# Explicit solution of a Cauchy problem

I have troubles finding the solution for the Cauchy problem $$u''(t) + u(t) = \mid t\mid$$ with conditions $$u(0) = 1, \, u'(0) = -1$$ for $$t \in \mathbb{R}$$

I tried to solve it with the formula for the solution of a linear system, reducing the order of the equation and ending up with a two dimensional system, but I get stuck. Is there another way to do it? Thanks in advance for any help!

Here is the standard procedure: the general solution of the homogeneous equation is $$a\sin\, x + b\cos \, x$$ There is particular solution $$u_0$$ of the form $$u_o(x)=a(x)\sin\, x + b(x)\cos \, x$$ where $$a(x)$$ and $$b(x)$$ are now functions. Plug this into the equation and determine the functions $$a(x)$$ and $$b(x)$$. The general solution to the given non-homogeneous equation is $$c\sin\, x + d\cos \, x+u_0$$ where $$c$$ and $$d$$ are constants. Find the values of these constants using the initial conditions.

This is called the method of variation of parameters and you can find more details in Wikipedia.

You should get, for instance by the method of unknown parameters for the two simpler equations $$u''+u=\pm t$$, $$u(t)=\begin{cases} A\cos(t)+B\sin(t)+t&\text{ for }t\ge 0\\ C\cos(t)+D\sin(t)-t&\text{ for }t<0 \end{cases}$$ and now have to determine the integration constants to match the initial conditions from both sides of $$t=0$$, $$A=1,~ B+1=-1\text{ and } C=1,~ D-1=-1$$ so that $$u(t)=\cos(t)-\sin(t)+[|t|-\sin(|t|)].$$

• Thanks! We can do this in general for a linear equation of order two, or just in case $u'$ doesn't appear? – astrobarrel Aug 22 '19 at 13:21
• It depends on "what" means. You can in general find a particular solution for $L[u]=|t|$ where $L$ is any linear differential operator by finding a solution $L[u_0]=t$ with initial conditions all zero, and then set $u_p(t)={\rm sign}(t)\,u_0(t)$. Here obviously $u_0(t)=t-\sin(t)=O(t^3)$ satisfies these conditions. – Lutz Lehmann Aug 22 '19 at 13:27

You could use Laplace transform to solve a Cauchy problem like this.

Start considering $$t\ge0$$, so $$|t|=t$$.

First apply the Laplace transform to the equation:

$$u''(t)+u(t)=t\Rightarrow \mathcal{L}\{u''(t)+u(t)\}(s)=\mathcal{L}\{t\}(s)$$

You will get $$s^2\mathcal{L}\{u(t)\}(s)-su(0)-u'(0)+\mathcal{L}\{u(t)\}(s)=\frac{1}{s^2}$$

Then solve the equation to $$\mathcal{L}\{u(t)\}(s)$$: $$\mathcal{L}\{u(t)\}(s)=\frac{1}{s^2}-\frac{2}{1+s^2}+\frac{s}{1+s^2}$$

Apply the Laplace transform inverse:

$$u(t)=\mathcal{L}^{-1} \left(\frac{1}{s^2}-\frac{2}{1+s^2}+\frac{s}{1+s^2}\right)$$

And you will get the solution:

$$u(t)=t-2\sin{t}+\cos{t}$$

• Thanks, nice solution! – astrobarrel Aug 22 '19 at 13:15