# Finding the maximum of a gradient vector

Im trying to find the maximum of my gradient vector G[x,y], I've tried several options including FindMaximumValue, FindMaximum etc. but i couldn't find it.

The full function is shown below, any help would be greatly appreciated

G[x_,y_]:= {-5.4 E^(-2.25 ((-4 + x)^2 + (-2.5 + y)^2)) (-4 + x) -
6.4 E^(-4 ((-3.5 + x)^2 + (-1.5 + y)^2)) (-3.5 + x) -
8 E^(-4 ((-3 + x)^2 + (-3.5 + y)^2)) (-3 + x) -
9. E^(-9 ((-2.5 + x)^2 + (-1.5 + y)^2)) (-2.5 + x) -
5.5125 E^(-3.0625 ((-2 + x)^2 + (-3 + y)^2)) (-2 + x) -
1.8 E^(-(-1 + x)^2 - (-1 + y)^2) (-1 + x) -
30. E^(-25 ((-0.75 + x)^2 + (-2 + y)^2)) (-0.75 +
x), -8 E^(-4 ((-3 + x)^2 + (-3.5 + y)^2)) (-3.5 + y) -
5.5125 E^(-3.0625 ((-2 + x)^2 + (-3 + y)^2)) (-3 + y) -
5.4 E^(-2.25 ((-4 + x)^2 + (-2.5 + y)^2)) (-2.5 + y) -
30. E^(-25 ((-0.75 + x)^2 + (-2 + y)^2)) (-2 + y) -
6.4 E^(-4 ((-3.5 + x)^2 + (-1.5 + y)^2)) (-1.5 + y) -
9. E^(-9 ((-2.5 + x)^2 + (-1.5 + y)^2)) (-1.5 + y) -
1.8 E^(-(-1 + x)^2 - (-1 + y)^2) (-1 + y)}

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G[x,y] is not scales quantity. What do you mean by maximum of gradient vector? – Algohi Jul 6 '14 at 5:57
If you plot Plot3D[G[x, y], {x, 0, 5}, {y, 0, 5}] you get a bunch of lumps, i want to find the "peaks" – Lebouski Jul 6 '14 at 6:17
note that, Plot3D[G[x, y], {x, 0, 5}, {y, 0, 5}] is actually a 3D plot if two functions. you can see it clearly from the plot. in this case you can find the max of each function separately, Maximize[G[x, y][[1]], {x, y}], Maximize[G[x, y][[2]], {x, y}]. – Algohi Jul 6 '14 at 6:28

I hope this is helpful (if I interpret your aim correctly): This is a little more challenging for your particular function.

g[x_, y_] := {-5.4 E^(-2.25 ((-4 + x)^2 + (-2.5 + y)^2)) (-4 + x) -
6.4 E^(-4 ((-3.5 + x)^2 + (-1.5 + y)^2)) (-3.5 + x) -
8 E^(-4 ((-3 + x)^2 + (-3.5 + y)^2)) (-3 + x) -
9. E^(-9 ((-2.5 + x)^2 + (-1.5 + y)^2)) (-2.5 + x) -
5.5125 E^(-3.0625 ((-2 + x)^2 + (-3 + y)^2)) (-2 + x) -
1.8 E^(-(-1 + x)^2 - (-1 + y)^2) (-1 + x) -
30. E^(-25 ((-0.75 + x)^2 + (-2 + y)^2)) (-0.75 +
x), -8 E^(-4 ((-3 + x)^2 + (-3.5 + y)^2)) (-3.5 + y) -
5.5125 E^(-3.0625 ((-2 + x)^2 + (-3 + y)^2)) (-3 + y) -
5.4 E^(-2.25 ((-4 + x)^2 + (-2.5 + y)^2)) (-2.5 + y) -
30. E^(-25 ((-0.75 + x)^2 + (-2 + y)^2)) (-2 + y) -
6.4 E^(-4 ((-3.5 + x)^2 + (-1.5 + y)^2)) (-1.5 + y) -
9. E^(-9 ((-2.5 + x)^2 + (-1.5 + y)^2)) (-1.5 + y) -
1.8 E^(-(-1 + x)^2 - (-1 + y)^2) (-1 + y)}


For simplicity maximizing $g(x,y).g(x,y)$ yields:

ma = Maximize[g[x, y].g[x, y], {x, y}]


yields: {0.595965, {x -> 1.1506, y -> 0.309118}}

However, you function has a lot of local maxima and minima. After looking at plot and choosing region of interest:

nma = NMaximize[{g[x, y].g[x, y], 0 < x < 1 && 0 < y < 2}, {x, y}]


yields: {8.14402, {x -> 0.607423, y -> 2.}}

Showing this interactively (the red mesh lines intersect at the first maximum and the green at the second):

gr = Manipulate[
Plot3D[g[x, y].g[x, y], {x, -1, 4}, {y, -1, 4}, PlotRange -> {0, r},
MeshFunctions -> {#1 &, #2 &, #1 &, #2 &},
Mesh -> {{ma[[2, 1, 2]]}, {ma[[2, 2, 2]]}, {nma[[2, 1,
2]]}, {nma[[2, 2, 2]]}},
MeshStyle -> {Red, Red, Green, Green}], {r, 1, 9}]


You can also visualize working backwards. A function whose gradient field is $g(x,y)$:

exp = (f[x, y] /.
First[DSolve[Thread[{D[f[x, y], x], D[f[x, y], y]} == g[x, y]],
f[x, y], {x, y}]]) /. C[1] -> 0;


Or perhaps more instructively with ContourPlot and StreamPlot (reassuringly exp is consistent with gradient):

pt = Point[{x, y} /. ma[[2]]]
ptn = Point[{x, y} /. nma[[2]]]
cp = ContourPlot[exp, {x, -1, 5}, {y, -1, 5}];
sp = StreamPlot[g[x, y], {x, -1, 5}, {y, -1, 5}];
Show[cp, sp,
Graphics[{{PointSize[0.02], Red, pt}, {PointSize[0.02], Green,
ptn}}]]


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Nice answer. Any reason why the contour plot is over $[-1,5]\times[-1,5]$ while the streamline plot is only over $[-1,4]\times[-1,4]$? – Rahul Jul 6 '14 at 18:52
@RahulNarain...tiredness...sorry...flat outside the mountains and fiddled to get them on....does not change the interpretation but will check when I get time – ubpdqn Jul 6 '14 at 22:40
@RahulNarain have edited... – ubpdqn Jul 7 '14 at 7:45