Determining polynomial coefficients by known derivatives

Hello community, I'm trying to create a cubic polynomial function, I know these: y(0)=1 y(30)=0.5 dy(0)/dx=-1 dy(30)/dx=-1 Can I determine the values for coefficients ?

Answer 1

[Dr. Seis]

Yes. Just in case this is homework, I will give an example to get you started.

Let's say you have the case below (I will start with the case where we do not have derivatives at a location).

We are trying to solve for A, B, and C

y(x) = A*x^2 + B*x + C

Knowns:

x1 = 0; y1 = 1;
x2 = 3; y2 = 6;
x3 = 7; y3 = 7;

So we can construct an inverse problem of the form G * m = d, where:

G = [x1^2,x1,1; x2^2,x2,1; x3^2,x3,1];
d = [y1; y2; y3];

We determine m (equal to [A; B; C] in our example) by:

m = G\d;

And we get:

plot(-1:10,polyval(m,-1:10),'r-', [x1,x2,x3],[y1,y2,y3],'k+')

[Dr. Seis]

In your case you have 2 known derivative points... based on the example/hint above, how would you define the G matrix to account for derivative information?

[Omur]

Maybe I need do define an another matrix as dg=diff(G); But how it will be calculated in sync with main matrix, also I only know start and end of the function, I have missing points which require some sort of interpolation. I have written dynamic analysis FEM codes but this thing struggled me.

[Dr. Seis]

Nope... you can put the derivative information right in the same G matrix. Another hint is below:

In your case you have information associated with:

y(x) = A*x^2 + B*x + C

and

y'(x) = 2*A*x + B

So... if you have y(x) info and y'(x) info you would set up your matrix equation like:

| x1^2,  x1,  1 | | A |      | y1    |
| x2^2,  x2,  1 | | B |  =   | y2    |
|               | | C |      | dydx1 |
|               |            | dydx2 |

How would you fill in the last 2 rows of G ?

[Dr. Seis]

Incidentally, I think you meant your dy(30)/dx = +1 (not -1). Once you factor that in and fill out your G matrix you should obtain something like:

[Omur]

Like 2x1 0 0 2x2 0 0 ? (How you do line down and write codes?) No, it's derivative is -1, therefore function is cubic. I guess ? Elige, I'm confused.

[Dr. Seis]

Almost... since you will need to expand to a 3-order polynomial, I will give you the G for a 2-order (which would give you the plot immediately above if dydx2 = +1). In your case, your y and y' info are at the same 2 x locations... so:

| x1^2,  x1,  1 | | A |      | y1    |
| x2^2,  x2,  1 | | B |  =   | y2    |
| 2*x1,   1,  0 | | C |      | dydx1 |
| 2*x2,   1,  0 |            | dydx2 |

You still need to have contribution from your "B", which is why the second column of the last two rows have the 1. Once you expand to a 3-order polynomial for your data, you should get something like:

[Dr. Seis]

To learn more about using mark-up to format your questions/answers/comments you can check out this page:

http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup

[Omur]

Thank you so much !

Answer 2

[Suzie Oman]

This answer doesn't incorporate the derivative. How would this be solved taking into account the derivative?

What about in 3 dimensions?

[Star Strider]

See this Comment (link).