MATLAB: How to solve the two equations numerically in which tabular data is also to be loaded

numerical solution

I have two equations as;

(D-0.025)^1.5 * J(c)  = 0.0025;  
 c = 0.5 - 0.01/D;

Where J(c) is the particular value of J corresponding to value of c.

There is table (c vs J) to be loaded in which values of "J" (2nd column) corresponding to "c" (1st column) are given as under

 Table = [0.1 0.1156; 0.2 0.1590; 0.3 0.1892; 0.4 0.2117; 0.5 0.2288; 0.6 0.2416; 0.7 0.2505]

My problem: 1. I want to find optimal value of "D" and "J".

Also Note that its not important that the value of D exist in the table; so i also need to interpolate the value of "D"(existing in between two cells) and corresponding "J" value.

Best Answer

You can rewrite the interpolation as a polynomial of degree 6; it works out as

- (25*x^6)/12 + (145*x^5)/24 - (359*x^4)/48 + (499*x^3)/96 - (239*x^2)/100 + (10583*x)/12000 + 117/2500

You can then substitute that into your equations, converting all of your floating point values to rationals.

The system you get can then be solved in terms of a value that is the primary root of a degree 15 polynomial,

R = root([512 -15616 222208 -1962496 12118592 -56160800 203296128 -579497504 1277765482 -2115195465 2460287016 -1208021246 11455984710 -86217800113 215987875056 -180007884864])
D = (1/10)*(86016*R^14-2476544*R^13+32687360*R^12-257504256*R^11+1311552000*R^10-4612323392*R^9+12775515552*R^8-33673493056*R^7+86104656976*R^6-190723241666*R^5+393335706409*R^4+1248605641188*R^3-12283484659202*R^2+31089547519966*R-25920766050455)/(136192*R^14-3095552*R^13+19046656*R^12+161221632*R^11-3087622016*R^10+19694850944*R^9-63486597472*R^8+88752422720*R^7+101264091972*R^6-913631183772*R^5-936427888987*R^4+34659208386904*R^3-152565699173910*R^2+279861195719556*R-190086479566579)
c = (1/10)*(15616*R^14-444416*R^13+5887488*R^12-48474368*R^11+280804000*R^10-1219776768*R^9+4056482528*R^8-10222123856*R^7+19036759185*R^6-24602870160*R^5+13288233706*R^4-137471816520*R^3+1120831401469*R^2-3023830250784*R+2700118272960)/(3840*R^14-109312*R^13+1444352*R^12-11774976*R^11+66652256*R^10-280804000*R^9+914832576*R^8-2317990016*R^7+4472179187*R^6-6345586395*R^5+6150717540*R^4-2416042492*R^3+17183977065*R^2-86217800113*R+107993937528)

The values are approximately D = 0.07782594605604250, c = .3715081472598230

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That means it is a sparse matrix based on the format of the output. If you want a full matrix format, simply do this:

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MATLAB: How to find the jacobian of the system of nonlinear equations

Hi,.

try:

syms x1 x2 x3
h=4;
gma=18.4;
ka1=0.2;
kp1=8.76
ka2=0.2;
kp2=8.76;
sindel=0.437;
cosdel=0.9;
pa1=ka1*gma*(h*x1+0.5*x1^2);
pp1=kp1*gma*0.5*x1^2;
pa2=ka2*gma*(x1*x2*+0.5*x2^2);
pp2=kp2*(x2*(h+x1)+0.5*x2^2);
za1=(0.5*h*x1^2+(x1^3)/6)/(h*x1+0.5*x1^2);
zp2=(0.5*(h+x1)*x2^2+((x2^3)/3))/((h+x1)*x2+0.5*x2^2);
za2=(0.5*x1*x2^2+((x2^3)/3))+(x1*x2+0.5*x2^2);
e1=pp1*sindel-pa1*sindel-pp2*sindel-pa2*sindel;
e2=pp1*cosdel+pa2*cosdel-pa1*cosdel-pp2*cosdel-x3;
e3=pp1*cosdel*(x1/3)+pp2*cosdel*zp2-pa1*cosdel*za1-pa2*cosdel*za2-x3*(x1+(h/3));
g=[e1; e2; e3];
J=jacobian([e1, e2, e3], [x1, x2, x3])

results in:

J =
[                                                                                                                                                                                                                                                                                                                                                                                                                                                                     (1075457*x1)/15625 - (95703*x2)/25000 - (10051*x2^3)/12500 - 20102/3125,                                                                                                                                                                                                                                                                                                                                              - (95703*x1)/25000 - (95703*x2)/25000 - (30153*x1*x2^2)/12500 - 95703/6250,          0]
[                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  (207*x2^3)/125 - (1971*x2)/250 + (88596*x1)/625 - 1656/125,                                                                                                                                                                                                                                                                                                                                                            (621*x1*x2^2)/125 - (1971*x2)/250 - (1971*x1)/250 - 3942/125,         -1]
[ (44298*x1^2)/625 - x3 - (207*x2^3*(x1*x2 + (x1*x2^2)/2 + x2^2/2 + x2^3/3))/125 - (1656*x1)/125 + (x2^2*((1971*x2*(x1 + 4))/250 + (1971*x2^2)/500))/(2*(x2^2/2 + (x1 + 4)*x2)) - (((414*x1)/125 + 1656/125)*(x1^3/6 + 2*x1^2))/(x1^2/2 + 4*x1) + (1971*x2*(x2^3/3 + (x1/2 + 2)*x2^2))/(250*(x2^2/2 + (x1 + 4)*x2)) - (207*x1*x2^3*(x2^2/2 + x2))/125 + (((207*x1^2)/125 + (1656*x1)/125)*(x1^3/6 + 2*x1^2)*(x1 + 4))/(x1^2/2 + 4*x1)^2 - (x2*(x2^3/3 + (x1/2 + 2)*x2^2)*((1971*x2*(x1 + 4))/250 + (1971*x2^2)/500))/(x2^2/2 + (x1 + 4)*x2)^2, ((x2^3/3 + (x1/2 + 2)*x2^2)*((1971*x1)/250 + (1971*x2)/250 + 3942/125))/(x2^2/2 + (x1 + 4)*x2) - (207*x1*x2^3*(x1 + x2 + x1*x2 + x2^2))/125 - (621*x1*x2^2*(x1*x2 + (x1*x2^2)/2 + x2^2/2 + x2^3/3))/125 + ((2*x2*(x1/2 + 2) + x2^2)*((1971*x2*(x1 + 4))/250 + (1971*x2^2)/500))/(x2^2/2 + (x1 + 4)*x2) - ((x2^3/3 + (x1/2 + 2)*x2^2)*((1971*x2*(x1 + 4))/250 + (1971*x2^2)/500)*(x1 + x2 + 4))/(x2^2/2 + (x1 + 4)*x2)^2, - x1 - 4/3]

Best regards

Stephan

Best Answer

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