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I was trying to solve a second order PDE and inside of the Lagrange-Charpit system, a first order coupled system popped out.

You can find the problem here: http://www.math.ucla.edu/~yanovsky/handbooks/PDEs.pdf on page 86, 87 if you want to get an idea of the whole problem. Here is the part I am interested in: $$ \left\{ \begin{array}{c} x'=p_1 \\ p_1'=-x \end{array} \right. $$

This is the coupled system, and they say it resolves to these two second-order equations: $$x'' + x =0 ,x ( s, 0) = s, x ' ( s, 0) = p ( s, 0) = 1 , \\p '' + p =0 ,p ( s, 0) = 1 ,p ' ( s, 0) = − x ( s, 0) = − s $$ And also they state the solutions to be: $$x ( s, t )= s · cos t +sin t,\\ p ( s, t )=cos t − s · sin t.$$

The method I know for solving such system in based on eigenvalues & eigenvectors to build a decoupling matrix. In the book, they use another method, as it is stated - they break the system into 2 second order equations and then solve them.

Can someone explain in more detail, how the method is applied?

EDIT:

I Tried to solve the equation with the eigenvalue & eigenvector approach, but I got to the result:

$$ \left\{ \begin{array}{c} x(s,t)=C_1cos(t) + iC_2sin(t) \\ p_1(s,t)=iC_2cos(t) - C_1sin(t) \end{array} \right. $$

Which can give me the result, which is stated there exactly if the integration constants had the values: $$ \left\{ \begin{array}{c} C_1= s \\ C_2= -i \end{array} \right. $$ Is there anything that suggests these values?

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    $\begingroup$ Why don't you edit your post to include the problem so people can see it without going to another website. $\endgroup$ Commented Jun 27, 2017 at 10:24
  • $\begingroup$ @Mattos Please accept my apologies, I edited the question. $\endgroup$
    – victor175
    Commented Jun 27, 2017 at 11:10
  • $\begingroup$ All good mate, no need to apologise. The author got those ODEs by differentiating $x$ and $p_{1}$ respectively i.e $$x' = p_{1} \implies x'' = p_{1}' = -x \implies x'' + x = 0$$ Similarly for the ODE in $p_{1}$. Now, you already got the same general solution as the authors did. Just remember that $i$ is a constant of sorts so $iC_{2}$ is just another constant, call it $C_{3}$. So you just need to apply your conditions. You have $x(s,t) = C_{1} \cos(t) + C_{3} \sin(t) \implies x(s,0) = C_{1} = s$ and $x'(s,0) = C_{3} = 1$. $\endgroup$ Commented Jun 27, 2017 at 11:41
  • $\begingroup$ @Mattos Very clever.. Please arrange this into an answer, so I can accept it. Thank you! $\endgroup$
    – victor175
    Commented Jun 27, 2017 at 12:36

2 Answers 2

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You have a coupled system of first order ODEs

\begin{align} x' &= p \quad (1) \\ p' &= -x \quad (2) \end{align}

This can be turned into two second order ODEs by differentiating both $x'$ and $p'$ again i.e

\begin{align} x'' &= p' \quad \text{from (1)} \\ &= -x \quad \text{from (2)} \\\\ p'' &= -x' \quad \text{from (2)} \\ &= -p \quad \text{from (1)} \end{align}

The boundary conditions can also be derived using your ODE system and the initial conditions for the PDE problem

\begin{align} x(s,0) &= s, \quad x'(s,0) = 1 \\ p(s,0) &= 1, \quad p'(s,0) = -s \end{align}

They are the same ODE, so we will just solve the one in $x$ here. There are two ways to solve this apart from using the matrix formulation. First, the ODE is linear so making an ansatz $x = e^{\lambda t}$ gives us $\lambda^{2} = -1 \implies \lambda = \pm i$, hence

\begin{align} x &= Ae^{it} + Be^{-it} \\ &= A(\cos(t) + i\sin(t)) + B(\cos(t) - i \sin(t)) \\ &= C_{1} \cos(t) + C_{2} \sin(t) \end{align}

where $C_{1} = A + B, C_{2} = i(A - B)$. The second way is to multiply the ODE by $x'$ and integrate i.e

\begin{align} x' x'' &= -x x' \\ \implies \frac{1}{2} x'^{2} &= - \frac{1}{2} x^{2} + k_{1} \\ \implies x' &= \pm \sqrt{k_{1} - x^{2}} \\ \implies \int \frac{dx}{\sqrt{k_{1} - x^{2}}} &= \pm \int dt \\ \implies \arcsin \left( \frac{x}{k_{1}} \right) &= \pm t + k_{2} \\ \implies x &= k_{1} \sin(k_{2} \pm t) \\ &= c_{1} \cos(t) + c_{2} \sin(t) \end{align}

where $c_{1} = k_{1} \sin(\pm k_{2}), c_{2} = k_{1} \cos(\pm k_{2})$. Now, applying your conditions we find

\begin{align} x(0) &= s \\ &= C_{1} \cos(0) + C_{2} \sin(0) \\ &= C_{1} \\ x'(0) &= 1 \\ &= -s \sin(0) + C_{2} \cos(0) \\ &= C_{2} \end{align}

Hence the particular solution is given by $x(s,t) = s \cos(t) + \sin(t)$. The same approach is used to solve the ODE in $p$.

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As to the added question:

You can determine the constants in the eigenvalue solution simply by evaluating the solution formula at $t=0$ $$ s=x(s,0)=C_1\cos(0)+iC_2\sin(0)=C_1\\ 1=p(s,0)=iC_2\cos(0)-C_1\sin(0)=iC_2 $$ giving exactly the same result as the cited solution.

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