Finding a Solution for a Differential System with Initial Conditions

[skrat]

Homework Statement

Find a solution of the system ${x}'''=2x+y$ and ${y}'''=x+2y$ for which $x(0)={x}'(0)=0$ and ${x}''(0)=1$ also for $y(0)={y}'(0)=0$ and ${y}''(0)=1$.

Homework Equations

The Attempt at a Solution

I must be doing something wrong:

$\begin{bmatrix} {x}'''\\ {y}''' \end{bmatrix}=\begin{bmatrix} 2 & 1\\ 1&2 \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}$

Eigenvalues are $\lambda _1=1$ and $\lambda _2=3$ so eigenvectors $v_1=(-1,1)$ and $v_2=(1,1)$.

Therefore matrix $D=\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}$ and matrix $P=\begin{bmatrix} -1 & 1\\ 1&1 \end{bmatrix}$

So general solution should be $\begin{bmatrix} x\\ y \end{bmatrix}=Pe^{Dx}\vec{c}=\begin{bmatrix} -1 & 1\\ 1&1 \end{bmatrix}\begin{bmatrix} e^x & 0\\ 0& e^{3x} \end{bmatrix}\begin{bmatrix} A\\ B \end{bmatrix}$

But all these conditions $x(0)={x}'(0)=0$ and ${x}''(0)=1$ also for $y(0)={y}'(0)=0$ and ${y}''(0)=1$... What do I do? :/

Thank you for your help!

 

[ehild]

Homework Statement

Find a solution of the system ${x}'''=2x+y$ and ${y}'''=x+2y$ for which $x(0)={x}'(0)=0$ and ${x}''(0)=1$ also for $y(0)={y}'(0)=0$ and ${y}''(0)=1$.

Homework Equations

The Attempt at a Solution

I must be doing something wrong:

$\begin{bmatrix} {x}'''\\ {y}''' \end{bmatrix}=\begin{bmatrix} 2 & 1\\ 1&2 \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}$

Eigenvalues are $\lambda _1=1$ and $\lambda _2=3$ so eigenvectors $v_1=(-1,1)$ and $v_2=(1,1)$.

Therefore matrix $D=\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}$ and matrix $P=\begin{bmatrix} -1 & 1\\ 1&1 \end{bmatrix}$

So general solution should be $\begin{bmatrix} x\\ y \end{bmatrix}=Pe^{Dx}\vec{c}=\begin{bmatrix} -1 & 1\\ 1&1 \end{bmatrix}\begin{bmatrix} e^x & 0\\ 0& e^{3x} \end{bmatrix}\begin{bmatrix} A\\ B \end{bmatrix}$

But all these conditions $x(0)={x}'(0)=0$ and ${x}''(0)=1$ also for $y(0)={y}'(0)=0$ and ${y}''(0)=1$... What do I do? :/

Thank you for your help!

λ2 is not 3, but third root of 3. Both x and y are functions of the same variable, say t. And you tried to find the solutions as x(t)=ae^λt and y(t)=be^λt. You cannot write exponents of x into the solution.

What does the last matrix equation mean for x(t) and y(t)? ( Expand the product.) You have to find the constants A and B so as the initial conditions are fulfilled.

ehild

 

[skrat]

$det\begin{bmatrix} 2-\lambda &1 \\ 1 & 2-\lambda \end{bmatrix}=(2-\lambda )^2-1=\lambda ^2-4\lambda +3=(\lambda -1)(\lambda -3)=0$ ?

am... it means that $\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} -Ae^x+Be^{3x}\\ Ae^x+Be^{3x} \end{bmatrix}$ but i have 6 conditions for only two constants...

 

[ehild]

It is not a first order but third order system of equations. And your matrix equation means that x= -Ae^x+Be^3x. Do you think it true? ehild

 

[skrat]

It is not a first order but third order system of equations.

Am, I can see the problem but I have to admit that I don't know what to do? Should I calculate the eigen values of matrix $A^3$ ?

And your matrix equation means that x= -Ae^x+Be^3x. Do you think it true?

Uf, good point, I guess the $'$ mean derivation by t and not by x.

 

[ehild]

Am, I can see the problem but I have to admit that I don't know what to do? Should I calculate the eigen values of matrix $A^3$ ?

No, why ? Still, yo can assume the solutions of form x=ae^λt, y=be^λt). Substitute into the original equations. You get values 1 and 3 for λ³. But λ is a complex number, and you have the complex third roots, 6 values altogether.

Or you can introduce dummy variables u=dx/dt, v=du/dt and p=dy/dt, q=dp/dt to transform the original equations into first order ones. Then you have a system of six first order equations and you can solve it with the standard method.

ehild

 

[skrat]

Aaa, ok, i get it now... Is this a standard procedure for systems that are second orders or higher?

So I will get a polynomial $\lambda ^6-4\lambda ^3+3=(\lambda -1)(\lambda ^3-3)(\lambda ^2+\lambda +1)$

This gives me $\lambda _1=1$ and $\lambda _{1,2}=\frac{1}{2}(-1\pm i\sqrt{3})$ and $\lambda _3= \sqrt[3]{3}$

How exactly do I get the other two?

 

[ehild]

λ⁶-4λ³+3=0 is a quadratic equation in λ³, with solutions λ³=1 and λ³=3.

So the roots are 1, (-1/2±√3/2 i) and 3^1/3, 3^1/3(-1/2±√3/2 i)

(You can factorize λ³-3 in the same way you did with λ³. If a=3^1/3, λ³-3=(λ-a)(λ²+aλ+a²).)

ehild

 

[skrat]

Thank you, now I know how to finish this!