# Finding $L^2$ norm of solution of ODE

I have a linear differential equation with real constant coefficients $$\sum\limits_{i=0}^3 a_i y^{(i)}(x)=0$$ with initial conditions $y^{(i)}(0)=y_i\in\mathbb{R}$ where $i=0,1,2$. I need to find $L^2(\mathbb{R}_+)$ norm of $y(x)$ assuming that $\lim\limits_{x\to+\infty}y(x)=0$.

How can I solve it? The chracteristic equation is of the third order with arbitrary coefficients!

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Any suggestions? –  user34574 Jun 27 '12 at 17:15
Now, is it clear? –  user34574 Jun 27 '12 at 22:11

For all $k,l\in\{0,1,2,3\}$ we denote $$I_{k,l}=\int\limits_{\mathbb{R}_+}y^{(k)}(x)y^{(l)}(x)d\mu(x)$$ Since $y(+\infty)=0$, and we are solving linear differetial equation with constant coefficients then $y^{(k)}(+\infty)=0$ for $k\in\{0,1,2,3\}$.

Note the following simple idenities $$I_{k,l}=-y_{k-1}y_{l}-I_{k-1,l+1}\quad\text{ for }\quad k\geq 1\\$$ $$I_{k+1,k}=-0.5 y_k^2\quad\text{ for }\quad k\geq 0\\$$ $$I_{k,l}=I_{l,k}\quad\text{ for }\quad k,l\geq 0\\$$ They allow us to write down $I$ matrix $$I= \begin{Vmatrix} I_{0,0} & -0.5 y_0^2 & -y_1y_0-I_{1,1} & -y_2y_0+0.5 y_1^2\\ -0.5 y_0^2 & I_{1,1} & -0.5 y_1^2 & -y_2 y_1-I_{2,2} \\ -y_1y_0-I_{1,1} & -0.5 y_1^2 & I_{2,2} & -0.5 y_2^2 \\ -y_2y_0+0.5 y_1^2 & -y_2 y_1-I_{2,2} & -0.5 y_2^2 & I_{3,3} \end{Vmatrix}$$ Now multiply our differential equation by $y^{(k)}(x)$ with $k\in\{0,1,2\}$, and integrate over $\mathbb{R}_+$. Then we get $$\sum\limits_{l=0}^3 a_l I_{k,l}=0\quad\text{where}\quad k\in\{0,1,2\}$$ This is a system of linear equations with unknowns $I_{0,0}$, $I_{1,1}$, $I_{2,2}$. Its solution is very difficult to get by hand but possible $$I_{0,0}= -\frac{-a_1 y_0^2-2 a_2 y_1 y_0-2 a_3 y_2 y_0+a_3 y_1^2}{2 a_0}-\frac{a_2 \left(-a_0 a_2 y_0^2-2 a_0 a_3 y_1 y_0-a_2^2 y_1^2-a_1 a_3 y_1^2-a_3^2 y_2^2-2 a_2 a_3 y_1 y_2\right)}{2 a_0 \left(a_1 a_2-a_0 a_3\right)},$$ $$I_{1,1}= -\frac{-a_0 a_2 y_0^2-2 a_0 a_3 y_1 y_0-a_2^2 y_1^2-a_1 a_3 y_1^2-a_3^2 y_2^2-2 a_2 a_3 y_1 y_2}{2 \left(a_1 a_2-a_0 a_3\right)},$$ $$I_{2,2}= -\frac{-a_0^2 y_0^2-2 a_0 a_1 y_1 y_0-a_1^2 y_1^2-a_0 a_2 y_1^2-a_1 a_3 y_2^2-2 a_0 a_3 y_1 y_2}{2 \left(a_1 a_2-a_0 a_3\right)}$$ I must confess, but I used Mathematica to solve this system. Here is the code

i[k_, l_] := -Subscript[y, k - 1] Subscript[y, l] - i[k - 1, l + 1] /;
k > l + 1 && l >= 0
i[k_, l_] := -1/2 Subscript[y, l]^2 /; k == l + 1 && l >= 0
i[k_, l_] := Subscript[i, k, l] /; k == l
i[k_, l_] := i[l, k] /; k < l

n := 3

Solve[
Table[Sum[Subscript[a, l] i[k, l], {l, 0, n}] == 0, {k, 0, n - 1}],
Table[Subscript[i, k, k], {k, 0, n - 1}]]


The answer to your question is $I_{0,0}^{1/2}$.

P.S. This method can be generalized to linear differential equations of arbitrary order.

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Presumably the $a_j$ are real. Let $p(z) = \sum_{j=0}^3 a_j z^j$. Let $r_j$ be the roots of this polynomial (counted by multiplicity).

If all $r_j$ have real part $\ge 0$, the only solution $y$ with $y \to 0$ at $+\infty$ is $0$.

If two $r_j$ have real part $\ge 0$ and one (say $r_3$) is negative, the only solution with $y(0) = y_0$ and $y \to 0$ at $\infty$ is $y = y_0 e^{r_3 t}$, and the $L^2({\mathbb R}_+)$ norm of this is $\|y_0|/\sqrt{-2 r_3}$.

If a complex conjugate pair of $r_j$ have real part $< 0$ (say $\alpha \pm \beta i$ with $\beta > 0$ and $\alpha < 0$)and the other root is nonnegative, the real solutions with $y \to 0$ at $\infty$ are $e^{\alpha t} \left(y_0 \cos(\beta t) + \dfrac{y_1 - \alpha y_0}{\beta} \sin(\beta t)\right)$, and their $L^2({\mathbb R}_+)$ norms are $$\sqrt{\frac {5\,{{\it y_0}}^{2}{\alpha}^{2}+{{\it y_0}}^{2}{\beta}^{2}-4\,\alpha\,{\it y_0}\,{\it y_1}+{{\it y_1}}^{2}}{ -4\left( {\alpha}^{2}+{ \beta}^{2} \right) \alpha}}$$

Other cases may be more complicated.

EDIT: if all $r_j$ have real part $< 0$, so all solutions go to $0$ as $t \to \infty$, then (with help from Maple) I get $$\|y\|^2 = (-2a_{{1}}{a_{{2}}}^{2}y_{{0}}y_{{1}}-2a_{{3}}y_{{1}}y_{{2}}{a_{{2} }}^{2}+ \left( -a_{{0}}{a_{{2}}}^{2}-{a_{{1}}}^{2}a_{{2}}+a_{{0}}a_{{1 }}a_{{3}} \right) {y_{{0}}}^{2}- \left( {a_{{2}}}^{3}+{a_{{3}}}^{2}a_{ {0}} \right) {y_{{1}}}^{2}-{a_{{3}}}^{2}{y_{{2}}}^{2}a_{{2}}+2\,a_{{3} } \left( -a_{{2}}a_{{1}}+a_{{0}}a_{{3}} \right) y_{{0}}y_{{2}})/ (2 a_{{0}} \left( -a_{{2}}a_{{1}}+a_{{0}}a_{{3}} \right) )$$

EDIT: This was assuming real $y_i$.

The general solution of the differential equation is $y(t) = \sum_{r} c_r e^{rt}$, the sum being over the roots of $p(z)$. The initial conditions are $y(0) = \sum_{r} c_r = y_0$, $y'(0) = \sum_{r} r c_r = y_1$, $y''(0) = \sum_r r^2 c_r = y_2$. Write these three equations as $Y = V C$ where $$V = \pmatrix{1 & 1 & 1\cr r_1 & r_2 & r_3\cr r_1^2 & r_2^2 & r_3^2\cr}$$ If $y_i$ are real, so is $y(t)$, and the square of its $L^2$ norm is $$\int_0^\infty y(t)^2\ dt = \sum_r \sum_s c_r c_s \int_0^\infty e^{(r+s)t}\ dt = \sum_r \sum_s \dfrac{-c_r c_s}{r+s} = - C^T M C$$ where $$M = \pmatrix{\dfrac{1}{2r_1} & \dfrac{1}{r_1+r_2} & \dfrac{1}{r_1+r_3}\cr \dfrac{1}{r_2+r_1} & \dfrac{1}{2r_2} & \dfrac{1}{r_2+r_3}\cr \dfrac{1}{r_3+r_1} & \dfrac{1}{r_3+r_2} & \dfrac{1}{2r_3}\cr}$$ Now $-C^T M C = -Y^T (V^{-1})^T M V^{-1} Y$, so what we need to do is express each entry of $(V^{-1})^T M V^{-1}$ in terms of the coefficients $a_j$. These entries are symmetric rational functions in $r_1$, $r_2$, $r_3$, so that should be possible: every symmetric polynomial in $k$ variables can be expressed as a polynomial in the elementary symmetric polynomials. But doing it by hand looks rather daunting.

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I was given a hint: $\Vert y\Vert_{L^(\mathbb{R}_+)}^2$ is a rational function of $a_i$ and $y_i$. –  user34574 Jun 28 '12 at 12:04
Since $\lim\limits_{x\to+\infty} y(x)=0$ can we conclude that $\text{Re}(r_j)<0$ for all $j$? How to get the last formula without Maple? –  user34574 Jun 29 '12 at 12:05
If all solutions have $\lim_{x \to \infty} y(x)=0$ we can conclude that all $\text{Re}(r_j) < 0$. –  Robert Israel Jun 29 '12 at 17:54
your expression for $y$ in last edit is incorrect if roots have multiplicity –  user34574 Jun 29 '12 at 20:40
The final answer should be OK if there are multiple roots, as it can be considered as a limit of the non-degenerate case. –  Robert Israel Jun 29 '12 at 21:22
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