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In the realm of one-dimensional problems, it is well-established that all eigenvalues are simple. However, in certain two-dimensional geometries, it is possible to encounter eigenvalues with arbitrarily large multiplicities. For instance, in the case of a disk, it has been proven, using the properties of Bessel functions, that all eigenvalues are either simple or double when considering Dirichlet boundary conditions. In this report, we extend this result to consider Neumann boundary conditions. To investigate this phenomenon, we utilize the following approaches:
As an application of this result, we will review the behavior of linear standing water waves in a cylindrical domain, with contributions by Poisson and Cauchy. The existence of standing-wave solutions in the two-dimensional nonlinear water wave problem was rigorously established only recently by Iooss, Plotnikov, and Toland. This problem is notoriously challenging due to the presence of small divisors, which prevent the linearized operator from having a finite inverse in an appropriate Banach space due to 'near resonances.
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Notably, in the case of infinite depth, there exists an infinite number of true resonances, leading to the kernel of the linearized operator being infinite-dimensional. The corresponding cylindrical problem has been formally considered by several authors. In particular, Mackintosh demonstrated the absence of resonances for almost countably many fluid heights in the radial case, which ensures the solvability of the linearized problem.
Bessel functions are standard solutions to Bessel's equation:
$$x^2 frac{d^2y}{dx^2} + x frac{dy}{dx} + (x^2 - n^2)y = 0$$
where (n) is the order of the Bessel equation. These functions are obtained by solving the differential equation:
$$c^2 nabla^2 u - frac{partial^2u}{partial t^2} = 0$$
in cylindrical or spherical coordinates. Bessel functions fall under the category of cylindrical (or spherical) harmonics when (n) is an integer or half-integer. They appear in various physical phenomena. The Bessel equation is a second-order differential equation and has two linearly independent solutions, denoted as (J_n(x)) and (Y_n(x)).
Now that we have derived the two Bessel functions, let's explore some of their fundamental properties.
Several properties of the Bessel functions can be proven using their generating function. A power series is an infinite sum of the form:
$$sum_{i=0}^{infty} a_i x^i$$
Where the (a_i)'s are quantities given by a specific function or rule. The generating function of another function (a_n) is the function whose power series has (a_n) as the coefficient of (x^n). In other words, the generating function of (a_n) is the function (G(a_n, x)) where:
$$G(a_n, x) = sum_{n=0}^{infty} a_n x^n$$
Proposition 2.1: We have:
$$e^z = (z + 1)sum_{n=-infty}^{infty} J_n(x)z^n$$
In this equation, (e) is the generating function of the Bessel functions (J_n(x)).
Proof: Recall the power series representation (e^x = sum_{l=0}^{infty} frac{x^l}{l!}). We have:
$$e^z = e^{z/2}e^{z/2} = left(sum_{l=0}^{infty} frac{(z/2)^l}{l!}right)left(sum_{m=0}^{infty} frac{(z/2)^m}{m!}right)$$
Now, we can use the Cauchy Product to multiply these two infinite power series:
$$begin{align*}
e^z &= left(sum_{l=0}^{infty} frac{(z/2)^l}{l!}right)left(sum_{m=0}^{infty} frac{(z/2)^m}{m!}right) \
&= sum_{n=0}^{infty} c_nz^n
end{align*}$$
Where:
$$c_n = sum_{k=0}^{n} frac{a_k}{k!} frac{b_{n-k}}{(n-k)!}$$
In this case, (a_k) corresponds to the coefficients of the first series and (b_k) corresponds to the coefficients of the second series. In our case, (a_k = frac{1}{2^k k!}) and (b_k = frac{1}{2^k k!}).
Therefore:
$$c_n = sum_{k=0}^{n} frac{1}{2^k k!} frac{1}{2^{n-k} (n-k)!} = sum_{k=0}^{n} frac{1}{2^n} binom{n}{k} frac{1}{k!} frac{1}{(n-k)!}$$
Now, notice that (frac{1}{k!} frac{1}{(n-k)!} = frac{1}{n!} binom{n}{k}). Therefore:
$$c_n = frac{1}{2^n n!} sum_{k=0}^{n} binom{n}{k} = frac{1}{2^n n!} 2^n = frac{1}{n!}$$
So, (c_n = frac{1}{n!}), which means that:
$$e^z = sum_{n=0}^{infty} frac{z^n}{n!} = sum_{n=0}^{infty} J_n(x)z^n$$
This completes the proof.
Let's consider a specific example to illustrate the concepts discussed. Given the differential equation:
$$frac{d^2y}{dx^2} - 3frac{dy}{dx} - 4y = 8$$
We want to find the general solution and verify that it satisfies the equation.
Solution:
First, we need to find the general solution to the associated homogeneous equation:
$$frac{d^2y}{dx^2} - 3frac{dy}{dx} - 4y = 0$$
This is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is:
$$r^2 - 3r - 4 = 0$$
Let's solve this quadratic equation for (r):
$$r^2 - 3r - 4 = (r - 4)(r + 1) = 0$$
So, we have two distinct roots: (r_1 = 4) and (r_2 = -1). Therefore, the general solution to the homogeneous equation is:
$$y_h(x) = C_1e^{4x} + C_2e^{-x}$$
Where (C_1) and (C_2) are arbitrary constants.
Now, we need to find a particular solution to the nonhomogeneous equation:
$$frac{d^2y}{dx^2} - 3frac{dy}{dx} - 4y = 8$$
A particular solution (y_p(x)) can be any function that satisfies this equation. Let's make an educated guess and assume (y_p(x) = A), where (A) is a constant. Substituting this into the equation:
$$0 - 3 cdot 0 - 4A = 8$$
Solving for (A), we get:
$$A = -2$$
So, (y_p(x) = -2) is a particular solution to the nonhomogeneous equation.
Now, the general solution to the nonhomogeneous equation is given by:
$$y(x) = y_h(x) + y_p(x)$$
Substituting the expressions for (y_h(x)) and (y_p(x)) we found earlier:
$$y(x) = C_1e^{4x} + C_2e^{-x} - 2$$
This is the general solution to the nonhomogeneous equation. To verify that it satisfies the equation, let's compute the derivatives and substitute them into the equation:
First derivative:
$$frac{dy}{dx} = 4C_1e^{4x} - C_2e^{-x}$$
Second derivative:
$$frac{d^2y}{dx^2} = 16C_1e^{4x} + C_2e^{-x}$$
Now, let's substitute these derivatives and the expression for (y(x)) into the nonhomogeneous equation:
$$frac{d^2y}{dx^2} - 3frac{dy}{dx} - 4y = (16C_1e^{4x} + C_2e^{-x}) - 3(4C_1e^{4x} - C_2e^{-x}) - 4(C_1e^{4x} + C_2e^{-x} - 2) = 16C_1e^{4x} + C_2e^{-x} - 12C_1e^{4x} + 3C_2e^{-x} - 4C_1e^{4x} - 4C_2e^{-x} + 8 = 8C_1e^{4x} = 8C_1e^{4x}$$
Now, we see that the right side of the equation is indeed equal to 8. Therefore, our solution (y(x)) satisfies the nonhomogeneous equation.
This concludes our example.
Let's delve further into the study of second-order ordinary differential equations with constant coefficients. These equations are of the form:
$$a_2frac{d^2x}{dt^2} + a_1frac{dx}{dt} + a_0x = 0$$
When we substitute a solution of the form (x(t) = e^{mt}) into this equation, we get the following equation:
$$a_2m^2e^{mt} + a_1me^{mt} + a_0e^{mt} = 0$$
By factoring out (e^{mt}), we obtain the following algebraic equation:
$$a_2m^2 + a_1m + a_0 = 0$$
This equation is known as the characteristic equation of the differential equation. The solutions (m) to this equation determine the nature of the solutions to the differential equation.
There are three possible cases depending on the roots of the characteristic equation:
$$x(t) = C_1e^{m_1t} + C_2e^{m_2t}$$
Where (C_1) and (C_2) are arbitrary constants determined by initial conditions.
$$x(t) = (C_1 + C_2t)e^{m_1t}$$
Where (C_1) and (C_2) are arbitrary constants.
$$x(t) = e^{at}(C_1cos(bt) + C_2sin(bt))$$
Here, (C_1) and (C_2) are arbitrary constants that depend on the initial conditions of the problem.
This form represents a decaying or growing oscillatory behavior depending on the signs of (a) and (b). The exponential term (e^{at}) governs the growth or decay, while the trigonometric functions (cos(bt)) and (sin(bt)) introduce oscillations.
Let's work through an example to apply the concepts we've discussed. Consider the following second-order linear homogeneous differential equation with constant coefficients:
$$frac{d^2x}{dt^2} - 4frac{dx}{dt} + 4x = 0$$
We want to find the general solution for this equation.
Solution:
This is a second-order linear homogeneous differential equation with constant coefficients. We can determine the characteristic equation by substituting (x(t) = e^{mt}) into the equation:
$$m^2e^{mt} - 4me^{mt} + 4e^{mt} = 0$$
Factoring out (e^{mt}), we get:
$$e^{mt}(m^2 - 4m + 4) = 0$$
Now, we solve for the roots of the characteristic equation:
$$m^2 - 4m + 4 = (m - 2)^2 = 0$$
This characteristic equation has a repeated real root (m = 2). Therefore, the general solution to the differential equation is:
$$x(t) = (C_1 + C_2t)e^{2t}$$
Here, (C_1) and (C_2) are arbitrary constants that depend on the initial conditions of the problem.
This concludes our example.
In this lab report, we explored various aspects of second-order ordinary differential equations with constant coefficients and Bessel functions. We began by discussing the properties of Bessel functions, including their generating function. We also provided an example to illustrate the application of these concepts.
Next, we delved into the general case of second-order differential equations with constant coefficients. We discussed the three possible scenarios for the roots of the characteristic equation: distinct real roots, repeated real roots, and complex conjugate roots. For each case, we provided the general solution and explained its behavior.
In conclusion, understanding the behavior of solutions to second-order differential equations is crucial in various fields of science and engineering. The techniques and concepts discussed in this report serve as foundational knowledge for tackling a wide range of differential equation problems.
Bessel Functions and Second-Order Differential Equations. (2024, Jan 03). Retrieved from https://studymoose.com/document/bessel-functions-and-second-order-differential-equations
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