Properties of the Laplace Transform

The Laplace transform has a number of interesting properties.

Property 1. Linearity of the Laplace transform: $$${L}{\left({a}{f{{\left({t}\right)}}}+{b}{g{{\left({t}\right)}}}\right)}={a}{L}{\left({f{{\left({t}\right)}}}\right)}+{L}{\left({g{{\left({t}\right)}}}\right)}$$$, where $$${a}$$$ and $$${b}$$$ are some constants.

The proof is straightforward through the definition:

$$${L}{\left({a}{\left({f{{\left({t}\right)}}}+{b}{g{{\left({t}\right)}}}\right)}\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{\left({a}{f{{\left({t}\right)}}}+{b}{g{{\left({t}\right)}}}\right)}{d}{t}={\int_{{0}}^{\infty}}{\left({a}{f{{\left({t}\right)}}}{{e}}^{{-{s}{t}}}+{b}{g{{\left({t}\right)}}}{{e}}^{{-{s}{t}}}\right)}{d}{t}$$$.

Now, using linearity, we can write that $$${\int_{{0}}^{\infty}}{\left({a}{f{{\left({t}\right)}}}{{e}}^{{-{s}{t}}}+{b}{g{{\left({t}\right)}}}{{e}}^{{-{s}{t}}}\right)}{d}{t}={a}{\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{d}{t}+{b}\int{{e}}^{{-{s}{t}}}{g{{\left({t}\right)}}}{d}{t}={a}{L}{\left({f{{\left({t}\right)}}}\right)}+{b}{L}{\left({g{{\left({t}\right)}}}\right)}$$$.

This completes the proof.

Note that there can be more than 2 terms, and the property will still hold.

Example 1 . Calculate $$${L}{\left({5}{{e}}^{{{3}{t}}}-{2}{\sin{{\left({7}{t}\right)}}}+{\cos{{\left({t}\right)}}}\right)}$$$.

From the table of Laplace transforms, it is known that $$${L}{\left({{e}}^{{{3}{t}}}\right)}=\frac{{1}}{{{s}-{3}}}$$$, $$${L}{\left({\sin{{\left({7}{t}\right)}}}\right)}=\frac{{7}}{{{{s}}^{{2}}+{{7}}^{{2}}}}$$$, and $$${L}{\left({\cos{{\left({t}\right)}}}\right)}=\frac{{s}}{{{{s}}^{{2}}+{{1}}^{{2}}}}$$$; so, using Property 1, we obtain the following:

$$${L}{\left({5}{{e}}^{{{3}{t}}}-{2}{\sin{{\left({7}{t}\right)}}}+{\cos{{\left({t}\right)}}}\right)}={5}{L}{\left({{e}}^{{{3}{t}}}\right)}-{2}{L}{\left({\sin{{\left({7}{t}\right)}}}\right)}+{L}{\left({\cos{{\left({t}\right)}}}\right)}=$$$

$$$=\frac{{5}}{{{s}-{3}}}-{2}\frac{{7}}{{{{s}}^{{2}}+{49}}}+\frac{{s}}{{{{s}}^{{2}}+{1}}}=\frac{{5}}{{{s}-{3}}}-\frac{{14}}{{{{s}}^{{2}}+{49}}}+\frac{{s}}{{{{s}}^{{2}}+{1}}}$$$.

Property 2. Scaling by constant: $$${L}{\left({f{{\left({a}{t}\right)}}}\right)}=\frac{{1}}{{s}}{F}{\left(\frac{{s}}{{a}}\right)}$$$, where $$${a}$$$ is a positive constant.

The proof is straightforward:

$$${L}{\left({f{{\left({a}{t}\right)}}}\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{f{{\left({a}{t}\right)}}}{d}{t}$$$.

Using the change of variable technique, we make the substituion $$${u}={a}{t}$$$; then, $$${d}{u}={a}{d}{t}$$$, which yields $$${d}{t}=\frac{{{d}{u}}}{{a}}$$$. Since $$${t}$$$ is changing from $$${0}$$$ to $$$\infty$$$, we have that $$${u}$$$ is changing from $$${a}\cdot{0}={0}$$$ to $$${a}\cdot\infty=\infty$$$.

So,

$$${\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{f{{\left({a}{t}\right)}}}{d}{t}={\int_{{0}}^{\infty}}{{e}}^{{-{s}\frac{{u}}{{a}}}}{f{{\left({u}\right)}}}\frac{{{d}{u}}}{{a}}=\frac{{1}}{{a}}{\int_{{0}}^{\infty}}{{e}}^{{-\frac{{s}}{{a}}{u}}}{f{{\left({u}\right)}}}{d}{u}=\frac{{1}}{{a}}{F}{\left(\frac{{s}}{{a}}\right)}$$$.

This completes the proof.

Property 3. $$${L}{\left({t}{f{{\left({t}\right)}}}\right)}=-\frac{{d}}{{{d}{s}}}{F}{\left({s}\right)}$$$, where $$${L}{\left({f{{\left({t}\right)}}}\right)}={F}{\left({s}\right)}$$$.

Proof.

From the definition, it follows that

$$${F}{\left({s}\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{d}{t}$$$.

Differentiating this equality with respect to $$${s}$$$ gives (note that the limits of integration are constant):

$$$\frac{{d}}{{{d}{s}}}{F}{\left({s}\right)}={\int_{{0}}^{\infty}}\frac{{d}}{{{d}{s}}}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{d}{t}$$$,

$$$\frac{{d}}{{{d}{s}}}{F}{\left({s}\right)}={\int_{{0}}^{\infty}}-{t}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{d}{t}$$$,

which finally gives

$$$-\frac{{d}}{{{d}{s}}}{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{t}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{d}{t}={L}{\left({t}{f{{\left({t}\right)}}}\right)}$$$.

This completes the proof.

Example 2. Calculate $$${L}{\left({t}{\sin{{\left({t}\right)}}}\right)}$$$.

Since $$${L}{\left({\sin{{\left({t}\right)}}}\right)}=\frac{{1}}{{{{s}}^{{2}}+{1}}}$$$, using property 3, we obtain that

$$${L}{\left({t}{\sin{{\left({t}\right)}}}\right)}=-\frac{{d}}{{{d}{s}}}\frac{{1}}{{{{s}}^{{2}}+{1}}}=\frac{{1}}{{{\left({{s}}^{{2}}+{1}\right)}}^{{2}}}\cdot\frac{{{d}{\left({{s}}^{{2}}+{1}\right)}}}{{{d}{s}}}=\frac{{{2}{s}}}{{{\left({{s}}^{{2}}+{1}\right)}}^{{2}}}$$$.

Property 4. $$${L}{\left(\frac{{f{{\left({t}\right)}}}}{{t}}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}$$$, where $$${F}{\left({s}\right)}={L}{\left({f{{\left({t}\right)}}}\right)}$$$.

Proof.

This property follows from property 3.

Let $$${g{{\left({t}\right)}}}=\frac{{f{{\left({t}\right)}}}}{{t}}$$$ and $$${L}{\left({g{{\left({t}\right)}}}\right)}={G}{\left({s}\right)}$$$. We need to prove that $$${G}{\left({s}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}$$$.

So, $$${f{{\left({t}\right)}}}={t}{g{{\left({t}\right)}}}$$$.

Taking the Laplace transform of both sides gives:

$$${L}{\left({f{{\left({t}\right)}}}\right)}={L}{\left({t}{g{{\left({t}\right)}}}\right)}$$$.

Or:

$$${F}{\left({s}\right)}=-\frac{{d}}{{{d}{s}}}{G}{\left({s}\right)}$$$ (using property 3).

Integrating this equality with respect to $$${s}$$$ and taking the limits $$${s}$$$ and $$$\infty$$$ yields:

$$${\int_{{s}}^{\infty}}-\frac{{d}}{{{d}{s}}}{G}{\left({s}\right)}{d}{s}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}$$$.

$$$-{\left(\lim_{{{a}\to\infty}}{G}{\left({a}\right)}-{G}{\left({s}\right)}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}$$$.

Since $$${G}{\left({s}\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{g{{\left({t}\right)}}}{d}{t}$$$, we have that $$$\lim_{{{s}\to\infty}}{G}{\left({s}\right)}={0}$$$.

So,

$$$-{\left({0}-{G}{\left({s}\right)}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}$$$.

$$${G}{\left({s}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}$$$.

This completes the proof.

Example 3. Calculate $$${L}{\left(\frac{{{\sin{{\left({t}\right)}}}}}{{t}}\right)}$$$.

Since $$${L}{\left({\sin{{\left({t}\right)}}}\right)}=\frac{{1}}{{{{s}}^{{2}}+{1}}}$$$, using property 4, we obtain that

$$${L}{\left(\frac{{{\sin{{\left({t}\right)}}}}}{{t}}\right)}={\int_{{s}}^{\infty}}\frac{{1}}{{{{p}}^{{2}}+{1}}}{d}{p}=\lim_{{{m}\to\infty}}{\int_{{s}}^{{m}}}\frac{{1}}{{{{p}}^{{2}}+{1}}}{d}{p}=\lim_{{{m}\to\infty}}{\operatorname{atan}{{\left({p}\right)}}}{{\mid}_{{s}}^{{m}}}=$$$

$$$=\lim_{{{m}\to\infty}}{\operatorname{atan}{{\left({m}\right)}}}-{\operatorname{atan}{{\left({s}\right)}}}=\frac{\pi}{{2}}-{\operatorname{atan}{{\left({s}\right)}}}={\operatorname{acot}{{\left({s}\right)}}}$$$.