Linear ODEs with Constant Coefficients: Repeated Roots

[georgesadler]

Professor MathTheBeautiful, thank you for finding the Null Space of Linear ODEs with Constant Coefficients and Repeated Roots. This is an error free video/lecture on YouTube.

[xiasuyang]

3:35 "I'll just linearly dependent my ass" cracked me so hard, lol

[andrewszymczak]

Okay I figured out one way to show where the t exp(-3t) comes from. Take our characteristic polynomial (λ+3)(λ+3) and offset it by a small parameter ε so that it becomes (λ+(3+ε))(λ+(3-ε)). Then the solution is of the form

u(t) = A exp(-(3+ε)t) + B exp(-(3-ε)t).

If we solve for A and B using the initial conditions, the expression becomes Cexp(-3t) + D(exp(εt) - exp(-εt))/ε for some messy constants C and D. Taking the limit as ε-->0 gives us Cexp(-3t) + D t exp(-3t) as desired. I'm pretty sure you can generalize this to roots of higher multiplicity. For example offsetting the characteristic polynomial by (λ+3+ε+δ)(λ+3+ε-δ)(λ+3-ε) and then inducting.

Aposteriori I can show that t^m exp(αt) is in the null space for a characteristic polynomial that factors (λ-α)^n (using a comnbinatorial argument), but that doesn't answer the question of discovery.

There is a strong analogy between this and solving linear recurrences: (1 + n + n^2 + ...) α^n as compared to (1 + t + t^2 + ...) exp(αt). You can use generating functions to show the appearance of (1 + n + n^2 + ...) so I wonder if we can use a similar technique here. Probably just need another level of abstraction.

[andreemcaldas]

6:30 I would not have guessed either. But I think it is a good "linear algebra" reasoning if you notice that you do not have two eigen vectors (functions), so, you could try an "extended eigen function". If U is an eigen function with eigen value a, then tU is an "extended eigen function", since its derivative is U + atU. It is like a triangular matrix with first column (a, 0) and second column (1, a). Jordan form!!! In other words, if you use the notation D to represent the operation of deriving then, (D-a)U = 0, and (D-a)(tU) = U. If you apply (D-a) twice, you send both U and tU to 0. This is what the homogeneous system asks you to find! Vectors such that if you apply (D-a) twice, you get 0. Sorry for my bad math typesetting!

[mirijason]

To come up with the second solution, just use the variation of constants. Take the solution as u(t)=C(t)*exp(-3*t). Plugging in the ODE, one gets after cancellation C ' ' (t) = 0. This equation has the general solution C(t)=a+b*t whose constants a and b are the two linearly independent parameters.

[elael]

In control the te^t appears naturally from the matrix exponential.
we would consider the column vectors x = [u'; u] and x' = [u''; u], from the equation we have u'' = -6u' -9u
Thus the system Ax = x' with A = [-6, -9; 1, 0]
has solution x(t) = e^(At) x(0)
where x(0) = [u'(0); u(0)] are the initial conditions.

[pawej]

6:38 istn't t*e^(-3*t) analogous to generalised eigenvector?
This situation in which we have repeated root of the characteristic polynomial looks similar to defective matrix problem.
I think that idea behind this extra t is just looking for "generalised eigenfunction"(?) of this ODE.

Let v be an eigenfunction of this ODE and g be generalised eigenfuction

(A-Iλ)g=v
Ag-Iλg=v
Ag=v+λg
so we're looking for a function g that, when ODE is applied, becomes not itself*constant (like an eigenfunction) but
Because we know that eigenfunction is e^(ct) we can write:

Ag=e^(ct)+λg

and now, using the product rule, we know that g=t*e^(ct)

we can also check it for 1st order ODE:
[t*e^(ct)]'=e^(ct)+c*te^(ct)
for 2nd order:

[rishabhgarg]

These videos are really helpful. Thanks for these. Can you tell which of the topics of Maths you gonna cover in near future ?

[debendragurung]

I wonder whoever thought of these first werent humans