin(nb) = g(x) - AC\cosh(nx)\), and we can find \(D\) with integration: $$ BD\sin(nb) = \frac{1}{\pi}\int_{0}^{a} (g(x) - AC\cosh(nx))\sinh(nx) \,\mathrm{d}x$$

Write the final solution considering all terms

Our final solution for \(u(x,y)\) will be a sum of all terms that satisfy the boundary conditions: $$u(x,y)=\sum_{n=1}^{\infty}A_n\cosh(n\pi x/a) \cos(n\pi y/a)$$ Where: $$ A_n = \frac{4}{ab} \int_{0}^{a} h(x)\cosh(n\pi x/a) \,\mathrm{d}x \int_{0}^{b} k(y)\cos(n\pi y/b) \,\mathrm{d}y $$ And that's the solution of Laplace's equation, considering the given boundary conditions.

Key concepts

Separation of Variables

When tackling Laplace's equation, which is vital in many physics and engineering problems, we often use the separation of variables technique. This method involves expressing a multi-variable function as the product of single-variable functions. In our case, we assume the solution to be in the form \(u(x, y) = X(x)Y(y)\). By substituting into the equation \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\), and dividing by the product \(X(x)Y(y)\), we arrive at two separate equations. Each equation depends solely on one variable, simplifying our problem significantly.
By solving these equations independently, we can find solutions that, when multiplied together, satisfy the original partial differential equation. This separation reduces a complex problem into more manageable ordinary differential equations, making it a powerful method in mathematical physics.

Boundary Conditions

Boundary conditions are constraints necessary in solving partial differential equations like Laplace's equation. They determine particular solutions by prescribing the behavior of the solution at the edges of the domain. In this problem, specific conditions are given for all four boundaries of a rectangle. These include:

• \( u(0, y) = k(y) \)
• \( u(a, y) = f(y) \)
• \( u(x, 0) = h(x) \)
• \( u(x, b) = g(x) \)

Applying these conditions allows us to gain information about the constants in the general solution. By substituting these into our separated solutions, we ensure the entire function adheres to the boundaries described. This ensures uniqueness and physically meaningful solutions within the context of the problem.

Fourier Series

Incorporating Fourier series into the solution helps break down complex functions using a combination of sines and cosines. This is particularly effective in boundary value problems where periodicity arises naturally (e.g., problems set in rectangular domains as here).
Fourier series aid in determining coefficients that ensure the solution satisfies the boundary conditions. In this exercise, terms of the form \(\cos(n\pi y/b)\) and \(\sin(n\pi y/b)\) appear, related to the conditions across \(y=0\) and \(y=b\). The series enables us to express the solution as an infinite sum, accommodating the infinite modes necessary to satisfy every point within the specified boundary conditions.

Ordinary Differential Equations

Once variables are separated, we deal with ordinary differential equations (ODEs). Specifically, you end up solving two ODEs reflecting each spatial dimension:

• \(X''(x) - \lambda^2 X(x) = 0\)
• \(Y''(y) + \lambda^2 Y(y) = 0\)

The solutions to these ODEs are combinatory functions (e.g., hyperbolic cosines and sines for \(X(x)\), and regular sines and cosines for \(Y(y)\)). These are foundational in constructing the solution for \(u(x,y)\).
Solving these ODEs means finding expressions that respond well to boundary conditions and correctly capture the necessary physical or geometrical properties dictated by these equations. These functions, encoded with constants, illuminate the structure of solutions in contexts like thermal distribution, electrostatics, or fluid flow.