5.2 Methods and Laplace transforms

Methods

D'Alembert's Theorem

Find missing solution: Try $y\left(x\right)=$old solution$\times u\left(x\right)$.

Variation of Parameters (for non-homogeneous ODEs)

Looking for solution in the form:

$y_{\mathrm{par}}={\sum }_{j=1}^n(u_j\left(x\right)y_j\left(x\right))$

For the case where $n=3$:

${\sum }_{j=1}^3(u_j^{\prime }{y}_j)=0$

${\sum }_{j=1}^3(u_j^{\prime }{y}_j^{\prime })=0$

${\sum }_{j=1}^3(u_j^{\prime }{y}_j^{\prime \prime })=g\left(x\right)$

Solve this system, then integrate each $u_j^{\prime }$ and multiply by $y_j$ to get:

$y_{\mathrm{par}}={\sum }_{j=1}^n(u_j\left(x\right)y_j\left(x\right))$

Alternatively:

$y_{\mathrm{par}}={\sum }_j(y_j\left(x\right){\int }_{x_0}^x(g\left(s\right)\frac{W_j\left[s\right]}{W_{}[y_1\left(s\right),\cdots y_n\left(s\right)]}d\left(s\right)))$

Diagonalisation

Introduce change of variables $x=Ty$:

$T{y}^{\prime }=ATy+g\left(t\right)\Rightarrow \frac{d\left(y\right)}{d\left(t\right)}=Dy+T^{-1}g=Dy+h$

This leads to a system of $n$ decoupled equations solved by direct integration:

$y_i^{\prime }=r_i{y}_i+h_i\Rightarrow y_i\left(t\right)=c_i{e}^{r_{i}t}+e^{r_{i}t}{\int }_{t_0}^t(e^{-r_{i}s}{h}_i\left(s\right)d\left(s\right))$

General solution in the original variables:

$x\left(t\right)=Ty\left(t\right)$

If the matrix is not diagonalisable, find its Jordan form $J$ and proceed similarly:

$P^{-1}AP=D,$ $J=\left[\begin{array}{cc}\lambda & 1\\ 0& \lambda \end{array}\right]$

Laplace Transform

Map $y\left(x\right)\to {{{{F_{}}_{}}_{}^{}}^{}}^{}\left(s\right)$ through:

${{{{F_{}}_{}}_{}^{}}^{}}^{}\left(s\right)={\int }_{\mathrm{\alpha }}^{\mathrm{\beta }}(K(s,x)y\left(x\right)d\left(x\right))$

The Laplace Transform of $ƒ\left(x\right)$ is defined for $x\in [0,\infty )$:

${{{{F_{}}_{}}_{}^{}}^{}}^{}\left(s\right)=L\left\{ƒ\right\}\left(s\right)\equiv {\int }_0^{\infty }({ℇ}^{-sx}ƒ\left(x\right)d\left(x\right))={\lim }_{T\to \infty }\left({\int }_0^T({ℇ}^{-sx}ƒ\left(x\right)d\left(x\right))\right)$

The Laplace Transform is linear:

$L\{c_1{ƒ}_1\left(x\right)+c_2{ƒ}_2\left(x\right)\}\left(s\right)=c_{1}L\{{ƒ}_1\left(x\right)\}+c_{2}L\{{ƒ}_2\left(x\right)\}$

If $ƒ,{ƒ}^{\prime },\cdots ,{ƒ}^{n-1}$ are continuous on $[0,\infty )$ and $\in E$, then:

$L\{{ƒ}^n\left(x\right)\}=s^{n}L\{ƒ\left(x\right)\}-s^{n-1}ƒ\left(0\right)-\cdots -s{ƒ}^{n-2}\left(0\right)-{ƒ}^{n-1}\left(x\right)$

Let $L\{ƒ\left(x\right)\}\left(s\right)={{{{F_{}}_{}}_{}^{}}^{}}^{}\left(s\right)$. For $ƒ\in E$:

• s-shift: $L\{{ℇ}^{-cx}ƒ\left(x\right)\}\left(s\right)={{{{F_{}}_{}}_{}^{}}^{}}^{}(s+c)$

• x-shift:$L\{ƒ(x-c)\}\left(s\right)={ℇ}^{-sc}{{{{F_{}}_{}}_{}^{}}^{}}^{}\left(s\right)$ if $c\geqq 0$ and $ƒ\left(x\right)=0$ for $x<0$

• Derivative:$L\{xƒ\left(x\right)\}\left(s\right)=-{{{{F_{}}_{}}_{}^{\prime }}^{}}^{}\left(s\right)$

• Scaling:$L\{ƒ(cx)\}\left(s\right)=\frac{1}{c}{{{{F_{}}_{}}_{}^{}}^{}}^{}\left(\frac{s}{c}\right)$, ${{{{F_{}}_{}}_{}^{}}^{}}^{}(sc)=\frac{1}{c}L\{ƒ\left(\frac{x}{c}\right)\}$ if $c>0$

Laplace transform of $x{ℇ}^x$:

$L\{x{ℇ}^x\}\left(s\right)=-{\left(\frac{1}{s-1}\right)}^{\prime }=\frac{1}{{(s-1)}^2}$

Change variable from $t$ to $t^{\prime }=t-c$:

$L\{u_c\left(t\right)ƒ(t-c)\}={ℇ}^{-cs}{{{{F_{}}_{}}_{}^{}}^{}}^{}\left(s\right)$

The unit step function:

$u_c\left(t\right)=\{\begin{array}{c}0,t<c\\ 1,t\geqq c\end{array}$

Laplace transform of the unit step function $c\ge 0$:

$L\{u_c\left(t\right)\}\left(s\right)={\int }_0^{\infty }({ℇ}^{-st}{u}_c\left(t\right)d\left(t\right))={\int }_c^{\infty }({ℇ}^{-st}d\left(t\right))=\frac{e^{-sc}}{s},s>0$

For 'nice' functions $ƒ\left(t\right)$:

${\int }_{-\infty }^{\infty }(\delta (t-t_0)ƒ\left(t\right)d\left(t\right))=ƒ\left(t_0\right)$

$L\{\delta (t-t_0)\}\left(s\right)={ℇ}^{-s{t}_0}$

$L\{\delta \left(t\right)\}\left(s\right)={{\lim }_{a\to 0}\left(ℇ\right)}^{-as}$

$L^{-1}\{{\lim }_{a\to 0}\left(\frac{{ℇ}^{-as}}{s^2-a^2}\right)\}=\frac{1}{a}{u}_0\left(t\right)\sinh (at)$

Laplace Proofs

$L\left\{y^{\prime }\right\}\left(s\right)={\int }_0^{\infty }({ℇ}^{-st}{y}^{\prime }d\left(t\right))=({ℇ}^{-st}y\left(t\right)){\mid }_0^{\infty }+sY\left(s\right)=sY\left(s\right)-y\left(0\right)$

With $L\{{(-t)}^nƒ\left(t\right)\}\left(s\right)=\frac{d^n({{{{F_{}}_{}}_{}^{}}^{}}^{}\left(s\right))}{{d\left(s\right)}^n}$, where ${{{{F_{}}_{}}_{}^{}}^{}}^{}\left(s\right)=L\{ƒ\left(t\right)\}\left(s\right)$, we get:

$L\{t^n{ℇ}^{at}\}\left(s\right)={(-1)}^n\frac{d^n}{{d\left(s\right)}^n}{(s-a)}^{-1}=\frac{\mathrm{n!}}{{(s-a)}^{n+1}}$

Convolution

$(ƒ\ast g)\left(t\right)\equiv {\int }_0^t(ƒ\left(t_1\right)g(t-t_1)d\left(t_1\right))$

$L\left\{ƒ\right\}\cdot L\left\{g\right\}=L\{ƒ\ast g\}$

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