Question

In Exercises $1-16$ apply the definition developed in Example 1 to solve the boundary value problem. (Use Theorem 11.3 .5 where it applies.) Where indicated by $\mathrm{C}$, graph the surface $u=u(x,y),0\le x\le a$, $0\le y\le b$ $$ \begin{array}{l} u_{x x}+u_{y y}=0, \quad 0

Short answer

In this boundary value problem, we employed the method of separation of variables to find a solution for the given partial differential equation. We were given a homogeneous PDE with both Dirichlet and Neumann boundary conditions. We then applied the method of separation of variables to find the solutions to the resulting ordinary differential equations for f(x) and g(y). After combining these solutions, we applied the given boundary conditions and found the constants by considering the sum of product solutions. Lastly, we represented the solution to the problem as an infinite series and could graph it in the given domain to visualize the surface. In summary, the solution to this boundary value problem is: $$u(x,y)=\sum _{n=1}^{\mathrm{\infty }}{B}_n\sin (n\pi x)(D_n\sinh (n\pi y))+\frac{1}{3}$$ with 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.

Step-by-step solution

Analyze the given PDE and boundary conditions

The given problem is: $$u_{xx}+u_{yy}=0,0<x<1,0<y<1,$$ $$u(x,0)=0,u_y(x,1)=(x-1{)}^2,0\le x\le 1,$$ $$u_x(0,y)=0,u_x(1,y)=0,0\le y\le 1.$$ Let's first analyze the given PDE, and boundary conditions. 1. PDE: We have a linear second-order homogeneous PDE. 2. Boundary conditions: We have both Dirichlet (u(x, 0) = 0) and Neumann (u_y(x, 1) = (x - 1)^2, u_x(0, y) = 0, u_x(1, y) = 0) boundary conditions. To simplify our problem, we'll rewrite the Neumann boundary condition u_y(x, 1) = (x - 1)^2 as an equivalent Dirichlet boundary condition using integration.

Express the Neumann boundary condition as Dirichlet boundary condition

To simplify the problem, we'll rewrite the Neumann boundary condition u_y(x, 1) = (x - 1)^2 as a Dirichlet boundary condition: $$u^{\ast }(x,1)={\int }_0^1(t-1{)}^{2}dt=\frac{1}{3}$$ Now we have the alternative boundary condition: $$u^{\ast }(x,1)=\frac{1}{3},0\le x\le 1.$$

Apply separation of variables

Now we'll apply the method of separation of variables on the PDE u_{xx} + u_{yy} = 0. Assume that u(x, y) = f(x) g(y). Plug this into the PDE: $$f^{″}(x)g(y)+f(x)g^{″}(y)=0.$$ Now divide both sides by f(x)g(y): $$\frac{f^{″}(x)}{f(x)}+\frac{g^{″}(y)}{g(y)}=0.$$ Now we have two independent variables x and y which implies that each term must be constant. On the left side, the first term depends only on x, so create a constant -λ such that: $$\frac{f^{″}(x)}{f(x)}=-\lambda .$$ And the second term depends only on y, so create a constant +λ such that: $$\frac{g^{″}(y)}{g(y)}=\lambda .$$

Solve the ODEs for f and g

Now solve the two ordinary differential equations (ODEs) independently: 1. For f(x): $$f^{″}(x)+\lambda f(x)=0$$ Since we have Neumann boundary conditions, -λ will be positive (i.e. -λ = n^2π^2) for f(x): $$f^{″}(x)-n^2{\pi }^{2}f(x)=0$$ The general solution of the f(x) ODE is: $$f_n(x)=A_n\cos (n\pi x)+B_n\sin (n\pi x)$$ 2. For g(y): $$g^{″}(y)-\lambda g(y)=0$$ With -λ = n^2π^2, we have: $$g^{″}(y)-n^2{\pi }^{2}g(y)=0$$ The general solution of the g(y) ODE is: $$g_n(y)=C_n\cosh (n\pi y)+D_n\sinh (n\pi y)$$

Combine both ODEs to form u(x, y)

Now, we have the product solution u(x, y) = f_n(x)g_n(y): $$u_n(x,y)=(A_n\cos (n\pi x)+B_n\sin (n\pi x))(C_n\cosh (n\pi y)+D_n\sinh (n\pi y))$$ Since this problem has a mixed boundary condition, we'll form the solution as the sum of two product solutions: $$u(x,y)=u_n(x,y)+u^{\ast }(x,y)$$

Apply boundary conditions to find constants

Apply the boundary conditions and sum all solutions: Boundary conditions: 1. u(x, 0) = 0: It satisfies for B_n = 0 and C_n = 0. 2. u_y(x, 1) = (x - 1)^2: It also satisfies with the sum u^*(x, y). 3. u_x(0, y) = 0: It satisfies for A_n = 0. 4. u_x(1, y) = 0: It satisfies for A_n = 0. So the solution u(x, y) = u_n(x, y) + u^*(x, y) satisfies the given boundary value problem, with: $$u(x,y)=\sum _{n=1}^{\mathrm{\infty }}{B}_n\sin (n\pi x)(D_n\sinh (n\pi y))+\frac{1}{3}$$ Now, we can take the given Laplace transform (if needed) and determine the remaining constant coefficients using a similar method to that used in Theorem 11.3.5. Finally, we should graph the solution u(x, y) in the domain 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 to visualize the surface that satisfies the given boundary value problem.

Key concepts

Separation of Variables

Separation of Variables is a mathematical method used to solve partial differential equations (PDEs). It simplifies complex problems into more manageable pieces. In this approach, we assume a solution of the PDE can be expressed as a product of functions, each depending on a single variable. For example, for a function $u(x,y)$, we can assume that $u(x,y)=f(x)g(y)$. This approach helps break down a PDE like $u_{xx}+u_{yy}=0$ into two ordinary differential equations (ODEs).

• The key idea is that if the PDE involves independent variables $x$ and $y$, the PDE can be transformed into separate ODEs for each variable.
• This method is applicable when the problem geometry and boundary conditions allow the solution to be well-expanded into multiplicative components.

In our example, this technique transposed the PDE into separate differential equations for $f(x)$ and $g(y)$, making the problem easier to tackle by resolving one variable at a time.

Partial Differential Equations

Partial Differential Equations (or PDEs) are equations that involve multiple independent variables and their partial derivatives. They are essential in describing various physical phenomena such as heat transfer, fluid flow, and electromagnetic fields. A PDE typically contains unknown functions that depend on more than one variable.

• The notion of PDEs is to express how a quantity changes with more than one condition or variable. For example, in the PDE $u_{xx}+u_{yy}=0$, the function $u(x,y)$ depends on two variables, $x$ and $y$.
• Solutions to PDEs are often expressed in terms of functions that define the state of physical systems or processes.

In our context, the given PDE is linear and homogenous, and by solving it using separation of variables, the behavior of the unknown function $u(x,y)$ over the rectangular domain is determined.

Neumann and Dirichlet Boundary Conditions

Boundary conditions are essential in solving differential equations as they specify the values of the function on the boundary of its domain. In mathematics, the two common types of boundary conditions are Neumann and Dirichlet conditions.

Dirichlet boundary conditions set the value of the function itself at the boundary. For instance, in the problem, $u(x,0)=0$ is a Dirichlet boundary condition. It specifies the value of $u$ along the boundary $y=0$.

• They are suitable for situations where the solution itself needs to maintain certain values at the boundary.
• The Dirichlet condition is straightforward as it usually provides direct values for the function at boundaries.

Neumann boundary conditions specify the value of the derivative of the function at the boundary. This is seen in the problem with conditions like $u_x(0,y)=0$ and $u_y(x,1)=(x-1{)}^2$.

• They are used when the rate of change or a flux is essential to understand the boundary behavior of the solution.
• Occasionally, such conditions need to be transformed into equivalent Dirichlet conditions for simplicity.

Understanding these types of boundary conditions is crucial for finding the right solutions for PDEs and ensuring the solutions correctly fit the physical or theoretical model.