I need to compute the inverse Laplace transformation of the following equation.

\begin{align*} f(s)&=\frac{A}{\prod_{i=1}^{L}(s+a_i)^m} \\ &=\frac{A}{(s+a_1)^m\,(s+a_2)^m\cdots (s+a_L)^m}. \end{align*}

What I managed to get is this: $$\mathcal{L}^{-1}\!\left(\frac{1}{(s+a)^m}\right)=\frac{t^{m-1}e^{-at}}{{(m-1)!}}.$$

How do I find the Laplace inverse of the series though? Someone suggested me using partial fraction decomposition and gave me this result: $$f(s)=\sum_{1\le k\le m\atop 1\le j\le L} {A_{jk}\over(s+a_j)^k},$$ but failed to tell how we achieved it.

  • That's not a solution, though. That's an intermediate result. Partial fractions certainly seems like a good approach, assuming the $a_i$ are unique. – Adrian Keister Sep 14 at 13:13
  • Yea I didnt get to the solution, I need help with understanding how we get the equation using partial fraction but I have little knowledge about the subject, if you can explain a bit it would be really helpful. Or if there is another approach to this problem – Viplav Prakash Sep 14 at 13:15
up vote 1 down vote accepted

As you've mentioned, $$\mathcal{L}^{-1}\!\left(\frac{1}{(s+a)^m}\right)=\frac{t^{m-1}e^{-at}}{{(m-1)!}}$$ is the key to the whole business. If we can find the partial fraction decomposition of $f(s),$ we can invoke this result to get the final answer. Now, it is known in the theory of partial fraction decomposition that if you have a function like $$g(s)=\frac{1}{(s-1)^2(s+1)^2},$$ you can decompose it with the ansatz (an ansatz is an initial guess that we take as a working hypothesis): $$g(s)=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{s+1}+\frac{D}{(s+1)^2}. $$ If you do all the algebra, find common denominators, add everything up, and solve the resulting equations for $A,B,C,$ and $D,$ you find that $$g(s)=-\frac{1}{4(s-1)}+\frac{1}{4(s-1)^2}+\frac{1}{4(s+1)}+\frac{1}{4(s+1)^2}.$$ So, we build up our solution as follows, bit-by-bit: \begin{align*} \frac{1}{(s+a_1)^m}&=\sum_{j=1}^{m}\frac{A_{1j}}{(s+a_1)^j} \\ \frac{1}{(s+a_2)^m}&=\sum_{j=1}^{m}\frac{A_{2j}}{(s+a_2)^j} \\ &\vdots \\ \frac{1}{(s+a_L)^m}&=\sum_{j=1}^{m}\frac{A_{Lj}}{(s+a_L)^j}. \end{align*} What we really want, of course, is all of these to be additive. We need to change the product in the original function entirely to a sum - which is what partial fraction decomposition is all about. So, we're going to write this (note the $A$ in the original can easily be absorbed into the $A_{ij}$): \begin{align*} f(s)&=\sum_{j=1}^{m}\frac{A_{1j}}{(s+a_1)^j}+\sum_{j=1}^{m}\frac{A_{2j}}{(s+a_2)^j}+\cdots + \sum_{j=1}^{m}\frac{A_{Lj}}{(s+a_L)^j} \\ &=\sum_{i=1}^L\sum_{j=1}^m\frac{A_{ij}}{(s+a_i)^j}. \end{align*} This is precisely what your friend you mentioned suggested to you. Now the final solution, once you find the $A_{ij},$ will then be $$f(t)=\sum_{i=1}^L\sum_{j=1}^m\frac{A_{ij}\,t^{j-1}e^{-a_{i}t}}{(j-1)!}. $$ The major part of the work here, of course, is to find the $A_{ij}.$ I would recommend a computer algebra system for doing that, like Mathematica.

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