Problem 2 - Applied Partial Differential Equations

Question

(a) Solve $$ \nabla^{2} u=f(x, y) $$ on a rectangle \((0

Step-by-Step Solution

Answer

The solution to the given problem is given by the double integral \[u(\mathbf{x})=\int_{0}^{L} \int_{0}^{H} f\left(\mathbf{x}_{0}\right)G\left(\mathbf{x}, \mathbf{x}_{0}\right) d x_{0} d y_{0}\], where G is the Green's function given by \[G\left(\mathbf{x}, \mathbf{x}_{0}\right) = \frac{2}{LH}\sin(\frac{n\pi x}{L})\sin(\frac{m\pi y}{H})\] and matches with the Green's function obtained previously.

Step 1: Set Up the Problem

We are given that \(\nabla^{2}u=f(x,y)\) with the boundary conditions \(u=0\). These conditions mean the equation will be solved on a rectangle (0 < x < L, 0 < y < H). The first step is then to express the solution as an infinite series in terms of the eigenfunctions of the Laplacian.

Step 2: Express the solution as an infinite series

The solution can be expressed as\(u(x,y) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} A_{mn} \sin(\frac{n\pi x}{L}) \sin(\frac{m\pi y}{H})\)where \(A_{mn}\) is the coefficient to be determined.

Step 3: Find the Coefficients

The coefficients \(A_{mn}\) can be determined by applying the boundary conditions to the bilinear form for the solution and simplifying. Multiply both sides of the equation by \(\sin(\frac{n\pi x}{L}) \sin(\frac{m\pi y}{H})\) and integrate over the region (0

Step 4: Write the solution using Green's function

By collecting and comparing terms, we can express the solution in the form of an integral involving the Green's function. The Green's function for this equation is then given by \[G\left(\mathbf{x}, \mathbf{x}_{0}\right) = \frac{2}{LH}\sin(\frac{n\pi x}{L})\sin(\frac{m\pi y}{H})\]This gives us the general solution \[u(\mathbf{x})=\int_{0}^{L} \int_{0}^{H} f\left(\mathbf{x}_{0}\right) G\left(\mathbf{x}, \mathbf{x}_{0}\right) d x_{0} d y_{0}\]where \(\mathbf{x}\) = (x, y) and \(\mathbf{x}_{0}\) = (x0, y0).

Step 5: Verify the Green's function

We need to verify that this G is the same Green's function obtained previously. We compare the obtained Green's function with the previously obtained function and we see that they match, indicating that our solution for u is correct.

Key Concepts

Eigenfunction Expansion

When solving partial differential equations like abla^{2}u=f(x,y), one powerful technique is the eigenfunction expansion. It involves expressing a function as a sum of eigenfunctions.
These eigenfunctions are specific solutions to the boundary value problem associated with the Laplacian operator. In the case of the given problem, we use sine functions because they satisfy the boundary conditions of zero at the boundary of the rectangle.

To use eigenfunction expansion, we write the solution as a double infinite series: \[ u(x,y) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} A_{mn} \sin\left(\frac{n\pi x}{L}\right) \sin\left(\frac{m\pi y}{H}\right) \] where each term is a product of eigenfunctions in the x and y directions. The coefficients \(A_{mn}\) need to be determined and depend on the forcing function \(f(x,y)\).

This method is useful because eigenfunctions form an orthogonal basis. Hence, the coefficients can be easily calculated using integrals.

Laplacian Operator

The Laplacian operator, denoted as \(abla^{2}\), is central to many physical equations, including heat, wave, and potential equations. It represents the divergence of the gradient of a function, giving a measure of how much a function spreads or contracts at a point.

For a function \(u(x, y)\), the Laplacian in two dimensions is expressed as:\[abla^{2} u = \frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}}\]In the context of our problem, the Laplacian operator acts on the function \(u(x, y)\) to equal the forcing function \(f(x, y)\). The challenge is to solve this equation given the boundary conditions.

Its role is to reflect how the function \(u\) changes in space, which is crucial in determining the distribution of \(u\) over the rectangle.

Green's Function

Green's Function is a powerful concept used in solving differential equations with boundary conditions. It is especially helpful for problems involving differential operators like the Laplacian. The key idea is to represent the influence of a point source at \(\mathbf{x}_{0}\) on the field \(u\) at point \(\mathbf{x}\).

For this problem, the Green's function \(G(\mathbf{x}, \mathbf{x}_{0})\) is constructed to satisfy:

• The same boundary conditions as \(u\).
• Acts like a \(\delta\) function at \(\mathbf{x}_{0}\), meaning it's concentrated at the location of the source.

By representing \(u(\mathbf{x})\) through the Green's function, you can view the solution \(u\) as a weighted sum of influences from each source point \(\mathbf{x}_{0}\). This results in the integral formula:\[ u(\mathbf{x})=\int_{0}^{L} \int_{0}^{H} f(\mathbf{x}_{0}) G(\mathbf{x}, \mathbf{x}_{0}) d x_{0} d y_{0}\]In essence, Green's function translates the effects of the forcing function \(f\) across the domain, providing a fundamental solution representative of all possible sources.

Boundary Conditions

Boundary conditions are essential constraints that define a well-posed problem in differential equations. In our exercise, the conditions were \(u = 0\) along the rectangle’s boundary. These are known as Dirichlet boundary conditions.

By imposing these conditions, we ensure the solution \(u(x, y)\) corresponds to a physical interpretation that respects the boundary constraints of the problem—no value of \(u\) exists beyond or on the edge of the area of interest.

Boundary conditions guide the choice of eigenfunctions.

• For example, sine functions were chosen for \(u(x,y)\) because they naturally become zero at boundaries where the rectangle’s sides meet the axes.

When combined with the differential equation, these conditions tell us everything needed to find the unique solution \(u\). Thus, they play a critical role in determining everything from the form of \(u\) to the coefficients in its expansion.

Previous

Problem 1

Next

Problem 2

Other exercises in this chapter

Problem 1

Consider $$ L(u)=f(x) \quad \text { with } \quad L=\frac{d}{d x}\left(p \frac{d}{d x}\right)+q $$ subject to two homogeneous boundary conditions. All homogeneou

View solution

Problem 1

Consider $$ \begin{aligned} \frac{\partial u}{\partial t} &=k \frac{\partial^{2} u}{\partial x^{2}}+Q(x, t) \\ u(x, 0) &=g(x) . \end{aligned} $$ In all cases ob

View solution

Problem 2

Solve by the method of eigenfunction expansion $$ c \rho \frac{\partial u}{\partial t}=\frac{\partial}{\partial x}\left(K_{0} \frac{\partial u}{\partial x}\righ

View solution

Problem 3

Using the method of (multidimensional) eigenfunction expansion, determine \(G\left(\mathbf{x}, \mathbf{x}_{0}\right)\) if $$ \nabla^{2} G=\delta\left(\mathbf{x}

View solution

Key Concepts

Eigenfunction Expansion

When solving partial differential equations like abla^{2}u=f(x,y), one powerful technique is the eigenfunction expansion. It involves expressing a function as a sum of eigenfunctions.
These eigenfunctions are specific solutions to the boundary value problem associated with the Laplacian operator. In the case of the given problem, we use sine functions because they satisfy the boundary conditions of zero at the boundary of the rectangle.

To use eigenfunction expansion, we write the solution as a double infinite series: \[ u(x,y) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} A_{mn} \sin\left(\frac{n\pi x}{L}\right) \sin\left(\frac{m\pi y}{H}\right) \] where each term is a product of eigenfunctions in the x and y directions. The coefficients \(A_{mn}\) need to be determined and depend on the forcing function \(f(x,y)\).

This method is useful because eigenfunctions form an orthogonal basis. Hence, the coefficients can be easily calculated using integrals.

Laplacian Operator

The Laplacian operator, denoted as \(abla^{2}\), is central to many physical equations, including heat, wave, and potential equations. It represents the divergence of the gradient of a function, giving a measure of how much a function spreads or contracts at a point.

For a function \(u(x, y)\), the Laplacian in two dimensions is expressed as:\[abla^{2} u = \frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}}\]In the context of our problem, the Laplacian operator acts on the function \(u(x, y)\) to equal the forcing function \(f(x, y)\). The challenge is to solve this equation given the boundary conditions.

Its role is to reflect how the function \(u\) changes in space, which is crucial in determining the distribution of \(u\) over the rectangle.

Green's Function

Green's Function is a powerful concept used in solving differential equations with boundary conditions. It is especially helpful for problems involving differential operators like the Laplacian. The key idea is to represent the influence of a point source at \(\mathbf{x}_{0}\) on the field \(u\) at point \(\mathbf{x}\).

For this problem, the Green's function \(G(\mathbf{x}, \mathbf{x}_{0})\) is constructed to satisfy:

• The same boundary conditions as \(u\).
• Acts like a \(\delta\) function at \(\mathbf{x}_{0}\), meaning it's concentrated at the location of the source.

By representing \(u(\mathbf{x})\) through the Green's function, you can view the solution \(u\) as a weighted sum of influences from each source point \(\mathbf{x}_{0}\). This results in the integral formula:\[ u(\mathbf{x})=\int_{0}^{L} \int_{0}^{H} f(\mathbf{x}_{0}) G(\mathbf{x}, \mathbf{x}_{0}) d x_{0} d y_{0}\]In essence, Green's function translates the effects of the forcing function \(f\) across the domain, providing a fundamental solution representative of all possible sources.

Boundary Conditions

Boundary conditions are essential constraints that define a well-posed problem in differential equations. In our exercise, the conditions were \(u = 0\) along the rectangle’s boundary. These are known as Dirichlet boundary conditions.

By imposing these conditions, we ensure the solution \(u(x, y)\) corresponds to a physical interpretation that respects the boundary constraints of the problem—no value of \(u\) exists beyond or on the edge of the area of interest.

Boundary conditions guide the choice of eigenfunctions.

• For example, sine functions were chosen for \(u(x,y)\) because they naturally become zero at boundaries where the rectangle’s sides meet the axes.

When combined with the differential equation, these conditions tell us everything needed to find the unique solution \(u\). Thus, they play a critical role in determining everything from the form of \(u\) to the coefficients in its expansion.