1.5: General First Order PDEs
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We have spent time solving quasilinear first order partial differential equations. We now turn to nonlinear first order equations of the form
\[F(x, y, u, u_x, u_y) = 0,\nonumber\]
for \(u = u(x, y)\).
If we introduce new variables, \(p = u_x\) and \(q = u_y\), then the differential equation takes the form
\[F(x, y, u, p, q) = 0.\nonumber\]
Note that for \(u(x, t)\) a function with continuous derivatives, we have
\[p_y = u_{xy} = u_{yx} = q_x.\nonumber\]
We can view \(F = 0\) as a surface in a five dimensional space. Since the arguments are functions of \(x\) and \(y\), we have from the multivariable Chain Rule that
\[\begin{align}\frac{dF}{dx}&=F_x+F_u\frac{\partial u}{\partial x}+F_p\frac{\partial p}{\partial x}+F_q\frac{\partial q}{\partial x}\nonumber \\ 0&=F_x+pF_u+p_xF_p+p_yF_q.\label{eq:1}\end{align}\]
Similarly, from \(\frac{dF}{dy} = 0\) we have that
\[\frac{dx}{F_p}=\frac{dy}{F_q}=-\frac{dq}{F_y+qF_u}.\nonumber\]
Combining these results we have the Charpit Equations
\[\label{eq:3}\frac{dx}{F_p}=\frac{dy}{F_q}=\frac{du}{pF_p+qF_q}=-\frac{dp}{F_x+pF_u}=-\frac{dq}{F_y+qF_u}.\]
These equations can be used to find solutions of nonlinear first order partial differential equations as seen in the following examples.
The Charpit equations were named after the French mathematician Paul Charpit Villecourt, who was probably the first to present the method in his thesis the year of his death, 1784. His work was further extended in 1797 by Lagrange and given a geometric explanation by Gaspard Monge (1746-1818) in 1808. This method is often called the Lagrange-Charpit method.
Find the general solution of \(u_x^2 + yu_y − u = 0\).
Solution
First, we introduce \(u_x = p\) and \(u_y = q\). Then,
\[F(x, y, u, p, q) = p^2 + qy − u = 0.\nonumber\]
Next we identify,
\[F_p = 2p,\quad F_q = y,\quad F_u = −1,\quad F_x = 0,\quad F_y = q.\nonumber\]
Then,
\[\begin{aligned}pF_p+qF_q&=2p^2+qy, \\ F_x+pF_u&=-p, \\ F_y+qF_u&=q-q=0.\end{aligned}\]
The Charpit equations are then
\[\frac{dx}{2p}=\frac{dy}{y}=\frac{du}{2p^2+qy}=\frac{dp}{p}=\frac{dq}{0}.\nonumber\]
The first conclusion is that \(q = c_1 =\) constant. So, from the partial differential equation we have \(u = p^2 + c_1y\).
Since \(du = pdx + qdy = pdx + c_1dy\), then
\[du-cdy=\sqrt{u-c_1y}dx.\nonumber\]
Therefore,
\[\begin{align}\int\frac{d(u-c_1y)}{\sqrt{u-c_1y}}&=\int dx\nonumber \\ \int\frac{z}{\sqrt{z}}&=x+c_2\nonumber \\ 2\sqrt{u-c_1y}&=x+c_2.\label{eq:4}\end{align}\]
Solving for \(u\), we have
\[u(x,y)=\frac{1}{4}(x+c_2)^2+c_1y.\nonumber\]
This example required a few tricks to implement the solution. Sometimes one needs to find parametric solutions. Also, if an initial condition is given, one needs to find the particular solution. In the next example we show how parametric solutions are found to the initial value problem.
Solve the initial value problem \(u_x^2 + u_y + u = 0\), \(u(x, 0) = x\).
Solution
We consider the parametric form of the Charpit equations,
\[\label{eq:5}dt=\frac{dx}{F_p}=\frac{dy}{F_q}=\frac{du}{pF_p+qF_q}=-\frac{dp}{F_x+pF_u}=-\frac{dq}{F_y+qF_u}.\]
This leads to the system of equations
\[\begin{aligned}\frac{dx}{dt}&=F_p=2p. \\ \frac{dy}{dt}&=F_q=1. \\ \frac{du}{dt}&=pF_p+qF_q=2p^2+q. \\ \frac{dp}{dt}&=-(F_x+pF_u)=-p. \\ \frac{dq}{dt}&=-(F_y+qF_u)=-q.\end{aligned} \]
The second, fourth, and fifth equations can be solved to obtain
\[\begin{aligned} y&=t+c_1. \\ p&=c_2e^{-t}. \\ q&=c_3e^{-t}.\end{aligned}\]
Inserting these results into the remaining equations, we have
\[\begin{aligned}\frac{dx}{dt}&=2c_2e^{-t}. \\ \frac{du}{dt}=2c_2^2e^{-2t}+c_3e^{-t}.\end{aligned}\]
These equations can be integrated to find Inserting these results into the remaining equations, we have
\[\begin{aligned}x&=-2c_2e^{-t}+c_4. \\ u&=-c_2^2e^{-2t}-c_3e^{-t}+c_5.\end{aligned}\]
This is a parametric set of equations for \(u(x, t)\). Since
\[e^{-t}=\frac{x-c_4}{-2c_2},\nonumber\]
we have
\[\begin{align} u(x,y)&=-c_2^2e^{-2t}-c_3e^{-t}+c_5.\nonumber \\ &=-c_2^2\left(\frac{x-c_4}{-2c_2}\right)^2-c_3\left(\frac{x-c_4}{-2c_2}\right)+c_5\nonumber \\ &=\frac{1}{4}(x-c_4)^2+\frac{c_3}{2c_2}(x-c_4).\label{eq:6}\end{align}\]
We can use the initial conditions by first parametrizing the conditions. Let \(x(s, 0) = s\) and \(y(s, 0) = 0\), Then, \(u(s, 0) = s\). Since \(u(x, 0) = x\), \(u_x(x, 0) = 1\), or \(p(s, 0) = 1\).
From the partial differential equation, we have \(p^2 + q + u = 0\). Therefore,
\[q(s, 0) = −p^2 (s, 0) − u(s, 0) = −(1 + s).\nonumber\]
These relations imply that
\[\begin{aligned}y(x,t)|_{t-0}=0\Rightarrow c_1&=0. \\ p(s,t)|_{t-0}=1\Rightarrow c_2&=1. \\ q(s,t)|_{t-0}=-(1+s)&=c_3.\end{aligned}\]
So,
\[\begin{aligned}y(s,t)&=t. \\ p(s,t)&=e^{-t}. \\ q(s,t)&=-(1+s)e^{-t}.\end{aligned}\]
The conditions on \(x\) and \(u\) give
\[\begin{aligned}x(s,t)&=(s+2)-2e^{-t}, \\ u(s,t)&=(s+1)e^{-t}-e^{-2t}.\end{aligned}\]