Solving ODE first order by Laplace transformation?

Question

given is the following first order ODE:

$\dot\epsilon(t) = \frac{1}{\eta}\cdot\sigma(t) + \frac{1}{E_1}\cdot\dot\sigma(t)$,

where $\eta$ and $E_1$ are constants. The initial conditions are: $\sigma(t = 0) = \sigma_0$; $\dot\sigma(t = 0) = 0$ and $\epsilon(t = 0) = \epsilon_0 = \frac{\sigma_0}{E_1}$ as well as $\dot\epsilon(t = 0) = 0$

The "function" $\sigma(t)$ is also given as $\sigma(t) = \sigma_0 = \text{const.}$

It's quite simple to solve this Equation by separation with the usage of the initial conditions. This gives:

$\epsilon(t) = \frac{\sigma_0}{E_1}+\frac{\sigma_0}{\eta}\cdot t$

Actually I want to solve the problem with Laplace-transformation. This leads me to (with $p_0 = \frac{1}{\eta}$ and $p_1 = \frac{1}{E_1}$):

$\mathcal{L}\{\epsilon\}(s)\cdot s-\epsilon_0 = p_0 \cdot \mathcal{L}\{\sigma\}(s) + p_1\cdot\bigl[\mathcal{L}\{\sigma\}(s)\cdot s - \sigma_0\bigr]$

... and now I don't know the way back; I tried it like the following:

$\int_0^t\epsilon(\xi)\delta(\xi)\text{d}\xi - \epsilon_0\delta(t) = p_0\cdot\int_0^t\sigma(\xi)\text{d}\xi + p_1\cdot\int_0^t\sigma(\xi)\delta(\xi)\text{d}\xi-p_1\cdot\sigma_0\delta(t)$

... but there's something wrong, 'cause this leads not to the same solution like given above.

Maybe someone could tell me my mistake. That would be nice!

Thanks in advance!

Answer 1

You have just one differential equation with two dependent variables $\epsilon(t)$ and $\sigma(t)$. That's not enough: you need another differential equation if you want to specify a solution uniquely.

EDIT: If you know $\sigma$ and want $\epsilon$, you get $$ {\mathcal L}\{\epsilon\}(s) = \left(\dfrac{1}{s\eta} + \dfrac{1}{E_1}\right) {\mathcal L}\{\sigma\}(s) + \dfrac{\epsilon_0 E_1 - \sigma_0}{E_1 s}$$ You said $\epsilon_0 = \sigma_0/E_1$ so the second term on the right is $0$.

You might note that the Laplace transform of $\displaystyle\int_0^t g(\xi) \; d\xi$ is $\dfrac{1}{s} {\mathcal L}\{g\}(s)$ so indeed $$ \epsilon(t) = \dfrac{1}{\eta} \int_0^t \sigma(\xi)\; d\xi + \dfrac{1}{E_1} \sigma(t)$$ which is what you would have obtained by integration without detouring through Laplace transforms.