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 \(u=u(x, y), 0 \leq x \leq a\), \(0 \leq y \leq b\) $$ \begin{array}{l} u_{x x}+u_{y y}=0, \quad 0

Short answer

Answer: The final solution for \(u(x, y)\) is given by $$u(x, y) = \sum_{n=1}^{\infty}\left(\frac{2}{\pi}\int_{0}^{\pi} x\sin x \sin(nx) dx\right)\sin(nx)\sinh(ny)$$

Step-by-step solution

Apply Separation of Variables

Assume that the solution \(u(x, y)\) can be represented as a product of functions of \(x\) and \(y\) respectively, i.e., \(u(x, y) = X(x)Y(y)\). Substituting \(u(x, y)\) into the given equation, we get $$X''(x)Y(y) + X(x)Y''(y) = 0$$

Separate the Variables

Divide both sides by \(X(x)Y(y)\) to separate the variables: $$\frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0$$

Introduce Constants and Get Two ODEs

Since the left side is a function of \(x\) while the right side is a function of \(y\), these two functions must be equal to each other and equal to a constant. Let's denote it by \(-\lambda\), this gives rise to two ordinary differential equations (ODEs): $$X''(x) = -\lambda X(x)$$ and $$Y''(y) = \lambda Y(y)$$

Solving the ODE for \(X(x)\)

To solve the first ODE, we will consider three cases: \(\lambda > 0\), \(\lambda = 0\), and \(\lambda < 0\). After solving the ODEs, we will apply the boundary conditions to find which lambda values are valid. If \(\lambda > 0\), we can let \(\lambda = k^2\) for some \(k > 0\). Then the ODE for \(X(x)\) becomes: $$X''(x) = -k^2 X(x)$$ The general solution is given by: $$X(x) = A\cos(kx) + B\sin(kx)$$ Applying boundary conditions: $$X(0) = A\cos(0) = 0 \Rightarrow A = 0$$ $$X(\pi) = B\sin(k\pi) = 0 \Rightarrow k = n$$ Where \(n\) is a positive integer. So \(\lambda = n^2\), and the solution is \(X_n(x) = \sin(nx)\)

Solving the ODE for \(Y(y)\)

We know \(\lambda = n^2\) for \(n = 1, 2, 3, \dots\), so the ODE for \(Y(y)\) is $$Y''(y) = n^2 Y(y)$$ The general solution is given by: $$Y_n(y) = C_n\cosh(ny) + D_n\sinh(ny)$$ Applying the boundary condition \(Y(0) = 0\), we get: $$C_n \cosh(0) = 0 \Rightarrow C_n = 0$$ So $$Y_n(y) = D_n\sinh((ny)$$

Formulating the General Solution

The general solution for \(u(x, y)\) is given by the sum of products of separated solutions over all \(n\): $$u_n(x, y) = X_n(x)Y_n(y) = (\sin(nx))(D_n\sinh(ny))$$ and $$u(x, y) = \sum_{n=1}^\infty X_n(x)Y_n(y) = \sum_{n=1}^{\infty}D_n\sin(nx)\sinh(ny)$$

Finding Coefficients \(D_n\)

We now need to find the coefficients \(D_n\) by applying the final boundary condition \(u(x, 0) = x\sin x\). From our general solution, we get: $$u(x, 0) = \sum_{n=1}^\infty D_n\sin(nx)\sinh(0) = 0$$ Which means $$x\sin x = \sum_{n=1}^{\infty}D_n\sin(nx)$$ To find \(D_n\), we can take the inner product (or use orthogonality) on both sides with respect to \(\sin(mx)\) for \(m = 1, 2, 3, \dots\): $$\int_{0}^{\pi} x\sin x \sin(mx) dx = \sum_{n=1}^{\infty}D_n \int_{0}^{\pi} \sin(nx)\sin(mx) dx$$ The right side of the equation simplifies due to orthogonality: $$\sum_{n=1}^{\infty} D_n \int_{0}^{\pi} \sin(nx)\sin(mx) dx = D_m \frac{\pi}{2}$$ So $$\int_{0}^{\pi} x\sin x \sin(mx) dx = D_m \frac{\pi}{2}$$ And finally: $$D_m = \frac{2}{\pi}\int_{0}^{\pi} x\sin x \sin(mx) dx$$

Final Solution

Now that we have \(D_m\), we can write down the final solution for \(u(x, y)\): $$u(x, y) = \sum_{n=1}^{\infty}\left(\frac{2}{\pi}\int_{0}^{\pi} x\sin x \sin(nx) dx\right)\sin(nx)\sinh(ny)$$

Key concepts

Differential Equations

Differential equations are mathematical tools that describe the relationship between a function and its derivatives. They express how the rate of change of a quantity is affected by the quantity itself and are essential in modeling various physical phenomena, ranging from the motion of planets to the flow of heat. An example of a differential equation is Newton's second law of motion that relates force, mass, and acceleration, which is the second derivative of position with respect to time.

In our textbook example, we deal with a partial differential equation (PDE) known as the Laplace equation, represented as \(u_{xx} + u_{yy} = 0\), where \(u_{xx}\) and \(u_{yy}\) denote the second partial derivatives of the function \(u(x, y)\) with respect to \(x\) and \(y\), respectively. The equation indicates that the sum of the second partial derivatives in both spatial dimensions \(x\) and \(y\) is zero, a situation commonly encountered in steady-state heat distribution and electrostatics.

Separation of Variables

Separation of variables is a method to solve differential equations where the solution is expressed as the product of functions, each in independent variables. The method works by transforming a PDE into simpler, separate ordinary differential equations (ODEs) that are easier to solve.

In our exercise's step-by-step solution, the given PDE \(u_{xx} + u_{yy} = 0\) gets simplified to \(X''(x)Y(y) + X(x)Y''(y) = 0\) by assuming \(u(x, y) = X(x)Y(y)\). The separation process involves dividing everything by \(X(x)Y(y)\), leading to an equation where each side only depends on one variable, allowing for the integration of terms involving \(x\) and \(y\) separately.

Ordinary Differential Equations (ODEs)

Ordinary differential equations (ODEs) are equations involving derivatives of a function of a single independent variable. In comparison to PDEs, which involve partial derivatives with respect to multiple independent variables, ODEs are generally simpler to solve.

The separation of variables in our exercise reduces the original PDE to two ODEs for the functions \(X(x)\) and \(Y(y)\), namely \(X''(x) = -\lambda X(x)\) and \(Y''(y) = \lambda Y(y)\), with \(\lambda\) being a separation constant. The ODEs are solved considering different cases for the constant \(\lambda\), and solutions are determined using given boundary conditions.

Fourier Series

Fourier series is a powerful mathematical tool used for expressing a periodic function as a sum of simple sine and cosine functions. It is extensively used in solving boundary value problems where solutions are required to satisfy certain conditions at the boundaries of a domain.

In the context of our boundary value problem, once the separate solutions for \(X_n(x)\) and \(Y_n(y)\) are identified, a Fourier series is used to construct the overall solution \(u(x, y)\). Each term of the series is a product of a sine function in \(x\) and a hyperbolic sine function in \(y\), multiplied by a coefficient \(D_n\). These coefficients are found by fitting the series to the boundary condition \(u(x, 0) = x\sin(x)\). The application of the Fourier series leads to the final form of the solution, taking into account the orthogonality of \(\sin(nx)\) functions, which simplifies the computation of \(D_n\) coefficients.