First Order Partial Differential Equations

Let us first consider the simplest case of the first order PDE of the form
$$\boxed{au_x(x,y)+bu_y(x,y)=0,\quad a^2+b^2\neq 0}$$
with constant coefficients $a,b\in\mathbb{R}$.


Example 1.
Solve
$$au_x(x,y)+bu_y(x,y)=0,\quad a^2+b^2\neq 0$$
where $a,b\in\mathbb{R}$.

We solve it by Geometric Method first. Note that
$$\begin{aligned} 0 &= au_x+bu_y\\ &=(a,b)\cdot (u_x,u_y)\\ &=\mathbf{v}\cdot \nabla u\\ &=\Vert\mathbf{v}\Vert D_\mathbf{v}u \end{aligned}$$
where $\mathbf{v}=(a,b)$ and $D_\mathbf{v}u$ is the directional derivative of $u$ along the direction $\mathbf{v}$. It follows that $u$ must be a constant along the direction $\mathbf{v}$. Then we get
$$u(x,y)=f(bx-ay)$$
for some $C^1$ function $f$. Here we call
$$bx-ay=C$$
for some arbitrary constant $C$ the characteristic curves which parallel to $\mathbf{v}$. On the other hand, we solve it by Coordinate Method. Let us consider the change of variables given by
$$\left\{\begin{array}{ll} s=ax+by\\ t={bx-ay}. \end{array} \right.$$
Then we have
$$u_x=\frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial x}=au_s+bu_t$$
and
$$u_y=\frac{\partial u}{\partial s}\frac{\partial s}{\partial y}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial y}=bu_s-au_t.$$
Plugging these terms into the equation, we get
$$\left(a^2+b^2\right)u_s=0,$$
to get
$$u=f(t)=f(bx-ay)$$
for some $C^1$ function $f$.


Next, we consider the general linear first order PDE of the form
$$\boxed{ a(x,y)u_x(x,y)+b(x,y)u_y(x,y)=c(x,y)u+d(x,y)}$$
with variable coefficients.


Example 2.
Solve
$$u_x(x,y)+yu_y(x,y)=0.$$

Note that
$$0=u_x+yu_y=(1,y)\cdot(u_x,u_y)$$
which implies the directional derivative of $u$ in the direction of $(1,y)$ is zero. The characteristic curves $(x(t),y(t))$ has the tangent vector $(1,y)$ given by
$$\frac{dx}{dt}=1\quad\text{and}\quad\frac{dy}{dt}=y$$
or
$$\frac{dy}{dx}=\frac{y}{1}.$$
Then we get
$$y=Ce^x$$
for some constant $C$. Since the solution of the PDE is constant along the characteristic curves, we conclude that
$$u(x,y)=u(x,Ce^x)=u(0,C)=u(0,ye^{-x})=f(ye^{-x})$$
for some $C^1$ function $f$.

Remark.

Note that every solution of the PDE is constant on the solution curves (characteristic curves) of the ODE.

For more examples, please refer the attached files.

Finally, we consider the general quasilinear first order PDE of the form
$$\boxed{a(x,y,u)u_x(x,y)+b(x,y,u)u_y(x,y)=c(x,y,u)}$$
In this case, if we represent the function $u(x,y)$ by a surface $z=u(x,y)$ in $xyz$-space, then the PDE can be written as
$$(a(x,y,u),b(x,y,u),c(x,y,u))\cdot(u_x(x,y),u_y(x,y),-1)=0$$
where $(u_x,u_y,-1)$ is the normal vector of the (solution) surface $z=f(x,y)$ and $(a,b,c)$ is called the characteristic direction. One sees that the solution surface is tangent to the characteristic direction. Then it suffices to solve the system of ODE
$$\left\{\begin{array}{ll} \displaystyle\frac{dx}{dt}=a(x,y,u)\\ \\ \displaystyle\frac{dy}{dt}=b(x,y,u)\\ \\ \displaystyle\frac{du}{dt}=c(x,y,u). \end{array} \right.$$

Example 3.

Solve
$$u(x,y)u_x(x,y)+u_y(x,y)=1$$
with side condition
$$u(2s^2,2s)=0,\quad s>0.$$

Consider the system of ODE
$$\left\{\begin{array}{ll} \displaystyle\frac{dx}{dt}=u\\ \\ \displaystyle\frac{dy}{dt}=1\\ \\ \displaystyle\frac{du}{dt}=1 \end{array} \right.$$
with the conditions at $t=0$ given by
$$x(s,0)=2s^2,\quad y(s,0)=2s,\quad u(s,0)=0,\quad s>0.$$
A direct computation yields that
$$x=\frac{1}{2}t^2+2s^2,\quad y=t+2s,\quad u=t.$$
Eliminating $t$ and $s$, we get
$$\left(u-\frac{1}{2}y\right)^2=x-\frac{1}{4}y^2,$$
to get
$$u=\frac{1}{2}y-\left(x-\frac{1}{4}y^2\right)^{1/2}$$
as $u(2s^2,2s)=0,\,s>0$. Note that the solution is not defined for $x<\frac{1}{4}y^2$ and is not differentiable when $x=\frac{1}{4}y^2$.

                                           
*詳細內容及更多範例可參考附件 First Order PDEs.pdf