# Prove y'' + exp(x)y'+y=0 has unique solution

1. Jun 24, 2012

### john_84

y''+exy'+y=0
y(0)=1
y('(0)=4
show the solution to this equation is unique

2. Jun 24, 2012

### Dick

You must have theorems that will allow you conclude a differential equation has a unique solution. What are they?

3. Jun 25, 2012

### christoff

We're going to compare two (possibly different) solutions of your initial value problem. Let $y_1$, $y_2$ be two solutions to your IVP. Then consider their difference $f:=y_1-y_2$. Direct verification shows that $f$ is a solution to the ODE with initial conditions $f(0)=0$, $f'(0)=0$. (check this)

However, by that same token, since $f$ satisfies the ODE, we can rearrange things and find $f''(x)=-\exp(x)f'(x)-f(x)$. Try evaluating $f''$ at zero. What does this tell you about $f$?

4. Jun 25, 2012

### john_84

christoff this is the way that the professor start with but he wants to use Gronwall’s Inequality to prove f is identically zero i multiplay the equation by f' but i end up with this
f'' f' + exp(x) (f') ^2 + f f' =0
d/dx[ (f')^2/2 + (f)^2/2]=-exp(x) (f')^2
at this point i couldn't use Gronwall’s Inequality because it states:
d/dx[f]<=kf
f(x)<=f(0)exp(kx)
in the RHS i have exp(x) which is not a constant what i should do?

5. Jun 25, 2012

### christoff

I'm not sure how you are supposed to use Gronwall's Identity to prove this. Your ODE is second order, and Gronwall's Identity only proves exponential boundedness of solutions of first-order equations. The problem isn't your exponential, your problem is that you have (f')^2/2+(f)^2/2 in the LHS, but you have (f')^2 in the RHS (these need to be equal, somehow).

You can reduce it to a first-order equation in two variables, but then the standard Gronwall's identity doesn't apply.

6. Jun 28, 2012

### tt2348

Alright so from what I see, we have $\frac{d}{dx}(\frac{f'^{2}}{2}+\frac{f^{2}}{2})=-e^{x}f'^{2}$... unpleasant looking.. but i have an idea.
multiply the entire thing by $2e^{x}$
Giving $e^{x}\frac{d}{dx}(f'^{2}+f^{2})=-2e^{2x}f'^{2}$
And for notation purposes... call $u(x)=f'(x)^{2}+f(x)^{2}$
then we have $e^{x}u'(x)=-2e^{x}f'^{2}$
using $\frac{d}{dx}(e^{x}u(x))=e^{x}u(x)+e^{x}u'(x)$
gives $\frac{d}{dx}(e^{x}u(x))-e^{x}u(x)=-2e^{2x}f'^{2}$
implies
$\frac{d}{dx}(e^{x}u(x))=e^{x}u(x)-2e^{2x}f'^{2}$
we know $2e^{2x}f'^{2}≥0$ so it follows
$\frac{d}{dx}(e^{x}u(x))=e^{x}u(x)-2e^{2x}f'^{2}≤e^{x}u(x)$
Which I believe should allow you to use the inequality proposed?
since f(0)=0, and f'(0)=0 , u(0)=0 I think proceeding from there shouldn't be a problem

7. Jun 28, 2012

### christoff

Brilliant solution, tt. That substitution really cleans things up a lot.

8. Jun 28, 2012

### HallsofIvy

Yes, very nice solution but it is not really necessary to solve the differential equation. Instead, letting v= y', the second equation can be written $v'= -y- e^xfv$ and together with $y'= v$ we can write as the first order vector equation
$$\frac{d\begin{pmatrix}y \\ v\end{pmatrix}}{dx}= \begin{pmatrix}0 & 1\\ -1 & -e^x\end{pmatrix}\begin{pmatrix}y \\ v\end{pmatrix}$$
That's a linear equation and the determinant of the coeffient matrix is 1, not 0, for all x and y and in particular in a neighborhood of (x, y)= (0, 4). Therefore, by the standard "existence and uniqueness" theorem for differential equations, there exist a unique solution.

Last edited by a moderator: Jun 28, 2012
9. Jun 28, 2012

### tt2348

very true! But what is f in $v'=-y-e^{x}fv$ ?

10. Jun 28, 2012

### john_84

thanks alot tt2348 that's a very clever solution