Problem 17 - Advanced Engineering Mathematics

Question

\(x=0\) is a regular singular point of the given differential equation. Show that the indicial roots of the singularity do not differ by an integer. Use the method of Frobenius to obtain two linearly independent series solutions about \(x=0\). Form the general solution on the interval \((0, \infty)\). $$ 4 x y^{\prime \prime}+\frac{1}{2} y^{\prime}+y=0 $$

Step-by-Step Solution

Answer

Roots are complex and do not differ by an integer; general solution is a combination of two series solutions.

Step 1: Rewrite the Differential Equation

First, we need to bring the differential equation into the standard form for checking a regular singular point. The given equation is \[4x y'' + \frac{1}{2} y' + y = 0.\]Divide through by the leading term coefficient (4x) to standardize it:\[y'' + \frac{1}{8x}y' + \frac{1}{4x}y = 0.\]

Step 2: Identify the Function Coefficients

Now we identify the functions:- \(P(x) = \frac{1}{8x}\)- \(Q(x) = \frac{1}{4x}\).

Step 3: Check for Regular Singular Point

A regular singular point requires that \((x - x_0)P(x)\) and \((x - x_0)^2Q(x)\) are analytic at \(x = x_0 = 0\).Calculate:- \((x - x_0)P(x) = x \times \frac{1}{8x} = \frac{1}{8}\)- \((x - x_0)^2Q(x) = x^2 \times \frac{1}{4x} = \frac{x}{4}.\)Both are analytic at \(x = 0\), proving \(x = 0\) is a regular singular point.

Step 4: Obtain the Indicial Equation

We use the Frobenius method by assuming a solution of the form \(y = x^r \sum_{n=0}^{\infty}a_nx^n\).Substitute into the equation and equate powers of \(x\).Analyzing terms, the indicial equation arises from the lowest power of \(x\), usually calculated by plugging the series into the differential equation. Solve the equations that stem from the lowest powers to find \(r\).Calculating, we find:\[4r(r-1)a_0 + \frac{1}{2} ra_0 + a_0 = 0.\]

Step 5: Solve the Indicial Equation

Simplify the indicial equation:\[4r^2 - 3r + 1 = 0.\]Use the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve,where \(a = 4, b = -3, c = 1\).Calculate:\[r = \frac{3 \pm \sqrt{9 - 16}}{8} = \frac{3 \pm \sqrt{-7}}{8}.\]Since the discriminant is negative, the roots are complex, and thus they do not differ by an integer.

Step 6: Frobenius Solution Construction

Since the roots do not differ by an integer, we can construct two series solutions from the general form. Each solution will begin with a different \(r\) from the roots found. Compute the corresponding power series.Let \(r_1 = \frac{3 + i\sqrt{7}}{8}\) and \(r_2 = \frac{3 - i\sqrt{7}}{8}\) and write:\[y_1(x) = x^{r_1} \sum_{n=0}^{\infty} a_n x^n,\]\[y_2(x) = x^{r_2} \sum_{n=0}^{\infty} b_n x^n.\]

Step 7: Form the General Solution

The general solution is formed by combining the two linearly independent solutions. Thus, we have:\[y(x) = C_1 y_1(x) + C_2 y_2(x),\]where \(C_1\) and \(C_2\) are constants determined by boundary conditions, valid on the interval \((0, \infty)\).

Key Concepts

Regular Singular Point

In differential equations, a regular singular point is where the behavior of solutions may undergo a shift. However, unlike an irregular singular point where things become a bit chaotic, a regular singular point tends to keep things under control but introduces interesting features in the solutions. For an equation having a regular singular point, we typically express it in the form:\[y'' + P(x)y' + Q(x)y = 0,\]where the functions - \((x-x_0)P(x)\) is analytic at \(x_0\), and- \((x-x_0)^2Q(x)\) is also analytic at \(x_0\).This ensures we categorize \(x_0\) as a regular singular point. The key is changing the equation's form to match these conditions, allowing the use of the Frobenius method to explore solutions. For example, with the equation \(4xy'' + \frac{1}{2}y' + y = 0\), transforming it into the needed format helped us see \(x = 0\) is a regular singular point.

Indicial Equation

When working with a regular singular point, the next step is defining the indicial equation. This equation is crucial as it helps determine the exponents for the series solutions in Frobenius's method. It comes from substituting a trial solution of the form:\[y = x^r \sum_{n=0}^{\infty} a_n x^n,\]and focusing on the lowest power of \(x\) terms in the differential equation. These terms usually involve the initial coefficient \(a_0\) aiding in framing the indicial equation, something like \[4r(r-1)a_0 + \frac{1}{2}ra_0 + a_0 = 0.\]Solving this grants the values of \(r\), which guides the rest of the series solution. The solutions to this indicial equation not only give us insights into the nature of solutions but also into whether we'll face complex roots or not, affecting the type of series solutions we can develop.

Series Solution

Once the indicial equation is solved, we use its roots to craft the series solution, crucial in handling differential equations at a regular singular point special case. The Frobenius method supports building this solution even when dealing with complex situations. We start with roots found from the indicial equation. Construct series solutions of the form:\[y_1(x) = x^{r_1} \sum_{n=0}^{\infty} a_n x^n,\]\[y_2(x) = x^{r_2} \sum_{n=0}^{\infty} b_n x^n.\]Here, \(r_1\) and \(r_2\) are roots computed earlier. We calculate different terms around these roots to build up power series solutions. These solutions involve adding up various power terms and often demand computations of coefficients like \(a_n\) or \(b_n\), ensuring each series satisfies the original differential equation. Usually, two linearly independent solutions arise unless the roots differ by an integer, opening up general solutions to these equations.

Complex Roots

Sometimes when solving the indicial equation, we encounter complex roots. These complex roots bring a fascinating twist to the solution process. In the case of the differential equation provided, the quadratic we dealt with led to:\[r = \frac{3 \pm i\sqrt{7}}{8}.\]These roots are indeed complex due to the negative discriminant, \(b^2 - 4ac = -7\).Complex roots imply solutions are inherently different. They mean we can't simply express our series solutions using real exponents but need a more creative approach. In many situations, this leads to involving complex numbers where we need- Euler's formula,- properties of exponential functions,and more mathematical tools to express the final solutions fully. While complex roots make the process intricate, they encourage a deeper dive into analysis techniques, unraveling richer solution patterns not present with simple real roots.