# Inconsistency when solving IVP using Laplace Transform with Dirac Delta

solving $$\dot{x}(t) + x(t) = \delta (t)$$ Using Laplace transform for $$x(0) = 1$$, we get:

$$sX(s)-1 + X(s) = 1$$

$$X(s) = \frac{2}{s+1}$$

so, $$x(t) = 2e^{-t}$$

However, evaluating at t=0, $$x(0) = 2 \neq 1$$ This disagrees with the initial condition. What went wrong here?

• The initial condition makes no sense. Because of the delta, the solution will have a jump discontinuity at $t=0$. But what side is the $1$ on? Also, what values has $x$ for $t<0$? What your calculus shows is a jump from $x(t)=0$ for $t<0$ to $1$ for the initial condition and on top of that a jump from $1$ to $2$ due to the Dirac delta. Commented Oct 20, 2021 at 4:52

Too long for a comment

There are two important points in this problem.

If you check the solution that you found, you will see that it does not satisfy the equation: $$\Big(\frac{d}{dt}+1\Big)2e^{-t}=0\neq \delta(t)$$

The general solution of an inhomogeneous equation (i.e. equation with non-zero RHS) is the general solution of a homogeneous equation, plus a particular solution of an inhomogeneous equation.

1. A particular solution of the inhomogeneous equation. We don't set any boundary condition (to meet the boundary condition we will use the arbitrary factor at a the general solution of a homogeneous equation). Applying the Laplace transformation $$sX(s)+X(s)=1 \,\Rightarrow \,X(s)=\frac{1}{1+s}\,\Rightarrow \, x_1(t)=h(t)e^{-t}$$ where $$h(t)$$ is a step-function ($$h(t)=0$$ for $$t<0$$, and $$h(t)=1$$ for $$x\geqslant0$$). Given that $$h'(t)=\delta(t)$$ and $$\,\delta(t)f(t)=\delta(t)f(0)$$, this is a particular solution of the initial equation.
2. The general solution of the homogeneous equation $$x'(t)+x(t)=0\,$$ is $$\,x_2(t)=Ce^{-t}$$, where $$C$$ is an arbitrary constant. So, the general solution of the initial equation is $$x(t)=h(t)e^{-t}+Ce^{-t}$$ Applying the boundary condition $$x(0)=1$$ we see that $$C=0$$. Therefore, the answer, valid for $$t\geqslant 0\,$$ $$x(t)=h(t)\,e^{-t}$$

If we put the boundary condition, for example, $$x(0)=2$$, we get the answer $$x(t)=(h(t)+1)\,e^{-t}$$

The reason that you get $$x(0)=2$$ is that you both set $$x(0^-)=1$$ and apply a Dirac $$\delta$$ that adds a unit step making $$x(0^+)-x(0^-)=1$$. Thus, $$x(0^+)=x(0^-)+\left(x(0^+)-x(0^-)\right) = 1 + 1 = 2.$$

This will be clearer if you displace the $$\delta$$ somewhat and take the differential equation as $$\dot{x}(t) + x(t) = \delta(t-\epsilon)$$ with $$x(0^-)=x_0.$$

When $$\epsilon>0$$ this gives $$\left(sX(s)-x_0\right)+X(s)=e^{-\epsilon s}$$ i.e. $$X(s) = \frac{x_0+e^{-\epsilon s}}{s+1} = \frac{x_0}{s+1} + \frac{e^{-\epsilon s}}{s+1}$$ so $$x(t) = x_0 e^{-t} H(t) + e^{-(t-\epsilon)} H(t-\epsilon) .$$ This function first has a step of size $$x_0$$ at $$t=0$$ and later another step of size $$1$$ (from the $$\delta$$ term) at $$t=\epsilon$$. If you take $$x_0=1$$ and let $$\epsilon\to 0$$ then you will get a total step of size $$2$$ at $$t=0$$ which is what you have seen.