# Solution of a First Order ODE with Discontinuous RHS

by H. R.   Last Updated May 11, 2018 21:20 PM

Assume that $\sigma:\mathbb{R}-\{0\}\to\mathbb{R}$ is function defined as \begin{align*} \sigma(t)= \begin{cases} \sigma_0 & t\gt0 \\ 0 & t\lt0 \end{cases} \tag{1} \end{align*} with $\sigma_0\in\mathbb{R}$. Suppose that we are looking for a function $\varepsilon$ such that \begin{align*} &q_1 \dot{\varepsilon}(t)=\sigma+p_1\dot{\sigma}(t),\quad \forall t \in \mathbb{R}-\{0\}, \\ &\varepsilon(0^-):=\lim_{t\to0^-}\varepsilon(t)=0, \tag{2} \end{align*} where the over dot denotes differentiation with respect to $t$ and $q_1\in\mathbb{R}-\{0\},\,p_1\in\mathbb{R}$.

Question 1. Does Eq.$(2)$ uniquely determines $\varepsilon(t)$ over $\mathbb{R}-\{0\}$?

I suppose that the answer to Question 1 is yes. I am not sure about this, please correct me if I am wrong. Bearing this in mind, let us find this function.

Direct Method

If $t<0$ then the ODE in $(2)$ implies that $\varepsilon(t)=A$. But the initial condition forces $A=0$ and we have \begin{align*} \varepsilon(t)=0,\quad t<0. \tag{3} \end{align*} If $t>0$ then the ODE reduces to $q_1\dot{\varepsilon}(t)=\sigma_0$ which has the solution \begin{align*} \varepsilon(t)=\frac{\sigma_0}{q_1}t+B,\quad t>0 \tag{4} \end{align*}

Question 2. How can I find the constant $B$?

In my text, the final answer is given as \begin{align*} \varepsilon(t)= \begin{cases} 0 & t<0 \\ \frac{\sigma_0}{q_1}t+\frac{p_1\sigma_0}{q_1} & t>0 \end{cases} \tag{5} \end{align*}

Laplace Transform Method

I checked the answer in $(5)$ via the Laplace transform method. If we define our Laplace transform as $$\mathcal{L}(f)(s):=\int_{0^-}^{+\infty}f(t)\exp(-st)dt\tag{6}$$ (note $0^-$ in the lower limit). Then we can easily obtain the above answer. Denoting that $\bar f(s):=\mathcal{L}(f)(s)$, using $\bar{\dot{f}}(s)=-f(0^-)+s\bar f(s)$ and $\bar\sigma(s)=\frac{\sigma_0}{s}$ we obtain \begin{align*} q_1\big(-\varepsilon(0^-)+s\bar\varepsilon(s)\big)&=\bar\sigma(s)+p_1\big(-\sigma(0^-)+s\bar\sigma(s)\big) \\ q_1s\bar\varepsilon(s)&=(1+p_1s)\bar\sigma(s)\\ \bar\varepsilon(s)&=\frac{(1+p_1s)}{q_1s^2}\sigma_0\\ \bar\varepsilon(s)&=\frac{1}{q_1}\Big(\frac{1}{s^2}+\frac{p_1}{s}\Big)\sigma_0\\ \varepsilon(t)&=\frac{1}{q_1}\big(t+p_1\big)\sigma_0\\ \varepsilon(t)&=\frac{\sigma_0}{q_1}t+\frac{p_1\sigma_0}{q_1} \tag{7} \end{align*}

which is the desired answer. But, I think that some part of this calculation is not allowed as I am sure that the direct method is completely OK!

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Integrate the entire equation like this

$$q_1\int_{-\delta}^\delta \dot\varepsilon(t)\ dt = \int_{-\delta}^\delta\sigma(t)\ dt + p_1 \int_{-\delta}^\delta \dot\sigma(t)\ dt$$

Then take the limit as $\delta \to 0$. You'll get

$$q_1\big[\varepsilon(0^+) - \varepsilon(0^-)\big] = p_1\sigma_0$$

Since $\varepsilon(0^-) = 0$, we have $\varepsilon(0^+) = \frac{p_1\sigma_0}{q_1}$ which is also the value of $B$

Dylan
May 11, 2018 18:55 PM