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:

Derivative in the time domain 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

\[ {{d^2}\over{d t^2}} x-{{d}\over{d t}} x-6 x=\cos(2t)\]

with initial conditions \(x(0)=0\) and \(x'(0)=1\) by using the Laplace transform.

#x(t) =# # -{{5\cdot \cos \left(2 t \right)}\over{52}} -{{\sin \left(2 t \right)}\over{52}} -{{3\cdot \euler^ {- 2 t }}\over{20}} + {{16\cdot \euler^{3 t }}\over{65}} #

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

\[\begin{array}{rcl}
\mathcal{L}(x') (s)&=& s\cdot y(s)-x(0) =s\cdot y(s)\\
\mathcal{L}(x'') (s)&=& s^2\cdot y(s)-2s =s^2\cdot y(s)-1 \\
\mathcal{L}(-\cos \left(2\cdot t\right)) (s)&=&\displaystyle -{{s}\over{s^2+4}} \\
\end{array}\]

Therefore, after all terms are moved to the left, the Laplace transform applied to the differential equation \( {{d^2}\over{d t^2}} x-{{d}\over{d t}} x-6 x=\cos(2t)\) gives

\[s^2\cdot y-s\cdot y-6\cdot y-{{s}\over{s^2+4}}-1=0\]

This can be rewritten to

\[\left(s^2-s-6\right)\cdot y-{{s}\over{s^2+4}}-1=0\]

Solving this equation with unknown #y# gives

\[ y (s)= {{s^2+s+4}\over{s^4-s^3-2\cdot s^2-4\cdot s-24}} \]

Partial fraction decomposition of the right-hand side leads to

\[ \begin{array}{rcl}y(s)
&=&\displaystyle {{s^2+s+4}\over{\left(s-3\right)\cdot \left(s+2\right)\cdot \left(s^2+4\right)}}\\
&=&\displaystyle {{-5\cdot s-2}\over{52\cdot \left(s^2+4\right)}}-{{3}\over{20\cdot \left(s+2\right)}}+{{16}\over{65\cdot \left(s-3\right)}}\\
&=&\displaystyle -{{5\cdot s}\over{52\cdot \left(s^2+4\right)}} -{{1}\over{26\cdot \left(s^2+4\right)}} -{{3}\over{20\cdot \left(s+2\right)}} + {{16}\over{65\cdot \left(s-3\right)}} \end{array}\]

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

\[ x (t)= -{{5\cdot \cos \left(2 t \right)}\over{52}} -{{\sin \left(2 t \right)}\over{52}} -{{3\cdot \euler^ {- 2 t }}\over{20}} + {{16\cdot \euler^{3 t }}\over{65}} \]

New example