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I'm interested in solving a Cauchy problem for the iterated singular EPD.

Weinstein (On a class of PDEs of even order, 1955) showed how the decomposition formula leads to the solution of the Cauchy problem for the iterated wave equation (so, for $k=0$).

My question, how one can use this decomposition to solve a Cauchy problem for the iterated EPD equation?

$L_kL_m u(t,x)=0$,

$D_t^i u(0,x)=u_i(x) , 0\leq i\leq 3 \\ $,

where $L_k := D^2_t-D^2_x+ \frac{k}{t}D_t$

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  • $\begingroup$ I assume EDP-> PDE? $\endgroup$
    – Igor Rivin
    Commented Oct 22, 2012 at 13:46
  • $\begingroup$ EPD : Euler-Poisson-Darboux $\endgroup$
    – MAK
    Commented Oct 22, 2012 at 13:51

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Is it not simply a conversion of an iterated homogeneous equation into a non-iterated non-homogeneous equation? By hypothesis, $L_m u = v$, where $L_k v = 0$. Every such $v$ can be obtained from some initial data at $t=0$. On the other hand, every every such $u$ can be obtained by solving the inhomogeneous linear equation, using advanced and retarded Green functions for $L_m$, plus some solution $w$ of the homogeneous equation $L_m w = 0$. Then the freedom in the choice of $w$ is fixed by matching the initial conditions for $u$ at $t=0$. Thus, to solve the iterated problem, you need only know how to solve the homogeneous and inhomogeneous non-iterated problems.

Are you perhaps looking for some explicit formulas for the Green functions that match initial conditions to the solutions that result from this procedure?

Update. About the choice of initial conditions. BTW, I didn't immediately pay attention to the fact that you want the initial data specified at $t=0$, which is the location of the singularity of the coefficients of your PDE. This is a kind of Fuchsian equation that requires extra care when imposing boundary conditions. I haven't examined this example in detail yet, but you may not be able to specify all the derivatives, $0\le i \le 3$, while requiring the solution to stay bounded at $t=0$. First, let me illustrate how to determine all the initial conditions when $t=0$ is a regular value for the coefficients. I may come back to the singular case later.

Write your solution as $u = u_v + w$, where $L_m w = 0$, $L_m u_v = v$ and $L_k v = 0$. You are free to set the initial conditions $D_t^{0,1}w(0,x) = u_{0,1}(0,x)$ and $D_t^{0,1} u_v(0,x) = 0$. Then the first two initial conditions are met, $D_t^{0,1}u(0,x) = u_{0,1}(0,x)$. On the other hand, just by taking time derivatives you find that $D_t^{0,1}v(0,x) = D_t^{0,1}L_m u_v(0,x) = D_t^{0,1}L_m u(0,x)$, where the right hand side can be determined from the knowledge of the derivatives $D_t^i u(0,x)$, $0\le i \le 3$. This gives you the initial conditions for solving $L_k v = 0$.

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  • $\begingroup$ @Igor, thank you very much. Indeed, it is easy to solve the non-homogeneous iterated equation. But, how shoold we choose the initial condition for $L_m u=v$ such that the solution of (auxiliary) second order EDP satisfies both the iterated EDP and the four initial conditions? $\endgroup$
    – MAK
    Commented Oct 23, 2012 at 6:53
  • $\begingroup$ To be honest, I'm not sure I understand your initial onditions. How are you specifying the values of the 0th-3rd time derivatives at $t=0$ needed to fix a unique solution? $\endgroup$
    – Igor Khavkine
    Commented Oct 23, 2012 at 7:27
  • $\begingroup$ for the sake of brevity, let me take $u_0(x)=x^n$, $n$ is a positive integer, and $u_1(x)=u_2(x)=u_3(x)=0$. $\endgroup$
    – MAK
    Commented Oct 23, 2012 at 8:21
  • $\begingroup$ Ah, I see now. The $D^i u(0,x)$ should have been $D^i_t u(0,x)$ in your question. $\endgroup$
    – Igor Khavkine
    Commented Oct 23, 2012 at 8:25
  • $\begingroup$ yes, as you said $\endgroup$
    – MAK
    Commented Oct 23, 2012 at 8:30

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