# Classify the given second-order linear PDE

Gold Member
Homework Statement:
Classify the second order linear pde, given by;

##U_t +2U_{tt}+3U_{xx}=0##
Relevant Equations:
use of discriminant
Now i learned how to use discriminant i.e ##b^2-4ac## and in using this we have;
##b^2-4ac##=##0-(4×3×2)##=##-24<0,## therefore elliptic.

The textbook has a slight different approach, which i am not familiar with as i was trained to use the discriminant at my undergraduate studies...
see textbook approach here; and the textbook solution is here; and i also checked wikipedia on this and they have used the discriminant approach. See the approach here; ( i am comfortable with this approach) Both approaches should be fine, one can use either approach...correct?

Staff Emeritus
Gold Member
The textbook is the same thing with some math modifications

1.) Instead of ##b^2-4ac## they have ##4ac-b^2## so the sign is flipped

2.) They write ##2b## as the coefficient instead of ##b##. So given that the coefficient of the cross term is called ##2b##, their discriminant becomes
##4ac-(2b)^2=4(ac-b^2)##. Since 4 is positive, the sign of the discriminant is determined by ##ac-b^2##.

• sysprog
Gold Member
The textbook is the same thing with some math modifications

1.) Instead of ##b^2-4ac## they have ##4ac-b^2## so the sign is flipped

2.) They write ##2b## as the coefficient instead of ##b##. So given that the coefficient of the cross term is called ##2b##, their discriminant becomes
##4ac-(2b)^2=4(ac-b^2)##. Since 4 is positive, the sign of the discriminant is determined by ##ac-b^2##.
ooooh, Thanks am trying to refresh on pde's cheers mate... 