Laplace transforms of differential equations

Differential equations and Laplace transforms: Laplace-transformations

Laplace transforms of differential equations

The Laplace transform of the derivative $f'$ of a function $f$ can be expressed in the Laplace transform of $f$ without the use of derivatives:

If $f$ is a differentiable function on $\ivco{0}{\infty}$ , the Laplace transform $\laplace{f}$ exists, and $\lim_{t\to\infty}\ee^{-st} f(t) = 0$ , then $\mathcal{L} f' (s) = s\cdot (\mathcal{ L}f)(s)-f(0)$

More generally, if $n$ is a natural number, $f$ is a piecewise $n$ -fold differentiable function, and, for all $k$ with $0\le k\lt n$ , the limit $\lim_{t\to\infty}\ee^{-st} f^{(k)}(t) = 0$ exists, then $\laplace{\left(f^{(n)}\right)}(s) = s^n\cdot \laplace{f}(s)-s^{n-1}f(0)-s^{n-2}f'(0)-\cdots-s\cdot f^{(n-2)}(0)-f^{(n-1)}(0)$

We begin by deriving the equality for the first derivative. We will make use of partial integration.

$\begin{array}{rcl}\mathcal{L} f' (s) &=& \displaystyle\int_0^{\infty}f'(t)\cdot\ee^{-st}\,\dd t\\ &&\phantom{xx}\color{blue}{\text{definition of Laplace transform}}\\&=& \displaystyle\left.f(t)\cdot\ee^{-st}\right|_{t=0}^{t=\infty}+s\cdot\int_0^{\infty}f(t)\cdot\ee^{-st}\,\dd t\\&&\phantom{xx}\color{blue}{\text{partial integration}}\\ &=& \displaystyle-f(0)+s\cdot(\mathcal{L}f(s))\\ &&\phantom{xx}\color{blue}{\lim_{t\to\infty}\ee^{-st} f(t) = 0}\\\end{array}$

For the proof of equality for any $n$ we use mathematical induction. The case $n=1$ has just been treated. Suppose $n\gt1$ . It follows from the induction hypothesis that

$\begin{array}{rcl} \mathcal{L} f^{(n)}(s) &=& \mathcal{L} (f^{(n-1)})'(s)\\ &=&s\cdot (\mathcal{ L}f^{(n-1)}(s))-f^{(n-1)}(0) \\ &=&s\cdot \left(s^{n-1}\mathcal{ L}f(s)-s^{n-2}f(0)-s^{n-3}f'(0)-\cdots-f^{(n-2)}(0) \right)-f^{(n-1)}(0) \\ &=&s^n\cdot \laplace{f}(s)-s^{n-1}f(0)-s^{n-2}f'(0)-\cdots-s\cdot f^{(n-2)}(0)-f^{(n-1)}(0)\end{array}$

Piecewise continuous functions require a separate treatment. For example, for the Heaviside function $u_a$ with $a\gt0$ we cannot just use as derivative the piecewise continuous function $u_a' = 0$ with a jump at $a$ take because

$\begin{array}{rcl}\left(\mathcal{L} u_a'\right) (s) &=& 0\\ s\cdot \left(\mathcal{L}u_a\right)(s)-f(0)&=&s\cdot \int_a^{\infty}\ee^{-ts}\dd t\\ &=& s\cdot\left. \frac{-\e^{-ts}}{s}\right|_a^{\infty}-u_a(0)\\ &=& \ee^{-as} \end{array}$

The jump at $a$ should be handled carefully: delta function $u_a' = \delta_a$ ,

$\mathcal{L} u_a' (s) = \int_0^\infty\delta_a(t)\cdot\e^{-st}\dd t = \e^{-at}$

Thanks to this property we can convert a linear differential equation with constant coefficients using Laplace transform into an algebraic equation. By calculating the inverse Laplace transform we can solve the differential equation. For a second order ODE are the operations shown below in a diagram

$\begin{array}{lcr}f''(t)+p\cdot f'(t)+q = r(t)&{\laplace{}\atop\longrightarrow}& P(s)\cdot\laplace{f}(s) = \laplace{r}(s)\\ \text{solution of }\big\uparrow&&\big\downarrow\text{ solve}\\ f(t) =\mathcal{L}^{-1}(F)(t)&{\mathcal{L}^{-1}\atop \longleftarrow}& \laplace{f}(s) = F(s)\end{array}$

Below are examples of this solution method.

Solve the differential equation

${{d^2}\over{d t^2}} x+x=t \euler^{t}$

with initial conditions $x(0)=0$ and $x'(0)=0$ by using the Laplace transform.

$x(t) =$ ${{\cos \left(t\right)}\over{2}} -{{\euler^{t}}\over{2}} + {{t\cdot \euler^{t}}\over{2}}$

In order to find the solution, we write $y=\mathcal{L}(x)$ . Then

$\begin{array}{rcl} \mathcal{L}(x'') (s)&=& s^2\cdot y(s)-2s =s^2\cdot y(s) \\ \mathcal{L}(-t\cdot \euler^{t}) (s)&=&\displaystyle -{{1}\over{\left(s-1\right)^2}} \\ \end{array}$

Therefore, after all terms are moved to the left, the Laplace transform applied to the differential equation ${{d^2}\over{d t^2}} x+x=t \euler^{t}$ gives

$s^2\cdot y+y-{{1}\over{\left(s-1\right)^2}}=0$

This can be rewritten to

$\left(s^2+1\right)\cdot y-{{1}\over{\left(s-1\right)^2}}=0$

Solving this equation with unknown $y$ gives

$y (s)= {{1}\over{s^4-2\cdot s^3+2\cdot s^2-2\cdot s+1}}$

Partial fraction decomposition of the right-hand side leads to

$\begin{array}{rcl}y(s) &=&\displaystyle {{1}\over{\left(s-1\right)^2\cdot \left(s^2+1\right)}}\\ &=&\displaystyle {{s}\over{2\cdot \left(s^2+1\right)}}-{{1}\over{2\cdot \left(s-1\right)}}+{{1}\over{2\cdot \left(s-1\right)^2}}\\ \end{array}$

so linearity of the inverse Laplace transform and determination of the inverse Laplace tansforms of the terms gives:

$x (t)= {{\cos \left(t\right)}\over{2}} -{{\euler^{t}}\over{2}} + {{t\cdot \euler^{t}}\over{2}}$

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