# Laplace transform

1. Jun 5, 2009

### KFC

I am reading a text about Laplace transform in solving differential equations. Seems that this method is so powerful. To practice how it works, I makeup a very simple problem

$$\frac{dy}{dt} = e^{wt}y$$

This equation could be solved by variable seperation and then intergrate LHS and RHS. But I would like to check that Laplace transform works. Let's

$$Y = \mathcal{L}[y]$$

is the Laplace transform of y. Note that

$$\mathcal{L}[dy/dt] = sY(s) - y(0)$$

and

$$\mathcal{L}[e^{wt}y] =Y(s-w)$$

where s is the parameter for Laplace transform. Hence, in Laplace domain, above equation becomes

$$sY(s) = Y(s-w)$$

(assuming y(0)=0) My question is : there is a shift in the variable s, so how to solve this equation to get Y?

2. Jun 5, 2009

### Mute

You don't. The problem with your simple example is that you chose a problem with a coefficient that is not constant, and the Laplace transform typically isn't applied to such problems.

Generally the Laplace transform is applied to linear ODEs with constant coefficients and a non-zero, time dependent right hand side. The Laplace transform is especially helpful when the RHS is discontinuous. For such problems you'll be able to solve for Y(s) and then do the inverse transform.

For your problem, you've gotten yourself a functional equation for Y(s), that I'm not sure how to solve. That doesn't mean it's not possible to solve it, but I'm not familiar with solving such problems.

Your relation almost looks like it's a Gamma function: recall the Gamma function has the functional relation

$$\Gamma(z+1) = z\Gamma(z)$$,

so for w = -1, $Y(s) = \Gamma(s)$ (this is apparently the unique solution, but I don't know how you'd show that). Of course, then you'd have to find the inverse laplace transform of a Gamma function, which usually isn't in the common tables!

So, generally, you wouldn't use the Laplace transform to solve that particular problem.

Last edited: Jun 5, 2009
3. Jun 10, 2009

### matematikawan

$$\frac{dy}{dt} = e^{wt}y$$

This DE is linear. It is perfectly alright to use Laplace transform method as long as you can solve the transformed equation.

The solution for $$sY(s) = Y(s-w)$$ is Y(s)=0

But this seem to contradict the solution given by Mute if w=-1.
So in order that to be able to solve $$\frac{dy}{dt} = e^{wt}y$$
using Laplace transform, I suspect w must be negative and less than -1.

But the original equation can be solve for all value of w

4. Jun 10, 2009

### KFC

Thanks for reply. Follow your reply, I still have some questions. First of all, we know that y(t=0) is the initial condition, but in what reason to make Y(s)=0?

5. Jun 11, 2009

### Mute

The reason, which I didn't notice before, is that if you solve your original equation, you find

$$\ln |y(t)| = \frac{e^{wt}}{w} + C \Rightarrow y(t) = A\exp\left[\frac{e^{wt}}{w}\right]$$

The initial condition y(0) = 0 is satisfied only for A = 0, which means y(t) = 0 for all t, and the laplace transform of zero is zero.

So, for w = -1 the gamma function solution for Y(s) seems to be erroneous, since Y(s) should be zero...

Of course, the trivial solution, Y(s) = 0 for all s also satisfies that relation, so I guess it's probably just the case that although the functional equation has non-trivial solutions, it's the trivial solution that needs to be picked to solve the ODE in this case. (I suppose it could also be the case that the inverse laplace transform of the non-trivial solution for Y(s) is zero, but that seems unlikely; they can't be analytic everywhere, as the Gamma function certainly isn't and we don't get a different solution for y(t) if w = -1).

Edit: Hm, maybe this is part of what's going on: in the other Laplace transform thread, Bobbybear mentioned that in order for something to be Laplace transformable, y(t) and its derivative cannot grow faster than the Laplace transform kernal, $\exp[-st]$, which as you can see, the nonzero solutions of this ODE don't! An exponential to an exponential grows way too fast for the Laplace transform to yield a finite result, unless w < 0, in which case the solution itself decays to zero as t -> infinity. (there could be other conditions on w, like matematikawan predicts, but that doesn't come out of this basic analysis)

So, part of the problem here is that the Laplace transform of y(t) only exists for w < 0. This still doesn't tell us how to pick which solution to the functional equation to use - though I suspect there is only the trivial solution and then a unique solution, so perhaps it's just a matter of checking if the inverse transform of the nontrivial solution satisfies the given initial conditions.

Last edited: Jun 11, 2009