image equation

In solving an initial value problem leading to an ordinary differentialequation, the Laplace transform offers often a way to simplify the equation:both sides are Laplace transformed.  The transformed equation, the so-calledimage equation, is in many cases simplier than the original differentialequation, since it does not contain the derivatives of the unknown function$y(t)$.  From the image equation one may solve the Laplace transform $Y(s)$ of$y(t)$ and then inverse transform $Y(s)$ getting $y(t)$.

Let’s consider e.g. the ordinary $n$’th order lineardifferential equation

$a_0{\displaystyle \frac{d^{n}y}{dt}}+a_1{\displaystyle \frac{d^{n-1}y}{d{x}^{n-1}}}+\mathrm{\dots }+a_{n-1}{\displaystyle \frac{dy}{dt}}+a_{n}y(t)=f(t)$(1)

subject to the initial conditions

$y(0)=y_0,y^{\prime }(0)=y_0^{\prime },\mathrm{\dots },y^{n-1}(0)=y_0^{(n-1)}.$(2)

Due to the linearity of the Laplace transform the image equationof (1) is

$a_0ℒ\{{\displaystyle \frac{d^{n}y}{d{x}^n}}\}+a_1ℒ\{{\displaystyle \frac{d^{n-1}y}{d{x}^{n-1}}}\}+\mathrm{\dots }+a_nℒ\{y(t)\}=ℒ\{f(t)\}.$(3)

Denote  $ℒ\{y(t)\}=:Y(s)$  and $ℒ\{f(t)\}=:F(s)$.  We put into (3) the expressionsof the Laplace transforms of the derivatives on the left hand side (see “Laplace transforms of derivatives (http://planetmath.org/LaplaceTransformsOfDerivatives)”)getting

$a_0[s^{n}Y(s)-(s^{n-1}{y}_0+s^{n-2}{y}_0^{\prime }+\mathrm{\dots }+y_0^{(n-1)})]$

$+a_1[s^{n-1}Y(s)-(s^{n-2}{y}_0+s^{n-3}{y}_0^{\prime }+\mathrm{\dots }+y_0^{(n-2)})]$

$+\mathrm{\dots }\mathit{\hspace{1em}\hspace{1em}}\mathrm{\dots }\mathit{\hspace{1em}\hspace{1em}}\mathrm{\dots }\mathit{\hspace{1em}\hspace{1em}}\mathrm{\dots }$

$+a_{n-1}[sY(s)-(y_0)]$

$=\mathit{\hspace{1em}}F(s).$

This equation is simplified to

$(a_0{s}^n+a_1{s}^{n-1}+\mathrm{\dots }+a_{n-1}s+a_n)Y(s)=$

$a_0[y_0{s}^{n-1}+y_0^{\prime }{s}^{n-2}+\mathrm{\dots }+y_0^{(n-1)}]+$

$+a_1[y_0{s}^{n-2}+y_0^{\prime }{s}^{n-3}+\mathrm{\dots }+y_0^{(n-2)}]+$

$+\mathrm{\dots }\mathit{\hspace{1em}\hspace{1em}}\mathrm{\dots }\mathit{\hspace{1em}\hspace{1em}}\mathrm{\dots }\mathit{\hspace{1em}\hspace{1em}}\mathrm{\dots }+$

$+a_{n-2}[y_{0}s+y_0^{\prime }]+a_{n-1}[y_0]+F(s).$

For brevity, denote in the last equation the polynomialmultiplier of $Y(s)$ by $\phi (s)$ and the sum preceding $F(s)$by $\psi (s)$.  Then the equation can be written as

$$\phi (s)Y(s)=\psi (s)+F(s),$$

i.e.

$Y(s)={\displaystyle \frac{\psi (s)}{\phi (s)}}+{\displaystyle \frac{F(s)}{\phi (s)}}.$(4)

The function $Y(s)$ defined by (4) is the Laplace transform ofthe solution $y(t)$ of the differential equation (1) whichsatisfies the initial conditions (2).  If we now find a function$y^{*}(t)$ the Laplace transform of which is the function $Y(s)$defined by (4), then $y^{*}(t)$ will do for $y(t)$ due to theuniqueness property of Laplace transform expressed in the entry“Mellin’s inverse formula (http://planetmath.org/MellinsInverseFormula)”.
If we seek the solution of (1) satisfying the zero initialconditions

$$x_0=x_0^{\prime }=x_0^{\prime \prime }=\mathrm{\dots }=x_0^{(n-1)}=0,$$

then  $\psi (s)\equiv 0$  and

$$Y(s)=\frac{F(s)}{\phi (s)},$$

i.e.

$$Y(s)=\frac{F(s)}{a_0{s}^n+a_1{s}^{n-1}+\mathrm{\dots }+a_n}.$$

Example.  The 4’th order differential equation

$y^{\prime \prime \prime \prime }(t)+y(t)=\mathrm{\hspace{0.33em}0}$(5)

should be solved with the initial conditions

$$y(0)=y^{\prime \prime \prime }(0)=1,y^{\prime }(0)=y^{\prime \prime }(0)=0.$$

The image equation of (5) is

$$s^{4}Y(s)-s^{3}y(0)-s^2{y}^{\prime }(0)-s{y}^{\prime \prime }(0)-y^{\prime \prime \prime }(0)+Y(s)=\mathrm{\hspace{0.33em}0},$$

i.e.

$$(s^4+1)Y(s)=s^3+1.$$

Thus one needs to determine the inverse Laplace transform of

$Y(s)={\displaystyle \frac{1}{4}}\cdot {\displaystyle \frac{4s^3}{s^4+1}}+{\displaystyle \frac{1}{s^4+1}}.$(6)

The zeroes of the numerator $s^4+1$ are the eighth roots of unity$e^{i\frac{\pi }{4}}$, $e^{i\frac{3\pi }{4}}$, $e^{i\frac{5\pi }{4}}$,$e^{i\frac{7\pi }{4}}$, in other words the complex numbers$\frac{\pm 1\pm i}{\sqrt{2}}$.  By the special case (3) ofthe Heaviside formula, the first addend of (6)corresponds the original function

$$\frac{1}{4}\sum _{\pm }{e}^{\frac{\pm 1\pm i}{\sqrt{2}}t}=\frac{e^{\frac{t}{\sqrt{2}}}+e^{-\frac{t}{\sqrt{2}}}}{2}\cdot \frac{e^{i\frac{t}{\sqrt{2}}}+e^{-i\frac{t}{\sqrt{2}}}}{2}=\cosh \frac{t}{\sqrt{2}}\cos \frac{t}{\sqrt{2}}.$$

Utilizing also the generalHeaviside formula (http://planetmath.org/HeavisideFormula) (1), one canget from (6) the result

$$y(t):=\cosh \frac{t}{\sqrt{2}}\cos \frac{t}{\sqrt{2}}+\frac{1}{\sqrt{2}}(\cosh \frac{t}{\sqrt{2}}\sin \frac{t}{\sqrt{2}}-\sinh \frac{t}{\sqrt{2}}\cos \frac{t}{\sqrt{2}}).$$