Finding the Value of c for 4u"+cu'+6u=0

[mathgirl2007]

Find the value of the constant c so that solutions of the equation 4u"+cu'+6u=0 tend to zero as fast as possible.


I think that c must be positive in order for this to tend towards zero but I cannot figure out what c has to be.


Thanks for your help!

 

[gabbagabbahey]

What is the characteristic polynomial of this ODE? What are the solutions (in terms of c) of that polynomial? What does that make the general solution for u?

 

[mathgirl2007]

this is all i was given.
would it be that m=4 lambda=c and k=6?

 

[gabbagabbahey]

Have you studied ODE's yet?? Are you familiar with the term characteristic polynomial?

For example, the characteristic polynomial of 2u''+5u'-3=0 is 2 \lambda^2+5\lambda-3=0 which has roots \lambda_1= -3 and \lambda_2=\frac{1}{2} and a general solution of u(x)=c_1e^{\lambda_1 x} +c_2e^{\lambda_2 x}

So, what is the characteristic polynomial of 4u" + cu'+6u=0?

 

[mathgirl2007]

So the equation would be 4\lambda^2+C\lambda+6=0

 

[mathgirl2007]

and the roots would be (-C+/- the square root(c^2 - 96))/8

 

[gabbagabbahey]

Yes, so what is the general solution u(x) then?

 

[mathgirl2007]

u(x) = c1e^(-C + the square root(c^2 - 96))/8)x + C2e^(-C - the square root(c^2 - 96))/8

 

[gabbagabbahey]

Good, now what does it mean for a function to tend to zero? (in terms of the derivative of the function)

 

[mathgirl2007]

its going to have to be getting infinitely smaller

 

[gabbagabbahey]

Yes, it gets smaller as x-gets larger and so du/dx is negative correct?

What is du/dx of your function?

 

[mathgirl2007]

im not sure how to find that here. is that the derivative in regards to x?

 

[mathgirl2007]

im really stuck. would it be that C must b be 1 to tend towards zero?

 

[gabbagabbahey]

Yes, du/dx is the derivative with regard to x...You know what u(x) is, so calculate u'.

 

[mathgirl2007]

okay i did that. now where in there can i find what c is?

 

[gabbagabbahey]

Well u' should be as large of a negative number as possible if u tends to zero as fast as possible correct?

In other words you want to find the minumum value of u'(x) with respect to c. The derivative of u'(x) with respect to c should be zero at its minimum, so solve the equation

\frac{d u'}{dc}=0

for c.

 

[klondike]

"(the solution) tends to zero as fast as possible"

My interpretation is that the transient response is "as short as possible".
given a second order system ay''+by'+cy=dx'+ex under zero input excitation or ay''+by'+cy=0, the response is due to non-zero initial condition(s). In order for the solution to converge (stable system and non oscillating), it requires either
1) a>0,b>0, and c>0, OR
2) a<0, b<0, and c<0
such that all poles are on the left half of s-plane (excluding j \omega) axis. In the other word, the Fourier transform must exist.

since the OP mentioned m, c, and k, I'm assuming she is dealing with a spring-mass-damper system (my&#039;&#039;+cy&#039;+ky=0 with non-equilibrium initial displacement
), I bet the answer is when \zeta=1 or critically damped. Note that when system is critically damped (double real roots), the solution doesn't take the form as other poster mentioned.

 

[gabbagabbahey]

"(the solution) tends to zero as fast as possible"

My interpretation is that the transient response is "as short as possible".
given a second order system ay&#039;&#039;+by&#039;+cy=dx&#039;+ex under zero input excitation or ay&#039;&#039;+by&#039;+cy=0, the response is due to non-zero initial condition(s). In order for the solution to converge (stable system and non oscillating), it requires either
1) a>0,b>0, and c>0, OR
2) a<0, b<0, and c<0
such that all poles are on the left half of s-plane (excluding j \omega) axis. In the other word, the Fourier transform must exist.

since the OP mentioned m, c, and k, I'm assuming she is dealing with a spring-mass-damper system (my&#039;&#039;+cy&#039;+ky=0 with non-equilibrium initial displacement
), I bet the answer is when \zeta=1 or critically damped. Note that when system is critically damped (double real roots), the solution doesn't take the form as other poster mentioned.

Yes, that makes more sense :0)