2
$\begingroup$

I have the following set of equations:

1)

$\text{$b_1$} \cos (\text{$\beta_1$} u)-\text{$b_1$} \cosh (\text{$\beta_1$} u)-\text{$b2$} \cos (\text{$\beta_1$} \theta y)-\text{$d_2$} \cosh (\text{$\beta_1$} \theta y)\sin (\text{$\beta_1$} u)-\sinh (\text{$\beta_1$} u)=0$

2)

$-\text{$b_1$} \sin (\text{$\beta_1$} u)-\text{$b_1$} \sinh (\text{$\beta_1$} u)+\text{$b_2$} \theta \sin (\text{$\beta $1} \theta y)-\text{$d_2$} \theta \sinh (\text{$\beta_1$} \theta y)+\cos (\text{$\beta_1$} u)-\cosh (\text{$\beta_1$} u)=0$

3)

$-\text{$b_1$} \cos (\text{$\beta_1$} u)-\text{$b_1$} \cosh (\text{$\beta_1$} u)+\alpha ^4 \text{$b_2$} \theta ^2 \cos (\text{$\beta_1$} \theta y)-\alpha ^4 \text{$d_2$} \theta ^2 \cosh (\text{$\beta_1$} \theta y)-\sin (\text{$\beta_1$} u)-\sinh (\text{$\beta_1$} u)=0$

... and I would like to find the constants $b_1$, $b_2$ and $d_2$.

I already know the answers:

$b_1 =\frac{\sin \left(\beta _1 u\right)-\sinh \left(\beta _1 u\right)}{\cosh \left(\beta _1 u\right)-\cos \left(\beta _1 u\right)}$

$b_2 =\frac{2 \cos \left(\beta _1 u\right) \left(\cos \left(\beta _1 u\right) \cosh \left(\beta _1 u\right)-1\right)}{\theta \left(\cosh \left(\beta _1 u\right)-\cos \left(\beta _1 u\right)\right) \left(\cos \left(\beta _1 \theta y\right) \sinh \left(\beta _1 \theta y\right)+\sin \left(\beta _1 \theta y\right) \cosh \left(\beta _1 \theta y\right)\right)}$

$d_2 = -b_2 \frac{\cos \left(\beta _1 \theta y\right)}{\cosh \left(\beta _1 \theta y\right)}$

I want to find those expressions myself, but so far I only have that (using Mathematica):

Click on it, to see it full-scale...

enter image description here

Any help would be very much appreciated !

$\endgroup$
1
$\begingroup$

Writing the system as $$Ab_1+Bb_2+Cd_2+D=0\tag1$$ $$Eb_1+Fb_2+Gd_2+A=0\tag2$$ $$Hb_1+Ib_2+Jd_2+E=0\tag3$$ should make it easy to solve where

$\qquad\begin{cases}A=\cos (\text{$\beta_1$} u)-\cosh (\text{$\beta_1$} u)& \\\\B=-\cos (\text{$\beta_1$} \theta y)& \\\\C=-\cosh (\text{$\beta_1$} \theta y)\sin (\text{$\beta_1$} u)& \\\\D=-\sinh (\text{$\beta_1$} u)& \\\\E=-\sin (\text{$\beta_1$} u)-\sinh (\text{$\beta_1$} u)\end{cases}$ $\qquad\begin{cases}F=\theta \sin (\text{$\beta $1} \theta y)& \\\\G=-\theta \sinh (\text{$\beta_1$} \theta y)& \\\\H=-\cos (\text{$\beta_1$} u)-\cosh (\text{$\beta_1$} u)& \\\\I=\alpha ^4 \theta ^2 \cos (\text{$\beta_1$} \theta y)& \\\\J=-\alpha ^4 \theta ^2 \cosh (\text{$\beta_1$} \theta y)\end{cases}$


$E\times (1)-A\times (2)$ gives $$Kb_2+Ld_2+M=0\tag4$$

$H\times (1)-A\times (3)$ gives $$Nb_2+Pd_2+Q=0\tag5$$ where $$K=BE-AF,\quad L=CE-AG,\quad M=DE-A^2$$ $$N=BH-AI,\quad P=CH-AJ,\quad Q=DH-AE$$

$P\times (4)-L\times (5)$ gives $$(PK-NL)b_2+PM-QL=0,$$ i.e. $$b_2=\frac{QL-PM}{PK-NL}$$

From $(4)$, $$d_2=\frac{MN-KQ}{PK-NL}$$

Finally, from $(1)$, we get $$b_1=\frac{-B(QL-PM)-C(MN-KQ)-D(PK-NL)}{A(PK-NL)}$$

$\endgroup$
  • 1
    $\begingroup$ Thank you so much ! This is a very helpful answer ! :) $\endgroup$ – james Oct 6 '18 at 9:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.