I am having some difficulty with a beam equation and I need to convert the equation below into a 1st order PDE with respect to time in order to compare with my code calculations.
In case the picture does not load, the equation is (rearranged to equal $\frac{d^2w}{dt^2}$ ):
$$\frac{d^2 w}{dt^2}= -\left(\frac{EI}{pA}\right)\cdot \left(\frac{d^4w}{dx^4}\right)$$
The original equation:
I want to turn the 2nd degree PDE in terms of $t$ into a 1st degree PDE in terms of t. I'm not quite sure how to proceed from here.
If you assume that: $$w(x,t)=X(x)T(t)$$ then you will see your equation reduces to: $$X(x)T''(t)=-CX''''(x)T(t)$$ multiplying both sides by $T'$ $$X(x)T''T'=-CX''''TT'$$ and now integrating wrt $t$. Notice that: $$\left[\frac12(T')^2\right]'=T''T'$$ $$\left[\frac12T^2\right]'=TT'$$ and so we get: $$X(x)\times\frac12(T')^2=-CX''''\times\frac12T^2$$ $$X(T')^2=-CX''''T^2$$ now you could rearrange: $$T'=j\sqrt{\frac{CX''''T^2}{X}}$$ However this is quite complicated
A common way used is to convert from 1 second order equation to 2 first order coupled equations, say: $$z_1(t)=T(t),\,z_2(t)=T'(t)$$ now it is clear that: $$z_1'(t)=T'(t)=z_2,\,z_2'(t)=T''(t)=-\frac{CX''''(x)T(t)}{X(x)}=-\frac{CX''''(x)z_1}{X(x)}$$ now letting: $$\mathbf{z}=\begin{pmatrix}z_1\\z_2\end{pmatrix}$$ you obtain: $$\mathbf{z}'=\underbrace{\begin{pmatrix}0 & 1\\-\frac{CX''''(x)}{X(x)} & 0\end{pmatrix}}_{\mathbf{M}}\mathbf{z}$$ and so you now have the first order equation: $$\mathbf{z}'=\mathbf{Mz}$$