Undergrad Solve second order linear differential equation

[lriuui0x0]

Consider the second order linear ODE with parameters $a, b$:

$$
xy'' + (b-x)y' - ay = 0
$$

By considering the series solution $y=\sum c_mx^m$, I have obtained two solutions of the following form:

$$
\begin{aligned}
y_1 &= M(x, a, b) \\
y_2 &= x^{1-b}M(x, a-b+1, 2-b) \\
\end{aligned}
$$

Where $M$ is a function after solving the recurrence coefficients, please feel free to check if my result is correct:

$$
M(x, a, b) = \sum_{m=0}^\infty \frac{\prod_{i=0}^{m-1}(a+i)}{m!\prod_{i=0}^{m-1}(b+i)}x^m
$$

The question is if $b=1$, we get repeated root and $y_1 = y_2$ hence linearly dependent. How to obtain a second independent solution? The problem has a hint of considering the following limit:

$$
\lim_{b \to 1} \frac{y_2 - y_1}{b - 1}
$$

 

[anuttarasammyak]

I observe in case $a=b$ the solution, other than trivial y=const,. is $y=Ce^x$ which meets with your result. It shows any constant A could be multiplied. For general a, b by transforming
y=e^{z(x)} to see differential equation of z, could you check/correct
y=Ce^{\frac{a}{b}x}
is a solution ?

 

[lriuui0x0]

I observe in case $a=b$ the solution, other than trivial y=const,. is $y=Ce^x$ which meets with your result. It shows any constant A could be multiplied. For general a, b by transforming
y=e^{z(x)} to see differential equation of z, could you check/correct
y=Ce^{\frac{a}{b}x}
is a solution ?

It's not a solution:

let $y=e^{\frac{a}{b}x}$, then the LHS of the differential equation is:

$$
\frac{a^2}{b^2}xe^{\frac{a}{b}x} + (b-x) \frac{a}{b}e^{\frac{a}{b}x} - ae^{\frac{a}{b}x} \ne 0
$$

 

[anuttarasammyak]

Thank you so much for correction. Following your way I got
c_{m+1}=\frac{a+m}{(m+1)(b+m)}c_m
for m=0,1,2,.. with $c_0$ given which is your $y_1$. I do not find the way to get $y_2$.

 

[pasmith]

Consider the second order linear ODE with parameters $a, b$:

$$
xy'' + (b-x)y' - ay = 0
$$

By considering the series solution $y=\sum c_mx^m$, I have obtained two solutions of the following form:

$$
\begin{aligned}
y_1 &= M(x, a, b) \\
y_2 &= x^{1-b}M(x, a-b+1, 2-b) \\
\end{aligned}
$$

Where $M$ is a function after solving the recurrence coefficients, please feel free to check if my result is correct:

$$
M(x, a, b) = \sum_{m=0}^\infty \frac{\prod_{i=0}^{m-1}(a+i)}{m!\prod_{i=0}^{m-1}(b+i)}x^m
$$

Applying Frobenius' Method and setting y = \sum_{n=0}^\infty c_nx^{n+r} with c_0 \neq 0 I obtained the indicial equation r(r - 1 + b) = 0 and the recurrence relation <br /> c_{n+1} = \frac{n + r + a}{(n + r + 1)(n + r + b)}c_n which are consistent with your result, althouh I would have used the property of the Gamma function that \Gamma(z + 1) = z \Gamma(z) to evaluate the products.

(@anuttarasammyak: Just using an integer power series, which to be fair is what @lriuui0x0 said they did, won't pick up the r = 1 - b solution.)

The question is if $b=1$, we get repeated root and $y_1 = y_2$ hence linearly dependent. How to obtain a second independent solution? The problem has a hint of considering the following limit:

$$
\lim_{b \to 1} \frac{y_2 - y_1}{b - 1}
$$

y_1 and (y_2 - y_1)/(b-1) are linearly independent for any b \neq 1; for b = 1 you can use l'Hopital's rule to calculate <br /> \lim_{b \to 1} \frac{x^{1-b}M(x,a-b+1, 2-b) - M(x,a,b)}{b-1} = <br /> - M(x,a,1)\log x - \left.\frac{\partial M}{\partial a}\right|_{(x,a,1)} - 2\left.\frac{\partial M}{\partial b}\right|_{(x,a,1)}<br /> and show that the result is a solution of the ODE which is linearly independent of M(x,a,1).

 

[lriuui0x0]

Applying Frobenius' Method and setting y = \sum_{n=0}^\infty c_nx^{n+r} with c_0 \neq 0 I obtained the indicial equation r(r - 1 + b) = 0 and the recurrence relation <br /> c_{n+1} = \frac{n + r + a}{(n + r + 1)(n + r + b)}c_n which are consistent with your result, althouh I would have used the property of the Gamma function that \Gamma(z + 1) = z \Gamma(z) to evaluate the products.

(@anuttarasammyak: Just using an integer power series, which to be fair is what @lriuui0x0 said they did, won't pick up the r = 1 - b solution.)
y_1 and (y_2 - y_1)/(b-1) are linearly independent for any b \neq 1; for b = 1 you can use l'Hopital's rule to calculate <br /> \lim_{b \to 1} \frac{x^{1-b}M(x,a-b+1, 2-b) - M(x,a,b)}{b-1} =<br /> - M(x,a,1)\log x - \left.\frac{\partial M}{\partial a}\right|_{(x,a,1)} - 2\left.\frac{\partial M}{\partial b}\right|_{(x,a,1)}<br /> and show that the result is a solution of the ODE which is linearly independent of M(x,a,1).

Just to double check on these steps. I got slightly different result compared with yours that doesn't involve partial derivative of $a$. Does it look correct?

Expand the limit with l'Hopital's rule:

$$

\begin{aligned} &\phantom{{}={}}\lim_{b\to 1}\frac{y_2-y_1}{b-1} \\ &= \biggl[\frac{\partial}{\partial b}(x^{1-b}M(x,a-b+1,2-b) - M(x,a,b))\biggr]_{b=1} \\ &= \biggl[-x^{1-b}\ln x M(x,a-b+1,2-b) + x^{1-b}\frac{\partial M(x,a-b+1,2-b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x + \biggl[\frac{\partial M(x,a-b+1,2-b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x + \biggl[\frac{\partial M(x,a,b)}{\partial b}(-2) - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x - 3\frac{\partial M}{\partial b}(x,a,1) \\ \end{aligned}

$$Check the Wronskian of $y_1 = M(x,a,1)$ and $y_2 = -M(x,a,1)\ln x - 3\frac{\partial M}{\partial b}(x,a,1)$:$$

\begin{aligned} &\phantom{{}={}} y_1y_2'-y_2y_1' \\ &= M(x,a,1)(-\frac{\partial M}{\partial x}(x,a,1)\ln x - M(x,a,1)\frac{1}{x} - 3\frac{\partial^2M}{\partial x\partial b}(x,a,1)) - \frac{\partial M}{\partial x}(x,a,1)(-M(x,a,1)\ln x - 3\frac{\partial M}{\partial b}(x,a,1)) \\ &= -M(x,a,1)^2 \frac{1}{x} -3M(x,a,1)\frac{\partial^2M}{\partial x\partial b}(x,a,1) + 3\frac{\partial M}{\partial x}(x,a,1)\frac{\partial M}{\partial b}(x,a,1) \end{aligned}

$$When $x \to 0$, the Wronskian diverges because of the $\frac{1}{x}$ term, so it's non-zero. Hence $y_1, y_2$ are independent.

 

[pasmith]

Just to double check on these steps. I got slightly different result compared with yours that doesn't involve partial derivative of $a$. Does it look correct?

Expand the limit with l'Hopital's rule:

$$

\begin{aligned} &\phantom{{}={}}\lim_{b\to 1}\frac{y_2-y_1}{b-1} \\ &= \biggl[\frac{\partial}{\partial b}(x^{1-b}M(x,a-b+1,2-b) - M(x,a,b))\biggr]_{b=1} \\ &= \biggl[-x^{1-b}\ln x M(x,a-b+1,2-b) + x^{1-b}\frac{\partial M(x,a-b+1,2-b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x + \biggl[\frac{\partial M(x,a-b+1,2-b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x + \biggl[\frac{\partial M(x,a,b)}{\partial b}(-2) - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\ &= -M(x,a,1)\ln x - 3\frac{\partial M}{\partial b}(x,a,1) \\ \end{aligned}

$$

You haven't checked that -M(x,a,1) \ln x - 3\frac{\partial M}{\partial b} actually solves the ODE; it doesn't. If L(y) = xy&#039;&#039; + (b-x)y&#039; - ay then differentiating partially with respect to b results in \frac{\partial}{\partial b}L(y) = L\left(\frac{\partial y}{\partial b}\right) + y&#039; and you can also show that L(y \ln x) = L(y) \ln x + 2y&#039; - y + \frac{(b-1)y}{x}. Since L(M) = 0 for all a and b you can see that L\left(M \ln x + 3 \frac{\partial M}{\partial b}\right) = -M&#039; - M when b = 1.

The problem is that you haven't calculated the derivative of M(x,a-b+1, 2-b) with respect to b correctly. By \frac{\partial M}{\partial b} we mean the derivative of M(x,a,b) with respect to b with x and a constant. But you are looking at f(x,a,b) = M(x, g(a,b),h(a,b)) with g(a,b) = a-b+1 and h(a,b) = 2 - b rather than M(x,a,b). The multivariate chain rule then gives \frac{\partial f}{\partial b} = -\frac{\partial M}{\partial a} - \frac{\partial M}{\partial b}, and we must evaluate the derivatives of M at (x, a-b+1, 2-b) rather than at (x,a,b). In the limit, of course, these are both (x,a,1).

 

[lriuui0x0]

You haven't checked that -M(x,a,1) \ln x - 3\frac{\partial M}{\partial b} actually solves the ODE; it doesn't. If L(y) = xy&#039;&#039; + (b-x)y&#039; - ay then differentiating partially with respect to b results in \frac{\partial}{\partial b}L(y) = L\left(\frac{\partial y}{\partial b}\right) + y&#039; and you can also show that L(y \ln x) = L(y) \ln x + 2y&#039; - y + \frac{(b-1)y}{x}. Since L(M) = 0 for all a and b you can see that L\left(M \ln x + 3 \frac{\partial M}{\partial b}\right) = -M&#039; - M when b = 1.

The problem is that you haven't calculated the derivative of M(x,a-b+1, 2-b) with respect to b correctly. By \frac{\partial M}{\partial b} we mean the derivative of M(x,a,b) with respect to b with x and a constant. But you are looking at f(x,a,b) = M(x, g(a,b),h(a,b)) with g(a,b) = a-b+1 and h(a,b) = 2 - b rather than M(x,a,b). The multivariate chain rule then gives \frac{\partial f}{\partial b} = -\frac{\partial M}{\partial a} - \frac{\partial M}{\partial b}, and we must evaluate the derivatives of M at (x, a-b+1, 2-b) rather than at (x,a,b). In the limit, of course, these are both (x,a,1).

Thanks @pasmith ! I found the error according to your correction, could you please check now? Especially if the argument on linear independence of the two solutions is valid?Expand the limit with l'Hopital's rule:
$$
\begin{aligned}
&\phantom{{}={}}\lim_{b\to 1}\frac{y_2-y_1}{b-1} \\
&= \biggl[\frac{\partial}{\partial b}(x^{1-b}M(x,a-b+1,2-b) - M(x,a,b))\biggr]_{b=1} \\
&= \biggl[-x^{1-b}\ln x M(x,a-b+1,2-b) + x^{1-b}\frac{\partial M(x,a-b+1,2-b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\
&= -M(x,a,1)\ln x + \biggl[\frac{\partial M(x,a-b+1,2-b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\
&= -M(x,a,1)\ln x + \biggl[-\frac{\partial M(x,a,b)}{\partial a} -\frac{\partial M(x,a,b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\
&= -M(x,a,1)\ln x - \frac{\partial M}{\partial a}(x,a,1) - 2\frac{\partial M}{\partial b}(x,a,1) \\
\end{aligned}
$$
Consider the differential operator $L(y) = xy'' + (b-x)y' -ay$, then consider $L(y\ln x)$, $\frac{\partial}{\partial b}L(y)$ and $\frac{\partial}{\partial a}L(y)$:
$$
\begin{aligned}
&\phantom{{}={}} L(y\ln x) \\
&= x(y\ln x)'' + (b-x)(y\ln x)' - a(y\ln x) \\
&= xy''\ln x +2y' - y\frac{1}{x} + (b-x)y' \ln x + (b-x)y\frac{1}{x} - ay\ln x \\
&= \ln x(xy'' + (b-x)y' -ay) + 2y' -y + (b-1)y\frac{1}{x} \\
&= \ln x L(y) + 2y' -y + (b-1)y\frac{1}{x} \\
\end{aligned}
$$

$$
\begin{aligned}
&\phantom{{}={}} \frac{\partial}{\partial b}L(y) \\
&= \frac{\partial}{\partial b}(xy'' + (b-x)y' - ay) \\
&= x\frac{\partial }{\partial b}y'' + y' +(b-x)\frac{\partial}{\partial b}y' - a\frac{\partial}{\partial b}y \\
&= x(\frac{\partial y}{\partial b})'' + (b-x) (\frac{\partial y}{\partial b})' - a\frac{\partial y}{\partial b} + y' \\
&= L(\frac{\partial y}{\partial b}) + y'
\end{aligned}
$$

$$
\begin{aligned}
&\phantom{{}={}} \frac{\partial}{\partial a}L(y) \\
&= \frac{\partial}{\partial a}(xy'' + (b-x)y' - ay) \\
&= x\frac{\partial }{\partial a}y'' +(b-x)\frac{\partial}{\partial b}y' - y -a\frac{\partial}{\partial a}y \\
&= x(\frac{\partial y}{\partial a})'' + (b-x) (\frac{\partial y}{\partial a})' - a\frac{\partial y}{\partial a} -y \\
&= L(\frac{\partial y}{\partial a}) - y
\end{aligned}
$$
Check $y_2 = -M(x,a,1)\ln x - \frac{\partial M}{\partial a}(x,a,1) - 2\frac{\partial M}{\partial b}(x,a,1)$ is indeed a solution to the differential equation:
$$
\begin{aligned}
&\phantom{{}={}} L(y_2) \\
&= -L(M \ln x) - L(\frac{\partial M}{\partial a}) - 2L(\frac{\partial M}{\partial b}) \\
&= -\ln xL(M) - 2M' +M -\frac{\partial}{\partial b}L(M) - M -2\frac{\partial}{\partial a}L(M) + 2M' \\
&= 0
\end{aligned}
$$
Check the Wronskian of $y_1 = M(x,a,1)$ and $y_2 = -M(x,a,1)\ln x - \frac{\partial M}{\partial a}(x,a,1) - 2\frac{\partial M}{\partial b}(x,a,1)$:
$$
\begin{aligned}
&\phantom{{}={}} y_1y_2'-y_2y_1' \\
&= M(-M'\ln x - M\frac{1}{x} - (\frac{\partial M}{\partial a})' -2(\frac{\partial M}{\partial b})') - M'(-M\ln x -\frac{\partial M}{\partial a} -2\frac{\partial M}{\partial b}) \\
&= -M^2 \frac{1}{x} -M(\frac{\partial M}{\partial a})' + M'\frac{\partial M}{\partial a} -2M(\frac{\partial M}{\partial a})' + 2M'\frac{\partial M}{\partial b}
\end{aligned}
$$
When $x \to 0$, the Wronskian diverges because of the $\frac{1}{x}$ term, so it's non-zero. Hence $y_1, y_2$ are independent.