📂Odinary Differential Equations

Inverse Laplace Transform of F(ks)

Formulas¹

Assuming that the Laplace transform $\mathcal{L} \left\{ f(t) \right\} = \displaystyle \int _{0} ^\infty e^{-st}f(t)dt = F(s)$ of the function $f(t)$ exists and is $s>a \ge 0$ for a positive number $k> 0$ then the inverse Laplace transform of $F(ks)$ is as follows.

$$\mathcal{L^{-1}} \left\{ F(ks) \right\} =\dfrac{1}{k}f\left(\frac{t}{k}\right),\quad s>\frac{a}{k}$$

Derivation

1

The Laplace transform of $f(ct)$

$$\mathcal{L} \left\{ f(ct) \right\} =\dfrac{1}{c}F\left(\dfrac{s}{c}\right), \quad s>ca$$

By substituting $\dfrac{1}{k}$ for $c$ in the above equation

$$\begin{align*} \mathcal{L} \left\{ f\left(\frac{t}{k}\right) \right\} &= kF(ks) \\ \implies F(ks) &= \dfrac{1}{k} \mathcal{L} \left\{ f\left(\frac{t}{k}\right) \right\} =\mathcal{L} \left\{\dfrac{1}{k} f\left(\frac{t}{k}\right) \right\} \end{align*}$$

Therefore

$$\mathcal{L^{-1}} \left\{ F(ks) \right\} =\dfrac{1}{k}f\left(\frac{t}{k}\right)$$

Since the condition when $c$ was $s>ca$, the condition naturally changes to $s>\dfrac{a}{k}$.

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2

By the definition of the Laplace transform,

$$\mathcal{L} \left\{ \frac{1}{k} f\left( \frac{t}{k} \right) \right\} =\dfrac{1}{k}\int _{0} ^\infty e^{-st}f \left( \frac{t}{k} \right) dt$$

Let us replace $\dfrac{t}{k}=\tau$ with. Then $st=sk\tau$ and $dt=kd\tau$ are, so

$$\begin{align*} \mathcal{L} \left\{ \frac{1}{k} f\left( \frac{t}{k} \right) \right\} &=\int _{0} ^\infty e^{-sk\tau}f \left(\tau \right) d\tau \\ &= F(ks) \end{align*}$$

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See Also

• Table of Laplace Transforms