## Loser66 one year ago On attachment Please, help

1. Loser66

2. Loser66

I would like to know from where we have "Hence $$x(t) = e^{tA}x_0$$ at 1.4, 1.5

3. freckles

4. freckles

like how did they get that equation?

5. Loser66

It didn't indicate to x(t) before. It worked on $$e^{tA}$$ and derivative of it. Suddenly, I have :" Hence x(t) =...." why?

6. ganeshie8

plug x(t) into the IVP $\dfrac{dx}{dt} = Ax~;~x(0)=x_0$ and see if it is really a solution

7. Loser66

If "plug x(t) into .." $$\dfrac{d}{dt}e^{tA}= Ae^{tA}$$ then $$\dfrac{d}{dt}x(t)= Ax(t)$$, right?

8. anonymous

since we have that $$\dfrac{d}{dt}e^{At}=Ae^{At}$$, it follows that $$x=e^{At}$$ is clearly a solution to $$\dfrac{dx}{dt}=Ax$$ (this is just a restatement of the previous fact)

9. ganeshie8

$$x(t) = e^{tA}x_0$$ right ?

10. Loser66

@ganeshie8 That is what I asked!! How and why we have that equation?

11. Loser66

and @oldrin.bataku gave the answer, but I didn't get how "it follows that x = e^(At) :(

12. ganeshie8

they are saying that function is a solution to the given IVP i was asking you to check if it is really the case by plugging x(t) into the differential eqn

13. ganeshie8

btw, $$x_0$$ is just a constant recall that if $$f(x)$$ is a solution , then a constant multiple of it is also a solution

14. anonymous

in fact, since the derivative is linear, we actually have $$\dfrac{d}{dt}\left(ke^{At}\right)=kAe^{At}$$ so the general solution is actually a whole parameterized family of solutions $$x(t)=kAe^{At}$$ for an $$n$$-by-$$1$$ constant matrix $$k$$

15. anonymous

oops, i meant $$x(t)=ke^{At}$$

16. anonymous

so if $$x(0)=x_0$$ then $$ke^{A\cdot0}=x_0\implies kI=x_0\implies k=x_0$$ so $$x(t)=x_0e^{At}=e^{At}x_0$$ is the unique solution to $$\dot x=Ax$$ subject to $$x(0)=x_0$$

17. Loser66

I got it. Thank you very much. I am confused with x_0. But now, I am ok with it.

18. ganeshie8

In the textbook, $$x_0$$ is a specific constant for the IVP. Maybe think of it like the usual $$c_0$$ if the past experience with simple ODE's helps..

19. Empty

I think it's cute how you can kinda pretend these aren't vectors and matrices and just kinda like $\frac{dx}{dt} = Ax$ "separate" $\frac{dx}{x} = Adt$ $\ln x = At +c$ $x=ke^{At}$

20. anonymous

note the idea here is actually very general and underlies what we call flows of vector fields -- $$d/dt$$ is an infinitesimal generator of the evolution of $$x$$ through time, i.e. it gives rise to flows

21. Loser66

I have another question, please explain me Ex 1

22. anonymous

which part is giving you trouble?

23. Loser66

1.31

24. anonymous

recall that for a matrix $$A$$ with columns $$A_1,A_2,A_3,\dots,A_n$$ the image of a column vector $$u=(u_1,\dots,u_n)^T$$ is $$Au=u_1A_1+\dots+u_nA_n$$

25. anonymous

i.e. the $$n$$-th column of the matrix is the action of the matrix on the $$n$$-th basis vector $$e_n$$, represented by a matrix of zeros with a $$1$$ in row $$n$$: $$e_1=\begin{bmatrix}1\\0\end{bmatrix},e_2=\begin{bmatrix}0\\1\end{bmatrix}$$so the matrix $$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$represents the transformation that takes $$e_1\to \begin{bmatrix}a\\c\end{bmatrix}$$ and $$e_2\to \begin{bmatrix}b\\d\end{bmatrix}$$

26. anonymous

so to figure out the first column of the matrix representing $$e^{At}$$, we just need to see how it behaves on $$\begin{bmatrix}1\\0\end{bmatrix}$$, and similarly for the second column

27. Loser66

I got what you meant . How about 1.30?

28. Loser66

1.33 is for $$e^{tA}(1,0)^T$$ that is the first column of $$e^{tA}$$ 1.34 is what? why not $$e^{tA}(0,1)^T$$ which is the second column of $$e^{tA}$$ Why it calculate $$e^{tA}(1,1)^T$$ ??

29. anonymous

consider that if $$v\ne 0$$ is an eigenvector of $$A$$, with $$Av=\lambda v$$, it follows that $$A^2v=\lambda^2 v$$ and more generally $$A^nv=\lambda^n v$$. so we have: $$e^{A}=\sum_{n=0}^\infty\frac1{n!} A^n\\e^Av=\left(\sum_{n=0}^\infty\frac1{n!} A^n\right)v=\sum_{n=0}^\infty\frac1{n!}\left(A^nv\right)=\sum_{n=0}^\infty\frac{\lambda^n}{n!}v=\left(\sum_{n=0}^\infty\frac{\lambda^n}{n!}\right)v=e^{\lambda}v$$

30. anonymous

if $$v$$ is an eigenvector of $$A$$ it's also an eigenvector of $$tA$$ as $$(tA)v=tAv=t\lambda v=(t\lambda )v$$ so:$$e^{tA}=e^{t\lambda}v$$

31. anonymous

so now we know the eigenvectors of $$e^{tA}$$ are those of $$A$$ and the eigenvalues $$\lambda$$ become $$e^{t\lambda }$$, which is what happened there

32. anonymous

now, since we know the eigenvectors $$v_1,\dots,v_n$$ of $$e^{tA}$$ and we have a vector $$v\in\operatorname{span}\{v_1,\dots,v_n\}$$ then we can decompose $$v_1=c_1v_1+\dots+c_nv_n$$ and it follows $$e^{tA}v=c_1 e^{tA}v_1+\dots+c_2 e^{tA}v_n=c_1 e^{t\lambda_1}v_1+\dots+c_n e^{t\lambda_n} v_n$$. this gives us a way to determine how $$e^{tA}$$ behaves on linear combinations of the eigenvectors

33. anonymous

so to figure out how $$e^{tA}$$ acts on $$e_1\begin{bmatrix}1\\0\end{bmatrix}$$ to figure out its first column we can decompose $$e_1=\dfrac12v_1+\dfrac12v_2$$ and $$e^{tA}e_1=\dfrac12 e^{tA}v_1+\dfrac12 e^{tA}v_2=\dfrac12e^{t\lambda_1}v_1+\dfrac12e^{t\lambda_2}v_2$$

34. Loser66

35. anonymous

since we have $$\lambda_1=1,\lambda_2=-1$$ and $$v_1=\begin{bmatrix}1\\1\end{bmatrix},v_2=\begin{bmatrix}1\\-1\end{bmatrix}$$ this says: $$e^{tA}\begin{bmatrix}1\\0\end{bmatrix}=\frac12e^t\begin{bmatrix}1\\1\end{bmatrix}+\frac12 e^{-t}\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}\frac12 e^t+\frac12 e^{-t}\\\frac12 e^t-\frac12 e^{-t}\end{bmatrix}$$ is our first column of $$e^{tA}$$

36. Loser66

Yes, that is 1.33

37. anonymous

in 1.34 they made a typo, they actually decomposed $$e_2=\frac12 v_1-\frac12 v_2$$ and then compute similarly to the above $$e^{tA}\begin{bmatrix}0\\1\end{bmatrix}=\frac12 e^t\begin{bmatrix}1\\1\end{bmatrix}-\frac12e^{-t}\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}\frac12 e^t-\frac12e^{-t}\\\frac12e^t+\frac12e^{-t}\end{bmatrix}$$

38. anonymous

so we've figured out the columns of $$e^{tA}$$ and can write it now as $$e^{tA}=e^{tA}\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}\frac12(e^t+e^{-t})&\frac12(e^t-e^{-t})\\\frac12(e^t-e^{-t})&\frac12(e^t+e^{-t})\end{bmatrix}$$

39. Loser66

Yes, I guessed it is a typo but it tortured me a lot since I didn't understand it. Now I got it. Thank you so so so much.

40. Loser66

Can I have another answer for another problem? The same topic, the same problem, just want to know the trick. Please.

41. Loser66

1.41 . Is there any way to quickly figure out $$-\dfrac{i+1}{4}$$ tem? and $$-\dfrac{i}{2}$$

42. anonymous

the easy way is to compute the inner products, since the eigenvectors $$\{v_j\}$$ are orthogonal we have $$\langle v_i,v_j\rangle=0\text{ where }i\ne j,\text{ or }|v_i|^2\text{ where }i=j$$

43. anonymous

so: $$\begin{bmatrix}1\\0\end{bmatrix}=c_1\begin{bmatrix}-2\\1+i\end{bmatrix}+c_2\begin{bmatrix}-2\\1-i\end{bmatrix}\\\begin{bmatrix}-2&1+i\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=c_1\begin{bmatrix}-2&1+i\end{bmatrix}\begin{bmatrix}-2\\1+i\end{bmatrix}+c_2\begin{bmatrix}-2&1+i\end{bmatrix}\begin{bmatrix}-2\\1-i\end{bmatrix}\\-2\cdot1+(1+i)\cdot0=c_1((-2)^2+(1+i)^2)+c_2((-2)^2+(1+i)(1-i))\\-2=c_1(4+1+2i-1)+c_2(4+1+1)\\-2=(4+2i)c_1+6c_2$$similarly we get$$\begin{bmatrix}-2&1-i\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=c_1\begin{bmatrix}-2&1-i\end{bmatrix}\begin{bmatrix}-2\\1+i\end{bmatrix}+c_2\begin{bmatrix}-2&1-i\end{bmatrix}\begin{bmatrix}-2\\1-i\end{bmatrix}\\-2\cdot1+(1-i)\cdot0=c_1((-2)^2+(1-i)(1+i))+c_2((-2)^2+(1-i)^2)\\-2=c_1(4+1+1)+c_2(4+1-2i-1)\\-2=6c_1+(4-2i)c_2$$now you can do substitution to solve $$c_1,c_2$$

44. Loser66

I got it. Again, thank you so much.

45. anonymous

oops they aren't orthonormal or even orthogonal here but they are linearly independent and we can still solve for unique $$c_1,c_2$$

46. Loser66

I did as usual and got a different solutions from the paper. I am checking.

47. anonymous
48. Loser66

Wow!! it takes a long time to work with :)