Proving that a solution to an IVP is unique and infinitely differentiable

[JPaquim]

Homework Statement

$\frac{d^2y}{dt^2} + t\frac{dy}{dt} + t^3y = e^t;\ \ \ y(0) = 0, \ \ y'(0) = 0$

Show that the solution is unique and has derivatives of all orders. Determine the fourth derivative of the solution at t = 0.2. The attempt at a solution

I'm somewhat lost here... Trying to Laplace Transform it produces a third degree ODE, which doesn't really seem any simpler...

I guess I can calculate the fourth derivative at 0 by first calculating the second, by directly substituting the initial conditions, differentiating the quation and finding the third derivative, and differentiate it again to find the fourth... Doing it like so gives me $y^{(4)} = 1$

However, I don't really know how to prove uniqueness nor C^∞ness... Any suggestions are welcome.

Cheers

 

[Melon Knight]

I'm pretty sure that your equation has a unique solution as it's a second-order linear ODE with initial conditions. It's a direct consequence of the Existence and Uniqueness Theorem.

As for showing that the solution is infinitely differentiable, I don't know if this is a valid argument but you already found the fourth derivative at t=0. Maybe you can find a pattern in successive derivatives and put down an argument.

I'm not sure if the existence of n derivatives at t=0 implies that the solution is n times differentiable, but maybe someone else can shed some light on this.

 

[JPaquim]

Thank you for your answer. My teacher's suggestion was to turn the equation into an equivalent system of linear differential equations, and then use Picard's theorem for systems to prove uniqueness. I guess I wasn't supposed to directly invoque the existence and uniqueness theorem for linear equations of nth order...

And the suggestion he gave for proving infinite differentiability was exactly to recursively differentiate the original equations a sufficient number of times to obtain a differential equation for the nth derivative, for any n. And then use the same argument for existence and uniqueness.

Cheers

 

[HallsofIvy]

It's fairly easy to show that a solution to such a d.e is infinitely differentiable.

If $$\frac{d^2y}{dt^2}+ t\frac{dy}{dt}+ t^3y= e^t$$ the y is obviously twice differentiable!

We can rewrite this as $$\frac{d^2y}{dt^2}= e^t- t\frac{dy}{dt}- t^3y$$. Since y is at least twice differentiable, the right side is differentiable and therefore, so is the left side: $$\frac{d^3y}{dt^3}= e^t- t\frac{d^2y}{dt^2}- \frac{dy}{dt}- t^3\frac{dy}{dx}- 3t^2y$$.

Now every function on the right side of that is differentiable and so
$$\frac{d^4y}{dt^4}= e^t- t\frac{d^3y}{dt^3}- \frac{d^2y}{dt^2}$$$$- \frac{d^2y}{dt^2}-$$$$t^3\frac{d^2y}{dt^2}- 3t^2\frac{dy}{dt}- 3t^3\frac{dy}{dt}- 6ty$$.

 

[JPaquim]

Exactly, that's what I was trying to say with recursively differentiating the original equation to obtain an equation for the nth derivative. But thank you, still ;)

Cheers