Initial-value problem for non-linear partial differential equation $y_x^2=k/y_t^2-1$

Question

For this problem, $y$ is a function of two variables: one space variable $x$ and one time variable $t$.
$k > 0$ is some constant.
And $x$ takes is value in the interval $[0, 1]$ and $t \ge 0$.
At the initial time, $y$ follows a parabolic profile, like $y(x, 0) = 1 - (x-\frac{1 }{2})^2$.

Finally, $y$ satisfies this PDE: $$ \left(\frac{\partial y} {\partial x}\right)^2 = \frac{k}{\left(\frac{\partial y} {\partial t}\right)^2} - 1.$$

Does anyone have an idea how to solve this problem (and find the expression of $y(x,t)$) ?

About: The problem arise in physics, when studying the temporal shift of a front of iron particles in a magnetic field.

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Edit: I solved it numerically on a (badly-designed) 1st-order numerical scheme with a small space & time discretization, with the initial condition I wanted (in Octave/Matlab, in Python and in OCaml + GNUplot). The numerical result was enough to confirm the theory and the experiment (the observation done in the lab), so I did not try any further to solve it analytically. See here for an animation of the front of iron matter, and here for more details (in French).

Answer 1

According to Mathematica, if this helps:

$y(x,t) = \pm \frac{x \sqrt{k-C_2^2}}{C_2}+C_2 t+C_1$

Mathematica won't solve it with the boundary condition.

Answer 2

Your equation can be written as follows:

$$F(p,q,x,t,y) = (p^2+1)q^2 -k = 0, \quad p = y_x, \quad q = y_t,$$

and hence:

\begin{align} F_p & = 2pq^2, \\ F_q & = 2(1+p^2) q,\\ F_t = F_x = F_y & = 0, \end{align}

so the Lagrange-Charpit equations read:

$$ \frac{\mathrm{d}x}{2 pq^2 }= \frac{\mathrm{d}t}{2(1+p^2)q} = \frac{\mathrm{d}y}{2p^2q^2 + 2(1+p^2)q^2 } = -\frac{\mathrm{d}p}{0} = - \frac{\mathrm{d}q}{0},$$

which tells you that $\mathrm{d}p = \mathrm{d}q = 0$ and, thus, $p = A$ (or $q = B$) constant. Since $p = y_x$ we have that $y(x,t) = Ax+f(t)$. Plug this into the original PDE to find:

$$f(t) = \pm \frac{\sqrt{k} t}{\sqrt{1+A^2}} + C, \quad C \in \mathbb{R}, $$

so the solution is finally given by:

$$ \color{blue}{y(x,t) = Ax \pm \frac{\sqrt{k} t}{\sqrt{1+A^2}} + C },$$

this is known as a complete integral of your PDE. It remains to set the initial condition $y(x,0)$. Can you take it from here?

Cheers!

Answer 3

$y_x^2=\dfrac{k}{y_t^2}-1$

$y_t^2=\dfrac{k}{y_x^2+1}$

$y_t=\pm\dfrac{\sqrt k}{\sqrt{y_x^2+1}}$

$y_{tx}=\mp\dfrac{\sqrt ky_xy_{xx}}{(y_x^2+1)^\frac{3}{2}}$

Let $u=y_x$ ,

Then $u_t=\mp\dfrac{\sqrt kuu_x}{(u^2+1)^\frac{3}{2}}$ with $u(x,0)=-2x+1$

$u_t\pm\dfrac{\sqrt kuu_x}{(u^2+1)^\frac{3}{2}}=0$ with $u(x,0)=-2x+1$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$

$\dfrac{dx}{ds}=\pm\dfrac{\sqrt ku}{(u^2+1)^\frac{3}{2}}=\pm\dfrac{\sqrt ku_0}{(u_0^2+1)^\frac{3}{2}}$ , letting $x(0)=f(u_0)$ , we have $x=\pm\dfrac{\sqrt ku_0s}{(u_0^2+1)^\frac{3}{2}}+f(u_0)=\pm\dfrac{\sqrt kut}{(u^2+1)^\frac{3}{2}}+f(u)$ , i.e. $u=F\left(x\mp\dfrac{\sqrt kut}{(u^2+1)^\frac{3}{2}}\right)$

$u(x,0)=-2x+1$ :

$F(x)=-2x+1$

$\therefore u=-2\left(x\mp\dfrac{\sqrt kut}{(u^2+1)^\frac{3}{2}}\right)+1$

$y_x=\pm\dfrac{2\sqrt kty_x}{(y_x^2+1)^\frac{3}{2}}-2x+1$

$(y_x+2x-1)(y_x^2+1)^\frac{3}{2}=\pm2\sqrt kty_x$

You will get $y_x(x,t)$ which is difficult to express explicitly, so does for $y(x,t)$ .

Answer 4

Like @DMoreno, let's rewrite the PDE as $$ F(x,t,p,q,y) := (p^2+1)q^2 -k = 0, \tag{1} $$ where $p:=y_x$ and $q:=y_t$. The characteristic curves obey the following differential equations: \begin{align} \dot{p}(s)&=-F_x-pF_y=0 \implies p(s)=p_0, \tag{2} \\ \dot{q}(s)&=-F_t-qF_y=0 \implies q(s)=q_0, \tag{3}\\ \dot{x}(s)&=F_p=2pq^2=2p_0q_0^2 \implies x(s)=x_0+2p_0q_0^2s, \tag{4} \\ \dot{t}(s)&=F_q=2q(p^2+1)=2q_0(p_0^2+1) \implies t(s)=t_0+2q_0(p_0^2+1)s, \tag{5} \\ \dot{y}(s)&=pF_p+qF_q=2p^2q^2+2q^2(p^2+1)=2q_0^2(2p_0^2+1) \\ &\implies y(s)=y_0+2q_0^2(2p_0^2+1)s. \tag{6} \end{align} In addition, we have the following relations among the constants $x_0,t_0,p_0,q_0,$ and $y_0$:

1. The initial condition $y(x,0)=1-\left(x-\frac{1}{2}\right)^2$ implies \begin{align} t(0)&=0 \implies t_0=0, \tag{7} \\ y(0)&=1-\left(x(0)-\frac{1}{2}\right)^2 \implies y_0=1-\left(x_0-\frac{1}{2}\right)^2; \tag{8} \end{align}
2. The PDE $(1)$ implies $$ (p_0^2+1)q_0^2-k=0; \tag{9} $$
3. Finally, Eq. $(8)$ implies $$ p_0=\frac{\partial y_0}{\partial x_0}=-2x_0+1. \tag{10} $$

Equations $(7)$ - $(10)$ allows us to rewrite $x,t,$ and $y$ in terms of a single constant ($q_0$, for instance) and the parameter $s$: \begin{align} x(s)&\overset{(4,10)}{=}\frac{1-p_0}{2}+2p_0q_0^2s\overset{(9)}{=}\frac{1}{2}\pm\sqrt{\frac{k}{q_0^2}-1} \left(-\frac{1}{2}+2q_0^2s\right), \tag{11} \\ t(s)&\overset{(5,7)}{=}2q_0(p_0^2+1)s\overset{(9)}{=}\frac{2ks}{q_0}, \tag{12} \\ y(s)&\overset{(6,8)}{=}1-\left(x_0-\frac{1}{2}\right)^2+2q_0^2(2p_0^2+1)s \overset{(10)}{=}1-\frac{p_0^2}{4}+2q_0^2(2p_0^2+1)s \\ &\overset{(9)}{=}1-\frac{1}{4}\left(\frac{k}{q_0^2}-1\right)+(4k-2q_0^2)s. \tag{13} \end{align} Ideally, one would solve Eqs. $(11)$ - $(12)$ in order to express $s$ and $q_0$ as functions of $x$ and $t$; Eq. $(13)$, then, would give $y$ as an explicit function of $x$ and $t$. Unfortunately, those equations are too complicated to be solved in closed form. It is possible, however, to present a solution in parametric form: Eq. $(12)$ yields $s=\frac{q_0t}{2k}$, which, when substituted in Eqs. $(11)$ and $(13)$, gives \begin{align} x(q_0,t)&=\frac{1}{2}\pm\sqrt{\frac{k}{q_0^2}-1} \left(-\frac{1}{2}+\frac{q_0^3t}{k}\right), \tag{14} \\ y(q_0,t)&=1-\frac{1}{4}\left(\frac{k}{q_0^2}-1\right)+\left(2q_0-\frac{q_0^3}{k}\right)t. \tag{15} \end{align} At any given time $t$, the set $\{(x(q_0,t),y(q_0,t))\in\mathbb{R}^2|q_0^2\leq k\}$ defines a curve in $\mathbb{R}^2$.