Using Charpit's method to solve a nonlinear PDE

Question

I have the nonlinear PDE $$p^2 - q^2 = 3u$$ with the initial condition $u(x, 0) = \cfrac{-x^2}{4}.$

Here's what I have done so far:

I defined the function $F$ to be equal $$F(x, y, p, q, u) = p^2 - q^2 - 3u,$$

and got

$$(F_x, F_y, F_p, F_q, F_u) = (0, 0, 2p, -2q, -3).$$

Therefore I managed to write down the characteristic

$$\frac{dx}{-F_p} = \frac{dy}{-F_q} = \frac{dp}{F_x + pF_u} = \frac{dq}{F_y + qF_u} = \frac{du}{-pF_p - qF_q},$$

or

$$\frac{dx}{-2p} = \frac{dy}{2q} = \frac{dp}{-3p} = \frac{dq}{-3q} = \frac{du}{-2p^2 + 2q^2}.$$

Since

$$\frac{dx}{-2p} = \frac{dp}{-3p}$$

and

$$\frac{dy}{2q} = \frac{dq}{-3q},$$

I got

$$p = \frac{3x}{2} + a$$

and

$$q = \frac{-3y}{2} + b$$

for some constants $a, b$. Also, $du = pdx + qdy$. Thus I got

$$du = \left(\frac{3x}{2} + a\right)dx + \left(\frac{-3y}{2} + b\right)dy.$$

Integrating both sides yields

$$u = \frac{3x^2}{4} + ax - \frac{3y^2}{4} + by + c,$$

where $c$ is a constant.

If this is a solution, then it must satisfy the initial condition. Putting $y = 0$ and $u = \cfrac{-x^2}{4}$ in the solution above gives us

$$\frac{-x^2}{4} = \frac{3x^2}{4} + ax + c.$$

However, there are no constants $a, b$ such that the equation above holds for all $x$. Then I concluded that there is no solution to the given PDE satisfying the given initial condition.

But, the function

$$u = \frac{-1}{4}(x - 2y)^2$$

is a solution to the given PDE, and it also satisfies the given initial condition.

I can't even guess what my mistake was. Any help would be highly appreciated.

Answer 1

For equations of the form $f(u,u_x,u_y)=0$ it is easily shown that $u_x=au_y$, $f(u,au_y,u_y)=0$ is the solution. For yours we then have that \begin{align} du=\pm\sqrt{\frac{3u}{1-a^2}}\left(a\mathrm dx+\mathrm dy\right)\quad\rightarrow\quad u(x,y)=\frac34\left(\frac{ax}{\sqrt{1-a^2}}+\frac{y}{\sqrt{1-a^2}}+b\right)^2. \end{align} Given that $u(x,0)=-x^2/4$: \begin{align} \frac34\left(\frac{ax}{\sqrt{1-a^2}}+b\right)^2+\frac{x^2}{4}=0, \end{align} for which we see that $a=\pm i/\sqrt2$ and $b=0$. We then find our two solutions \begin{align} u_{\pm}(x,y)=\frac{-(x\pm2y)^2}{4}. \end{align}

Answer 2

The Chappit's method is difficult to apply in case of non-linear PDEs. In the present case the method used by Eli Bartlett is simpler and more reliable. Nevertheless we try to see where is the mistake in the OP's calculus.

We must remember that the Charpit-Lagrange ODEs are not true everywhere but only on some particular lines. To makes this evident we will check the Charpit-Lagrange ODEs in the case of the PDE : $$u_x^2-u_y^2=3u,$$ knowing a particular solution, for example : $$u(x,y)=-\frac14(x+2y)^2$$ $p=u_x=-\frac12(x+2y)\quad\implies\quad dp=-\frac12 dx-dy$

$q=u_y=-(x+2y)\quad\implies\quad dq=-dx-2dy$

$du=-\frac12(x+2y)(dx+2dy)$

We put $p,q,dp,dq,du$ into the Charpit-Lagrange ODEs : $$\frac{dx}{-F_p} = \frac{dy}{-F_q} = \frac{dp}{F_x + pF_u} = \frac{dq}{F_y + qF_u} = \frac{du}{-pF_p - qF_q},$$ $$\frac{dx}{-2p} = \frac{dy}{2q} = \frac{dp}{-3p} = \frac{dq}{-3q} = \frac{du}{-2p^2 + 2q^2},$$ leading to : $$\frac{dx}{-2(-\frac12(x+2y))} = \frac{dy}{2(-(x+2y))} = \frac{-\frac12 dx-dy}{-3(-\frac12(x+2y))} = \frac{-dx-2dy}{-3(-(x+2y))} = \frac{-\frac12(x+2y)(dx+2dy)}{-2(-\frac12(x+2y))^2 + 2(-(x+2y))^2}.$$ After simplification : $$dx = -\frac12 dy = -\frac13(dx+2dy) = -\frac13(dx+2dy)=-\frac13(dx+2dy)$$ This is OK except $\quad dx = -\frac12 dy\quad\implies\quad y=2x+c$

This shows that the Charpit-Lagrange ODEs are satisfied on the curves $\quad y(x)=2x+c\quad$ but not elsewere.

Of course at the first start of the problem we don't know what are the curves on which the Charpit-Lagrange ODEs are satisfied. Thus if we have some integrations to do we have to take care.

This explains why integrating $\quad \frac{dx}{-2p} = \frac{dp}{-3p}\quad\implies\quad dp=\frac32 x\quad$ doesn't gives $\quad p = \frac{3x}{2} + a\quad$ which isn't correct. In other words $p$ isn't function of $x$ only but is also function of $y$ which is function of $x$ on a curve $y(x)$ which is unknown at this stage.

The same difficulty arises with $\frac{dy}{2q} = \frac{dq}{-3q}$.

Now we forget the particular solution used above and we come back to the original problem $$u_x^2-u_y^2=3u\quad;\quad u(x,0)=-\frac14 x^2 .$$

$$\frac{3dx}{2p} = \frac{3dy}{-4q} = \frac{dp}{p} = \frac{dq}{q} = \frac{du}{2u}$$ $\frac{dp}{p} = \frac{du}{2u} \quad\implies\quad p=C\:|u|^{1/2} \quad\implies\quad \frac{u_x}{|u|^{1/2}}=C \quad\implies\quad |u|^{1/2}=2Cx+f(y)$ $$u=\pm\big(ax+f(y)\big)^2$$ On the same manner $\frac{3dy}{-4q} = \frac{dq}{q}$ leads to $$u=\pm\big(by+g(x)\big)^2$$ with $|a^2-b^2|=\frac34$

Both together : $$u(x,y)=\pm(ax+by+c)^2 \quad;\quad a^2-b^2=\mp\frac34$$ $a,b,c$ constants real or complex.

With condition $u(x,0)=-\frac14 x^2$ it is easy to find the sign - and $c=0$ ; $a=\frac12$ then $b=\pm 1.$

$$\boxed{u(x,y)=-\left(\frac{x}{2}\pm y\right)^2 }$$

In case of non-linear PDE indeed this isn't the simplest method and full of traps.