Using Laplace Transforms to Solve Linear DEs

Section 5.6 Using Laplace Transforms to Solve Linear DEs

Laplace transforms turn derivatives into algebraic operations. Therefore, particularly for certain linear equations, I can expect Laplace transforms to turn DEs into algebraic equations. I’ll demonstrate the use of this technique in a series of examples. First, I’ll start with a very well known case: homogeneous second order constant coefficient linear equations. In all these examples, I’ll include initial condition, since the value of the initial conditions naturally arising in the process of solving with Laplace transforms.

Example 5.6.1.

\begin{align*} \amp y^{\prime \prime} + y = 0 \amp \amp y(0) = 1 \amp \amp y^\prime(0) = 0 \end{align*}

I apply the Laplace transform, making use of the initial values when I transform the derivative.

\begin{align*} \calL \{ y^{\prime \prime} + y \} \amp = 0\\ (s^2Y - sy(0) - y^\prime(0)) + Y \amp = 0\\ s^2Y - s + Y \amp = 0 \end{align*}

Now this is an algebraic equation. I solve for \(Y\text{.}\)

\begin{align*} (s^2+1)Y \amp = s\\ Y \amp = \frac{s}{s^2 + 1} \end{align*}

Then I invert the Laplace transform using the tables of Laplace transforms and inverse transforms.

\begin{align*} y \amp = \calL^{-1} \left\{ \frac{s}{s^2+1} \right\} = \cos t \end{align*}

I recover the expected \(y(t) = \cos t\text{,}\) but without any calculation of the characteristic equation or interpretation of complex roots. I also didn’t have to get the complete solution and solve for unknown constants, since I made use of the initial values in the process. If initial values were not given, I would have to use unknown constants in their place in the calculation; this is doable, but given how naturally the initial values show up in the calculation, it’s always preferable to work with initial values is possible.

Example 5.6.2.

Now consider a harmonic system with friction and assume that \(b^2 - 4mk \lt 0\text{,}\) so that I know to expect sinusoidal solutions.

\begin{align*} \amp my^{\prime \prime} + b y^\prime + ky = 0 \amp \amp y(0) = 1 \amp \amp y^\prime(0) = \frac{-b}{2m} \end{align*}

Then we can apply the Laplace transform to the entire equation.

\begin{align*} \calL \{ my^{\prime \prime} + b y^\prime + ky \} \amp = \calL \{ 0 \}\\ m (s^2Y - sy(0) - y^\prime(0)) + b(sY - y(0)) + kY \amp = 0\\ Y(ms^2 + bs + k) - ms - b + \frac{b}{2} \amp = 0 \end{align*}

This is an algebraic system, so I solve for \(Y\text{.}\)

\begin{align*} Y \amp = \frac{ms+\frac{b}{2} }{ms^2+bs+k} = \frac{s + \frac{b}{2m}}{s^2 + \frac{b}{m} s + \frac{k}{m}}\\ \amp = \frac{s + \frac{b}{2m}}{ \left( s^2 + \frac{b}{m} s + \frac{b^2}{4m^2} \right)+ \left( \frac{k}{m} - \frac{b^2}{4m^2} \right) }\\ \amp = \frac{s + \frac{b}{2m}}{ \left( s + \frac{b}{2m} \right)^2 + \left( \frac{4km - b^2}{4m^2} \right)} \end{align*}

Now I need to do the inverse transform. This looks a bit complicated, but it does fit this rule from the tables.

\begin{equation*} \calL^{-1} \left\{ \frac{s+\alpha}{(s + \alpha)^2 + \beta^2} \right\} = e^{-\alpha t}{\cos \beta t} \end{equation*}

Fiting to the patter, I need \(\alpha = \frac{b}{2m}\) and \(\beta = \frac{\sqrt{4mk - b^2}}{2m}\text{.}\) The assumption at the start that \(b^2 - 4mk \lt 0\) ensures that the square is defined. It also ensure that the denominator of \(Y(s)\) is an irreducible quadratic, meaning that I don’t have to do partial fractions to split it up into two linear fractions, which would have other Laplace inverses. Now I can finally proceed to do the inverse transform.

\begin{align*} y \amp = \calL^{-1} \left\{ \frac{s + \frac{b}{2m}}{ \left( s + \frac{b}{2m} \right)^2 + \left( \frac{4mk - b^2}{4m^2} \right)} \right\}\\ \amp = e^{-\frac{b}{2m}t} \cos \left( \frac{\sqrt{4mk-b^2}}{2m} t \right) \end{align*}

Example 5.6.3.

Now I’ll add a forcing term to the general harmonic form, again assuming that \(b^2 - 4acm \lt 0\text{.}\) Since I can take a Laplace transform of a \(\delta\)-function, we’ll use that for the forcing term. \(F\delta_0(t)\) is a sudden impact with force \(F\) (in appropriate units) at time \(t\text{.}\) I’ll use initial values of \(y(0) = y^\prime(0) = 0\text{,}\) so that the system is initially at rest and the only motion is due to the sudden impact. I apply the Laplace transform to the differential equation.

\begin{align*} my^{\prime \prime} + by^\prime + ky \amp = F \delta_0(t)\\ \calL \{ my^{\prime \prime} + by^\prime + ky \} \amp = \calL \{ F \delta_0(t) \}\\ m (s^2 Y - s y(0) - y^\prime(0)) + b(sY - y(0)) + kY \amp = F\\ (ms^2 + bs + k) Y \amp = F \end{align*}

This is now an algebraic equation, so I solve for \(Y\text{.}\)

\begin{align*} Y \amp = \frac{F}{ms^2 + bs + k}\\ Y \amp = \frac{\frac{F}{m}}{s^2 + \frac{b}{m}s + \frac{k}{m}}\\ Y \amp = \frac{\frac{F}{m}}{\left( s^2 + \frac{bs}{m} + \frac{b^2}{4m^2} \right) + \left( \frac{k}{m} - \frac{b^2}{4m^2} \right) }\\ Y \amp = \frac{\frac{F}{m}}{\left( s^2 + \frac{bs}{m} + \frac{b^2}{4m^2} \right) + \left( \frac{4km - b^2}{4m^2} \right) } \end{align*}

This looks like something that will transform back to an expoentially decaying sine function, since there is a constant in the numerator. I’ll adjust the constants to get a matching constant for the inverse transform.

\begin{align*} Y \amp = \frac{F}{m} \frac{2m}{\sqrt{4km-b^2}} \frac{\frac{\sqrt{4km-b^2}}{2m}}{\left( s + \frac{b}{2m} \right)^2 + \left( \frac{4km - b^2}{4m^2} \right) } \end{align*}

Now I apply the inverse transform that produces an exponentially decaying sine function, using \(\alpha = \frac{b}{2m}\) and \(\beta = \frac{\sqrt{4km - b^2}}{2m}\text{.}\)

\begin{align*} y(t) \amp = \frac{2m}{4km-b^2} \frac{F}{m} e^{\frac{-b}{2m}t} \sin \left( \frac{\sqrt{4km-b^2}}{2m}t \right)\\ y(t) \amp = \frac{2F}{4km-b^2} e^{\frac{-b}{2m}t} \sin \left( \frac{\sqrt{4km-b^2}}{2m}t \right) \end{align*}

This seems reasonable. The sudden impact causes a sine wave to start, which then decays. I could ask: what changes if I move the impact to a later time? If the impact is at \(t=4\text{,}\) then the forcing term is \(\delta_4(t)\text{.}\) I again proceed with the Laplace transform.

\begin{align*} my^{\prime \prime} + by^\prime + ky \amp = F \delta_4(t)\\ \calL \{ my^{\prime \prime} + by^\prime + ky \} \amp = \calL \{ F \delta_0(t) e^{-4s}\}\\ m (s^2 Y - s y(0) - y^\prime(0)) + b(sY - y(0)) + kY \amp = F e^{-4s}\\ (ms^2 + bs + k) Y \amp = F e^{-4s} \end{align*}

This is now an algebraic equation, so I solve for \(Y\text{.}\)

\begin{align*} Y \amp = \frac{Fe^{-4s}}{ms^2 + bs + k}\\ Y \amp = \frac{\frac{Fe^{-4s}}{m}}{s^2 + \frac{b}{m}s + \frac{k}{m}}\\ Y \amp = e^{-4s} \frac{\frac{F}{m}}{\left( s^2 + \frac{bs}{m} + \frac{b^2}{4m^2} \right) + \left( \frac{k}{m} - \frac{b^2}{4m^2} \right) }\\ Y \amp = e^{-4s} \frac{F}{m} \frac{2m}{\sqrt{4km-b^2}} \frac{\frac{\sqrt{4km-b^2}}{2m}}{\left( s + \frac{b}{2m} \right)^2 + \left( \frac{4km - b^2}{4m^2} \right) } \end{align*}

This is identitcal to the previous inverse transform except for multiplication by \(e^{-4s}\text{.}\) I can use the rules for shifts; multiplication by an exponential in the \(s\) domain is a shift in the \(t\) domain. I take the result from the previous calculation, shift it, and multipy by a unit step function.

\begin{align*} y(t) \amp = u_4(t) \frac{2m}{4km-b^2} \frac{F}{m} e^{\frac{-b}{2m}(t-4)} \sin \left( \frac{\sqrt{4km-b^2}}{2m} (t-4) \right)\\ y(t) \amp = u_4(t) \frac{2F}{4km-b^2} e^{\frac{-b}{2m} (t-4) } \sin \left( \frac{\sqrt{4km-b^2}}{2m} (t-4) \right) \end{align*}

The result is the same sine wave, just shifted \(4\) units to the right. This makes sense: the sine wave doesn’t start until after the sudden impact.

Example 5.6.4.

This example has an exponential forcing term. I’ll apply the Laplace transform to the differential equation.

\begin{align*} y^{\prime \prime} + 4y^\prime + 4y \amp = e^{-2t} \\ y(0) \amp = 0 \\ y^\prime(0) \amp = 2\\ s^2 Y - sy(0) - y^\prime(0) + 4(sY - y(0)) + 4Y \amp = \frac{1}{s+2} \end{align*}

This is now an algebraic equation, so I solve for \(Y\text{.}\)

\begin{align*} s^2 Y - 2 + 4sY + 4Y \amp = \frac{1}{s+2} \\ (s^2 + 4s + 4)Y \amp = \frac{1}{s+1} + 2 = \frac{2s+3}{s+2} \\ Y \amp = \frac{2s+3}{(s+2)(s+2)^2} = \frac{2s+3}{(s+2)^3} \end{align*}

I need to use partial fractions to split this into forms that are amenable to the inverse transform. (I do all the partial fractions steps by computer, not showing the work.)

\begin{align*} Y \amp = \frac{2}{(s+2)^2} + \frac{1}{(s+2)^3} \end{align*}

Then I can apply the inverse transform. This is a little tricky, but I treat both of these are derivatives. Looking at them as derivatives, I can use the derivative rule to find the original functions in the \(t\) domain.

\begin{align*} \frac{1}{(s+2)^2} \amp = \frac{d}{ds} \frac{-1}{(s+2)}\\ \frac{1}{(s+2)^3} \amp = \frac{1}{2} \frac{d^2}{ds^2} \frac{-1}{(s+2)}\\ Y \amp = -2 \frac{d}{ds} \frac{1}{(s+2)} + \frac{1}{2} \frac{d^2}{ds^2} \frac{1}{(s+2)} \\ y \amp = 2te^{-2t} + t^2e^{-2t} = e^{-2t}(2t+t^2) \end{align*}

Example 5.6.5.

This example has a linear forcing term. I apply the Laplace transform to the differential equation.

\begin{align*} y^{\prime \prime} + 4y \amp = 4t^2 - 4t + 10\\ y(0) \amp = 0 \\ y^\prime(0) \amp = 3\\ s^2 Y - sy(0) - y^\prime(0) + 4Y \amp = \calL \{ 4t^2 - 4t + 10 \} \end{align*}

This is now an algebraic equation, so I solve for \(Y\text{.}\)

\begin{align*} s^2 Y + 4sY -3 \amp = \frac{8}{s^3} - \frac{4}{s^2} + \frac{10}{s} \end{align*}

I need to use partial fractions to split this into forms that are amenable to the inverse transform.

\begin{align*} Y \amp = \frac{8-4s+10s^2 + 3s^3}{s^3(s^2+4)} \\ Y \amp = \frac{2s^2-s+2}{s^3} + \frac{-2s+4}{s^2+4} \\ \amp = \frac{2}{s} - \frac{1}{s^2} + \frac{2}{s^3} - \frac{2s}{s^2+4} + \frac{4}{s^2+4} \end{align*}

Then I apply the inverse transform.

\begin{align*} y \amp = 2 - t + t^2 - 2 \cos 2t + 2 \sin 2t \end{align*}

Example 5.6.6.

Since I can take Laplace transforms of piecewise-continuous functions, here is an example with a piecewise forcing term.

\begin{equation*} f(t) = \left\{ \begin{matrix} 0 \amp t \leq \pi \\ 3 \cos t \amp t > \pi \end{matrix} \right. \end{equation*}

This represents a sinusoidal forcing term which is turned of at time \(t=\pi\text{.}\) I calculate the Lapalce transform of this piecewise function and then apply the Laplace transform to the whole equation. I can treat this piecewise function as a product of a unit step function and a shifted cosine function, which allows me to use the tables to find an appropriate transform.

\begin{align*} y^{\prime \prime} + y \amp = f(t) \\ y(0) \amp = 0\\ y^\prime(0) \amp = 0\\ s^2Y + Y \amp = 3 \calL \{ u_\pi(t) \cos t\} = 3 \calL \{ -u_\pi(t) \cos (t-\pi) \} \\ \amp = -3 e^{-\pi s} \frac{s}{s^2+1} \end{align*}

This is now an algebraic equation, so I solve for \(Y\text{.}\)

\begin{align*} (s^2+1)Y \amp = \frac{-3se^{-\pi s}}{s^2+1} \\ Y \amp = \frac{-3se^{-\pi s}}{(s^2+1)^2} \\ Y \amp = -3e^{-\pi s} \frac{s}{(s^2+1)^2} \end{align*}

I’ve written this as a derivative, so that I can use the inverse rule for derivative. I apply the inverse transform, also using the fact that multiplying by the exponential in the \(s\) domain will produce a shift in the \(t\) domain.

\begin{align*} Y \amp = -3e^{-\pi s} \frac{1}{2} \frac{d}{dt} \frac{-1}{s^2+1} \\ y \amp = \frac{-3}{2} u_{\pi}(t) (t-\pi) \sin (t-\pi) \end{align*}

The forcing term is discontinuous, representing a sudden change, but the resulting solution is still continuous. The force suddenly changes, but the system still responds continuously, as is reasonable to expect.

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