Classify the given second-order linear PDE

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chwala
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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;

Textbook page on classifying second-order linear PDEs: definition of the coefficient matrix A = [a b; b c], eigenvalue test for elliptic, parabolic and hyperbolic types, and Example 1.2 with PDEs listed for classification.


and the textbook solution is here;
Handwritten solution listing four symbol matrices A with determinants and classifications: [[2,0],[0,3]] det 6 elliptic; [[0,0],[0,17]] det 0 parabolic; [[2,2],[2,1]] det -2 hyperbolic; [[-1,-1],[-1,1]] det -2 hyperbolic.
and i also checked wikipedia on this and they have used the discriminant approach. See the approach here; ( i am comfortable with this approach)
Text page titled Elliptic, Hyperbolic, and Parabolic PDEs, with the general second-order PDE form and the discriminant test b^2 - 4ac for elliptic, hyperbolic and parabolic cases, plus example equations.


Both approaches should be fine, one can use either approach...correct?
 
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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##.
 
Office_Shredder said:
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...:cool: