### Key Concepts

 Data Analytics: Summaries on Strang's Linear Algebra - Differential Equations
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Sunday, 24 February 2013 07:05

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We have $n$ equations $\frac{du}{dt}=Au$ starting from the vector $u(0)$ at $t=0$. Our main goal is to solve linear constant coefficient equations by exponentials $e^{\lambda t}x$, when $Ax=\lambda x$.

### Solution of du/dt=Au

Use $u=e^{\lambda t} x$ when $Ax=\lambda x$, $\frac{du}{dt}=\lambda e^{\lambda t}x$ agrees with $Au=Ae^{\lambda t} x$. All components of the special solution $u=e^{\lambda t} x$ share the same $e^{\lambda t}$
The steps for solving the $du/dt=Au$ are as follows:

1. Obtain eigenvectors $x_i$ and eigenvalues $\lambda_i$ of $A$.
2. Write $u(0)$ as a combination $c_1x_1 + \cdots + c_nx_n$ of the eigenvectors of $A$
3. Multiply each eigenvector $x_i$ by $e^{\lambda_i t}$.
4. The solution is the combination of pure solutions $e^{\lambda t}x$: $u(t) = c_1e^{\lambda_1 t}x_1 + \cdots + c_ne^{\lambda_n t}x_n$

### Conditions for Stability

Here are the conditions for stability, steady state and blowup:

1. Stability In this case we wish $e^{\lambda t}\text{ }\rightarrow \text{ }0$ as $t\text{ }\rightarrow\text{ }0$. When $\lambda$ is real, we must have $\lambda_i < 0$ for all $i$. If $\lambda$ is a complex no. we have $Re(\lambda_i) < 0$ for all $\lambda_i$
2. Steady State Here we have $\lambda_k = 0$ and $Re(\lambda_i)<0$ for all $i\neq k$
3. Blowup In this case blowup happens when any $Re(\lambda_i)>0$

#### Stability of 2x2 Matrices

$A$ is stable and $u(t)\text{ }\rightarrow\text{ }0$ when all eigenvalues have negative real parts.
The 2x2 matrix $A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$ must pass 2 tests:
$\lambda_1 + \lambda_2 < 0$. The trace $T=a+d$ must be -ve.
$\lambda_1\lambda_2 > 0$. The determinant $D=ad-bc$ must be +ve.

### Decoupling and Matrix Exponentials

We have $\frac{du}{dt}=Au$. Matrix $A$ couples the components of $u$. The whole point of eigenvectors is to uncouple. We can write $u=Sv$ where $S$ is the eigenvector matrix that uncouples or diagonalizes. Substituting, $S \frac{dv}{dt}=ASv$ => $\frac{dv}{dt}=S^{-1}ASv = \Lambda v$. No more coupling since $\frac{dv_1}{dt}=\lambda_1 v_1, \cdots$. The soln. is $v(t)=e^{\Lambda t} v(0)$ $u(t)=Se^{\Lambda t}S^{-1} v(0)$ But $e^{At}=Se^{\Lambda t}S^{-1}$
What is meant by a matrix exponential ? $u(t)=Se^{\Lambda t}S^{-1} v(0) =e^{At}u(0)$ The idea is that there is a power series for exponentials. Thus $e^{At}=I+At+\frac{1}{2}(At)^2+\frac{1}{6}(At)^3 + \cdots + \frac{(At)^n)}{n!} + \cdots$ Note that $e^{At}u(0)$ solves the differential eqn. quickly even if there is a shortage of eigenvectors. But how do we show that $e^{At}=Se^{\Lambda t}S^{-1}$ ?
We can again use the Taylor series. Thus we have that $u(t)=Se^{\Lambda t}S^{-1} v(0) =e^{At}u(0)$

### 2nd Order Equations

We have the following D.E : $y" + by' + ky = 0$ We can change the above 2nd order eqn. by a similar process for the difference eqn. case a s follows:
Let $u=\left[\begin{array}{c}y'\\y\end{array}\right]$. Then $u'=\left[\begin{array}{c}y"\\y'\end{array}\right] = \left[\begin{array}{cc}-b&&-k\\1&&0\end{array}\right]\left[\begin{array}{c}y"\\y'\end{array}\right]$ We then obtain the eigenvalues of $A$
$det(A-\lambda I)$ gives $\lambda^2+b\lambda+k=0$.

Last Updated on Tuesday, 26 February 2013 03:46