$\dot{X}(t)=A(t)X(t)$

Question

Ordinary, Linear, variable coefficient, 2nd order dif. eqn. Hoca ödev verdi. eigen value yi bulup çözebilirsiniz dedi. Fakat ne kadar uğraştıysam bir türlü sonuca ulaşamadım. yardımcı olursanız sevinirim. Teşekkür ederim. Kolay gelsin.

 

$\left[\begin{array}{c} x'_1(t)\\ x'_2(t)\\ \end{array} \right]=\left[ \begin{array}{cc}  \cos (2 t) & 1-\sin (2 t) \\  -1-\sin (2 t) & -\cos (2 t) \\ \end{array} \right]\left[ \begin{array}{c} x_1 \\ x_2\\ \end{array} \right]$

 

$X'(t)=A(t)X(t)$

Comments

$\dot{X}(t)$ olarak ifade ettiğiniz şey $X'(t)$ mi?

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evet hocam fiziksel gösterimi öyle

Answer 1

Eger soru soyle olsaydi (soruda typo olabilecegini dusunuyorum) biraz daha kolay olabilirdi. Not: Butun adimlar Mathematica ile hesaplandi. Malesef Mathematica analitik olarak cozemedi ama bazi baslangic degerleri icin numerik cozumler karsilastirildi.

 

$X'(t)=A(t)X(t)$

 

$A(t)=\left[ \begin{array}{cc} \cos (2 t) & 1-\sin (2 t) \\ 1-\sin (2 t) & \cos (2 t) \\ \end{array} \right]$

 

 

Bu durumda $\displaystyle\int_0^tA(s)ds=\displaystyle\int_0^t\left[ \begin{array}{cc} \cos (2 s) & 1-\sin (2 s) \\ 1-\sin (2 s) & \cos (2 s) \\ \end{array} \right]ds=\left[ \begin{array}{cc} \sin (t) \cos (t) & t-\sin ^2(t) \\ t-\sin ^2(t) & \sin (t) \cos (t) \\ \end{array} \right]$ olmak uzere

 

 

$A(t)\cdot\displaystyle\int_0^tA(s)ds=\left[ \begin{array}{cc} \cos (2 t) & 1-\sin (2 t) \\ 1-\sin (2 t) & \cos (2 t) \\ \end{array} \right]\left[ \begin{array}{cc} \sin (t) \cos (t) & t-\sin ^2(t) \\ t-\sin ^2(t) & \sin (t) \cos (t) \\ \end{array} \right]$

 

 

$=\left[ \begin{array}{cc} (\cos (t)-\sin (t)) (-t \sin (t)+\sin (t)+t \cos (t)) & \left[t-\sin ^2(t)\right] \cos (2 t)-\sin (t) (\sin (2 t)-1) \cos (t) \\ \left[t-\sin ^2(t)\right] \cos (2 t)-\sin (t) (\sin (2 t)-1) \cos (t) & (\cos (t)-\sin (t)) (-t \sin (t)+\sin (t)+t \cos (t)) \\ \end{array} \right]$

 

 

$\displaystyle\int_0^tA(s)ds\cdot A(t)=\left[ \begin{array}{cc} \sin (t) \cos (t) & t-\sin ^2(t) \\ t-\sin ^2(t) & \sin (t) \cos (t) \\ \end{array} \right]\left[ \begin{array}{cc} \cos (2 t) & 1-\sin (2 t) \\ 1-\sin (2 t) & \cos (2 t) \\ \end{array} \right]$

 

 

 

$=\left[ \begin{array}{cc} (\cos (t)-\sin (t)) (-t \sin (t)+\sin (t)+t \cos (t)) & \left[t-\sin ^2(t)\right] \cos (2 t)-\sin (t) (\sin (2 t)-1) \cos (t) \\ \left[t-\sin ^2(t)\right] \cos (2 t)-\sin (t) (\sin (2 t)-1) \cos (t) & (\cos (t)-\sin (t)) (-t \sin (t)+\sin (t)+t \cos (t)) \\ \end{array} \right]$

 

 

 

$\implies A(t)\cdot\displaystyle\int_0^tA(s)ds=\displaystyle\int_0^tA(s)ds\cdot A(t)$ olur ve bu durumda temel matris su sekilde verilir.

 

 

$\Phi(t)=e^{\displaystyle\int_0^tA(s)ds}=e^{\left[ \begin{array}{cc} \sin (t) \cos (t) & t-\sin ^2(t) \\ t-\sin ^2(t) & \sin (t) \cos (t) \\ \end{array} \right]}$

 

Eger matrisimiz kosegen olsaydi soyle bir kolayligimiz olurdu

 

$e^{\left[ \begin{array}{cc} d_1& 0\\ 0 & d_2 \\ \end{array} \right]}=\left[ \begin{array}{cc} e^{d_1}& 0\\ 0 & e^{d_2} \\ \end{array} \right]$

 
 

Sansliyiz ki $B=\left[ \begin{array}{cc} \sin (t) \cos (t) & t-\sin ^2(t) \\ t-\sin ^2(t) & \sin (t) \cos (t) \\ \end{array} \right]$ matrisimiz simetrik ve butun simetrik matrisler kosegenlestirilebilir. O zaman hadi kosegenlestirelim.

 Once ozdegerleri bulalim.

 

$\left|B-\lambda I \right|=\left|\begin{array}{cc} \sin (t) \cos (t) -\lambda& t-\sin ^2(t) \\ t-\sin ^2(t) & \sin (t) \cos (t)-\lambda \\ \end{array} \right|=0$

 

$\implies\lambda ^2-2 \lambda \sin (t) \cos (t)-\sin ^4(t)+2 t \sin ^2(t)+\sin ^2(t) \cos ^2(t)-t^2=0$

 

$\lambda _1=\frac{1}{2} (-2 t+\sin (2 t)-\cos (2 t)+1),\,\, \lambda_2= \frac{1}{2} (2 t+\sin (2 t)+\cos (2 t)-1)$

 

$\lambda _1=\frac{1}{2} (-2 t+\sin (2 t)-\cos (2 t)+1): \implies (B-\lambda_1I)V=0$

 

$\left[ \begin{array}{cc} \frac{1}{2} (2 t+\cos (2 t)-1) & t-\sin ^2(t) \\ t-\sin ^2(t) & \frac{1}{2} (2 t+\cos (2 t)-1) \\ \end{array} \right]\left[ \begin{array}{c} v_1 \\ v_2\\ \end{array} \right]=\left[ \begin{array}{c} 0 \\ 0\\ \end{array} \right]$

 

 

 

$\implies \left[ \begin{array}{c} v_1 \\ v_2\\ \end{array} \right]=\left[ \begin{array}{c} -1 \\ 1\\ \end{array} \right]$

 

$\lambda_2= \frac{1}{2} (2 t+\sin (2 t)+\cos (2 t)-1): \implies (B-\lambda_2I)V=0$

 

$\left[ \begin{array}{cc} \frac{1}{2} (-2 t-\cos (2 t)+1) & t-\sin ^2(t) \\ t-\sin ^2(t) & \frac{1}{2} (-2 t-\cos (2 t)+1) \\ \end{array} \right]\left[ \begin{array}{c} v_3 \\ v_4\\ \end{array} \right]=\left[ \begin{array}{c} 0 \\ 0\\ \end{array} \right]$

 
 

$\implies \left[ \begin{array}{c} v_3 \\ v_4\\ \end{array} \right]=\left[ \begin{array}{c} 1 \\ 1\\ \end{array} \right]$

 

 

Simdi $P=\left[ \begin{array}{cc} v_1&v_3 \\ v_2 &v_4 \\ \end{array} \right]=\left[ \begin{array}{cc} -1&1 \\ 1 &1 \\ \end{array} \right]$ olarak tanimlayalim ve

 

 

$P^{-1}=\dfrac12\left[ \begin{array}{cc} 1&1 \\ -1 &1 \\ \end{array} \right]$ olur.

 
 

$D=P^{-1}\left[\displaystyle\int_0^tA(s)ds\right]P$

 

 

$=\dfrac12\left[ \begin{array}{cc} 1&1 \\ -1 &1 \\ \end{array} \right]\left[ \begin{array}{cc} \sin (t) \cos (t) & t-\sin ^2(t) \\ t-\sin ^2(t) & \sin (t) \cos (t) \\ \end{array} \right]\left[ \begin{array}{cc} -1&1 \\ 1 &1 \\ \end{array} \right]$

 

 

$\implies D=\left[ \begin{array}{cc} t-\sin ^2(t)+\sin (t) \cos (t) & 0 \\ 0 & -t+\sin ^2(t)+\sin (t) \cos (t) \\ \end{array} \right]$

 
 

$PDP^{-1}=\left[\displaystyle\int_0^tA(s)ds\right]\implies Pe^DP^{-1}=e^{\displaystyle\int_0^tA(s)ds}$

 
 

$\Phi(t)=e^{\displaystyle\int_0^tA(s)ds}=Pe^DP^{-1}$

 

 

$=Pe^{\left[ \begin{array}{cc} t-\sin ^2(t)+\sin (t) \cos (t) & 0 \\ 0 & -t+\sin ^2(t)+\sin (t) \cos (t) \\ \end{array} \right]}P^{-1}$

 

 

$=\left[ \begin{array}{cc} -1&1 \\ 1 &1 \\ \end{array} \right]\left[ \begin{array}{cc} e^{t-\sin ^2(t)+\sin (t) \cos (t)} & 0 \\ 0 &e^{ -t+\sin ^2(t)+\sin (t) \cos (t)} \\ \end{array} \right]\dfrac12\left[ \begin{array}{cc} 1&1 \\ -1 &1 \\ \end{array} \right]$

 

 

$\Phi(t)=\left[ \begin{array}{cc} \frac{1}{2} \left[e^{2 t}+e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} & \frac{1}{2} \left[e^{2 t}-e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} \\ \frac{1}{2}\left[e^{2 t}-e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} & \frac{1}{2} \left[e^{2 t}+e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} \\ \end{array} \right]$

 

 

$X(t)=\Phi(t)C$

 
 

$\left[ \begin{array}{c} x_1(t)\\ x_2(t)\\ \end{array} \right]=\left[ \begin{array}{cc} \frac{1}{2} \left[e^{2 t}+e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} & \frac{1}{2} \left[e^{2 t}-e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} \\ \frac{1}{2} \left[e^{2 t}-e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} & \frac{1}{2} \left[e^{2 t}+e^{2 \sin ^2(t)}\right] e^{-t-\sin ^2(t)+\sin (t) \cos (t)} \\ \end{array} \right]\left[ \begin{array}{c} C_1 \\ C_2\\ \end{array} \right]$

 

 

$\left[ \begin{array}{c} C_1 \\ C_2\\ \end{array} \right]=\left[ \begin{array}{c} 1 \\ 2\\ \end{array} \right]$ baslangic degerleri icin cozumler.

 

Comments

umarım soru sahibi hayattadır ^^