I have the following set of equations:


$\text{$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$} u)=0$


$-\text{$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$} u)=0$


$-\text{$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 !


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 $$b_1=\frac{-B(QL-PM)-C(MN-KQ)-D(PK-NL)}{A(PK-NL)}$$

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

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