5.7 The Method Of Frobenius
Given the equation \(p(x)y'' + q(x)y' + r(x)y = 0\). If \(x_0\) is a singular point of the ODE and if the
limits \(\lim \limits _{x \rightarrow x_0}\big (x -x_0\big ) \dfrac {q(x)}{p(x)}\) and \(\lim \limits _{x \rightarrow x_0}\big (x -x_0\big )^2\hspace {0.1cm} \dfrac {r(x)}{p(x)}\) both exist and are finite, then we say that \(x_0\) is a singular point.
Theorem 5.7.1 (Method Of Frobenius).
Suppose \(p(x)y'' + q(x)y' + r(x)y = 0\) has a regular singular point \(x =0\), then there exists at least one solution of the form \[y(x) = x^r \sum ^{\infty } _{k = 0} a_k x^k\] In this case , the solution is \[y(x) = \sum ^{\infty }_{k=0} a_k x^{k+r}\] Substituting this \(y\) into the equation and setting the coefficient of the lowest power of \(x\) to zero gives the indicial equation, a quadratic in \(r\). Its roots decide which of three cases applies, and the three differ only in how the second solution has to be built when the first does not supply one.
- 1.
- If the indicial equation has two real roots \(r_1\) and \(r_2\) such that \(r_1 -r_2\) is not an integer, then \begin {align*} y_1 & = x^{r_1} \sum ^{\infty }_{k = 0} a_k x^k\\\\ y_2 & = x^{r_2} \sum ^{\infty }_{k=0} a_k x^k\\ \end {align*}
- 2.
- If the equation has a double root \(r_1\) then \begin {align*} y_1 & = x^r \sum ^{\infty }_{k = 0} a_k x^k\\\\ y_2 & = x^r \sum ^{\infty }_{k = 0} b_kx^k + \big (\ln x\big ) y_1\\ \end {align*}
- 3.
- If \(r_1 - r_2\) is a non-zero integer, with \(r_1 > r_2\), then \begin {align*} y_1 & = x^{r_1} \sum ^{\infty }_{k = 0} a_k x^k\\\\ y_2 & = x^{r_2} \sum ^{\infty }_{k=0} b_kx^k + C\big (\ln (x)\big ) y_1\\ \end {align*}
Solve the equation \(4x^2 y'' - 4x^2 y' + (1 -2x)y = 0\)
So we have that \(p(x) = 4x^2 , r(x) = 1-2x\) and \(q(x) = -4x^2\).
We check if \(x =0\) is a regular singular point.
\[\lim _{x \rightarrow 0} x \frac {q(x)}{p(x)} = \lim _{x \rightarrow 0} x\Bigg (\frac {-4x^2}{4x^2}\Bigg ) = 0\]
\[\lim _{x \rightarrow 0} x^2 \frac {r(x)}{p(x)} = \lim _{x \rightarrow 0} x^2 \Bigg ( \frac {1-2x}{4x^2}\Bigg ) = \frac {1}{4}\]
Both limits exist and so \(x = 0\) is a regular singular point. Hence the solution is \(\displaystyle {y(x) = x^r \sum ^{\infty }_{k=0} a_kx^k}\).
\begin {align*} y(x) & = \sum ^{\infty }_{k =0} a_kx^{k + 1}\\\\ y'(x) & = \sum ^{\infty }_{k = 0} a_k (k+r) x^{k+r -1}\\\\ y''(x) & = \sum ^{\infty }_{k = 0} a_k (k+r) (k+r -1) x^{k+r -2} \end {align*}
\[\implies \hspace {0.0cm} 4x^2 \sum ^{\infty }_{k = 0}a_k (k+r) (k + r - 1) x^{k + r -2} - 4x^2 \sum ^{\infty }_{k = 0} a_k (k+1) x^{k + r -1} + (1-2x) \sum ^{\infty }_{k = 0} a_k x^{k + r} = 0\]
\[\implies \hspace {0.0cm} \sum ^{\infty }_{k = 0}4(k+ r)(k+r - 1) a_k x^{k+r} - \sum ^{\infty }_{k = 0} 4 a_k (k+r) x^{k + r + 1} + \sum ^{\infty }_{k = 0} a_k x^{k+ r} - \sum ^{\infty }_{k = 0}2a_k x^{k + r + 1} = 0\]
\[ \sum ^{\infty }_{k = 0}\big [ 4(k+r)(k + r -1) + 1\big ] a_k x^{k + r} - \sum ^{\infty }_{k = 0}\big [4(k+r) + 2\big ] a_k x^{k+r + 1} = 0\]
\[ \sum ^{\infty }_{k = 0}\big [4(k+r) (k+r-1) + 1\big ] a_kx^{k +r} - \sum ^{\infty }_{k = 1}\big [4(k+r - 1 ) + 2\big ] a_{k -1}x^{k + r} = 0\]
\[\implies \hspace {0.0cm} \big [4r(r-1)+1\big ]a_0x^r + \sum ^{\infty }_{k = 1}\big [4(k+r) (k+r-1) + 1\big ] a_kx^{k +r} - \sum ^{\infty }_{k = 1}\big [4(k+r - 1 ) + 2\big ] a_{k -1}x^{k + r} = 0\]
\begin {align*} 4r(r-1) + 1 & = 0\\ 4r^2 - 4r + 1 & = 0\\ \implies r & = \frac {1}{2} \end {align*}
\[\big [4(k+r)(k+r-1) + 1\big ] a_k - \big [4(k + r -1) + 2\big ]a_{k-1} = 0\]
\[\implies \hspace {1cm} a_k = \frac {\big [4(k + r -1) + 2\big ]}{\big [4(k+r)(k+r-1) + 1\big ]}\hspace {0.2cm}a_{k-1}\]
And replacing for \(r= \dfrac {1}{2}\), we get \[a_k = \frac {4k + 2 - 4 + 2}{4 \Bigg (k+ \dfrac {1}{2}\Bigg )\Bigg ( k - \dfrac {1}{2}\Bigg ) + 1}\hspace {0.1cm} a_{k-1}\]
\[a_k = \frac {4k}{4k^2 -1 + 1}\hspace {0.1cm} a_{k-1}\]
\[a_k = \frac {1}{k}\hspace {0.1cm}a_{k-1}\hspace {0.1cm},\hspace {0.3cm} k = 1,2,3,\cdot \cdots \cdots \]
\begin {align*} a_1 = a_0\hspace {0.1cm}, \hspace {0.4cm} a_2 & = \frac {1}{2}a_1 = \frac {1}{2}a_0\hspace {0.2cm},\\\\ a_3 & = \frac {1}{3}a_2 = \frac {1}{2(3)}a_0\\\\ a_4 & = \frac {1}{(2)(3)(4)}a_0 \end {align*}
\begin {align*} y_1(x) & = x^{1/2} \bigg (a_0 + a_0x + \frac {1}{2!}a_0x^2 + \frac {1}{3!}a_0 x^3 + \cdots \cdots \cdot \bigg )\\ & = a_0x^{1/2}\sum ^{\infty }_{n =0} \frac {x^n}{n!}\\ & = a_0 x^{1/2}e^x \end {align*}
Set \(a_0 = 1,\hspace {0.4cm} y_1 = x^{1/2}e^x\)
\[y_2 = \sum ^{\infty }_{n=0} b_n x^{n+ 1/2} + \big (\ln x\big ) x^{1/2} e^x\]
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