Hermite differential equation












1














I'm currently reading Applied PDEs by Logan and he defines the Hermite differential equation as
$$-y''+x^2y=Ey$$
and we are to show that $v_n(x)=H_n(x)e^{-x^2/2}$ is a solution where $E=2n+1$ and $H_n(x) = (-1)^n e^{x^2} frac{d^n}{dx^n} e^{-x^2}$



I suppose I'm concerned that there isn't a solution because after working for awhile I checked around and the differential equation was always written as $y''-2xy'+2ny=0$. I'm also aware there is more than one way to define $H_n(x)$. In any case, I'd appreciate knowing that it can at least be done in principle and that this is not, in fact, a typo.... and a hint would be appreciated as well.






pde







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  • Have you tried direct substitution? The definition of Hermite polynomials you are using seems okay to me
    – Yuriy S
    Nov 24 at 15:56










  • It's not the definition of the polynomials I am concerned about.... but the definition of the DE. I was merely mentioning that the definition of the polynomials varies because this might explain why the DE I'm encountering is different.
    – Clclstdnt
    Nov 25 at 1:07
















1














I'm currently reading Applied PDEs by Logan and he defines the Hermite differential equation as
$$-y''+x^2y=Ey$$
and we are to show that $v_n(x)=H_n(x)e^{-x^2/2}$ is a solution where $E=2n+1$ and $H_n(x) = (-1)^n e^{x^2} frac{d^n}{dx^n} e^{-x^2}$



I suppose I'm concerned that there isn't a solution because after working for awhile I checked around and the differential equation was always written as $y''-2xy'+2ny=0$. I'm also aware there is more than one way to define $H_n(x)$. In any case, I'd appreciate knowing that it can at least be done in principle and that this is not, in fact, a typo.... and a hint would be appreciated as well.






pde







share|cite|improve this question






















  • Have you tried direct substitution? The definition of Hermite polynomials you are using seems okay to me
    – Yuriy S
    Nov 24 at 15:56










  • It's not the definition of the polynomials I am concerned about.... but the definition of the DE. I was merely mentioning that the definition of the polynomials varies because this might explain why the DE I'm encountering is different.
    – Clclstdnt
    Nov 25 at 1:07














1












1








1







I'm currently reading Applied PDEs by Logan and he defines the Hermite differential equation as
$$-y''+x^2y=Ey$$
and we are to show that $v_n(x)=H_n(x)e^{-x^2/2}$ is a solution where $E=2n+1$ and $H_n(x) = (-1)^n e^{x^2} frac{d^n}{dx^n} e^{-x^2}$



I suppose I'm concerned that there isn't a solution because after working for awhile I checked around and the differential equation was always written as $y''-2xy'+2ny=0$. I'm also aware there is more than one way to define $H_n(x)$. In any case, I'd appreciate knowing that it can at least be done in principle and that this is not, in fact, a typo.... and a hint would be appreciated as well.






pde







share|cite|improve this question













I'm currently reading Applied PDEs by Logan and he defines the Hermite differential equation as
$$-y''+x^2y=Ey$$
and we are to show that $v_n(x)=H_n(x)e^{-x^2/2}$ is a solution where $E=2n+1$ and $H_n(x) = (-1)^n e^{x^2} frac{d^n}{dx^n} e^{-x^2}$



I suppose I'm concerned that there isn't a solution because after working for awhile I checked around and the differential equation was always written as $y''-2xy'+2ny=0$. I'm also aware there is more than one way to define $H_n(x)$. In any case, I'd appreciate knowing that it can at least be done in principle and that this is not, in fact, a typo.... and a hint would be appreciated as well.






pde




pde






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share|cite|improve this question











share|cite|improve this question




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asked Nov 24 at 15:49









Clclstdnt

463313




463313












  • Have you tried direct substitution? The definition of Hermite polynomials you are using seems okay to me
    – Yuriy S
    Nov 24 at 15:56










  • It's not the definition of the polynomials I am concerned about.... but the definition of the DE. I was merely mentioning that the definition of the polynomials varies because this might explain why the DE I'm encountering is different.
    – Clclstdnt
    Nov 25 at 1:07


















  • Have you tried direct substitution? The definition of Hermite polynomials you are using seems okay to me
    – Yuriy S
    Nov 24 at 15:56










  • It's not the definition of the polynomials I am concerned about.... but the definition of the DE. I was merely mentioning that the definition of the polynomials varies because this might explain why the DE I'm encountering is different.
    – Clclstdnt
    Nov 25 at 1:07
















Have you tried direct substitution? The definition of Hermite polynomials you are using seems okay to me
– Yuriy S
Nov 24 at 15:56




Have you tried direct substitution? The definition of Hermite polynomials you are using seems okay to me
– Yuriy S
Nov 24 at 15:56












It's not the definition of the polynomials I am concerned about.... but the definition of the DE. I was merely mentioning that the definition of the polynomials varies because this might explain why the DE I'm encountering is different.
– Clclstdnt
Nov 25 at 1:07




It's not the definition of the polynomials I am concerned about.... but the definition of the DE. I was merely mentioning that the definition of the polynomials varies because this might explain why the DE I'm encountering is different.
– Clclstdnt
Nov 25 at 1:07










1 Answer
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Define $y$ as follows



begin{eqnarray}
y(x) &=& u(x)e^{-x/2} \
frac{{rm d}^2y}{{rm d}x^2} &=& e^{-x^2/2} left[left(x^2-1right) u(x)+frac{{rm d}^2u}{{rm d}x^2}-2 x frac{{rm d}u}{{rm d}x}right]
end{eqnarray}



Replacing we get



begin{eqnarray}
-frac{{rm d}^2y(x)}{{rm d}x^2} + x^2y &=& E y(x) \
Leftrightarrow e^{-x^2/2} left[-frac{{rm d^2}u}{{rm d}x^2}+2 x frac{{rm d}u}{{rm d}x}+u(x)right] &=& E e^{-x^2/2}u(x) \
Leftrightarrow frac{{rm d^2}u}{{rm d}x^2} -2x frac{{rm d}u}{{rm d}x} + (E-1)u(x) &=& 0
end{eqnarray}



From here you can conclude the solutions are of the form



$$
u(x) = H_n(x)
$$



where $n = (E - 1) / 2$, or equivalently $E = 2n + 1$






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    1 Answer
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    Define $y$ as follows



    begin{eqnarray}
    y(x) &=& u(x)e^{-x/2} \
    frac{{rm d}^2y}{{rm d}x^2} &=& e^{-x^2/2} left[left(x^2-1right) u(x)+frac{{rm d}^2u}{{rm d}x^2}-2 x frac{{rm d}u}{{rm d}x}right]
    end{eqnarray}



    Replacing we get



    begin{eqnarray}
    -frac{{rm d}^2y(x)}{{rm d}x^2} + x^2y &=& E y(x) \
    Leftrightarrow e^{-x^2/2} left[-frac{{rm d^2}u}{{rm d}x^2}+2 x frac{{rm d}u}{{rm d}x}+u(x)right] &=& E e^{-x^2/2}u(x) \
    Leftrightarrow frac{{rm d^2}u}{{rm d}x^2} -2x frac{{rm d}u}{{rm d}x} + (E-1)u(x) &=& 0
    end{eqnarray}



    From here you can conclude the solutions are of the form



    $$
    u(x) = H_n(x)
    $$



    where $n = (E - 1) / 2$, or equivalently $E = 2n + 1$






    share|cite|improve this answer


























      1














      Define $y$ as follows



      begin{eqnarray}
      y(x) &=& u(x)e^{-x/2} \
      frac{{rm d}^2y}{{rm d}x^2} &=& e^{-x^2/2} left[left(x^2-1right) u(x)+frac{{rm d}^2u}{{rm d}x^2}-2 x frac{{rm d}u}{{rm d}x}right]
      end{eqnarray}



      Replacing we get



      begin{eqnarray}
      -frac{{rm d}^2y(x)}{{rm d}x^2} + x^2y &=& E y(x) \
      Leftrightarrow e^{-x^2/2} left[-frac{{rm d^2}u}{{rm d}x^2}+2 x frac{{rm d}u}{{rm d}x}+u(x)right] &=& E e^{-x^2/2}u(x) \
      Leftrightarrow frac{{rm d^2}u}{{rm d}x^2} -2x frac{{rm d}u}{{rm d}x} + (E-1)u(x) &=& 0
      end{eqnarray}



      From here you can conclude the solutions are of the form



      $$
      u(x) = H_n(x)
      $$



      where $n = (E - 1) / 2$, or equivalently $E = 2n + 1$






      share|cite|improve this answer
























        1












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        1






        Define $y$ as follows



        begin{eqnarray}
        y(x) &=& u(x)e^{-x/2} \
        frac{{rm d}^2y}{{rm d}x^2} &=& e^{-x^2/2} left[left(x^2-1right) u(x)+frac{{rm d}^2u}{{rm d}x^2}-2 x frac{{rm d}u}{{rm d}x}right]
        end{eqnarray}



        Replacing we get



        begin{eqnarray}
        -frac{{rm d}^2y(x)}{{rm d}x^2} + x^2y &=& E y(x) \
        Leftrightarrow e^{-x^2/2} left[-frac{{rm d^2}u}{{rm d}x^2}+2 x frac{{rm d}u}{{rm d}x}+u(x)right] &=& E e^{-x^2/2}u(x) \
        Leftrightarrow frac{{rm d^2}u}{{rm d}x^2} -2x frac{{rm d}u}{{rm d}x} + (E-1)u(x) &=& 0
        end{eqnarray}



        From here you can conclude the solutions are of the form



        $$
        u(x) = H_n(x)
        $$



        where $n = (E - 1) / 2$, or equivalently $E = 2n + 1$






        share|cite|improve this answer












        Define $y$ as follows



        begin{eqnarray}
        y(x) &=& u(x)e^{-x/2} \
        frac{{rm d}^2y}{{rm d}x^2} &=& e^{-x^2/2} left[left(x^2-1right) u(x)+frac{{rm d}^2u}{{rm d}x^2}-2 x frac{{rm d}u}{{rm d}x}right]
        end{eqnarray}



        Replacing we get



        begin{eqnarray}
        -frac{{rm d}^2y(x)}{{rm d}x^2} + x^2y &=& E y(x) \
        Leftrightarrow e^{-x^2/2} left[-frac{{rm d^2}u}{{rm d}x^2}+2 x frac{{rm d}u}{{rm d}x}+u(x)right] &=& E e^{-x^2/2}u(x) \
        Leftrightarrow frac{{rm d^2}u}{{rm d}x^2} -2x frac{{rm d}u}{{rm d}x} + (E-1)u(x) &=& 0
        end{eqnarray}



        From here you can conclude the solutions are of the form



        $$
        u(x) = H_n(x)
        $$



        where $n = (E - 1) / 2$, or equivalently $E = 2n + 1$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 at 16:17









        caverac

        13.4k21029




        13.4k21029






























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