Differential equations – Laplace transform – Basic properties

The Laplace transform \(F(s)=\mathcal{L}\{f(t)\}(s)=\displaystyle\int_0^{\infty}e^{-st}f(t)\,dt\) exists for standard functions, such as exponential functions, polynomials, sines and cosines and combinations of these functions:

1) Let \(f(t)=1\), then we have: \(F(s)=\displaystyle\int_0^{\infty}e^{-st}f(t)\,dt=\int_0^{\infty}e^{-st}\,dt=-\frac{1}{s}e^{-st}\bigg|_0^{\infty}=\frac{1}{s}\) for \(s>0\).

2) Let \(f(t)=e^{at}\), then we have: \(F(s)=\displaystyle\int_0^{\infty}e^{-st}f(t)\,dt=\int_0^{\infty}e^{-st}e^{at}\,dt=\int_0^{\infty}e^{-(s-a)t}\,dt=\frac{1}{s-a}\) for \(s>a\geq0\).

3) Let \(f(t)=\cos(at)\), then we have for \(a>0\): \(F(s)=\displaystyle\int_0^{\infty}e^{-st}\cos(at)\,dt=\frac{1}{a}\int_0^{\infty}e^{-st}\,d\sin(at)\). Now we use integration by parts to obtain

\begin{align*} F(s)&=\frac{1}{a}e^{-st}\sin(at)\bigg|_0^{\infty}-\frac{1}{a}\int_0^{\infty}\sin(at)\,de^{-st}=0+\frac{s}{a}\int_0^{\infty}e^{-st}\sin(at)\,dt=-\frac{s}{a^2}\int_0^{\infty}e^{-st}\,d\cos(at)\\[2.5mm] &=-\frac{s}{a^2}e^{-st}\cos(at)\bigg|_0^{\infty}+\frac{s}{a^2}\int_0^{\infty}\cos(at)\,de^{-st}=\frac{s}{a^2}-\frac{s^2}{a^2}\int_0^{\infty}e^{-st}\cos(at)\,dt=\frac{s}{a^2}-\frac{s^2}{a^2}F(s),\quad s>0. \end{align*}

Hence we have: \(\displaystyle\left(1+\frac{s^2}{a^2}\right)F(s)=\frac{s}{a^2}\). This implies that \(F(s)=\displaystyle\frac{s}{s^2+a^2}\) for \(s>0\). Note that this formula also holds for \(a=0\).

Similarly we have for \(f(t)=\sin(at)\) that \(F(s)=\displaystyle\int_0^{\infty}e^{-st}\sin(at)\,dt=\frac{a}{s^2+a^2}\) for \(s>0\).

4) Let \(f(t)=t^n\) for \(n\in\{1,2,3,\ldots\}\), then we have

\[F(s)=\int_0^{\infty}e^{-st}t^n\,dt=-\frac{1}{s}\int_0^{\infty}t^n\,de^{-st}=-\frac{1}{s}t^ne^{-st}\bigg|_0^{\infty}+\frac{1}{s}\int_0^{\infty}e^{-st}\,dt^n =0+\frac{n}{s}\int_0^{\infty}e^{-st}t^{n-1}\,dt,\quad s>0.\]

This is a reduction formula. Using \(\mathcal{L}\{1\}(s)=\displaystyle\frac{1}{s}\) we obtain for \(n=1\) that \(F(s)=\displaystyle\frac{1}{s}\cdot\frac{1}{s}=\frac{1}{s^2}\). For \(n=2\) we obtain \(\displaystyle\frac{2}{s}\cdot\frac{1}{s^2}=\frac{2}{s^3}\) and so on. In general we obtain for \(n\geq1\):

\[F(s)=\frac{n}{s}\cdot\frac{n-1}{s}\cdots\frac{1}{s}\int_0^{-st}\,dt=\frac{n(n-1)\cdots1}{s^n}\cdot\frac{1}{s}=\frac{n!}{s^{n+1}},\quad s>0.\]

5) Let \(f(t)=\cosh(at)=\displaystyle\frac{e^{at}+e^{-at}}{2}\), then we have: \(F(s)=\displaystyle\frac{1}{2}\left(\frac{1}{s-a}+\frac{1}{s+a}\right)=\frac{1}{2}\cdot\frac{s+a+s-a}{(s-a)(s+a)}=\frac{s}{s^2-a^2}\) for \(s>a\geq0\).

Similarly, let \(f(t)=\sinh(at)=\displaystyle\frac{e^{at}-e^{-at}}{2}\), then: \(F(s)=\displaystyle\frac{1}{2}\left(\frac{1}{s-1}-\frac{1}{s+a}\right)=\frac{1}{2}\cdot\frac{s+a-s+a}{(s-a)(s+a)}=\frac{a}{s^2-a^2}\) for \(s>a\geq0\).

Theorem: If \(F(s)=\displaystyle\int_0^{\infty}e^{-st}f(t)\,dt=\mathcal{L}\{f(t)\}(s)\) is the Laplace transform of \(f(t)\) for \(s>0\), then

\[F(s-a)=\int_0^{\infty}e^{-(s-a)t}f(t)\,dt=\int_0^{\infty}e^{-st}\cdot e^{at}f(t)\,dt=\mathcal{L}\{e^{at}f(t)\}(s)\]

is the Laplace transform of \(e^{at}f(t)\) for \(s>a\).

This implies that for \(n\in\{0,1,2,\ldots\}\)

\[\mathcal{L}\{e^{at}\cos(bt)\}(s)=\frac{s-a}{(s-a)^2+b^2}\,\quad\mathcal{L}\{e^{at}\sin(bt)\}(s)=\frac{b}{(s-a)^2+b^2} \quad\text{and}\quad\mathcal{L}\{t^ne^{at}\}(s)=\frac{n!}{(s-a)^{n+1}},\quad s>a.\]

Let \(F(s)=\displaystyle\int_0^{\infty}e^{-st}f(t)\,dt=\mathcal{L}\{f(t)\}(s)\) be the Laplace transform of \(f(t)\) for \(s>0\), then

\[F'(s)=\int_0^{\infty}e^{-st}(-t)f(t)\,dt=\mathcal{L}\{(-t)f(t)\}(s),\quad s>0.\]

This implies:

Theorem: If \(F(s)=\displaystyle\int_0^{\infty}e^{-st}f(t)\,dt=\mathcal{L}\{f(t)\}(s)\) is the Laplace transform of \(f(t)\) for \(s>0\), then for \(s>0\)

\[\mathcal{L}\{t^nf(t)\}(s)=\int_0^{\infty}e^{-st}t^nf(t)\,dt=(-1)^nF^{(n)}(s),\quad n=0,1,2,\ldots.\]

Examples:

1) If \(F(s)=\mathcal{L}\{1\}(s)=\displaystyle\int_0^{\infty}e^{-st}\,dt=\frac{1}{s}\) for \(s>0\), then we have:

\[\mathcal{L}\{t^n\}(s)=\int_0^{\infty}e^{-st}t^n\,dt=(-1)^nF^{(n)}(s)=(-1)^n\cdot(-1)(-2)\cdots(-n)s^{-n-1}=\frac{n!}{s^{n+1}},\quad s>0.\]

2) If \(G(s)=\mathcal{L}\{\sin(t)\}(s)=\displaystyle\int_0^{\infty}e^{-st}\sin(t)\,dt=\frac{1}{s^2+1}\) for \(s>0\), then we have:

\[\mathcal{L}\{t\sin(t)\}(s)=-G'(s)=\frac{2s}{(s^2+1)^2},\quad s>0.\]

3) If \(H(s)=\mathcal{L}\{\cos(t)\}(s)=\displaystyle\int_0^{\infty}e^{-st}\cos(t)\,dt=\frac{s}{s^2+1}\) for \(s>0\), then we have:

\begin{align*} \mathcal{L}\{t^2\cos(t)\}(s)&=(-1)^2H''(s)=H''(s)=\frac{d}{ds}H'(s)=\frac{d}{ds}\left(\frac{s^2+1-2s^2}{(s^2+1)^2}\right) =\frac{d}{ds}\left(\frac{1-s^2}{(s^2+1)^2}\right)\\[2.5mm] &=\frac{-2s\cdot(s^2+1)^2-2(s^2+1)\cdot2s\cdot(1-s^2)}{(s^2+1)^4}=\frac{2s(-s^2-1-2+2s^2)}{(s^2+1)^3}=\frac{2s(s^2-3)}{(s^2+1)^3},\quad s>0. \end{align*}

Let \(F(s)=\displaystyle\int_0^{\infty}e^{-st}f(t)\,dt=\mathcal{L}\{f(t)\}(s)\) be the Laplace transform of \(f(t)\) for \(s>0\), then using integration by parts we obtain for \(s>0\)

\[\mathcal{L}\{f'(t)\}(s)=\int_0^{\infty}e^{-st}f'(t)\,dt=\int_0^{\infty}e^{-st}\,df(t)=e^{-st}f(t)\bigg|_0^{\infty}-\int_0^{\infty}f(t)\,de^{-st} =-f(0)+s\int_0^{\infty}e^{-st}f(t)\,dt=sF(s)-y(0).\]

This implies that \(\mathcal{L}\{f''(t)\}(s)=s\left(sF(s)-f(0)\right)-f'(0)=s^2F(s)-sf(0)-f'(0)\) and so on:

Theorem: If \(f\) is continuous and \(f'\) is piecewise continuous on every subinterval \([0,A]\) with \(A>0\) and \(|f(t)|\leq Ke^{at}\) for all \(t\geq M>0\), then we have for \(s>a\geq0\):

\[\mathcal{L}\{f'(t)\}(s)=\int_0^{\infty}e^{-st}f'(t)\,dt=sF(s)-f(0).\]

Theorem: If \(f,f',\ldots,f^{(n-1)}\) are continuous and \(f^{(n0}\) is piecewise continuous on every subinterval \([0,A]\) with \(A>0\) and \(|f^{(k)}(t)|\leq Ke^{at}\) for all \(t\geq M>0\) and \(k=0,1,2,\ldots,n-1\), then we have for \(s>a\geq0\):

\[\mathcal{L}\{f^{(n)}(t)\}(s)=\int_0^{\infty}e^{-st}f^{(n)}(t)\,dt=s^nF(s)-s^{n-1}f(0)-\cdots-sf^{(n-2)}(0)-f^{(n-1)}(0).\]

Example: \(y''(t)+y(t)=t\) with \(y(0)=1\) and \(y'(0)=0\). Using the characteristic equation \(r^2+1=0\) and the method of undetermined coefficients we easily obtain that the general solution is \(y(t)=t+c_1\cos(t)+c_2\sin(t)\). The initial values then lead to \(c_1=1\) and \(c_2=-1\). Hence the solution of the initial-value problem is: \(y(t)=t+\cos(t)-\sin(t)\).

Using the Laplace transform we obtain: let \(Y(s)=\mathcal{L}\{y(t)\}(s)\) be the Laplace transform of \(y(t)\), then using partial fractions we find

\[s^2Y(s)-sy(0)-y'(0)+Y(s)=\frac{1}{s^2}\quad\Longrightarrow\quad(s^2+1)Y(s)=s+\frac{1}{s^2}\quad\Longrightarrow\quad Y(s)=\frac{s^3+1}{s^2(s^2+1)}=\frac{As+B}{s^2}+\frac{Cs+D}{s^2+1}.\]

Then we have: \(s^3+1=As(s^2+1)+B(s^2+1)+Cs^3+Ds^2=(A+C)s^3+(B+D)s^2+As+B\) which implies that \(A=0\), \(B=C=1\) and \(D=-1\). Hence we have:

\[Y(s)=\frac{1}{s^2}+\frac{s}{s^2+1}-\frac{1}{s^2+1}\quad\Longrightarrow\quad y(t)=t+\cos(t)-\sin(t).\]

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Last modified on April 25, 2021