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

[john_84]

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


i tried to find the answer but iam stuck please help

 

[Dick]

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

 

[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 $f$?

 

[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?

 

[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.

 

[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

 

[christoff]

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

 

[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.

 

[tt2348]

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.

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

 

[john_84]

thanks a lot tt2348 that's a very clever solution