Computer assignment in MATLAB

Critical Points of Functions of Two and Three Variables

Assignment Participants

Andreas Karlsson

andrekar@hotmail.com

Daniel Lehtinen

dleh@kth.se

Contents

Problem 1 .............................................................................. 1Solution procedure .......................................................................MATLAB calculations .................................................................. 2

MATLAB code ......................................................................Calculation results ............................................................. 3

Plots ........................................................................................ 4Conclusion ............................................................................... 5

Problem 2 .............................................................................. 6Solution procedure .......................................................................MATLAB calculations .....................................................................

MATLAB code ......................................................................Calculation results ............................................................. 7

Conclusion ..................................................................................

Problem 3 .............................................................................. 8Solution procedure .......................................................................MATLAB calculations .................................................................. 9

MATLAB code ..................................................................... Calculation results ..............................................................

Plots ..................................................................................... 10Conclusion ............................................................................. 11

Problem 4 ............................................................................ 12Solution procedure .......................................................................MATLAB calculations .....................................................................

MATLAB code ......................................................................Calculation results ........................................................... 13

Conclusion ..................................................................................

Problem 5 ............................................................................ 14Solution procedure .......................................................................MATLAB calculations .....................................................................

MATLAB code ......................................................................Calculation results ........................................................... 15

Plots ...................................................................................... 16Conclusion ............................................................................. 17

Problem 6 ............................................................................ 18Solution procedure .......................................................................MATLAB calculations .....................................................................

MATLAB code ......................................................................Calculation results ........................................................... 19

Plots ...................................................................................... 20Conclusion ..................................................................................

Problem 1

Find and classify all critical points of the function

g x , y= x42x3 y−6x2 y2 y4

x4 y 41,

which you studied in the previous lesson, and compare your results with the contour and gradient plot of g that you have already made.

Solution procedure

To find critical points of g, we must set the partial derivatives equal to 0 and solve for x and y.With the coordinates in hand, our next step is to calculate the second derivatives. These second derivatives will make the Hessian matrix. Finally, we calculate the Hessian determinant and this value will classify the critical points for us.

MATLAB calculations

MATLAB code

syms x y zf=(x^4 + 2*(x^3)*y - 6*(x^2)*(y^2) + y^4)/(x^4 + y^4 +1)%figure(1);%ezsurf(f, [-3, 3, -3, 3])fx=diff(f,x);sfx = simplify(fx);fy=diff(f,y);sfy = simplify(fy);[xcr,ycr]=solve(fx,fy);[xcr,ycr] %punkterna i vilka derivatorna är nollfxx=diff(fx,x);sfxx=simplify(fxx);fxy=diff(fx,y);sfxy=simplify(fxy);fyy=diff(fy,y);sfyy=simplify(fyy);%hessdetf=fxx*fyy-fxy^2 %hessiandeterminantenhessdetf_simplified = simplify(hessdetf)hessdetf_simplified_inserted = subs(hessdetf_simplified, [x,y], [0,0])gradf = jacobian(f, [x, y])hessmatf = jacobian(gradf, [x, y])hessmatf_inserted = subs(hessmatf, [x,y], [0,0])hessdetf1 = simplify(det(hessmatf))%simplify(hessdetf - hessdetf1)

[xcr(1), ycr(1), subs(hessdetf, [x,y], [xcr(1), ycr(1)]), subs(fxx, [x,y], [xcr(1), ycr(1)])]

%syms h k%q = fxx*h^2 + 2*fxy*h*k + fyy*k^2%q1 = simplify(q)%q2 = subs(q, [x,y], [0,0])%subs(fx, [x,y], [0,0])%subs(fy, [x,y], [0,0])

[xx, yy] = meshgrid(-3:.1:3,-3:.1:3);ffun = inline(vectorize(f));fxfun = inline(vectorize(gradf(1)));fyfun = inline(vectorize(gradf(2)));%figure(2);%contour(xx, yy, ffun(xx,yy), 30)%hold on%[xx, yy] = meshgrid(-3:.25:3,-3:.25:3);%quiver(xx, yy, fxfun(xx,yy), fyfun(xx,yy), 0.6)%axis equal tight, hold off

Calculation results

Partial derivatives of g

Simplified using MATLABs built-in function simplify().

f ' x=−2x−2x2 x5 y−3xy5−3xy−6y2 x46y66y2

x4 y412

f ' y=2x7−3x3 y 4x3−6x6 y6x2 y5−6x2 y2y3

x4 y 412

Which resulted in one point (0, 0).

Second derivatives of g

Due to the length of the derivatives, we did not reproduce them here.

f ' xx=...

f ' xy= f ' yx=...

f ' yy=...

Hessian matrix (the square matrix of second partial derivatives)

H x , y = f ' xx f ' xyf ' yx f ' yyH 0,0 =0 0

0 0det H 0,0 =0

Plots

Figure 1. Contour plot

Figure 2. Surf plot

Conclusion

det(H) > 0 and any second derivate is positive => local minimum.det(H) > 0 and any second derivate is negative => local maximum.det(H) < 0 => saddlepoint.

Since the Hessian determinant equals 0, the second derivatives test to evaluate if the critical points is either max, min or saddle points fails. The figure in the plot looks like a pair of crossed saddle points with some kind of plane in the intersection.

Problem 2

Find and classify the critical points of the function

h x , y , z =3x24y2z 2−9xyz.

You may have trouble substituting the symbolic values of the coordinates of the critical points into the Hessian matrix; if so, convert them to numerical values, using double.

Solution procedure

We calculate the first partial derivatives and set them to 0 to receive the critical points.We evaluate the Hessian matrix at the critical points and compute the eigenvalues of the matrix. The values of the eigenvalues will determine the type of critical points.

MATLAB calculations

MATLAB code

syms x y zf=(3*x^2 + 4*y^2 + z^2 - 9*x*y*z)gradf2 = jacobian(f,[x,y,z])hessmat = jacobian(gradf2, [x,y,z])[xcr2, ycr2, zcr2] = solve(gradf2(1),gradf2(2),gradf2(3));[xcr2, ycr2, zcr2]

H1=subs(hessmat,[x,y,z],[xcr2(1), ycr2(1), zcr2(1)])H2=subs(hessmat,[x,y,z],[xcr2(2), ycr2(2), zcr2(2)])H3=subs(hessmat,[x,y,z],[xcr2(3), ycr2(3), zcr2(3)])H4=subs(hessmat,[x,y,z],[xcr2(4), ycr2(4), zcr2(4)])H5=subs(hessmat,[x,y,z],[xcr2(5), ycr2(5), zcr2(5)])

double(eig(H1))double(eig(H2))double(eig(H3))double(eig(H4))double(eig(H5))

Calculation results

Partial derivatives

f ' x=6x−9yz

f ' y=8y−9xz

f ' z=2z−9xy

Hessian matrix

H x , y , z= 6 −9z −9y−9z 8 −9x−9y −9x 2

Cordinates for critical points

0 0 04/9 2/9⋅3 4/9⋅3−4 /9 2/9⋅3 −4/9⋅3−4 /9 −2 /9⋅3 4/9⋅34/9 −2 /9⋅3 −4/9⋅3

Eigenvalues

H1 = (6, 8, 2)H2 = (14.0565, -4.3445, 6.2880)H3 = (14.0565, -4.3445, 6.2880)H4 = (14.0565, -4.3445, 6.2880)H5 = (14.0565, -4.3445, 6.2880)

Conclusion

The significance of the eigenvalues of the Hessian matrix is that if all of them are positive at a critical point, the function has a local minimum there; if all are negative, the function has a local maximum; if they have mixed signs, the function has a saddle point; and if at least one of them is 0, the critical point is degenerate.

According to this, our function results in a single local minimum and four saddle points.

Problem 3

Find and classify all critical points of the function

x4 y4xy−x2− y2 .

Relate your results to a simultaneous contour and gradient plot of the function.

Solution procedure

First we plot the f'x=0 and f'y=0 to find the points where f'x=f'y=0. We estimated the coordinates for when the curves intersected in the plot. We sent those estimated values into the newton2d function to get the exact values.

Then we calculate the second derivatives and use them to calculate the Hessian determinant, in which we use our coordinates.

MATLAB calculations

MATLAB code

syms x yk=(x^4 + y^4 + x*y -x^2 -y^2)kx=diff(k,x)ky=diff(k,y)[x1,y1]=meshgrid(-8:.1:8,-8:.1:8);kxfun=inline(vectorize(kx)); kyfun=inline(vectorize(ky));figure(1);contour(x1,y1,kxfun(x1,y1),[0,0],'r'), hold oncontour(x1,y1,kyfun(x1,y1),[0,0],'b'), hold off

[xsol,ysol]=newton2d(kx, ky, -0.8, 0.8); [xsol,ysol][xsol,ysol]=newton2d(kx, ky, 0.8, -0.8); [xsol,ysol][xsol,ysol]=newton2d(kx, ky, 0, 0); [xsol,ysol][xsol,ysol]=newton2d(kx, ky, -0.5, -0.5); [xsol,ysol][xsol,ysol]=newton2d(kx, ky, 0.5, 0.5); [xsol,ysol]

hessdetk = simplify(diff(kx, x)*diff(ky, y) - diff(kx, y)^2)

double(subs(hessdetk, [x,y], [-3^(1/2)/2, 3^(1/2)/2])) double(subs(hessdetk, [x,y], [3^(1/2)/2, -3^(1/2)/2])) double(subs(hessdetk, [x,y], [0, 0]))double(subs(hessdetk, [x,y], [-0.5, -0.5]))double(subs(hessdetk, [x,y], [0.5, 0.5]))

double(subs(diff(kx, x), [x,y], [-3^(1/2)/2, 3^(1/2)/2]))double(subs(diff(kx, x), [x,y], [3^(1/2)/2, -3^(1/2)/2]))double(subs(diff(kx, x), [x,y], [0,0]))

figure(2);ezcontour(k, [0.75, 0, 0.75, 0])

figure(3);ezsurf(k, [-1.5, 1.5, -1.5, 1.5])

Calculation results

Critical points (coordinates where kx = ky = 0)

(-0.8, 0.8) −3 /2 ,3/2(0.8, -0.8) 3/2 ,−3/2

(0,0) (0,0)

(-0.5, -0.5) (-0.5, -0.5)

(0.5, 0.5) (0.5, 0.5)

Left values are the values we estimated in the plot, and the right are the exact values according to newton2d function.

Hessian determinant

144x2 y2−24x2−24y23

Hessian determinants calculated for critical points

−3 /2 ,3/2 48

3/2 ,−3/2 48

(0,0) 3

(-0.5, -0.5) 0

(0.5, 0.5) 0

We have three local min- or maxpoints and two other points. To find out whether it is min or max, we have to look at the second derivatives.

Classified critical pointsWe put the first three points into any second derivate, to find out if they are minimum or maximum points.

We find the first two to be positive indicating that they're local minimum points. The third one has a negative secondary derivative, which means that it's a local maximum.

Plots

Figure 3. ezsurf

Figure 4. contour (kx=0: red ky=0: blue)

Figure 5. Ezcontour of one of the two undefined points (the other one is mirrored).

Conclusion

To define the two last points, we look at the contour plot above. We can see that they're saddle points.

−3 /2 ,3/2 min

3/2 ,−3/2 min

(0,0) max

(-0.5, -0.5) saddle

(0.5, 0.5) saddle

Problem 4

Find and classify all critical points of the function

x4 y4z 4xyz.

MATLAB will report many critical points, but only a few of them are real.

Solution procedure

To find critical points of the function, we must set the partial derivatives equal to 0 and solve for x, y and z. Then we calculate the Hessian matrix for the second derivatives.We discarded the non-real critical points.

MATLAB calculations

MATLAB code

syms x y zf=(x^4 + y^4 + z^4 + x*y*z)

gradf = jacobian(f,[x,y,z])hessmat = jacobian(gradf, [x,y,z])[xcr, ycr, zcr] = solve(gradf(1),gradf(2),gradf(3));[xcr, ycr, zcr]

H1=subs(hessmat,[x,y,z],[xcr(1), ycr(1), zcr(1)])H2=subs(hessmat,[x,y,z],[xcr(2), ycr(2), zcr(2)])H3=subs(hessmat,[x,y,z],[xcr(3), ycr(3), zcr(3)])H6=subs(hessmat,[x,y,z],[xcr(6), ycr(6), zcr(6)])H7=subs(hessmat,[x,y,z],[xcr(7), ycr(7), zcr(7)])

eigH1 = double(eig(H1))eigH2 = double(eig(H2))eigH3 = double(eig(H3))eigH6 = double(eig(H6))eigH7 = double(eig(H7))

Calculation results

Partial derivatives

f ' x=4x3 yz

f ' y=4y3 xz

f ' z=4z3 xy

Hessian matrix

H x , y , z =12x2 z yz 12y2 xy x 12z2

Critical points

x y z

0 0 0

-¼ ¼ ¼

-¼ -¼ -¼

¼ -¼ ¼

¼ ¼ -¼

Eigenvalues

0 0 0

0.250 1 1

0.250 1 1

0.250 1 1

0.250 1 1

Conclusion

Our eigenvalues determine the critical points as being minimum except the first one (0, 0, 0) which is degenerate.

Problem 5

Find and classify all critical points of the function

hx , y =y2⋅exp x2 −x−3y.

You will need the graphical/numerical method to find the critical points.

Solution procedure

We plot f'x = 0 and f'y = 0 and estimated the critical points where f'x = f'y = 0. We calculate the Hessian matrix, and puts in the values returned from the solve function. Then we calculate the eigenvalues to see which type of point it is.

MATLAB calculations

MATLAB code

syms x yh = y^2*exp(x^2) - x - 3*y

hx = diff(h, x)hy = diff(h, y)

[x1,y1]=meshgrid(0:.05:5,-.5:.05:1.5);hxfun=inline(vectorize(hx)); hyfun=inline(vectorize(hy));figure(1);contour(x1,y1,hxfun(x1,y1),[0,0],'r'), hold oncontour(x1,y1,hyfun(x1,y1),[0,0],'b'), hold off

gradh = jacobian(h,[x,y])hessmat = jacobian(gradh, [x,y])[xcr, ycr] = solve(gradh(1),gradh(2));[xcr, ycr]

H1=subs(hessmat,[x,y],[xcr(1), ycr(1)])eigH1 = double(eig(H1))H2=subs(hessmat,[x,y],[xcr(2), ycr(2)])eigH2 = double(eig(H2))

figure(2);ezsurf(h, [0.23, 0.26, 1.4, 1.45])

figure(3);ezsurf(h, [1.33, 1.35, 0.24, 0.26])

Calculation results

Partial derivatives

h ' x=2y2 x⋅exp x2−1

h ' y=2y⋅exp x2−3

Estimated coordinates for critical points

x y

0.25 1.42

1.34 0.25

Eigenvalues

H1 = (5.3429, 1.4987)H2 = (16.8761, -1.3836)

Where the first one is a minimum and the other a saddle point.

Plots

Figure 6. Contour (f'x = 0: red; f'y = 0: blue)

Figure 7. Minimum point

Figure 8. Saddle point.

Conclusion

The plots verifies that there is a minimum and a saddle point.

Problem 6

Find and classify all critical points of the function

h x , y = y2⋅exp x− x−3y.

Use both the analytical and the graphical/numerical methods to find the critical points, and compare the results.

Solution procedure

Analytical

To find critical points of h, we must set the partial derivatives equal to 0 and solve for x and y.We calculate the second derivatives and put them into the Hessian matrix. Then we calculate the eigenvalues which will classify our critical points.

Graphical/numerical

With a contour plot of the derivatives, we are able to narrow down the input range close to the cutting points. Then we calculate the exact values with the newton2d function.

MATLAB calculations

MATLAB code

syms x yh = y^2*exp(x) - x - 3*y

hx = diff(h, x)hy = diff(h, y)

[x1,y1]=meshgrid(-5:.01:10,-5:.01:5);hxfun=inline(vectorize(hx)); hyfun=inline(vectorize(hy));figure(1);contour(x1,y1,hxfun(x1,y1),[0,0],'r'), hold oncontour(x1,y1,hyfun(x1,y1),[0,0],'b'), hold off

[xsol,ysol]=newton2d(hx, hy, 0.81, 0.67);[xsol,ysol]

gradh = jacobian(h,[x,y])hessmat = jacobian(gradh, [x,y])[xcr, ycr] = solve(gradh(1),gradh(2));[xcr, ycr]

H1=subs(hessmat,[x,y],[xcr(1), ycr(1)])eigH1 = double(eig(H1))

figure(2);ezsurf(h, [log(9/4)-0.1, log(9/4)+0.1, 2/3-0.1, 2/3+0.1])

Calculation results

Critical point

(log(9/4), 2/3)

Hessian matrix

H x , y = y 2⋅ex 2y⋅ex

2y⋅ex 2⋅e x H log9/4 ,2/3=1 3

3 9/2Eigenvalues

6.2231-0.7231

Plots

Figure 9. Saddle point.

Conclusion

The graphical (newton2d) and the analytical (solve) methods returns the same results. Mixed signs in the eigenvalues results in a saddle point at the critical point, just as the plot show us.