Solve the following initial value problems: (a) (d 𝐮)/(d t)=( 0 2 2 0 ) 𝐮, 𝐮(1

Solve the following initial value problems: (a) (d 𝐮)/(d...

Solve the following initial value problems: $(a) \frac{d \mathbf{u}}{d t}=\left(\begin{array}{ll}0 & 2 \\ 2 & 0\end{array}\right) \mathbf{u}, \mathbf{u}(1)=\left(\begin{array}{l}1 \\ 0\end{array}\right)$; (b) $\frac{d \mathbf{u}}{d t}=\left(\begin{array}{rr}1 & -2 \\ -2 & 1\end{array}\right) \mathbf{u}, \mathbf{u}(0)=\left(\begin{array}{r}-2 \\ 4\end{array}\right)$; (c) $\frac{d \mathbf{u}}{d \ell}=\left(\begin{array}{rr}1 & 2 \\ -1 & 1\end{array}\right) \mathbf{u}, \mathbf{u}(0)=\left(\begin{array}{l}1 \\ 0\end{array}\right)$; (d) $\frac{d \mathbf{u}}{d t}=\left(\begin{array}{rrr}-1 & 3 & -3 \\ 2 & 2 & -7 \\ 0 & 3 & -4\end{array}\right) \mathbf{u}, \mathbf{u}(0)=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)$; (e) $\frac{d \mathbf{u}}{d t}=\left(\begin{array}{rrr}2 & 1 & -6 \\ -1 & 0 & 4 \\ 0 & -1 & -2\end{array}\right) \mathbf{u}, \mathbf{u}(\pi)=\left(\begin{array}{r}2 \\ -1 \\ -1\end{array}\right)$; (f) $\frac{d \mathbf{u}}{d t}=\left(\begin{array}{llll}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \\ 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0\end{array}\right) \mathbf{u}, \mathbf{u}(2)=\left(\begin{array}{l}1 \\ 0 \\ 0 \\ 1\end{array}\right) ; \quad(g) \frac{d \mathbf{u}}{d t}=\left(\begin{array}{rrrr}2 & 1 & -1 & 0 \\ -3 & -2 & 0 & 1 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 1 & -1\end{array}\right) \mathbf{u}, \quad \mathbf{u}(0)=\left(\begin{array}{r}1 \\ -1 \\ 2 \\ 1\end{array}\right)$.

Solve the following initial value problems: $(a) \frac{d \mathbf{u}}{d t}=\left(\begin{array}{ll}0 & 2 \\ 2 & 0\end{array}\right) \mathbf{u}, \mathbf{u}(1)=\left(\begin{array}{l}1 \\ 0\end{array}\right)$;
(b) $\frac{d \mathbf{u}}{d t}=\left(\begin{array}{rr}1 & -2 \\ -2 & 1\end{array}\right) \mathbf{u}, \mathbf{u}(0)=\left(\begin{array}{r}-2 \\ 4\end{array}\right)$;
(c) $\frac{d \mathbf{u}}{d \ell}=\left(\begin{array}{rr}1 & 2 \\ -1 & 1\end{array}\right) \mathbf{u}, \mathbf{u}(0)=\left(\begin{array}{l}1 \\ 0\end{array}\right)$;
(d) $\frac{d \mathbf{u}}{d t}=\left(\begin{array}{rrr}-1 & 3 & -3 \\ 2 & 2 & -7 \\ 0 & 3 & -4\end{array}\right) \mathbf{u}, \mathbf{u}(0)=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)$;
(e) $\frac{d \mathbf{u}}{d t}=\left(\begin{array}{rrr}2 & 1 & -6 \\ -1 & 0 & 4 \\ 0 & -1 & -2\end{array}\right) \mathbf{u}, \mathbf{u}(\pi)=\left(\begin{array}{r}2 \\ -1 \\ -1\end{array}\right)$;
(f) $\frac{d \mathbf{u}}{d t}=\left(\begin{array}{llll}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \\ 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0\end{array}\right) \mathbf{u}, \mathbf{u}(2)=\left(\begin{array}{l}1 \\ 0 \\ 0 \\ 1\end{array}\right) ; \quad(g) \frac{d \mathbf{u}}{d t}=\left(\begin{array}{rrrr}2 & 1 & -1 & 0 \\ -3 & -2 & 0 & 1 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 1 & -1\end{array}\right) \mathbf{u}, \quad \mathbf{u}(0)=\left(\begin{array}{r}1 \\ -1 \\ 2 \\ 1\end{array}\right)$.

Key Concepts

Jordan Canonical Form

When a matrix cannot be diagonalized due to a lack of sufficient independent eigenvectors, the Jordan canonical form offers a generalized framework. It organizes the matrix into Jordan blocks, allowing for the computation of the matrix exponential even in cases of defective matrices by taking into account the structure of these blocks.

Diagonalization

Diagonalization involves expressing the system matrix as PDP?¹, where D is a diagonal matrix containing the eigenvalues and P is the matrix of corresponding eigenvectors. This simplifies many operations, including finding the matrix exponential, as the exponential of a diagonal matrix is simply the exponential of its individual entries.

Initial Value Problems

An initial value problem specifies the value of the unknown function(s) at a particular starting point. The provided initial conditions ensure the uniqueness of the solution for the differential equation system, guiding the application of methods like the matrix exponential to yield the correct, particular solution.

Matrix Exponential

The matrix exponential is a method used to solve linear differential equations with constant coefficients by computing e^(At). This approach transforms the differential equation into an algebraic form and allows the solution to be expressed succinctly as u(t) = e^(A(t-t?))u(t?), making it a powerful tool for both theoretical and practical applications.

Linear Systems of Differential Equations

This concept refers to an ordered set of differential equations where the derivatives of several unknown functions are linked linearly through constant coefficients. It provides a framework for modeling and solving problems by translating them into matrix form, which leverages linear algebra tools.

Eigenvalues and Eigenvectors

These are fundamental in analyzing and solving linear systems. By determining the eigenvalues and eigenvectors of the system matrix, one can often decouple the system into independent components. They play a crucial role in understanding the behavior of the system and in simplifying the computation of the matrix exponential through diagonalization.

Transcript

00:01 So given the matrix a in our initial condition x not, we want to find the initial value problem given.

00:08 We want to solve that.

00:09 So first we need to find our eigenvalues and our eigenvectors.

00:16 So first let's find eigenvalues.

00:20 So in order to do that, we set determinant of a minus lambda i equal to zero.

00:28 So we have the determinant is equal to the determinant of 1 -5.

00:39 Half minus lambda because we're going to subtract a lambda along the diagonal.

00:45 That's what the lambda i means.

00:47 And then we have negative three halves on the off diagonals.

00:52 And then we have another one half minus lambda.

00:57 So we're going to take the determinant.

01:00 So we have one half minus lambda squared minus negative three halves times negative three halves gives us 9 over 4, and we want to set this equal to 0.

01:21 So 1 .5 minus lambda squared is equal to 9 over 4, so 1 half minus lambda is equal to plus or minus 3 halves.

01:40 So solving this, we get lambda is equal to 1 half plus or minus 3 halves.

01:48 So we have our lambda 1 is equal to 1 plus 3 over 2, which gives us 2, and lambda 2 is 1 minus 3, which is negative 2 over 2, so negative 1.

02:11 So these are our eigenvalues.

02:15 So now we need to find the corresponding eigenvectors.

02:23 So first let's find the 1 corresponding to lambda 1 equal to 2.

02:33 So we want to set a minus slame to i times x equal to 0.

02:40 So 1 1 1 .2 is negative 3 halves.

02:47 So we're going to plug that in for the diagonal.

02:50 And then the off diagonal, we're also negative 3 halves.

02:56 And then we're going to multiply our general vector x1, x2, and we set this equal to the 0 vector.

03:06 So this tells us negative three halves x1 minus three halves x2 equals zero and both equations are the same.

03:22 So what does this tell us? negative three halves x1 is equal to positive three halves times x2.

03:32 So x1 is equal to minus x2.

03:37 So x1 and x2 have to be opposites.

03:40 So that's all we're given.

03:42 So we can let x1 or x2 be anything.

03:47 They just have to satisfy this relation.

03:51 So let's just pick x1 equals 1, and then so our x2 has to be the opposite.

04:01 So this is our first eigenvector.

04:07 So similarly, let's find the eigenvector corresponding to lambda 2 equals minus 1.

04:20 So we have a minus lambda i.

04:24 So one half minus negative one is positive three halves...

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