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&#039;^{2}}{2}+\frac{f^{2}}{2})=-e^{x}f&#039;^{2}... unpleasant looking.. but i have an idea.
multiply the entire thing by 2e^{x}
Giving e^{x}\frac{d}{dx}(f&#039;^{2}+f^{2})=-2e^{2x}f&#039;^{2}
And for notation purposes... call u(x)=f&#039;(x)^{2}+f(x)^{2}
then we have e^{x}u&#039;(x)=-2e^{x}f&#039;^{2}
using \frac{d}{dx}(e^{x}u(x))=e^{x}u(x)+e^{x}u&#039;(x)
gives \frac{d}{dx}(e^{x}u(x))-e^{x}u(x)=-2e^{2x}f&#039;^{2}
implies
\frac{d}{dx}(e^{x}u(x))=e^{x}u(x)-2e^{2x}f&#039;^{2}
we know 2e^{2x}f&#039;^{2}≥0 so it follows
\frac{d}{dx}(e^{x}u(x))=e^{x}u(x)-2e^{2x}f&#039;^{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&#039;= -y- e^xfv and together with y&#039;= v we can write as the first order vector equation
\frac{d\begin{pmatrix}y \\ v\end{pmatrix}}{dx}= \begin{pmatrix}0 &amp; 1\\ -1 &amp; -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&#039;= -y- e^xfv and together with y&#039;= v we can write as the first order vector equation
\frac{d\begin{pmatrix}y \\ v\end{pmatrix}}{dx}= \begin{pmatrix}0 &amp; 1\\ -1 &amp; -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&#039;=-y-e^{x}fv ?

 

[john_84]

thanks a lot tt2348 that's a very clever solution