# $\frac{dx}{dt}=-\lambda x +\epsilon x(t-a)$ series solution via Laplace method

Consider the following equation $$\frac{dx}{dt}=-\lambda x +\epsilon x(t-a), \quad x(0)=1,\quad |\epsilon|\ll1$$ where $a$ and $\lambda$ are positive constants and $x(t-a)$ means the function $x(t)$ evaluated at $t-a$. The task is to obtain a solution as a series in powers of $\epsilon$. The book suggests to show that $x(t)=e^{-\lambda t} w(t)$ satisfies $$\frac{dw}{dt}=\delta w(t-a)\quad w(0)=1,\quad \delta=\epsilon e^{\lambda a}$$ and show that the perturbation expansion is given by the following expression: $$w(t)=1+t\sum_{n=1}^{\infty}\frac{\delta^n}{n!}\left(t-na\right)^{n-1}$$ hence $$x(t)=e^{-\lambda t}\left(1+t\sum_{n=1}^{\infty}\frac{\delta^n}{n!}\left(t-na\right)^{n-1}\right)$$ Since the method given in the solution is basically verification by substitution and proof by induction, I have been looking for a more systematic approach. The presence of time delay in RHS suggests that Laplace transform method could be appropriate here. So I proceed as follows. Apply Laplace transform to both sides of the original equation observing the initial conditions: $$x(t)\fallingdotseq X(s)$$ $$sX(s)-x(0)=-\lambda X(s) +\epsilon e^{-sa}X(s)$$ Solve for $X(s)$: $$X(s)=\frac{x(0)}{s+\lambda-\epsilon e^{-sa}}=\frac{1}{s+\lambda-\epsilon e^{-sa}}$$ Expand in powers of $\epsilon$: $$X(s)=\frac{1}{s+\lambda}\left(\frac{1}{1-\epsilon\frac{e^{-sa}}{s+\lambda}}\right)=\frac{1}{s+\lambda}\left(1+\sum_{n=1}^{\infty}\frac{\epsilon^n e^{-nas}}{(s+\lambda)^n}\right)=\frac{1}{s+\lambda}+\sum_{n=1}^{\infty}\frac{\epsilon^n e^{-nas}}{(s+\lambda)^{n+1}}$$ Taking inverse transforms $$\frac{1}{s+\lambda} \risingdotseq e^{-\lambda t}$$ $$\frac{e^{-nas}}{(s+\lambda)^{n+1}} \risingdotseq \frac{(t-na)^n e^{-\lambda(t-na)}}{n!}$$ But then I lose the factor of $t$ before the sum as compared to the book solution. Where am I making the slip? Thanks in advance.

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What is the \fallingdotseq $\fallingdotseq$ for? I've never seen this notation. –  Patrick Da Silva May 29 '12 at 22:44
@PatrickDaSilva For noting it is an inverse transform equality, if I'm not reading wrongly. –  Pedro Tamaroff May 29 '12 at 22:44
@Peter Tamaroff : I understand that on one side you have a function and on the other its Laplace Transform, but I wanted to know the precise meaning of the symbol. –  Patrick Da Silva May 29 '12 at 22:46
I was just telling anon, sometimes people choose a simbol to denote something. Such as $e^{ta} \sqsubseteq \frac{1}{s-a}$ –  Pedro Tamaroff May 29 '12 at 22:47
I realised that this is a possibly outdated symbol specific to old Russian textbooks for example this one. Could not find any I am aware of available in preview –  Valentin May 29 '12 at 23:02

Let $D=\frac{\mathrm{d}}{\mathrm{d}t}$, then we can rewrite your equation as $$(D+\lambda-\epsilon e^{-aD})x=0\tag{1}$$ To invert the operator, consider the expansion: \begin{align} \frac{1}{\lambda+x-\epsilon e^{-ax}} &=\frac{1}{\lambda+x}\frac{1}{1-\epsilon\frac{e^{-ax}}{\lambda+x}}\\ &=\frac{1}{\lambda+x}+\epsilon\frac{e^{-ax}}{(\lambda+x)^2}+\epsilon^2\frac{e^{-2ax}}{(\lambda+x)^3}+\epsilon^3\frac{e^{-3ax}}{(\lambda+x)^4}+\dots\tag{2} \end{align} Using integrating factors we get that $$(\lambda+D)^{-1}f=e^{-\lambda t}\int e^{\lambda t}f\,\mathrm{d}t\tag{3}$$ which leads us to $$(\lambda+D)^{-k-1}0=e^{-\lambda t}\sum_{j=0}^kc_j\frac{t^{k-j}}{(k-j)!}\tag{4}$$ and therefore the coefficient of $\epsilon^k$ would be $$e^{-kaD}(\lambda+D)^{-k-1}0=e^{-\lambda (t-ka)}\sum_{j=0}^kc_j\frac{(t-ka)^{k-j}}{(k-j)!}\tag{5}$$