# image equation

In solving an initial value problem leading to an ordinary differential equation, the Laplace transform offers often a way to simplify the equation: both sides are Laplace transformed.  The transformed equation, the so-called image equation, is in many cases simplier than the original differential equation, 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 linear differential equation

 $\displaystyle a_{0}\frac{d^{n}y}{dt}+a_{1}\frac{d^{n-1}y}{dx^{n-1}}+\ldots+a_{% n-1}\frac{dy}{dt}+a_{n}y(t)\;=\;f(t)$ (1)

subject to the initial conditions

 $\displaystyle y(0)=y_{0},\quad y^{\prime}(0)=y_{0}^{\prime},\quad\ldots,y^{n-1% }(0)=y_{0}^{(n-1)}.$ (2)

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

 $\displaystyle a_{0}\mathcal{L}\{\frac{d^{n}y}{dx^{n}}\}+a_{1}\mathcal{L}\{% \frac{d^{n-1}y}{dx^{n-1}}\}+\ldots+a_{n}\mathcal{L}\{y(t)\}\;=\;\mathcal{L}\{f% (t)\}.$ (3)

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

 $\displaystyle a_{0}[s^{n}Y(s)-(s^{n-1}y_{0}+s^{n-2}y_{0}^{\prime}+\ldots+y_{0}% ^{(n-1)})]$ $\displaystyle+a_{1}[s^{n-1}Y(s)-(s^{n-2}y_{0}+s^{n-3}y_{0}^{\prime}+\ldots+y_{% 0}^{(n-2)})]$ $\displaystyle+\ldots\qquad\ldots\qquad\ldots\qquad\ldots$ $\displaystyle+a_{n-1}[sY(s)-(y_{0})]$ $\displaystyle=\quad F(s).$

This equation is simplified to

 $\displaystyle(a_{0}s^{n}+a_{1}s^{n-1}+\ldots+a_{n-1}s+a_{n})Y(s)\;=$ $\displaystyle a_{0}[y_{0}s^{n-1}+y_{0}^{\prime}s^{n-2}+\ldots+y_{0}^{(n-1)}]+$ $\displaystyle+a_{1}[y_{0}s^{n-2}+y_{0}^{\prime}s^{n-3}+\ldots+y_{0}^{(n-2)}]+$ $\displaystyle+\ldots\qquad\ldots\qquad\ldots\qquad\ldots+$ $\displaystyle+a_{n-2}[y_{0}s+y_{0}^{\prime}]+a_{n-1}[y_{0}]+F(s).$

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

 $\varphi(s)Y(s)\;=\;\psi(s)+F(s),$

i.e.

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

The function $Y(s)$ defined by (4) is the Laplace transform of the solution $y(t)$ of the differential equation (1) which satisfies 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 the uniqueness 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 initial conditions

 $x_{0}=x_{0}^{\prime}=x_{0}^{\prime\prime}=\ldots=x_{0}^{(n-1)}=0,$

then  $\psi(s)\equiv 0$  and

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

i.e.

 $Y(s)\;=\;\frac{F(s)}{a_{0}s^{n}+a_{1}s^{n-1}+\ldots+a_{n}}.$

Example.  The 4’th order differential equation

 $\displaystyle y^{\prime\prime\prime\prime}(t)+y(t)\;=\;0$ (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

 $s^{4}Y(s)-s^{3}y(0)-s^{2}y^{\prime}(0)-sy^{\prime\prime}(0)-y^{\prime\prime% \prime}(0)+Y(s)\;=\;0,$

i.e.

 $(s^{4}\!+\!1)Y(s)\;=\;s^{3}\!+\!1.$

Thus one needs to determine the inverse Laplace transform of

 $\displaystyle Y(s)\;\;=\;\;\frac{1}{4}\!\cdot\!\frac{4s^{3}}{s^{4}+1}+\frac{1}% {s^{4}+1}.$ (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) of the 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 general Heaviside formula (http://planetmath.org/HeavisideFormula) (1), one can get from (6) the result

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

## References

• 1 N. Piskunov: Diferentsiaal- ja integraalarvutus kõrgematele tehnilistele õppeasutustele. Teine köide. Viies trükk.  Kirjastus Valgus, Tallinn (1966).

Title image equation ImageEquation 2014-03-20 20:16:56 2014-03-20 20:16:56 pahio (2872) pahio (2872) 13 pahio (2872) Topic msc 34A05 msc 44A10