6.2: Regular Singular Points - Cauchy-Euler Equations

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The value \(x = x_0\) is called a regular singular point of the ode \[\label{eq:1}(x-x_0)^2y''+p(x)(x-x_0)y'+q(x)y=0,\] if \(p(x)\) and \(q(x)\) have convergent Taylor series about \(x = x_0\), i.e., \(p(x)\) and \(q(x)\) can be written as a power-series in \((x − x_0)\):

\[\begin{array}{l}p(x)=p_0+p_1(x-x_0)+p_2(x-x_0)^2+\cdots , \\ q(x)=q_0+q_1(x-x_0)+q_2(x-x_0)^2+\cdots ,\end{array}\nonumber\] with \(p_n\) and \(q_n\) constants, and \(q_0\neq 0\) so that \((x − x_0)\) is not a common factor of the coefficients. Any point x = x0 that is not an ordinary point or a regular singular point is called an irregular singular point. Many important differential equations of physical interest have regular singular points, and their solutions go by the generic name of special functions, with specific names associated with now famous mathematicians like Bessel, Legendre, Hermite, Laguerre and Chebyshev.

Here, we will only consider the simplest ode with a regular singular point at \(x = 0\). This ode is called a Cauchy-Euler equation, and has the form \[\label{eq:2}x^2y''+\alpha xy'+\beta y=0,\] with \(\alpha\) and \(\beta\) constants. Note that \(\eqref{eq:1}\) reduces to a Cauchy-Euler equation (about \(x = x_0\)) when one considers only the leading-order term in the Taylor series expansion of the functions \(p(x)\) and \(q(x)\). In fact, taking \(p(x) = p_0\) and \(q(x) = q_0\) and solving the associated Cauchy-Euler equation results in at least one of the leading-order solutions to the more general ode \(\eqref{eq:1}\). Often, this is sufficient to obtain initial conditions for numerical solution of the full ode. Students wishing to learn how to find the general solution of \(\eqref{eq:1}\) can consult Boyce & DiPrima.

An appropriate ansatz for \(\eqref{eq:2}\) is \(y = x^r\), when \(x > 0\) and \(y = (−x)^r\) when \(x < 0\), (or more generally, \(y = |x|^r\) for all \(x\)), with \(r\) constant. After substitution into \(\eqref{eq:2}\), we obtain for both positive and negative \(x\) \[r(r-1)|x|^r+\alpha r|x|^r+\beta |x|^r=0,\nonumber\] and we observe that our ansatz is rewarded by cancelation of \(|x|^r\). We thus obtain the following quadratic equation for \(r\): \[\label{eq:3}r^2+(\alpha -1)r+\beta=0,\]

which can be solved using the quadratic formula. Three cases immediately appear: (i) real distinct roots, (ii) complex conjugate roots, (iii) repeated roots. Students may recall being in a similar situation when solving the second-order linear homogeneous ode with constant coefficients. Indeed, it is possible to directly transform the Cauchy-Euler equation into an equation with constant coefficients so that our previous results can be used.

The idea is to change variables so that the power law ansatz \(y = x^r\) becomes an exponential ansatz. For \(x > 0\), if we let \(x = e^ξ\) and \(y(x) = Y(ξ)\), then the ansatz \(y(x) = x^r\) becomes the ansatz \(Y(ξ) = e^{rξ}\), appropriate if \(Y(ξ)\) satisfies a constant coefficient ode. If \(x < 0\), then the appropriate transformation is \(x = −e^ξ\), since \(e^ξ > 0\). We need only consider \(x > 0\) here and subsequently generalize our result by replacing \(x\) everywhere by its absolute value.

We thus transform the differential equation \(\eqref{eq:2}\) for \(y = y(x)\) into a differential equation for \(Y = Y(ξ)\), using \(x = e^ξ\), or equivalently, \(ξ = \ln x\). By the chain rule, \[\begin{aligned}\frac{dy}{dx}&=\frac{dY}{d\xi}\frac{d\xi}{dx} \\ &=\frac{1}{x}\frac{dY}{d\xi} \\ &=e^{-\xi}\frac{dY}{d\xi},\end{aligned}\] so that symbolically, \[\frac{d}{dx}=e^{-\xi}\frac{d}{d\xi}.\nonumber\]

The second derivative transforms as \[\begin{aligned}\frac{d^2y}{dx^2}&=e^{-\xi}\frac{d}{d\xi}\left(e^{-\xi}\frac{dY}{d\xi}\right) \\ &=e^{-2\xi}\left(\frac{d^2Y}{d\xi^2}-\frac{dY}{d\xi}\right).\end{aligned}\]

Upon substitution of the derivatives of \(y\) into \(\eqref{eq:2}\), and using \(x = e^ξ\), we obtain \[\begin{aligned}e^{e\xi}\left(e^{-2\xi}(Y''-Y')\right)+\alpha e^{\xi}\left(e^{-\xi}Y'\right)+\beta Y&=Y''+(\alpha -1)Y'+\beta Y \\ &=0.\end{aligned}\]

As expected, the ode for \(Y = Y(ξ)\) has constant coefficients, and with \(Y = e^{rξ}\), the characteristic equation for \(r\) is given by \(\eqref{eq:3}\). We now directly transfer previous results obtained for the constant coefficient second-order linear homogeneous ode.

Distinct Real Roots

This simplest case needs no transformation. If \((\alpha − 1)^2 − 4\beta > 0\), then with \(r_±\) the real roots of \(\eqref{eq:3}\), the general solution is \[y(x)=c_1|x|^{r_+}+c_2|x|^{r_-}.\nonumber\]

Distinct Complex-Conjugate Roots

If \((\alpha − 1)^2 − 4\beta < 0\), we can write the complex roots of \(\eqref{eq:3}\) as \(r_± = \lambda ± i\mu\). Recall the general solution for \(Y = Y(ξ)\) is given by \[Y(\xi)=e^{\lambda\xi}(A\cos\mu\xi +B\sin\mu\xi );\nonumber\] and upon transformation, and replacing \(x\) by \(|x|\), \[y(x)=|x|^{\lambda}(A\cos (\mu\ln |x|)+B\sin (\mu\ln |x|)).\nonumber\]

Repeated Roots

If \((\alpha − 1)^2 − 4\beta = 0\), there is one real root \(r\) of \(\eqref{eq:3}\). The general solution for \(Y\) is \[Y(\xi )=e^{r\xi}(c_1+c_2\xi ),\nonumber\] yielding \[y(x)=|x|^r(c_1+c_2\ln |x|).\nonumber\]

We now give examples illustrating these three cases.

Example \(\PageIndex{1}\)

Solve \(2x 2y'' + 3xy' − y = 0\) for \(0 ≤ x ≤ 1\) with two-point boundary condition \(y(0) = 0\) and \(y(1) = 1\).

Solution

Since \(x > 0\), we try \(y = x^r\) and obtain the characteristic equation \[\begin{aligned}0&=2r(r-1)+3r-1 \\ &=2r^2+r-1 \\ &=(2r-1)(r+1).\end{aligned}\]

Since the characteristic equation has two real roots, the general solution is given by \[y(x)=c_1x^{\frac{1}{2}}+c_2x^{-1}.\nonumber\]

We now encounter for the first time two-point boundary conditions, which can be used to determine the coefficients \(c_1\) and \(c_2\). Since \(y(0)=0\), we must have \(c_2 = 0\). Applying the remaining condition \(y(1) = 1\), we obtain the unique solution \[y(x)=\sqrt{x}.\nonumber\]

Note that \(x = 0\) is called a singular point of the ode since the general solution is singular at \(x = 0\) when \(c_2\neq 0\). Our boundary condition imposes that \(y(x)\) is finite at \(x = 0\) removing the singular solution. Nevertheless, \(y'\) remains singular at \(x = 0\). Indeed, this is why we imposed a two-point boundary condition rather than specifying the value of \(y' (0)\) (which is infinite)

Example \(\PageIndex{2}\)

Solve \(x^2y'' + xy' + \pi^2y = 0\) with two-point boundary condition \(y(1) = 1\) and \(y(\sqrt{e}) = 1\).

Solution

With the ansatz \(y = x^r\), we obtain \[\begin{aligned}0&=r(r-1)+r+\pi^2 \\ &=r^2+\pi^2,\end{aligned}\] so that \(r=\pm i\pi\). Therefore, with \(\xi =\ln x\), we have \(Y(\xi )=A\cos\pi\xi +B\sin\pi\xi\), and the general solution for \(y(x)\) is \[y(x)=A\cos (\pi\ln x)+B\sin (\pi\ln x).\nonumber\]

The first boundary condition \(y(1) = 1\) yields \(A = 1\). The second boundary condition \(y(\sqrt{e}) = 1\) yields \(B = 1\).

Example \(\PageIndex{3}\)

Solve \(x^2y'' + 5xy' + 4y = 0\) with two-point boundary condition \(y(1) = 0\) and \(y(e) = 1\).

Solution

With the ansatz \(y = x^r\), we obtain \[\begin{aligned}0&=r(r-1)+5r+4 \\ &=r^2+4r+4 \\ &=(r+2)^2,\end{aligned}\] so that there is a repeated root \(r=-2\). With \(\xi =\ln x\), we have \(Y(\xi )=(c_1+c_2\xi )e^{-2\xi }\), so that the general solution is \[y(x)=\frac{c_1+c_2\ln x}{x^2}.\nonumber\]

The first boundary condition \(y(1) = 0\) yields \(c_1 = 0\). The second boundary condition \(y(e) = 1\) yields \(c_2 = e^2\). The solution is therefore \[y(x)=\frac{e^2\ln x}{x^2}.\nonumber\]