Please help me with this math problem?

Question

Let J=(λ, 0, 1, λ), where λ is an arbitrary real number. (2x2 matrix, λ and 0 on the left, 1 and λ on the right.)

 

(a) Find J^2, J^3, and J^4.

(b) Use an inductive argument to show that J^n=(λ^n, 0, nλ^(n-1), λ^n).

(c) Determine exp(Jt).

(d) Use exp(Jt) to solve the initial value problem x'=Jx, x(0)=x^0.

Answer 1

Let me continue the Andre's answer:

 

For e^Jt he obtained the following:

 

e^Jt=Σ_n=0^∞(1/n!){(λⁿ, 0);(nλ^n-1, λⁿ)}tⁿ=Σ_n=0^∞{( (λⁿtⁿ)/n!, 0);( (nλ^n-1tⁿ)/n!, (λⁿtⁿ)/n!)}=

=[Σ_n=1^∞{( (λt)ⁿ/n!, 0); ( (λt)^n-1t/(n-1)!, (λt)ⁿ/n!)}]+{(1,0);(0,1)}={ (e^λt, 0); (te^λt; e^λt) }

 

So, e^Jt={ (e^λt, 0); (te^λt; e^λt) }

 

The initial value problem x′=Jx; x(0)=x^0; has a solution in the form:

 

x=x^0*e^Jt; Since we determined e^Jt in the previous part, we can formally multiply matrix e^Jt by a vector x^0 to get a solution. I assume x^0 means (1,1) vector. Then the solution is:

 

x(t)=(e^λt,(t+1)e^λt)

Answer 2

I'm assuming you know how to multiply matrices and what a Taylor series is. If not, let me know.

 

(a)  J² just means multiply matrix J by itself, J² = J*J. Similarly, J³=J*J*J etc.

 

Let's do J² for your matrix:

 

J²=J*J=(λ, 0, 1, λ)*(λ, 0, 1, λ)=(λ², 0, 2λ, λ²)

 

Similarly,

 

J³=J²*J=(λ², 0, 2λ, λ²)*(λ, 0, 1, λ)=(λ³, 0, 3λ², λ³)

J⁴=J³*J=(λ³, 0, 3λ, λ³)*(λ, 0, 1, λ)=(λ⁴, 0, 4λ³, λ⁴)

 

(b) In part (a), we see a pattern emerging, so we claim

 

Jⁿ=(λⁿ, 0, nλ^n-1, λⁿ) for any positive integer n.

 

The inductive argument goes like this:

 

1. The statement is true for n=1, because

 

J¹=J=(λ, 0, 1, λ) by definition.

 

2. Let the statement be true for some arbitrary but fixed n=k:

 

J^k=(λ^k, 0, kλ^k-1, λ^k)

 

Then for n=k+1

 

J^k+1=J^k*J=(λ^k, 0, kλ^k-1, λ^k)* (λ, 0, 1, λ)=(λ^k+1, 0, (k+1)λ^k, λ^k+1).

 

Therefore, by the principle of mathematical induction, the statement is true for all n=1,2,3,...

 

(c) The matrix exponential exp(A) of a matrix A is defined as a Taylor series of matrices analogous to the Taylor series of the function e^t about t=0:

 

exp(A)= I + A + 1/2! A² + 1/3! A³ + .... =Σ( (1/n!) Aⁿ, n=0..infinity),

 

where I is the identity matrix and A⁰=I.

 

Therefore,

 

exp(Jt) = Σ( (1/n!) (Jt)ⁿ, n=0..infinity) = Σ( (1/n!) Jⁿtⁿ, n=0..infinity)

 

For our matrix J=(λ, 0, 1, λ), we showed in part (b) Jⁿ=(λⁿ, 0, nλ^n-1, λⁿ), so that

 

exp(Jt) =Σ( (1/n!) (λⁿ, 0, nλ^n-1, λⁿ) tⁿ, n=0..infinity)

 

We can distribute the summation and the tⁿ into each entry and turn the series of matrices into a matrix of series. We get

 

exp(Jt) = (Σ (1/n!)λⁿtⁿ, 0, Σ (1/n!) nλ^n-1tⁿ, Σ (1/n!) λⁿtⁿ)

 

Now use the definition of the Taylor series of the function e^λt about t=0 and get

 

exp(Jt) = ( e^λt, 0, t e^λt, e^λt).

 

(d) We have the initial value problem

 

x'=Jx, with x(0)=x₀

 

I claim that the solution is

 

x= exp(Jt) x₀.

 

Actually, this statement is true for all matrices J, but the proof is especially simple for the triangular matrix J given in this problem, J=(λ, 0, 1, λ). I will show that x' and Jx are equal to the same thing.

 

First,

 

x' = (exp(Jt) x₀)' =  d/dt (e^λt, 0, t e^λt, e^λt) x₀ =  ( λ e^λt, 0, λt e^λt + e^λt, λ e^λt) x₀

 

Second,

 

J x = (λ, 0, 1, λ)*(exp(Jt) x₀) = (λ, 0, 1, λ)*(e^λt, 0, t e^λt, e^λt) x₀

= ( λ e^λt, 0, λt e^λt+ e^λt, λ e^λt) x₀

 

Both sides of x'=Jx are equal to ( λ e^λt, 0, λt e^λt+ e^λt, λ e^λt) x₀, so

 

x= exp(Jt) x₀

 

is a solution to the equation. By the uniqueness theorem it is the only solution. qed