I Solving ODE with Laplace transform and Existence and Uniqueness Theorem

[lys04]

This is my answer to the ODE (I think it's correct) via Laplace transform, but I'm more concerned about whether or not my explanation for part b is correct or not?
Any help would be greatly appreciated!!

 

[anuttarasammyak]

I observe your solution (a) is continuous at t=$\pi$.

 

[lys04]

I observe your solution (a) is continuous at t=π.

Oh yeah you're right, what might be a good explanation for (b) then?

 

[lys04]

I graphed it and it seems like it’s twice differentiable as well

 

[pasmith]

I would expect y and y' to be continuous, and y'' to have a jump discontinuity.

 

[lys04]

I would expect y and y' to be continuous, and y'' to have a jump discontinuity.

Why?

 

[anuttarasammyak]

Because
If y has a jump, derivatives y' and y" would have "hyper" jump which does not coincide with RHS of the ODE.
If y' has a jump, derivative y" would have "hyper" jump which does not coincide with RHS of the ODE.
y" should have a jump as RHS of the ODE has. "hyper" jump here is actually delta function and its derivative.

 

[lys04]

If y has a jump, derivatives y' and y" would have "hyper" jump which does not coincide with RHS of the ODE.
If y' has a jump, derivative y" would have "hyper" jump which does not coincide with RHS of the ODE.
y" should have a jump as RHS of the ODE has. "hyper" jump here is actually delta function and its derivative.

By a jump do you mean a discontinuity? A heaviside function is discontinuous though?

 

[anuttarasammyak]

By a jump do you mean a discontinuity? A heaviside function is discontinuous though?

Yes. Is there anything you feel uneasy ?

 

[lys04]

Yes. Is there anything you feel uneasy ?

Yeah. By y do you mean my solution found through Laplace transform? But then you speak of the RHS, is that referring to the Heaviside function?
So if you’re talking about the Heaviside function then I’m not sure what you mean by y’ and y’’ Would have hyper jumps which do not coincide with the RHS of the ODE because y (Heaviside does have a jump)?
Also what does a hyper jump mean?

 

[anuttarasammyak]

OK. First could you calculate second derivative of your solution and show it to us?

 

[lys04]

OK. First could you calculate second derivative of your solution and show it to us?

Yeah sure I could do that but the question says without doing any explicit calculations?

 

[lys04]

Okay, the for $$t<=pi$$,

the first derivative is $$e^{t}sin(3t)+3e^{t}cos(3t)$$ and the second derivative is $$-8e^{t}sin(3t)+6e^{t}cos(3t)$$

and for $t>pi$,
the first derivative is $$e^{t}sin(3t)+3e^{t}cos(3t)+10e^{t-\pi}sin(3(t-\pi))$$ and the second derivative is $$-8e^{t}sin(3t)+6e^{t}cos(3t)+10e^{t-\pi}sin(3(t-\pi))+30e^{t-\pi}cos(3(t-\pi))$$

 

[anuttarasammyak]

Thanks for your effort. Say your sokution is y
y^&quot;=-8e^{t}sin(3t)+6e^{t}cos(3t)=8e^\pi at t=$\pi-0$
y^&quot;=-8e^{t}sin(3t)+6e^{t}cos(3t)+10e^{t-\pi}sin(3(t-\pi))+30e^{t-\pi}cos(3(t-\pi))=8e^\pi+30 at t=$\pi+0$. We see discotinuity of gap 30 at at t=$\pi$ in y".

You have already found that y is continuas at t=$\pi$. You can easily check that y' is continuous too there.

This is an example of post #7.

 

[lys04]

Thanks for your effort. Say your sokution is y
y^&quot;=-8e^{t}sin(3t)+6e^{t}cos(3t)=8e^\pi at t=$\pi-0$
y^&quot;=-8e^{t}sin(3t)+6e^{t}cos(3t)+10e^{t-\pi}sin(3(t-\pi))+30e^{t-\pi}cos(3(t-\pi))=8e^\pi+30 at t=$\pi+0$. We see discotinuity of gap 30 at at t=$\pi$ in y".

You have already found that y is continuas at t=$\pi$. You can easily check that y' is continuous too there.

This is an example of post #7.

Hm yeah, but how can I tell all of this without doing explicit calculations?

 

[pasmith]

Why?

Because the highest derivative is the most badly behaved. This accords with our physical intuition, and guarantees that the lower order derivatives will actually exist.

The way of solving this equation without transforms is as a particular integral (which is zero for t \leq \pi) plus a linear combination of complementary functions. The coefficients of the complementary functions in [0,\pi] are fixed by the conditions at zero; the cofficients in (\pi,\infty) are fixed by requiring y and y&#039; to be continuous at \pi.

This is also the solution obtained by the Laplace transform method, if you use the result <br /> \mathcal{L}(y&#039;) = p\mathcal{L}(y) - y(0). If y were discontinuous at \pi, you would instead have <br /> \mathcal{L}(y&#039;) = p\mathcal{L}(y) + (y(\pi^{-}) - y(\pi^{+}))e^{-p\pi} - y(0) with a similar result for \mathcal{L}(y&#039;&#039;).

 

[anuttarasammyak]

ODE is
y^&quot;-2y&#039;+10y=30H(t-\pi)
H is in either y,y' or y". If H is in y or y',
H&#039;(x)=\delta (x)
would appear from y' or y" which is calculated from such y or y'. But we do not find it in ODE.
This is just another rephrasing of post #7.

 

[anuttarasammyak]

This is also the solution obtained by the Laplace transform method, if you use the result L(y′)=pL(y)−y(0). If y were discontinuous at π, you would instead have L(y′)=pL(y)+(y(π−)−y(π+))e−pπ−y(0) with a similar result for L(y″).

I have been a lazy learner of Laplace transform so let me know it more.
Laplace transform of ODE in post #17 with the initial condition is
(p^2-2p+10) \mathcal{L}(y)=3+30\frac{e^{-\pi p}}{p}
$\mathcal{L}(y)$ is written as sum of
\frac{1}{p-1-3i}, \frac{1}{p-1+3i}, \frac{1}{p}
terms. Is it a right track to solve this ODE ?

 

[pasmith]

Once you have <br /> \mathcal{L}(y) = \frac{3}{(p- 1)^2 + 3^2}\left(1 + \frac{10e^{-\pi p}}{p}\right) you can look these up in transform tables.

 

[Kirti_Vardhan_1]

Hello. It is a second order linear ordinary differential equation with constant coefficients and initial values.

y(t) = y_{h}(t) + y_{p}(t)
In general one may ask themselves what happened to the transient (general, homogenous) and steady state response (particular, inhomogenous)?
\textbf{Transient Response: Solution Families/Categories Of Finite, Known Cases Of Solutions}
The transient response is the homogenous solution. The homogenous solution occurs for:
- Distinct real roots,
- Repeated real roots,
- Complex roots
\textbf{Steady State response: inhomogenous solution/forcing function }
The transient response is the homogenous solution. The homogenous solution occurs for:
The steady state response is the inhomogeneous solution.

The inhomogeneous solution requires a forcing function. Here, the forcing function is a unit step/Heaviside delayed by pi units.

h(t - \pi)
In general the unit step function scaled and shifted is defined as:
<br /> K h(t - T_{0}) =<br /> \begin{cases}<br /> 0 &amp; \text{for } t &lt; T_{0} \\<br /> K &amp; \text{for } t \geq T_{0}<br /> \end{cases}<br /> <br />
Whose derivative is an equally scaled and shifted delta pulse:
<br /> \frac{d}{dt} (K h(t - T_{0})) = K \delta(t - T_{0})<br /> <br />
The derivative of the dirac delta (an term included within and of the second derivative of y(t)) is a distribution defined over a test function phi:
<br /> \delta&#039;(t) = \int_{-\infty}^{\infty} \delta&#039;(t) \phi(t) \, dt = -\phi&#039;(0)<br /> <br />


For O.D.E, the transient response is the homogenous solution to the equation.
The steady state response is directly related to the forcing function.

it is fair to say that the transient response of a second-order linear differential equation is both differentiable everywhere, continuous in all cases of roots:
- real and distinct,
-real and repeated,
- and complex

Hence y(t) would be discontinuous (jump discontinuity)...and the nature of the y'(t) and y"(t) follows from the jump discontinuity of the step function, a term in y(t)...
....

Discontinuities in y(t) occur at the point where the unit step/heaviside function is applied

So for the solution for this family/category of these equations as the behavior of the homogenous parts are well defined across all finite and known cases, the discontinuity behavior arises from the forcing function, inhomogeneous solution.