Question

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

Short answer

Question: Find the solution u(x, y) of the given partial differential equation \(u_{xx} + u_{uy} = 0\) with boundary conditions \(u(0, y) = 0\), \(u(a, y) = 0\), and \(u_y(x, 0) = f(x)\), where \(a = \pi\), \(b = 1\), and \(f(x)\) is defined in the interval [0, \(\pi\)] as \(f(x) = x\) for \(0 \leq x \leq \frac{\pi}{2}\) and \(f(x) = \pi - x\) for \(\frac{\pi}{2} \leq x \leq \pi\). Answer: The solution u(x, y) satisfying the given partial differential equation and boundary conditions is: $$u(x, y) = \sum_{n=1}^{\infty} -\frac{f(x)}{n\pi} \sin(n\pi x)e^{-n\pi y},$$ where \(f(x)\) is the Fourier sine series obtained by substituting the given definition of \(f(x)\) and calculating the integral. The explicit solution can be found by computing the polynomial for the provided values of \(a\) and \(b\).

Step-by-step solution

Separation of Variables

Assume the solution of the PDE has the form \(u(x, y) = X(x)Y(y)\). Plug this into the given PDE, $$X''(x)Y(y) + X(x)Y'(y) = 0.$$ Now divide both sides of the equation by \(XY\). This gives $$\frac{X''(x)}{X(x)} = -\frac{Y'(y)}{Y(y)}.$$ Since the left side depends only on \(x\) and the right side depends only on \(y\), they must both be equal to some constant, say \(-k^2\). Hence, we have two ordinary differential equations (ODEs): 1. \(X''(x) + k^2X(x) = 0\) 2. \(Y'(y) + k^2Y(y) = 0\)

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

To solve ODEs for \(X(x)\) and \(Y(y)\), we can use the following general solutions for the above ODEs: 1. \(X(x) = A\cos(kx) + B\sin(kx)\) 2. \(Y(y) = Ce^{-ky}\) where \(A\), \(B\), and \(C\) are constants. Now, let's apply the given boundary conditions on the general solutions.

Apply Boundary Conditions for X(x) and Y(y)

Using the boundary conditions \(u(0, y) = 0\) and \(u(a, y) = 0\), we can find the values of the constants \(A\) and \(B\): 1. \(u(0, y) = 0 \Rightarrow X(0) = A\cos(0) + B\sin(0) \Rightarrow A = 0\) 2. \(u(a, y) = 0 \Rightarrow X(a) = 0 \Rightarrow B\sin(ka) = 0\) Ans since \(B\neq0\), we have \(\sin(ka) = 0\). Thus, \(ka=n\pi\), where n is a positive integer, meaning that \(k=\frac{n\pi}{a}\). Now obtain the series solution by summing over all possible values of \(k\): $$X(x) = B_n\sin(k_nx).$$ Next, consider the boundary condition \(u_y(x, 0) = f(x)\). Taking the derivative \(Y'(y)\), we have: $$Y'(y) = -k_nCe^{-k_ny}.$$ At \(y = 0\), we get, $$Y'(0) = -k_nC.$$ Recall that \(u(x, y) = X(x)Y(y)\), and therefore \(u_y(x, y) = X(x)Y'(y)\). So, for \(y=0\), we have: $$u_y(x, 0) = -k_nB_nC = f(x).$$ Now, we need to find Fourier sine coefficients for \(f(x)\) given its definition.

Compute Fourier Sine Coefficients for f(x)

We first define f(x) in its given interval as $$ f(x)= \begin{cases} x, & 0 \leq x \leq \frac{\pi}{2} \\ \pi - x, & \frac{\pi}{2} \leq x \leq \pi \end{cases} $$ Using the Fourier sine series formula, find the coefficients \(B_n\), where \(n\) is a positive integer: $$B_n = \frac{2}{a} \int_{0}^{a} f(x) \sin\left(\frac{n\pi x}{a}\right) dx.$$ After calculating the integral for the given boundary values \(a=\pi\) and \(b=1\), derive the Fourier sine series: $$f(x) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{a}\right).$$

Find U(x, y) from X(x) and Y(y)

We found the Fourier sine series for \(f(x)\). Now, derive the product solution from \(X(x)\) and \(Y(y)\). Recall that: $$u(x, y) = X(x)Y(y) = B_n\sin(\frac{n\pi x}{\pi})\left[Ce^{-\frac{n\pi y}{\pi}}\right].$$ Substitute the expression for \(C\) from step 3: $$u(x, y) = -\frac{f(x)}{n\pi} \sin(\frac{n\pi x}{\pi})e^{-\frac{n\pi y}{\pi}}.$$ Finally, sum the series solution for all possible values of \(n\) to get the general solution satisfying the boundary conditions: $$u(x, y) = \sum_{n=1}^{\infty} -\frac{f(x)}{n\pi} \sin(n\pi x)e^{-n\pi y}.$$ Now, find the explicit solution by substituting the Fourier sine series of \(f(x)\) (derived in step 4) and compute the polynomial. The obtained solution u(x, y) will be the desired answer to the exercise considering all provided boundary conditions.

Key concepts

Partial Differential Equations

Partial Differential Equations (PDEs) are equations that involve multiple independent variables and their partial derivatives. They are essential in describing phenomena in engineering, physics, and other fields. PDEs can model things like heat conduction, wave propagation, and fluid dynamics. In our exercise, we have a PDE given by:\[ u_{xx} + u_{yy} = 0 \]This is actually a standard form known as the Laplace's equation, which is crucial in potential theory and electrostatics. The general approach to solving PDEs is to look for solutions that satisfy given boundary conditions, leading to what we call boundary value problems (BVPs). The boundary conditions help us determine the specific solution from the infinite possibilities. In the exercise, we need to work with boundary constraints like \(u(x, b)=0\) and others to find the solution \(u(x, y)\).

Separation of Variables

Separation of Variables is a technique used to solve partial differential equations, which involves assuming that the solution can be decomposed into functions that depend on individual variables. In our case, the proposed solution form is:\[ u(x, y) = X(x)Y(y) \]This assumption allows us to split the original PDE into simpler ordinary differential equations (ODEs), one for each variable. The equation:\[ X''(x)Y(y) + X(x)Y''(y) = 0 \]is divided by \(XY\) to isolate the variables, resulting in:\[ \frac{X''(x)}{X(x)} = -\frac{Y''(y)}{Y(y)} = -k^2 \]Thus reducing a complex PDE problem into two ODEs, one for \(X(x)\) and another for \(Y(y)\). Applying this technique simplifies our task, making it easier to apply boundary conditions and find solutions for \(X(x)\) and \(Y(y)\).

Fourier Series

Fourier Series are used to represent a function as a sum of sine and cosine terms, making them highly useful for solving PDEs, especially with periodic boundary conditions. In this exercise, we need to find the Fourier sine series for the function \(f(x)\):\[ f(x) = \begin{cases} x, & 0 \leq x \leq \frac{\pi}{2} \ \pi - x, & \frac{\pi}{2} \leq x \leq \pi \end{cases} \]The coefficients \(B_n\) are determined using the integral:\[ B_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \sin\left(\frac{n\pi x}{\pi}\right) \, dx \]By computing these integrals, we can derive the series representation of \(f(x)\). This representation plays a crucial role in solving the PDE because it allows us to express the function in terms of simple trigonometric terms, which can be used to meet the boundary conditions effectively.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) involve functions of a single variable and their derivatives. They are simpler than PDEs and often result from using separation of variables. In this exercise, we encounter:1. \(X''(x) + k^2 X(x) = 0\)2. \(Y''(y) + k^2 Y(y) = 0\)These equations arise from the separation of variables, relating the spatial component \(x\) and the other variable \(y\) separately. The solution for equations of this form typically involves trigonometric and exponential functions:- The general solution for \(X(x)\) is \(\cos(kx)\) and \(\sin(kx)\).- For \(Y(y)\), the solution depends on the boundary conditions, often resulting in \(e^{-ky}\).The constants and specific form of the solutions are determined by applying boundary conditions, ensuring that the solution not only solves the ODEs but also fits the physical or geometrical constraints of the problem.