Question

In Exercises \(17-28\) define the formal solution of $$ u_{x x}+u_{u y}=0, \quad 0

Short answer

To find the formal solution for the given PDE \(u_{xx} + u_{uy} = 0\) with specific boundary conditions and values, we first separated the variables and solved the resulting ODEs for X(x) and Y(y). We then applied boundary conditions and obtained a series representation for the formal solution. Finally, we calculated specific coefficients \(A_n\) for the given values of \(a = 1, b = 1\) and \(g(y) = y(y^3 - 2y^2 + 1)\). The formal solution is given by: \(u(x, y) = \sum_{n=0}^\infty A_n \sin\left(\frac{(2n+1)\pi x}{2}\right)e^{-k^2 y}\), where \(k = \frac{(2n+1)\pi}{2}\) and the coefficients \(A_n\) can be found using: \(A_n = \frac{2}{b} \int_0^b g(y)\sin\left(\frac{(2n+1)\pi x}{2b}\right)e^{-k^2 y} dy\).

Step-by-step solution

Separation of Variables

We assume the solution is of the form \(u(x, y) = X(x) \cdot Y(y)\). Substituting this into the given PDE, we have: \(X''(x)Y(y)+ X(x)Y'(y)=0 \Longrightarrow \frac{X''(x)}{X(x)}= -\frac{Y'(y)}{Y(y)}\) Since the left part of the equation depends only on \(x\) and the right part depends only on \(y\), both expressions must equal a constant for it to hold. Let's denote that constant as \(k^2\): \(\frac{X''(x)}{X(x)} = -\frac{Y'(y)}{Y(y)} = k^2\) Now we have two ODEs to solve: \(X''(x) = k^2X(x)\) \(Y'(y) = -k^2Y(y)\)

Solving the ODEs for X(x) and Y(y)

The general form of solution for these ODEs is: \(X(x) = A\sin(kx)+B\cos(kx)\) \(Y(y) = Ce^{-k^2y}\) Now, we use the boundary conditions to determine the values of \(A,B,\) and \(C\).

Applying Boundary Conditions

Using the boundary conditions: \(u(x, 0)=0 \Rightarrow X(x)Y(0)=0 \Rightarrow Y(0)=0\). Since \(Y(y) \neq 0\) (trivial solution), we get: \(A\sin(kx)+B\cos(kx) = 0\) (1) \(u(x, b)=0 \Rightarrow X(x)Y(b)=0 \Rightarrow Y(b)=0\). Since \(X(x) \neq 0\): \(Ce^{-k^2b} = 0 \Rightarrow C=0\) (trivial solution, not taken) This suggests taking \(k=\frac{(2n+1)\pi}{2b}\) for \(n=0,1,2,\dots\) Now, we have: \(X(x)= A\sin\left(\frac{(2n+1)\pi x}{2b}\right)\). Since \(X'(0)=0\): \(u_x(x, y)= \left(X'(x)Y(y) \right)|_{x = 0} = 0\) (2) Solve this equation to find \(X'(x) = \frac{(2n+1)\pi}{2b}A\cos\left(\frac{(2n+1)\pi x}{2b}\right)\) (3) Now, we integrate (3) w.r.t \(x\) from \(0\) to \(a\), and using boundary condition \(u(a,y)=g(y)\), we get: \(\left(\sum_{n=0}^\infty A_n \sin\left(\frac{(2n+1)\pi x}{2b}\right)\right)e^{-k^2 y} = g(y)\)

Solve for A_n

Multiply both sides by \(\sin\left(\frac{(2m+1)\pi}{2b}\right)e^{-k^2 y}\), \(m=0,1,2,\dots\), and integrate w.r.t \(y\) from \(0\) to \(b\): \(\int_0^b \sum_{n=0}^\infty A_n\sin\left(\frac{(2n+1)\pi x}{2b}\right)\sin\left(\frac{(2m+1)\pi x}{2b}\right)e^{-2k^2 y} dy = \int_0^b g(y)\sin\left(\frac{(2m+1)\pi x}{2b}\right)e^{-k^2 y} dy\) We use orthogonality condition to get \(A_m\): \(A_m = \frac{2}{b} \int_0^b g(y)\sin\left(\frac{(2m+1)\pi x}{2b}\right)e^{-k^2 y} dy\) Now that we have \(A_m\), we can find the solution \(u(x, y)\):

Formal Solution

\(u(x, y) = \sum_{n=0}^\infty A_n \sin\left(\frac{(2n+1)\pi x}{2b}\right)e^{-k^2 y}\) Now we plug the specific values \(a = 1, b = 1,\) and \(g(y) = y\left(y^3 - 2y^2 + 1\right)\) into the above equation and find \(A_n\) using the formula mentioned in step 4 and finally, substitute this \(A_n\) back in the solution u(x,y) equation.

Key concepts

Separation of Variables

The technique of separation of variables is a method for solving partial differential equations (PDEs), in which the PDE is broken down into simpler, ordinary differential equations (ODEs). This approach assumes that the solution to a PDE can be expressed as the product of functions, where each function depends only on one of the independent variables.

Consider the PDE given by \(u_{xx} + u_{yy} = 0\) with certain boundary conditions. By assuming that \(u(x, y) = X(x) \cdot Y(y)\), we essentially separate the variables. This means we transform our PDE into two ODEs based on the premise that if the product of two functions is zero, then at least one of the functions must itself be zero. This net result is highly practical, as it simplifies the complex problem and allows us to solve the equation step by step for each variable separately.

Partial Differential Equations

Partial differential equations (PDEs) are equations that involve rates of change with respect to continuous variables. The PDE \(u_{xx} + u_{yy} = 0\) is an example where \(u_{xx}\) and \(u_{yy}\) indicate the second partial derivatives of \(u\) with respect to \(x\) and \(y\), respectively. PDEs are fundamental to the description of physical phenomena such as heat conduction, waves, and quantum mechanics.

Dealing with PDEs often requires specialized methods such as separation of variables, transform methods, or numerical analysis. By imposing boundary conditions, we can tailor general solutions to specific problems and accurately model various physical situations.

Solving Ordinary Differential Equations

Solving ordinary differential equations (ODEs) involves finding a function that satisfies the equation, often with given initial or boundary conditions. In the context of our problem, after applying separation of variables, we are left with two ODEs that depend solely on \(x\) and \(y\), separately.

The general solutions have known forms, but what remains is the task of determining the coefficients by applying the boundary conditions. This involves evaluating the solutions at the boundaries and adjusting parameters accordingly. ODEs can often be solved analytically, but in more complex cases, numerical methods might be used. The key point for students to understand here is the character of the behavior imposed by the boundary conditions on the ODE solution.

Eigenvalues and Eigenfunctions

In the context of differential equations, eigenvalues and eigenfunctions occur in problems that possess some form of symmetry which generally leads to a PDE having special solutions under certain conditions. When we perform separation of variables on a PDE, we often get a resultant expression in which each side is equal to a common constant value - the eigenvalue.

For each eigenvalue, there’s a corresponding function, called the eigenfunction, which satisfies the ODE along with the initial or boundary conditions. These eigenfunctions form the building blocks of the solution to the original PDE. The takeaway is that eigenvalues and eigenfunctions are integral to constructing solutions for PDEs with boundary conditions, as they encapsulate the quantitative measure and spatial patterns of these solutions, respectively.