# when to use frobenius method

\end{align} \label{ex1eq3}\], First, $$a_0 = 0$$. If r 1−r 2 ∈ Z, then only r = max{r Q4 (5 points) Use the method of Frobenius to find the general solution to the differential equation: z?y" + xy' – 2y = 0 Note: The power series involved will have only finitely many non-zero terms (i.e., it … We can use the method of Frobenius whenever we have a differential equation that meets the conditions described above. But perhaps we can combine the methods. It can be shown that the power series in a Frobenius solution of ( eq:7.5.1 ) converges on some open interval , where . An important class of functions that arises commonly in physics are the Bessel functions. Let y=Ún=0 ¥a xn+r. In the following we solve the second-order differential equation called the hypergeometric differential equation using Frobenius method, named after Ferdinand Georg Frobenius. Suppose that $p(x) y'' + q(x) y' + r(x) y = 0$ has a regular singular point at $$x=0$$, then there exists at least one solution of the form $y = x^r \sum_{k=0}^\infty a_k x^k .$ A solution of this form is called a Frobenius-type solution. Let us try, \begin{align} y &= x^r \sum_{k=0}^\infty a_k x^k \\ &= \sum_{k=0}^\infty a_k x^{k+r} , \label{ex2eq2} \end{align}, where $$r$$ is a real number, not necessarily an integer. (Notice that A 0 = 0 is a constant multiple of the indicial equation r(r 1) + p 0r + q 0 = 0). The Set-Up The Calculations and Examples The Main Theorems Inserting the Series into the DE Getting the Coe cients Observations Roots Di ering by a Positive Integer Here we have r 1 =r 2 +N for some positive integer N . First we must define the gamma function, $\Gamma(x) = \int_0^\infty t^{x-1} e^{-t} \, dt .$, Notice that $$\Gamma(1) = 1$$. Example $$\PageIndex{3}$$: Expansion around a regular singular point, Often, and for the rest of this section, $$x_0 = 0$$. Method of Frobenius. The gamma function also has a wonderful property. Method of Frobenius. Question: Using the Frobenius method, find a basis of solutions of the ODE {eq}\; x^2 {y}'' + 6x{y}' + (4x^2 + 6)y = 0 {/eq}. The Method of Frobenius Step 2: Set A 0 = A 1 = A 2 = = 0. For negative b there are no solutions. can be changed to $$x^2 y'' + x y' + \lambda^2 x^2 y = 0$$. In mathematics, the Method of Frobenius, named for Ferdinand Georg Frobenius, is a method to nd an innite series solution for a second-order ordinary dierential equation of the form x2y00+p(x)y0+q(x)y= 0 with y0dy dx and y00d2y dx2in the vicinity of the regular singular point x= 0. Supposing that $$a_0 \not= 0$$ we obtain, This equation is called the indicial equation. Let us now discuss the general Method of Frobenius. x^2 y double prime + 6x*y prime + (4x^2 + 6)y = 0. \], Bessel functions will be convenient constant multiples of $$y_1$$ and $$y_2$$. We obtain our two linearly independent solutions by taking the real and imaginary parts of $$y$$. .\], $y =\sum_{k=0}^\infty a_k x^{k+r}=\sum_{k=0}^\infty \dfrac{1}{k!} x��\M�7r��f����&�ym~,��Q����d�-ـ4�dyY寧��*������Y ��fU�Y�MV���fv��C��xs�웼���b�W���O�쥩\vvO��kk�u��KA���.�=�����/�_>��f/��;�����=�b��˗϶�/o/fs��V�f��s�ۻ�?n��=n���l�|�wb��lo.�l�/hzW}����� Ly = 2x2y00 ¡xy0 +(1+x)y = 0 x = 0 is a RSP. The constants are picked for convenience. y = X1 n=0 anx n+r (4.16) Then we write the general solution as $$y = A y_1 + B y_2$$. Consequently our solution---if we can find one---may only make sense for positive $$x$$. (v) If the indicial equation has two real roots such that $$r_1-r_2$$ is an integer, then one solution is, and the second linearly independent solution is of the form, \[y_2 = x^{r_2} \sum_{k=0}^\infty b_k x^{k} + C (\ln x) y_1 ,$. \label{ex2eq3} \end{align} \], Plugging Equations \ref{ex2eq2} - \ref{ex2eq3} into our original differential equation (Equation \ref{ex2eq1}) we obtain, \begin{align} 0 &= 4x^2y''-4x^2y'+(1-2x)y \\ &= 4x^2 \, \left( \sum_{k=0}^\infty (k+r)\,(k+r-1) \, a_k x^{k+r-2} \right)-4x^2 \, \left( \sum_{k=0}^\infty (k+r) \, a_k x^{k+r-1} \right)+(1-2x) \left( \sum_{k=0}^\infty a_k x^{k+r} \right) \\ &=\left( \sum_{k=0}^\infty 4 (k+r)\,(k+r-1) \, a_k x^{k+r} \right)-\left( \sum_{k=0}^\infty 4 (k+r) \, a_k x^{k+r+1} \right)+\left( \sum_{k=0}^\infty a_k x^{k+r} \right)-\left( \sum_{k=0}^\infty 2a_k x^{k+r+1} \right) \\ &=\left( \sum_{k=0}^\infty 4 (k+r)\,(k+r-1) \, a_k x^{k+r} \right)-\left( \sum_{k=1}^\infty 4 (k+r-1) \, a_{k-1} x^{k+r} \right)+\left( \sum_{k=0}^\infty a_k x^{k+r} \right)-\left( \sum_{k=1}^\infty 2a_{k-1} x^{k+r} \right)\\ &=4r(r-1) \, a_0 x^r + a_0 x^r +\sum_{k=1}^\infty \left( 4 (k+r)\,(k+r-1) \, a_k - 4 (k+r-1) \, a_{k-1}+a_k -2a_{k-1} \right) \, x^{k+r}\\ &=\left( 4r(r-1) + 1 \right) \, a_0 x^r +\sum_{k=1}^\infty\left( \left( 4 (k+r)\,(k+r-1) + 1 \right) \, a_k-\left( 4 (k+r-1) + 2 \right) \, a_{k-1} \right) \, x^{k+r} .\end{align}, To have a solution we must first have $$\left( 4r(r-1) + 1 \right) \, a_0 = 0$$. Finally, we can formulate the method of Frobenius series as follows. 2n 2, so Frobenius’ method fails. When $$p$$ is not an integer, $$J_p$$ and $$J_{-p}$$ are linearly independent. Does the Frobenius method work for all second order linear differential equations with only regular singular points? stream Previous question Next question Transcribed Image Text from this Question. On the other hand if we make the slight change, $\lim_{x \to 0} ~x \dfrac{q(x)}{p(x)} =\lim_{x \to 0} ~x \dfrac{(1+x)}{x^2} = \lim_{x \to 0} ~\dfrac{1+x}{x} =\text{DNE}.$. The method we will use to find solutions of this form and other forms that we’ll encounter in the next two sections is called the method of Frobenius, and we’ll call them Frobenius solutions. In this section we define ordinary and singular points for a differential equation. But what about a second solution? The main idea is the following theorem. Algebra-equation.com provides practical strategies on online solver frobenius, syllabus for intermediate algebra and multiplying and dividing rational and other math topics. We also show who to construct a series solution for a differential equation about an ordinary point. L. Nielsen, Ph.D. for arbitrary constants $$A$$ and $$B$$. (i)Given the equation (14) with a regular singular point at x= , solve the indicial equation (18) and nd possible values for r. Note that if we required the normalization ~a( ) = 1 from the beginning, the indicial equation would have been r2 + ~b( ) 1 Therefore we only get the trivial solution $$y=0$$. Have questions or comments? If, furthermore, the limits, $\lim_{x \to x_0} ~ (x-x_0) \dfrac{q(x)}{p(x)} \qquad \text{and} \qquad \lim_{x \to x_0} ~ (x-x_0)^2 \dfrac{r(x)}{p(x)}$. If we only get one, we either use the ideas above or even a different method such as reduction of order (Exercise 2.1.8) to obtain a second solution. also Fuchsian equation). But the power series solutions is limited to series whose exponents are non-negative integers. <> What we will do is to try a solution of the form, $4 x^2 y'' - 4 x^2 y' + (1-2x)y = 0, \label{ex2eq1}$, and again note that $$x=0$$ is a singular point. (i)Given the equation (14) with a regular singular point at x= , solve the indicial equation (18) and nd possible values for r. Note that if we required the normalization ~a( ) = 1 from the beginning, the indicial equation would have been r2 + ~b( ) 1 In this _|���E���'�4�к��p����p�@�cY�X5�ك�0�Q�#���Q{n,\��Q�Y ���%��{=&�����x�\�K� �~�.� ������ˮ�e�X��'8L,T OK, so we know what $$r$$ has to be. Using the first root, we plug in, $y_1 = x^{r_1} \sum_{k=0}^\infty a_k x^{k} ,$, and we solve for all $$a_k$$ to obtain the first solution. Suppose that $$p$$ is not an integer. 6Named after the German astronomer and mathematician Friedrich Wilhelm Bessel (1784 – 1846). This equation comes up for example when finding fundamental modes of vibration of a circular drum, but we digress. Why do I have to use Frobenius method in Bessel's equation? Use the Frobenius method to determine the general power series solution of the differential equation: – Let a trial solution be of the form y = a0xc + a 1x c+1 + a 2x c+2 + a 3x c+3 + … + a rx c+r + … – Differentiating equation to obtain y’, y’’ – Substituting y and y’’ into each term of the given equation That knowledge we obtained simply by looking at the coefficient of $$x^r$$. 0. results in a complex-valued function---all the $$a_k$$ are complex numbers. Furthermore, these functions oscillate, although they are not periodic. 5See Joseph L. Neuringera, The Frobenius method for complex roots of the indicial equation, International Journal of Mathematical Education in Science and Technology, Volume 9, Issue 1, 1978, 71–77. Watch the recordings here on Youtube! We have one solution, let us call it $$y_1 = x^{1/2} e^x$$. We define the Bessel functions of the first kind of order $$p$$ and $$-p$$ as, $J_p(x) = \dfrac{1}{2^p\Gamma(1+p)} y_1=\sum_{k=0}^\infty \dfrac{{(-1)}^k}{k! It also turns out that $$Y_n(x)$$ and $$J_n(x)$$ are linearly independent. both exist and are finite, then we say that $$x_0$$ is a regular singular point. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Let y=Ún=0 ¥a xn+r. The reader can (and should) try this to obtain for example the first three terms, \[b_1 = b_0 -1 , \qquad b_2 = \dfrac{2b_1-1}{4} , \qquad b_3 = \dfrac{6b_2-1}{18} , \qquad \ldots$. 2. Other equations can sometimes be solved in terms of the Bessel functions. This is a method that uses the series solution for a differential equation, where we assume the solution takes the form of a series. We then fix $$b_0$$ and obtain a solution $$y_2$$. For example, these functions appear when solving the wave equation in two and three dimensions. In the Frobenius Method we look for the solution in the form y = [[infinity].summation over (n=0)] [a.sub.n][x.sup.n+r]. So, \begin{align} 0 &= 2 x y' - y \\ &= 2 x r x^{r-1} - x^r\\&= (2r-1) x^r \end{align}., Therefore $$r= \dfrac{1}{2}$$, or in other words $$y = x^{1/2}$$. Let us try $$y=x^r$$ for some real number $$r$$. A similar method of solution can be used for matrix equations of the first order, too. (k-p)(k-1-p) \cdots (2-p)(1-p)}. In particular there are three questions in my text book that I have attempted. Answer to: Using the Frobenius method, determine a basis of solutions of the ODE. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Therefore when $$n$$ is an integer, we have the general solution to Bessel's equation of order $$n$$. Verify that the indicial equation of Bessel's equation of order $$p$$ is $$(r-p)(r+p)=0$$. L. Nielsen, Ph.D. Use Frobenius method to find the solutions of the following ODES 1. ry" +y' – 12y = 0 The point $$0$$ is a singular point, but not a regular singular point. It turns out we want to try for another solution of the form, $y_2 = \sum_{k=0}^\infty b_k x^{k+r} + (\ln x) y_1 ,$, $y_2 = \sum_{k=0}^\infty b_k x^{k+1/2} + (\ln x) x^{1/2} e^x .$, We now differentiate this equation, substitute into the differential equation and solve for $$b_k$$. These make sense only for integer orders $$n$$ and are defined as limits of linear combinations of $$J_p(x)$$ and $$J_{-p}(x)$$ as $$p$$ approaches $$n$$ in the following way: $Y_n(x) = \lim_{p\to n} \dfrac{\cos(p \pi) J_p(x) - J_{-p}(x)}{\sin(p \pi)} .$, As each linear combination of $$J_p(x)$$ and $$J_{-p}(x)$$ is a solution to Bessel's equation of order $$p$$, then as we take the limit as $$p$$ goes to $$n$$, $$Y_n(x)$$ is a solution to Bessel's equation of order $$n$$. Missed the LibreFest? Examples open all close all. A long computation ensues and we obtain some recursion relation for $$b_k$$. In the following we solve the second-order differential equation called the hypergeometric differential equation using Frobenius method, named after Ferdinand Georg Frobenius. (Compiled 27 January 2018) In this lecture we will consider the Frobenius series solution of the Bessel equation, which arises during the process of separation of … See Figure 7.4 for graphs of Bessel functions. The Frobenius equation is the Diophantine equation , where the a i are positive integers, b is an integer, and a solution must consist of non-negative integers. }{\left(\dfrac{x}{2}\right)}^{2k+n} .\]. If we are lucky and find two, we are done. Step 3: Use the system of equations For example, given a positive constant $$\lambda$$. The Method of Frobenius We now approach the task of actually finding solutions of a second-order linear dif ferential equation near the regular singular point x = 0. Picking $$a_0$$ to be a different constant only gets us a constant multiple of $$y_1$$, and we do not have any other $$r$$ to try; we only have one solution to the indicial equation. 0 m r m m x a y 0 1 m r m m x a r m y 0 2 1 m r m m x a r m r m y )o�Ʈvð��m���ծv� %��x��ծvK̡=����n����������/�0.����ߋ7����Ÿלcٞ���W����E�pT�+h�\�����cL����`wSk ��7�G���C��j��t{�����w��_�2��[ڝ1�O�o�x>㕉͒�0��s�ڙ�?�J����Rt;x}�ӟna��\$� �;���࡯���n�T0f��4L��,9�u�ݺ�����Vp��/n_���g�iw�|�>9�ː����o�p Setting the first coefficient (usually the coefficient of $$x^r$$) in the series to zero we obtain the indicial equation, which is a quadratic polynomial in $$r$$. Let us first look at a simple first order equation, Note that $$x=0$$ is a singular point. The method we will use to find solutions of this form and other forms that we’ll encounter in the next two sections is called the method of Frobenius, and we’ll call them Frobenius solutions. It can be shown that the power series in a Frobenius solution of ( eq:7.5.1 ) converges on some open interval , where . This is a method that uses the series solution for a differential equation, where we assume the solution takes the form of a series. (k+p)(k-1+p) \cdots (2+p)(1+p)}, \\ y_2= x^{-p} \sum_{k=0}^{\infty} \dfrac{(-1)^k x^{2k}}{2^{2k}k! The other solution is the so-called Bessel function of second kind. In the Frobenius method one examines whether the equation (2) allows a series solution of the form y ⁢ ( x ) = x s ⁢ ∑ n = 0 ∞ a n ⁢ x n = a 0 ⁢ x s + a 1 ⁢ x s + 1 + a 2 ⁢ x s + 2 + … , Finally, we can formulate the method of Frobenius series as follows. Let, be an ODE. Multiplying by a constant, the general solution for positive $$x$$ is. In section 4 we use the Frobenius Method to solve differential equations in the neighborhood of a singular regular point. Say you have (c-2)y'' +y' = 0 and you are asked to solve a) the equation about the point x=0 b) the equation about the point x=2 When do you use frobenius over power series. $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, 7.3: Singular Points and the Method of Frobenius, [ "article:topic", "targettag:lower", "method of Frobenius", "Bessel functions", "regular singular point", "gamma function", "Frobenius-type solution", "indicial equation", "authorname:lebl", "showtoc:no" ], https://math.libretexts.org/@app/auth/2/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FDifferential_Equations%2FBook%253A_Differential_Equations_for_Engineers_(Lebl)%2F7%253A_Power_series_methods%2F7.3%253A_Singular_Points_and_the_Method_of_Frobenius, $$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, The Frobenius method for complex roots of the indicial equation. y = X1 n=0 anx n+r (4.16) yields the solution y(x) = ce1=x, which could not be captured by a Frobenius expansion about x0 = 0. 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