How can I get the particular integral for resonance case for this partial differential equation with Dirac delta function?

Question

I was trying to solve the differential equation of an undamped simple beam with length $L$, mass per unit length $m$, and flexural rigidity $EI$, and subjected to a moving load $P$ wit speed $v$.
$$m\ddot u+EIu''''=P\delta(x-vt)$$ with these boundary and initial conditions:
$$u(0,t)=0,u(L,t)=0,u''(0,t)=0,u''(L,t)=0,u(x,0)=0,\dot u(x,0)=0$$ I first applied Laplace transform with respect to $x$, which yielded this equation:
$$m\ddot U+EIs^4u=EIs^2u'(0,t)+EIu'''(0,t)+Pe^{-vts}$$ For the particular integral part of the solution which came from the last part of the equation, it was
$$U=\frac {Pe^{-vts}}{mv^2s^2+EIs^4}$$ which gave the solution
$$u=\frac P{mv^2} \left[x-vt-\frac1v \sqrt\frac{EI}m \sin\sqrt \frac m{EI}v(x-vt) \right] H(x-vt)$$ The full soluton after applying the boundary and initial conditions was $$u=\left[-\frac{2L^4v}{\pi^3}\sqrt\frac m{EI}\sum\limits_{n=1}^\infty\frac 1{n^3(n^2\pi^2EI-mL^2v^2)}\sin\frac{n^2\pi^2}{L^2} \sqrt \frac{EI} mt \sin\frac {n\pi} L x+\frac 1{mv^2}\left(\frac{vt}L-1\right) x + \frac 1{mv^3} \sqrt\frac{EI}m\left(\cos\sqrt\frac m{EI}v^2t-\cot\sqrt\frac m{EI}vL\sin\sqrt\frac m{EI}v^2t\right) \sin\sqrt\frac m{EI}vx+\frac 1{mv^2}\left[x-vt-\frac1v \sqrt \frac{EI} m \sin\sqrt\frac m{EI}v(x-vt)\right]H(x-vt)\right]P$$ The problem now is when the velocity of the load causes what seems like resonance, matching one of the beam frequencies, which is this value
$$v=\frac{k\pi}L\sqrt\frac{EI}m$$ This caused two problems:

1. First, the particular integral $$\frac {PL^2}{k^2\pi^2EI} \left[x-\frac{k\pi}L\sqrt\frac{EI}mt-\frac L{k\pi} \sin \left( \frac{k\pi} Lx-\frac{k^2\pi^2}{L}\sqrt\frac{EI}mt\right) \right] H\left(x-\frac{k\pi}L \sqrt\frac{EI}mt\right)$$ will match part of the complementary function, which is not accepted.
2. The term $\cot\sqrt\frac m{EI}vL$ will be infinity, which is not possible.
   That means the particular solution much differ slightly for that case, maybe it should be multiplied by $t$ or something similar. I think the change should be in the first step of Laplace transform as it usually be, but I can't find how it will change. Can anyone help me with this?

Answer 1

I managed, somehow, to get the solution that I need. I recalled that one of my professors of dynamic analysis said that the solution for the resonance case can be obtained through the differentiation of both numerator and denominator with respect to the load frequency, this seems like L'Hôpital's rule. For example, for the equation $\ddot u+\alpha^2u=A\cos{\omega t}+B\sin{\omega t}$, the particular integral is $A\frac{\cos{\omega t}}{\alpha^2-\omega^2}+B\frac{\sin{\omega t}}{\alpha^2-\omega^2}$, but if $\omega=\alpha$, then the denominator will be zero, so both numerator and denominator are differentiated with respect to $\omega$, which produces $A\frac{t\sin{\omega t}}{2\omega}-B\frac{t\cos{\omega t}}{2\omega}$, which is the solution that can be obtained using other mathematical methods.
I used the same concept for the term that contains $\cot\sqrt\frac m{EI}vL$, by turning it into a fraction form like this: $$\frac{\cos\sqrt\frac m{EI}vL\sin\sqrt\frac m{EI}v^2t\sin\sqrt\frac m{EI}vx}{\sin\sqrt\frac m{EI}vL}$$ and differentiated both numerator and denominator with respect to $v$, which resulted in: $$\frac{-L\sin\sqrt\frac m{EI}vL\sin\sqrt\frac m{EI}v^2t\sin\sqrt\frac m{EI}vx+2vt\cos\sqrt\frac m{EI}vL\cos\sqrt\frac m{EI}v^2t\sin\sqrt\frac m{EI}vx+x\cos\sqrt\frac m{EI}vL\sin\sqrt\frac m{EI}v^2t\cos\sqrt\frac m{EI}vx}{L\cos\sqrt\frac m{EI}vL}$$ and since $\sin\sqrt\frac m{EI}vL=0$, it became: $$\frac{2vt}L\cos\sqrt\frac m{EI}v^2t\sin\sqrt\frac m{EI}vx+\frac xL\sin\sqrt\frac m{EI}v^2t\cos\sqrt\frac m{EI}vx$$ Starting from that discovery, I decided to assume that solution contains all terms of $\cos\sqrt\frac m{EI}v^2t\cos\sqrt\frac m{EI}vx$, $\cos\sqrt\frac m{EI}v^2t\sin\sqrt\frac m{EI}vx$, $\sin\sqrt\frac m{EI}v^2t\cos\sqrt\frac m{EI}vx$, and $\sin\sqrt\frac m{EI}v^2t\sin\sqrt\frac m{EI}vx$, each multiplied by $\frac{vt}L$ or $\frac xL$, which is total of eight terms, along with the Heaviside function part and polynomial part. Following the same steps I did for the original case, only two terms of the eight remained, and the were the same two I found using the differentiation method, so the final solution became: $$u=\left[-\frac{2L^4v}{\pi^3}\sqrt\frac m{EI}\sum\limits_{n=1,n\ne k}^\infty\frac 1{n^3(n^2\pi^2EI-mL^2v^2)}\sin\frac{n^2\pi^2}{L^2} \sqrt \frac{EI} mt \sin\frac {n\pi} L x+\frac 1{mv^2}\left(\frac{vt}L-1\right) x + \frac 1{mv^3} \sqrt\frac{EI}m\left(\cos\sqrt\frac m{EI}v^2t+\frac7{2Lv}\sqrt\frac{EI}m\sin\sqrt\frac m{EI}v^2t\right)\sin\sqrt\frac m{EI}vx-\frac1{mLv^3}\sqrt\frac{EI}mx\sin\sqrt\frac m{EI}v^2t\cos\sqrt\frac m{EI}vx-\frac2{mLv^2}\sqrt\frac{EI}mt\cos\sqrt\frac m{EI}v^2t\sin\sqrt\frac m{EI}vx+\frac 1{mv^2}\left[x-vt-\frac1v\sqrt\frac{EI}m\sin\sqrt\frac m{EI}v(x-vt)\right]H(x-vt)\right]P$$ The only term that seemed new to me was $\frac7{2Lv}\sqrt\frac{EI}m\sin\sqrt\frac m{EI}v^2t\sin\sqrt\frac m{EI}vx$, but I think this happened because it was excluded from the infinite summation for $n=k$.