Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to determinate the linearly independence between some special functions defined by ODE? For example:

  1. ${}_1F_1(a;b;x)$ , $x^{1-b}{}_1F_1(a-b+1;2-b;x)$ when $b$ is integer

  2. ${}_2F_1(a,b;c;x)$ , $x^{1-c}{}_2F_1(a-c+1,b-c+1;2-c;x)$ when $c$ is integer

  3. HeunC$(\alpha,\beta,\gamma,\delta,\eta;x)$ , $x^{-\beta}$HeunC$(\alpha,-\beta,\gamma,\delta,\eta;x)$ when $\beta$ is integer

share|cite|improve this question
${}_mF_n$ is hypergeometric function? – Frank Science May 17 '12 at 9:15
You can write a hypergeometric function like this: $\displaystyle F\left( \left. { a_1,a_2,\dots,a_m \atop b_1,b_2,\dots,b_n } \right| z \right)$, or $F(a_1,a_2,\dots,a_m;b_1,b_2,\dots,b_n;z)$, so you might lose a semicolon in $F(a;b;x)$ and $F(a,b;c;x)$. – Frank Science May 17 '12 at 9:17
Thanks to reminder. – doraemonpaul May 17 '12 at 21:09
Please note that this question base on the information of , and , so this question should be correctly asked. – doraemonpaul Jun 15 '12 at 4:35

The usual route to proving the linear independence of two solutions $f_1(z)$, $f_2(z)$ is to compute the Wronskian determinant,


If this determinant is identically zero, then your two functions are linearly dependent.

For your first pair, the Wronskian is $\dfrac{\exp\,z\sin\,\pi b}{\pi z^b}$; for integer $b$, this Wronskian is clearly zero, and thus your pair is linearly dependent. For your second, the Wronskian is $(1-c)\dfrac{(1-z)^{c-a-b-1}}{z^c}$; this zeroes out only if $c=0$, so apart from that case, your two functions are linearly independent.

Abel's identity is useful if you do not happen to have explicit expressions for your two solutions.

share|cite|improve this answer
It sounds great for using Abel's identity to find the Wronskian, but the constant part of Abel's identity still unavoidable for needing directly using the solutions from the ODE, so for the Wronskian of special functions Abel's identity still not useful to find it exactly. – doraemonpaul Jun 15 '12 at 4:56
Reminder the constant part of the Wronskian often act as the key point to distinguish the zero and non-zero cases, like the Wronskian of question $1$. – doraemonpaul Jun 15 '12 at 5:03
Careful about your statement "If this determinant is identically zero, then your two functions are linearly dependent" This is not true in general. For example, $x^2$ and $x|x|$ are not linearly dependent in any neighborhood of $0$, yet the Wronskian vanishes. – John Sep 24 '12 at 15:58
@John, you're right (though on the other hand, does a linear combination of your pair satisfy a certain second-order differential equation?) – J. M. Sep 24 '12 at 16:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.