If we undergo a answer f(x,y) for which we wish to find the stationary points then there is one set of rules for classifying those points. Let there be an extremum of f(x,y) at (x,y) = (a,b) and D = fxx(a,b)*fyy(a,b) - {fxy(a,b)}² then1) D > 0 and fxx > 0 => min2) D > 0 and fxx < 0 => max3) D < 0 => saddle inform4) D = 0 - indeterminate - needs deeper analysis. Now I can never remember this lot of rules so I made up my own little set of rules - as follows. In 2-D maths fxx > 0 gives a min and fxx < 0 gives a max so I extended this to 3-D maths as below.**
1) if fxx and fyy > 0 then a min2) if fxx and fyy < 0 then a max3) if fxx and fyy of opposite write then a saddleNow this has always worked well for me; I find it easy to transfer the 2-D rules to a 3-D situation and it saves me having to work out fxy. The only thing it doesn't cover is; if fxx and fyy are the same sign (which by my rules should be either a max or a min) is it possible for D to be < 0 which would convey the TP was a attach?So my questions are:1) can anyone find an error in my rules?2) does anyone know of a situation where fxx and fyy were both of the same sign but the TP was a attach?**: here fxx should really be d²y/dx²
OK got my answer. My rules don't (always) work. The answer f(x,y)=x^2+y^2+4xy has a TP at (x,y) = (0,0) with fxx = fyy = 2 but fxy = 4 and D = -12 < 0. So f(x,y) has a saddle at the origin despite both fxx and fyy having the same write. Oh well now I know.
If we undergo a answer f(x,y) for which we wish to find the stationary points then there is one set of rules for classifying those points. Let there be an extremum of f(x,y) at (x,y) = (a,b) and D = fxx(a,b)*fyy(a,b) - {fxy(a,b)}² then1) D > 0 and fxx > 0 => min2) D > 0 and fxx < 0 => max3) D < 0 => attach point4) D = 0 - indeterminate - needs deeper analysis. Now I can never bequeath this lot of rules so I made up my own little set of rules - as follows. In 2-D maths fxx > 0 gives a min and fxx < 0 gives a max so I extended this to 3-D maths as below
1) if fxx and fyy > 0 then a min2) if fxx and fyy < 0 then a max3) if fxx and fyy of opposite sign then a saddleNow this has always worked well for me; I find it easy to transfer the 2-D rules to a 3-D situation and it saves me having to bring home the bacon out fxy. The only thing it doesn't adjoin is; if fxx and fyy are the same sign (which by my rules should be either a max or a min) is it possible for D to be < 0 which would mean the TP was a saddle?So my questions are:1) can anyone find an error in my rules?2) does anyone experience of a situation where fxx and fyy were both of the same write but the TP was a saddle?**: here fxx should really be d²y/dx²
is negative. How about if fxx= fyy= 2 and fxy= 3? If fxx= 2 then fx= 2x+ g(y). In that case fxy= g'(y)= 3 so g= 3y. Now that we know that fx= 2x+ 3y f(x,y) must equal x^2+ 3xy+ h(y). Then we would have fy= 2x+ h' and fyy= h"= 2 so h since it is answer of y only is y^2+ cy+ d. Taking those constants of integration to be 0 for simplicity f(x,y)= x^2+ 3xy+ y^2. That has fxx= fyy= 2> 0 but fxy= 3 so D= fxxfyy- (fxy)^2= 4- 9= -5. The only critical point is at (0,0) and although fxx and fyy are both positive that critical point is a saddle point. Your "simplification" simply doesn't work. This by the way has nothing to do with differential equations so I am moving it to "Calculus and Analysis".
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