Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

In solving first order PDE's with solution $u(x,y)$, when constructing a graph of $u$ as a union of initial curves $C_s$ emanating from the initial curve $\Gamma$. My lecture notes say, for each $s \in (\alpha,\beta) \subset \mathbb{R}$, $$C_s:= \{(x(t;s),y(t;s),z(t;s)) | t \in (-\epsilon_s, \epsilon_s), \epsilon_s >0\}$$ and $$\mathrm{Graph}_u = \bigcup_{s\in (\alpha,\beta)\subset\mathbb{R}}C_s$$

I don't understand this 'semi-colon' notation given when defining $C_s$ as surely $x,y,z$ are functions of only one variable?

I appreciate any help!

share|cite|improve this question
It denotes the parameter $s$. Maybe better would be to write $x_s(t)$, etc... – David Mitra Jan 18 '13 at 16:36

For each value of $s$, the curve $C_s$ has a parametric equation in which $x,y,z$ are functions of $t$. To distinguish $x(t)$ from the parametric equation of $C_1$ and $x(t)$ from the parametric equation of $C_2$, one can use subscripts: let $(x_s,y_s,z_s)$ be the functions that parameterize $C_s$. Including the argument $t$, the notation becomes $(x_s(t),y_s(t),z_s(t))$.

For various reasons, subscripts and superscripts are not always good places to keep variables in. This is how $(x(t;s),y(t;s),z(t;s))$ becomes an attractive alternative. The semicolon has no mathematical meaning; it's just a separator, like a comma. Using the semicolon emphasizes that $t$ and $s$ are not considered to be "on equal footing". If we did not want to emphasize this, then writing $(x(t,s),y(t,s),z(t,s))$ would be just as correct: after all, $x,y,z$ are functions of two variables.

share|cite|improve this answer

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.