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

  1. Jun 24, 2012 #1
    show the solution to this equation is unique

    i tried to find the answer but iam stuck please help
  2. jcsd
  3. Jun 24, 2012 #2


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    You must have theorems that will allow you conclude a differential equation has a unique solution. What are they?
  4. Jun 25, 2012 #3
    We're going to compare two (possibly different) solutions of your initial value problem. Let [itex] y_1 [/itex], [itex] y_2 [/itex] be two solutions to your IVP. Then consider their difference [itex] f:=y_1-y_2 [/itex]. Direct verification shows that [itex] f [/itex] is a solution to the ODE with initial conditions [itex] f(0)=0 [/itex], [itex] f'(0)=0 [/itex]. (check this)

    However, by that same token, since [itex] f [/itex] satisfies the ODE, we can rearrange things and find [itex]f''(x)=-\exp(x)f'(x)-f(x) [/itex]. Try evaluating [itex] f'' [/itex] at zero. What does this tell you about [itex]f [/itex]?
  5. Jun 25, 2012 #4
    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:
    in the RHS i have exp(x) which is not a constant what i should do?
  6. Jun 25, 2012 #5
    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.
  7. Jun 28, 2012 #6
    Alright so from what I see, we have [itex]\frac{d}{dx}(\frac{f'^{2}}{2}+\frac{f^{2}}{2})=-e^{x}f'^{2}[/itex]... unpleasant looking.. but i have an idea.
    multiply the entire thing by [itex]2e^{x}[/itex]
    Giving [itex]e^{x}\frac{d}{dx}(f'^{2}+f^{2})=-2e^{2x}f'^{2}[/itex]
    And for notation purposes... call [itex]u(x)=f'(x)^{2}+f(x)^{2} [/itex]
    then we have [itex] e^{x}u'(x)=-2e^{x}f'^{2}[/itex]
    using [itex]\frac{d}{dx}(e^{x}u(x))=e^{x}u(x)+e^{x}u'(x)[/itex]
    gives [itex]\frac{d}{dx}(e^{x}u(x))-e^{x}u(x)=-2e^{2x}f'^{2}[/itex]
    we know [itex]2e^{2x}f'^{2}≥0 [/itex] so it follows
    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
  8. Jun 28, 2012 #7
    Brilliant solution, tt. That substitution really cleans things up a lot.
  9. Jun 28, 2012 #8


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    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 [itex]v'= -y- e^xfv[/itex] and together with [itex]y'= v[/itex] we can write as the first order vector equation
    [tex]\frac{d\begin{pmatrix}y \\ v\end{pmatrix}}{dx}= \begin{pmatrix}0 & 1\\ -1 & -e^x\end{pmatrix}\begin{pmatrix}y \\ v\end{pmatrix}[/tex]
    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
  10. Jun 28, 2012 #9
    very true! But what is f in [itex]v'=-y-e^{x}fv[/itex] ?
  11. Jun 28, 2012 #10
    thanks alot tt2348 that's a very clever solution
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