Solving System x'' + αx = 0: An Eigenform Approach

[c0der]

Homework Statement

Solve the system x'' + αx = 0

x = [ x1 x2 ]
α constant

Homework Equations

x'' + αx = 0 Where x is a vector [x1 x2]

The Attempt at a Solution

The solution of x'' + αx = 0 is x = Acosh(√α x) + Bsinh(√α x)

Differentiating x

x' = √α*Acosh(√α x) + √α*Bsinh(√α x)
x'' = α*Acosh(√α x) + α*Bsinh(√α x)

Plugging this back into the original ODE

α*Acosh(√α x) + α*Bsinh(√α x) - α(Acosh(√α x) + Bsinh(√α x)) = 0 is satisfied

I can't get this in Eigenform to solve for the system. What do I do in this case?

 

[HallsofIvy]

Homework Statement

Solve the system x'' + αx = 0

x = [ x1 x2 ]
α constant

Homework Equations

x'' + αx = 0 Where x is a vector [x1 x2]

The Attempt at a Solution

The solution of x'' + αx = 0 is x = Acosh(√α x) + Bsinh(√α x)

That makes no since at all. x is a two dimensional vector but you have it as both a single function and the independent variable!
IF we were given a single real function, x, satisfying x''+ αx= 0 THEN the solution would be either x= Asinh(√αt)+ Bcosh(√αt), x= Asin(√αt)+ Bcos(√αt), or x= At+ B, depending upon whether α was positive, negative, or 0 and t is the (unnamed in the equation) independent variable.

Are you given that α single real number? If so then the general solution would be
$$\begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix}= \begin{bmatrix}Asinh(√αt)+ Bcosh(√αt) \\ Csinh(√αt)+ Dcosh(√αt)\end{bmatrix}$$
if α> 0,
$$\begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix}= \begin{bmatrix}Asin(√αt)+ Bcos(√αt) \\ Csin(√αt)+ D cos(√αt)\end{bmatrix}$$
lf α< 0,
and
$$\begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix}= \begin{bmatrix}At+ B \\ Ct+ D\end{bmatrix}$$
if α= 0.

Differentiating x

x' = √α*Acosh(√α x) + √α*Bsinh(√α x)
x'' = α*Acosh(√α x) + α*Bsinh(√α x)

Plugging this back into the original ODE

α*Acosh(√α x) + α*Bsinh(√α x) - α(Acosh(√α x) + Bsinh(√α x)) = 0 is satisfied

I can't get this in Eigenform to solve for the system. What do I do in this case?

I'm not sure what you mean by "eigenform" but if α is a single number then the differential equation can be written in matrix form as
$$\begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix}'= \begin{bmatrix}α & 0 \\ 0 & α\end{bmatrix}\begin{bmatrix}x_1(t)\\ x_2(t)\end{bmatrix}$$
which has α as the only eigenvalue.

 

[c0der]

Sorry wasn't clear enough

Apologies, I should have been clear. This is the system:

d^2/dx^2(y₁) = α(y₁ - y₂)
d^2/dx^2(y₂) = α(y₂ - y₁)

Hence, why your set of solutions won't work. This is why I tried to put it into eigenform, like the harmonic oscillator problem, which is easily done to get the frequencies for an mdof system. This method doesn't work for this type of problem

 

[pasmith]

Apologies, I should have been clear. This is the system:

d^2/dx^2(y₁) = α(y₁ - y₂)
d^2/dx^2(y₂) = α(y₂ - y₁)

This is why I tried to put it into eigenform, like the harmonic oscillator problem, which is easily done to get the frequencies for an mdof system. This method doesn't work for this type of problem.

It is obvious from inspection that if $u = y_1 + y_2$ and $v = y_1 - y_2$ then
$$\frac{d^2 u}{dx^2} = 0 \\<br /> \frac{d^2 v}{dx^2} = 2av$$

 

[c0der]

Yes but how do you solve it formally? I use the eigenvalue method to get:

y1 = 0
y2 = Acosh(sqrt(2a)x) + Bsinh(sqrt(2a)x) but differentiating this twice and subbing this into the second equation and solving for y1 doesn't satisfy the boundary conditions