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

Consider the partial differential equation:

$$\frac{\partial u}{\partial t}=\frac{\partial^2u}{\partial x^2} + a \frac{\partial u}{\partial x} + bu$$

for the function $u (x; t)$ where $a$ and $b$ are constants.

By using substitution of the form $u(x,t) = \exp(\alpha x+\beta t)v(x,t)$;

And suitable choice of constants alpha and beta, show that the PDE can be reduced to the heat equation

$$\frac{\partial v}{\partial t}=\frac{\partial^2v}{\partial x^2}.$$

share|cite|improve this question

closed as off-topic by This is much healthier., Claude Leibovici, RecklessReckoner, Hanul Jeon, Carl Mummert Jul 15 '14 at 10:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Claude Leibovici, RecklessReckoner, Hanul Jeon, Carl Mummert
If this question can be reworded to fit the rules in the help center, please edit the question.

Just differentiate $v$ with respect to $x,x^2$ and $t$. Add the terms and you should see your reslut. Note that $v$ is determined by the substitution. Usually, you should show a bit more effort and tell us what you have tried so far. – Quickbeam2k1 Feb 4 '14 at 6:36

you could compare the equations $$\beta v+ \frac{\partial v}{\partial t} \\=\alpha^2v+\frac{\partial^2 v}{\partial x^2}+2\alpha\frac{\partial v}{\partial x}+a\alpha v+a\frac{\partial v}{\partial x}+bv$$ and the old one and choose the constant that fits your request.

share|cite|improve this answer

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