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Find the general solution to the linear system $\frac{du}{dt}=Au$ for the following matrices:

a.) $\pmatrix{2&1\\0&2}$

b.) $\pmatrix{-1&-1\\4&-5}$

  • I got the eigenvalues for a:

    $\lambda = 2,2$ repeated roots and the eigenvector, $v_1=\pmatrix{0\\1}$, so the general solution is $c_1e^{2t} + c_2te^{2t}$ but the answer is $\pmatrix{c_1e^{2t}+c_2te^{2t}\\c_2e^{2t}}$. How did they get that?

  • For b, $\lambda = -3,-3$ and eigenvector, $v_1=\pmatrix{1\\2}$ but the answer is $\pmatrix{c_1e^{-3t}+c_2(\frac{1}{2} +t)e^{-3t}\\2c_1e^{-3t}+2c_2te^{-3t}}$ how did they get that?

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  • $\begingroup$ What method are you using to compute the solutions? Are you trying to find $e^{At}$? $\endgroup$
    – JSchlather
    Commented Nov 26, 2012 at 2:42
  • $\begingroup$ @JacobSchlather yes, i am. I am taking the eigenvalues of each matrix and finding $e^{At}$ like you said. $\endgroup$
    – Q.matin
    Commented Nov 26, 2012 at 2:45

2 Answers 2

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For linear systems $\dot{u} = Au$, the general solution is $u = e^{At}u_0$.

The way to calculate this is by $e^{At}u_0 = e^{(A-\lambda I)t}e^{\lambda t}u_0 = \sum_{k=0}^{\infty} (A- \lambda I)^k \frac{t^k}{k!} e^{\lambda t}u_0 = \sum_{k=0}^{mult(\lambda) - 1} (A- \lambda I)^k \frac{t^k}{k!} e^{\lambda t}u_0$. Which is true if $u_0$ is in the eigenspace of $\lambda$.

So in general, $e^{At}u_0 = \sum_i \sum_{k=0}^{mult(\lambda_i) - 1} (A- \lambda_i I)^k \frac{t^k}{k!} e^{\lambda_i t}u_0^{(i)}$, where ${u_0}^{(i)}$ is the component of $u_0$ in the eigenspace of $\lambda_i$. And $mult(\lambda_i)$ is the geometric multiplicity of $\lambda_i$.

For (a), your eigenvalue is 2 with geometric multiplicty of 2.

The first term in the series will be $e^{2t}u_0 = e^{2t} \pmatrix{c_1 \\ c_2}$ And the second term in the series is $(A- 2 I) t e^{2 t}u_0 = te^{2t} \pmatrix{0 & 1 \\ 0 & 0} \pmatrix{c_1 \\ c_2} = te^{2t} \pmatrix{c_2 \\ 0}$

Adding the two yields solution $\pmatrix{e^{2t}c_1 + te^{2t}c_2 \\ e^{2t}c_2}$.

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  • $\begingroup$ Thanks a lot for that!!! And can you write out the computation please, so that I can understand the equation you just provided? $\endgroup$
    – Q.matin
    Commented Nov 26, 2012 at 3:42
  • $\begingroup$ Added computation for part (a) $\endgroup$ Commented Nov 26, 2012 at 3:48
  • $\begingroup$ Stuart, how come you did $(A- 2 I) t e^{2 t}u_0 = te^{2t} \pmatrix{0 & 1 \\ 0 & 0} \pmatrix{c_1 \\ c_2} = te^{2t} \pmatrix{c_2 \\ 0}$ for the second one but not for the first one $e^{2t}u_0 = e^{2t} \pmatrix{c_1 \\ c_2}$? $\endgroup$
    – Q.matin
    Commented Nov 28, 2012 at 6:06
  • $\begingroup$ Because that's what the series says to do xD $\endgroup$ Commented Nov 28, 2012 at 8:34
  • $\begingroup$ I am going to need to do a lot more practice then :) $\endgroup$
    – Q.matin
    Commented Nov 29, 2012 at 5:45
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For the first system you do not need any eigenvalues or eigenvectors. Note that the second equation has the form $$ \dot u_2=2u_2. $$ Hence the solution $$ u_2(t)=c_2e^{2t}. $$ You plug it into the first equation and get $$ \dot u_1=2u_1+c_2e^{2t}. $$ This is a first order linear equation and can be solved by any appropriate method, e.g., by the method of integrating factor: $$ \frac{d}{dt}(u_1e^{-2t})=c_2\Rightarrow u_1=c_2te^{2t}+c_1e^{2t}. $$ So the general solution is $$ \begin{bmatrix} u_1\\ u_2 \end{bmatrix}=\begin{bmatrix}c_1e^{2t}+c_2te^{2t}\\ c_2e^{2t} \end{bmatrix} $$ For the second problem you will need a generalized vector that satisfies $(A-\lambda I)^2v=0$. Any vector will satisfy this equation (why), and hence we can take the one which is simply looking and linearly independent of what you already have found. E.g., $v^{T}=(1,0)$ works.

Hence the general solution would be $$ \begin{bmatrix} u_1\\ u_2 \end{bmatrix}=c_1\begin{bmatrix}1\\ 2 \end{bmatrix} e^{-3t}+c_2(I+(A-\lambda I)t)ve^{-3t} $$ and you get your (almost, up to a mulplicative constant) answer.

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  • $\begingroup$ Thanks a lot!! Stuart answered it, but this is very useful too! $\endgroup$
    – Q.matin
    Commented Nov 26, 2012 at 3:58

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