Math 555: Differential Equations

6.2 Solving Initial Value Problems

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6.2.1 The Laplace Transform of a Derivative

Now that we know how to compute the Laplace transform of a function let us compute the Laplace transform of the derivative of a function. If we start with a function $f(t)$ that we know has a well-defined Laplace transform, then we have that

$$\mathcal{L}\left[f(t)\right] = \displaystyle\int_0^{\infty} f(t)e^{-st}\,dt \lt \infty, \quad s\gt c\ge 0.$$
This implies that

$$\displaystyle\lim_{s\to\infty} \displaystyle\int_s^{\infty} f(t)e^{-st}\,dt = 0,\quad s\gt c\ge 0.$$
This give us that
$$\displaystyle\lim_{t\to\infty} f(t)e^{-st} = 0,\quad s\gt c\ge 0.$$

Then the Laplace transform of the derivative of $f$ is

$$\mathcal{L}\left[f'(t)\right] = \displaystyle\int_0^{\infty} f'(t)e^{-st}\,dt,\quad s\gt c\ge 0.$$
We will compute a formula for the Laplace transform of a function using integration by parts .



$$\begin{align*} \mathcal{L}\left[f'(t)\right] &= \displaystyle\int_0^{\infty} f'(t)e^{-st}\,dt \\ \\ &= \left|f(t)e^{-st}\right|_0^{\infty} + \displaystyle\int_0^{\infty} f(t)se^{-st}\,dt \\ \\ &= \displaystyle\lim_{t\to\infty}f(t)e^{-st} - f(0)e^0 + s\displaystyle\int_0^{\infty} f(t)e^{-st}\,dt \\ \\ &= s\mathcal{L}\left[f(t)\right] - f(0). \end{align*}$$
It is easier to derive the Laplace transform of the second derivative using the previous formula for the first derivative.

$$\begin{align*} \mathcal{L}[f''(t)] &= s\mathcal{L}[f'(t)] - f'(0) \\ \\ &= s\left\{ s\mathcal[f(t)] - f(0)\right\} - f'(0) \\ \\ &= s^2\mathcal[f(t)] - sf(0) - f'(0). \end{align*}$$
We can now compute the Laplace transform of higher derivatives of $f(t)$ by following the pattern of derivatives.

$$\begin{align*} \mathcal{L}\left[f^{(n)}(t)\right] &= s\mathcal{L}\left[f^{(n-1)}(t)\right] - f^{(n-1)}(0) \\ \\ &= s\left\{ s\mathcal{L}\left[f^{(n-2)}(t)\right] - f^{(n-1)}(0) \right\} - f^{(n-1)}(0) \\ \\ &= s^2\mathcal{L}\left[f^{(n-2)}(t)\right] - sf^{(n-1)}(0) - f^{(n-1)}(0) \\ \\ &= s^2\left\{ s\mathcal{L}\left[f^{(n-3)}(t)\right] - f^{(n-2)}(0) \right\} - sf^{(n-2)}(0) - f^{(n-1)}(0) \\ \\ &= s^3\mathcal{L}\left[f^{(n-3)}(t)\right] - s^2f^{(n-3)}(0) - sf^{(n-2)}(0) - f^{(n-1)}(0) \\ &\ddots \\ &= s^n\mathcal{L}\left[f(t)\right] - s^{n-1}f(0) -\cdots- sf^{(n-2)}(0) - f^{(n-1)}(0) \end{align*}$$

The final expression

$$ \mathcal{L}\left[f^{(n)}(t)\right] = s^n\mathcal{L}\left[f(t)\right] - s^{n-1}f(0) -\cdots- sf^{(n-2)}(0) - f^{(n-1)}(0) $$
is very important and appears on the Laplace Transform Table .

6.2.2 Introductory Examples

Applying the above formula ties into the general usefulness of Laplace transforms. Let us start with some examples then discuss what we see.

Example 6.2.1

Solve the initial value problem

$$y'' + 2y' + 4y = 0;\qquad y(0) = 0,\ y'(0) = 1.$$

Solution

If we compute the Laplace transform of both sides we have

$$\begin{align*} \mathcal{L}\left[y'' + 2y' + 4y\right] &= \mathcal{L}[0] \\ \\ \mathcal{L}\left[y''\right] + 2\mathcal{L}\left[y'\right] + 4\mathcal{L}[y] &= \displaystyle\int_0^{\infty} 0\,e^{-st}\,dt \\ \\ s^2\mathcal{L}[y] - sy(0) - y'(0) + 2\left\{s\mathcal{L}[y] - y(0)\right\} + 4\mathcal{L}[y] &= \displaystyle\int_0^{\infty} 0\,dt \\ \\ \left(s^2 + 2s + 4\right)\mathcal{L}[y] - 1 &= 0 \\ \\ \left(s^2 + 2s + 4\right)Y(s) &= 1 \\ \end{align*}$$
From this we can use algebra to find an expression for the Laplace transform of $Y(s)$ of our solution.

$$Y(s) = \frac{1}{s^2 + 2s + 4}$$
We need to use more algebra to determine an algebraic expression in the right column of our Laplace Transform Table . If we complete the square in the denominator we obtain

$$\begin{align*} Y(s) &= \dfrac{1}{\left(s^2 + 2s + 1\right) + 3} \\ \\ &= \dfrac{1}{(s + 1)^2 + 3} \\ \\ &= \dfrac{1}{\sqrt{3}}\dfrac{\sqrt{3}}{(s + 1)^2 + 3} \\ \\ &= \dfrac{1}{\sqrt{3}}\dfrac{b}{(s-a)^2+b^2},\quad s>a, \\ \end{align*}$$
where $a=-1$ and $b-\sqrt{3}$. Looking for the Inverse Laplace Transform by selecting the function in the left column of entry 9 we have the solution

$$y(t) = \mathcal{L}^{-1}\left[\dfrac{1}{\sqrt{3}}\dfrac{\sqrt{3}}{(s+1)^2+3}\right] = \dfrac{1}{\sqrt{3}}\mathcal{L}^{-1}\left[\dfrac{\sqrt{3}}{(s+1)^2+3}\right] = \dfrac{1}{\sqrt{3}}e^{-t}\sin\left(\left(\sqrt{3}\right)\,t\right).$$

Example 6.2.2

Solve the initial value problem

$$y'' + \omega^2y = \cos(2t),\quad\omega^2\neq 4,\ y(0)=1,\ y'(0)=0.$$

Solution

If we compute the Laplace transform of both sides we have

$$\begin{align*} s^2Y(s) - s(1) - (0) + \omega^2Y(s) &= \dfrac{s}{s^2+4} \\ \\ \left(s^2 + \omega^2\right)Y(s) &= s + \dfrac{s}{s^2+4} \\ \\ Y(s) &= \dfrac{s}{s^2 + \omega^2} + \dfrac{s}{\left(s^2 + \omega^2\right)\left(s^2+4\right)} \\ \end{align*}$$
Using partial fraction decomposition

$$\begin{align*} \dfrac{s}{\left(s^2 + \omega^2\right)\left(s^2+4\right)} &= \dfrac{As + B}{s^2 + \omega^2} + \dfrac{Cs + D}{s^2 + 4} \\ \\ s &= (As + B)(s^2 + 4) + (Cs + D)(s^2 + \omega^2) \\ \\ &= As^3 + Bs^2 + 4As + 4B + Cs^3 + Ds^2 + C\omega^2s + D\omega^2 \\ \\ A + \ \ \ \ \ C &= 0 \qquad \ \ B + \ \ \ \ D = 0 \\ 4A + \omega^2C &= 1 \qquad 4B + \omega^2D = 0 \\ \\ 4A + \ \ \ 4C &= 0 \qquad \ \ 4B + \ \ 4D = 0 \\ 4A + \omega^2C &= 1 \qquad \ \ 4B + \omega^2D = 0 \\ \\ \left(\omega^2-4\right)C &= 1 \qquad \left(\omega^2-4\right)D = 0 \\ \\ C &= \dfrac{1}{\omega^2-4} \qquad D = 0 \\ \\ A &= \dfrac{-1}{\omega^2-4} \qquad B = 0 \\ \\ \dfrac{s}{\left(s^2 + \omega^2\right)\left(s^2+4\right)} &= \dfrac{-1}{\omega^2-4}\cdot\dfrac{s}{s^2 + \omega^2} + \dfrac{1}{\omega^2-4}\cdot\dfrac{s}{s^2 + 4} \end{align*}$$
Substituting the partial fraction decomposition into our equation for $Y(s)$ results in

$$\begin{align*} Y(s) &= \dfrac{s}{s^2 + \omega^2} + \dfrac{-1}{\omega^2-4}\cdot\dfrac{s}{s^2 + \omega^2} + \dfrac{1}{\omega^2-4}\cdot\dfrac{s}{s^2 + 4} \\ \\ &= \left(1 - \dfrac{1}{\omega^2-4}\right)\dfrac{s}{s^2 + \omega^2} + \dfrac{1}{\omega^2-4}\cdot\dfrac{s}{s^2 + 4} \\ \\ &= \left(\dfrac{\omega^2-4}{\omega^2-4} - \dfrac{1}{\omega^2-4}\right)\dfrac{s}{s^2 + \omega^2} + \dfrac{1}{\omega^2-4}\cdot\dfrac{s}{s^2 + 4} \\ \\ &= \dfrac{\omega^2-5}{\omega^2-4}\dfrac{s}{s^2 + \omega^2} + \dfrac{1}{\omega^2-4}\cdot\dfrac{s}{s^2 + 4}. \end{align*}$$
Using the Laplace Transform Table we have that the inverse Laplace transform of both sides of the equation yields

$$y(t) = \dfrac{\omega^2-5}{\omega^2-4}\cos(\omega t) + \dfrac{1}{\omega^2-4}\cos(2t).$$

6.2.2 Solving Initial Value Problems

Notice that when one solves an initial value problem using the Laplace transform, one determines the solution using algebraic methods instead of using differential and integral calculus . Notice also that we utilize the initial conditions when we convert our differential equation into an algebraic equation and the unique solution is found immediately. No variation of parameters or undetermined coefficients is required. While partial fraction decomposition results in a linear algebra problem similar to the one found in the method undetermined coefficients, one does not need to first determine linear independence or dependence.

In the rest of this chapter we are going to develop stronger methods based on this section to solve more complicated differential equations, including differential equation with discontinuous forcing functions. The Laplace Transform belongs to a class of operators called Integral Transforms . There is a lot of study of integral transforms in mathematics to solve differential equations that arise in science and engineering.

Exercise 6.2.1

Solve the initial value problem

$$y'' + 3y' + 2y = 0;\quad y(0)=1,\ y'(0)=-1.$$

View Solution

Computing the Laplace transform of both sides results in

$$\begin{align*} s^2Y(s) - s(1) - (-1) + 3\left(sY(s) - 1\right) + 2Y(s) &= 0 \\ \\ \left(s^2 + 3s + 2\right)Y(s) &= s + 2 \\ \\ Y(s) &= \dfrac{s+2}{s^2 + 3s + 2} \\ \end{align*}$$
Dividing like factors gives us

$$\dfrac{s+2}{(s+2)(s+1)} = \dfrac{1}{s+1}.$$
The algebraic expression for $Y(s)$ becomes

$$Y(s) = \dfrac{1}{s+1}.$$
Using our Laplace Transform Table to find the inverse Laplace transform results in the solution

$$y(t) = e^{-t}.$$

Exercise 6.2.2

Solve the initial value problem

$$ y'' - 2y' + 4y = 0;\quad y(0)=2,\ y'(0)=0. $$

Check Your Work

$$ y(t) = 2e^{-t}\cos\sqrt{3}t - \frac{2}{\sqrt{3}}e^{-t}\sin\sqrt{3}t $$

Video Solution

6.2.3 Laplace Transform of a Series

The Laplace transform of certain functions can be found from their Taylor series expressions. For example the Maclaurin series for $\sin(t)$ is

$$ \sin(t) = \displaystyle\sum_{n=0}^{\infty} (-1)^n\dfrac{x^{2n+1}}{(2n+1)!}.$$
Since the Laplace transform is linear we should have

$$\begin{align*} \mathcal{L}\left[\sin(t)\right] &= \mathcal{L}\left(\displaystyle\sum_{n=0}^{\infty} (-1)^n\dfrac{x^{2n+1}}{(2n+1)!}\right) \\ \\ &= \displaystyle\sum_{n=0}^{\infty}\mathcal{L}\left((-1)^n\dfrac{x^{2n+1}}{(2n+1)!}\right) \\ \\ &= \displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^n}{(2n+1)!}\mathcal{L}\left(x^{2n+1}\right) \\ \\ &= \displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^n}{(2n+1)!}\left(\dfrac{(2n+1)!}{s^{2n+2}}\right) \\ \\ &= \displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^n}{\left(s^2\right)^{n+1}} \\ \\ &= \dfrac{1}{s^2}\displaystyle\sum_{n=0}^{\infty}\left(-\dfrac{1}{s^2}\right)^n,\quad\text{a geometric series} \\ \\ &= \dfrac{1}{s^2}\cdot\dfrac{1}{1 - \left(-\dfrac{1}{s^2}\right)},\quad \left|\dfrac{1}{s^2}\right| < 1 \\ \\ &= \dfrac{1}{s^2}\cdot\dfrac{1}{1 + \dfrac{1}{s^2}},\quad s^2 > 1 \\ \\ &= \dfrac{1}{s^2}\cdot\dfrac{s^2}{s^2 + 1},\quad s > 1 \\ \\ &= \dfrac{1}{s^2 + 1},\quad s > 1. \end{align*}$$

6.2.4 Differentiation of the Laplace Transform of a Function

If $f(t)$ satisfies Theorem 6.1.2 and

$$F(s) = \mathcal{L}\left[f(t)\right] = \displaystyle\int f(t)e^{-st}\,dt, $$
then

$$\begin{align*} F'(s) &= \dfrac{dF}{ds} = \dfrac{d}{ds}\displaystyle\int f(t)e^{-st}\,dt \\ \\ &= \displaystyle\int \dfrac{d}{ds}\left(f(t)e^{-st}\right)\,dt \\ \\ &= \displaystyle\int f(t)\dfrac{d}{ds}\left(e^{-st}\right)\,dt \\ \\ &= \displaystyle\int f(t)\left(-te^{-st}\right)\,dt \\ \\ &= \displaystyle\int -tf(t)e^{-st}\,dt \\ \\ &= \mathcal{L}\left[-tf(t)\right]. \end{align*}$$

Exercise 6.2.3

Compute the Laplace transform of $te^{at}$.

View Solution

$$\begin{align*} \mathcal{L}\left[te^{at}\right] &= -\mathcal{L}\left[-te^{at}\right] \\ \\ &= -\dfrac{d}{ds}\left(\mathcal{L}\left[e^{at}\right]\right) \\ \\ &= -\dfrac{d}{ds}\left(\dfrac{1}{s-a}\right),\quad s > a \\ \\ &= \dfrac{1}{(s-a)^2},\quad s > a \end{align*}$$

6.2.5 Bessel's Equation

Consider Bessel's equation of order zero

$$ty'' + y' + ty = 0.$$
We know that $x=0$ is a regular singular point of the differential equation and that the solution may become unbounded as $t\to 0^+$. However if one of the two fundamental solutions remains bounded as $t\to 0^+$, we can use the Laplace transform to find a bounded solution. Computing the Laplace transform of both sides yields

$$\begin{align*} -\dfrac{d}{ds}\mathcal{L}\left[y''\right] + \mathcal{L}\left[y'\right] - \dfrac{d}{ds}Y(s) &= 0 \\ \\ -\dfrac{d}{ds}\left[s^2Y(s) - sy(0) - y'(0)\right] + sY(s) - y(0) - Y'(s) &= 0 \\ \\ -s^2Y'(s) - 2sY(s) + y(0) + sY(s) - y(0) - Y'(s) &= 0 \\ \\ -\left(1 + s^2\right)Y'(s) - sY(s) &= 0 \\ \\ \left(1 + s^2\right)Y'(s) + sY(s) &= 0 \end{align*}$$
We can solve this first-order separable differential equation for $Y(s)$.

$$\begin{align*} \left(1 + s^2\right)Y'(s) &= -sY(s) \\ \\ \dfrac{1}{Y}\dfrac{dY}{ds} &= -\dfrac{s}{1+s^2} \\ \\ \ln\left|Y\right| &= -\displaystyle\int\dfrac{s}{1+s^2}\,ds = -\dfrac{1}{2}\ln|1+s^2| + c \\ \\ Y(s) &= c\left(1 + s^2\right)^{-1/2} \end{align*}$$
Now we established in Section 5.7 that the solution is Bessel's function of the first kind of order zero,

$$J_0(t) = \displaystyle\sum_{n=0}^{\infty} (-1)^n\dfrac{t^{2n}}{4^n(n!)^2},\quad t \gt 0.$$
This tells us that

$$ \mathcal{L}\left[J_0(t)\right] = \dfrac{c}{\sqrt{1 + s^2}},\quad s \gt 0.$$
It is also true that
$$\displaystyle\lim_{s\to\infty} \mathcal{L}[J_0'(t)] = \displaystyle\lim_{s\to\infty}\displaystyle\int_0^{\infty}J_0'(t)e^{-st}\,dt = \displaystyle\int_0^{\infty}J_0'(t)\left(\displaystyle\lim_{s\to\infty}e^{-st}\right)\,dt = 0.$$
Thus we have that

$$0 = \displaystyle\lim_{s\to\infty} \mathcal{L}[J_0'(t)] = \displaystyle\lim_{s\to\infty}\left(s\mathcal{L}\left[J_0(t)\right] - J_0(0)\right) = \displaystyle\lim_{s\to\infty}\left(\dfrac{cs}{\sqrt{1+s^2}} - 1\right) = c - 1.$$
Therefore $c=1$ and

$$ \mathcal{L}\left[J_0(t)\right] = \dfrac{1}{\sqrt{1 + s^2}},\quad s \gt 0.$$

6.2.4 Final Value Theorem

We can use the Laplace Transform to compute the limit of $J_0(t)$ as $t\to\infty$ using the Final Value Theorem .

$$\begin{align*} \displaystyle\lim_{s\to 0}sF(s) &= \displaystyle\lim_{s\to 0} s\displaystyle\int_0^{\infty} f(t)e^{-st}\,dt \\ \\ &= \displaystyle\lim_{s\to 0} -\displaystyle\int_0^{\infty} f(t)\left(-se^{-st}\right)\,dt \\ \\ &= -\displaystyle\lim_{s\to 0} \displaystyle\int_0^{\infty} f(t)\dfrac{d}{dt}\left(e^{-st}\right)\,dt \end{align*}$$
Integrating by parts this expression becomes

$$\begin{align*} \displaystyle\lim_{s\to 0}sF(s) &= -\displaystyle\lim_{s\to 0}\left\{\left[ f(t)e^{-st} \right]_0^{\infty} - \displaystyle\int_0^{\infty} f'(t)e^{-st}\,dt\right\} \\ \\ &= -\displaystyle\lim_{s\to 0}\left\{\left[ 0 - f(0) \right] - \displaystyle\int_0^{\infty} f'(t)e^{-st}\,dt\right\} \\ \\ &= f(0) + \displaystyle\lim_{s\to 0}\displaystyle\int_0^{\infty} f'(t)e^{-st}\,dt \\ \\ &= f(0) + \displaystyle\int_0^{\infty} \displaystyle\lim_{s\to 0}\left(f'(t)e^{-st}\right)\,dt \\ \\ &= f(0) + \displaystyle\int_0^{\infty} f'(t)\displaystyle\lim_{s\to 0}\left(e^{-st}\right)\,dt \\ \\ &= f(0) + \displaystyle\int_0^{\infty} f'(t)\,dt \\ \\ &= f(0) + \displaystyle\lim_{t\to\infty} f(t) - f(0) \\ \\ &= \displaystyle\lim_{t\to\infty} f(t). \end{align*}$$
Thus the statement of the Final Value Theorem is

Final Value Theorem

If $f$ and its derivative $f'$ have Laplace transforms for $s\gt 0$ with $\mathcal{L}\left\{f(t)\right\} = F(s)$, then

$$ \lim_{t\to\infty} f(t) = \lim_{s\to 0}sF(s) $$
if both of these limits exist.

Replacing $f(t)$ with $J_0(t)$ we have

$$\displaystyle\lim_{t\to\infty} J_0(t) = \displaystyle\lim_{s\to 0}\dfrac{cs}{\sqrt{1 + s^2}} = 0.$$
Similarly,

$$\begin{align*} \displaystyle\lim_{t\to\infty} J_0'(t) &= \displaystyle\lim_{s\to 0}s\mathcal{L}\left[J_0'(t)\right] = \displaystyle\lim_{s\to 0}s\left\{s\mathcal{L}\left[J_0(t)\right] - J_0(0)\right\} \\ \\ &= \displaystyle\lim_{s\to 0}\left[\dfrac{cs^2}{\sqrt{1+s^2}} - sJ_0(0)\right] = 0. \end{align*}$$

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