📂Partial Differential Equations

Solution of Nonlinear First Order PDE Using Characteristic Equations

Explanation¹

• When emphasizing that x and p are variables of a partial differential equation, they are denoted in normal font as $x,p \in \mathbb{R}^{n}$, and when emphasizing them as functions of $s$, they are denoted in bold font as $\mathbf{x}, \mathbf{p} \in \mathbb{R}^{n}$.

Characteristic Equations

$$ \begin{cases} \dot{\mathbf{p}} (s) = -D_{x}F\big(\mathbf{p}(s),\ z(s),\ \mathbf{x}(s) \big)-D_{z}F\big(\mathbf{p}(s),\ z(s),\ \mathbf{x}(s) \big)\mathbf{p}(s) \\ \dot{z}(s) = D_{p}F\big(\mathbf{p}(s),\ z(s),\ \mathbf{x}(s) \big) \cdot \mathbf{p}(s) \\ \dot{\mathbf{x}}(s) = D_{p}F\big(\mathbf{p}(s),\ z(s),\ \mathbf{x}(s) \big) \end{cases} $$

The solution of nonlinear first-order partial differential equations using characteristic equations varies slightly depending on how the differential equation is given. This is distinguished by the linearity of the given differential equation and differs for cases of linear, quasi-linear, and fully nonlinear. The stronger the nonlinearity, the more challenging it is.

Solution

Homogeneous Linear

If the given partial differential equation is completely linear, it can be solved most easily. The condition for $\mathbf{p}(s)$ in the characteristic equation is so simple that it is unnecessary. Consider the following linear and homogeneous differential equation.

$$ \begin{equation} F(Du, u, x) = \mathbf{b}(x)\cdot Du(x)+c(x)u(x)=0 \quad (x\in \Omega \subset \mathbb{R}^{n}) \label{eq1} \end{equation} $$

Here, if we set the variables of $F$ as $p, z, x$, it would be as follows.

$$ \begin{equation} F(p,\ z,\ x)=\mathbf{b}(x)\cdot p +c(x)z=b_{1}p_{1}+\cdots +b_{n}p_{n}+cz = 0 \label{eq2} \end{equation} $$

If we calculate $D_{p}F$, it would be as follows.

$$ D_{p}F=(F_{p_{1}}, \dots, F_{p_{n}})=(b_{1}, \dots, b_{n})=\mathbf{b}(x) $$

Then, the characteristic equation is as follows.

$$ \begin{align*} \dot{\mathbf{x}}(s) &= \mathbf{b}(x) \\ \dot{z}(s) &= \mathbf{b}(\mathbf{x}(s))\cdot \mathbf{p}(s) \end{align*} $$

By $(2)$, $\dot{z}(s)$ is as follows.

$$ \dot{z}(s) = -c(\mathbf{x}(s))z $$

Therefore, the characteristic equation for a homogeneous linear first-order partial differential equation is as follows.

$$ \left\{ \begin{align*} \dot{\mathbf{x}}(s)&=\mathbf{b}(x) \\ \dot{z}(s) &= -c(\mathbf{x}(s))z \end{align*} \right. $$

Here, it can be seen through an example that the characteristic equation for $\mathbf{p}(s)$ is not necessary to solve the problem.

Example

Suppose the following differential equation is given.

$$ \left\{ \begin{align*} x_{1} u_{x_{2}} - x_{2} u_{x_{1}} &= u && \text{in } \Omega \\ u&=g && \text{on } \Gamma \end{align*} \right. $$

• $\Omega=\left\{ x_{1}>0,\ x_{2}>0 \right\}$
• $\Gamma=\left\{ x_{1}>0,\ x_{2}=0 \right\}$

Then in $(1)$, $\mathbf{b}=(-x_{2},\ x_{1}), c=-1$. Thus, the characteristic equation is as follows.

$$ \left\{ \begin{align*} \dot{x}^{1} &= -x^{2} \\ \dot{x}^{2} &=x^{1} \\ \dot{z}&=z \end{align*} \right. $$

Since this is a simple ordinary differential equation, it can be easily solved as follows.

$$ \left\{ \begin{align*} x^{1}(s) &=x^{0}\cos s \\ x^{2}(s)&=x^{0} \sin s \\ z(s)&=z^{0}e^s=g(x^{0})e^s \end{align*} \right. $$

Here, $x^{0}$ is a constant set to pass through the $x_{1}-$axis $(\Gamma)$ when $s=0$. Then, by the boundary condition $z=u=g\ \mathrm{on}\ \Gamma$, when $s=0$, $z(0)=z^{0}=g(x^{0})$. Now, fix the point $(x_{1},\ x_{2}) \in \Omega$.

$$ (x_{1},\ x_{2})=(x^{1}(s),\ x^{2}(s)) = (x^{0} \cos (s),\ x^{0} \sin (s)) $$

Then, for $s>0, x^{0}>0$, we obtain the following.

$$ x_{1}^{2} + x_{2}^{2} = (x^{0})^{2}\cos^{2}(s) + (x^{0})^{2}\sin^{2}(s) = (x^{0})^{2} \implies x^{0}=({x_{1}}^{2}+{x_{2}}^{2})^{1/2} \\ \dfrac{x_{2}}{x_{1}} = \dfrac{x^{0}\sin (s)}{x^{0} \cos (s)} = \tan (s) \implies s=\arctan \left( \frac{x_{2}}{x_{1}} \right) $$

Therefore, the solution of the equation is as follows.

$$ \begin{align*} u(x)&=u(x^{1}(s),\ x^{2}(s)) \\ &= z(s) \\ &=g(x^{0})e^s \\ &= g(({x_{1}}^{2}+{x_{2}}^{2})^{1/2})e^{\arctan \left(\frac{x_{2}}{x_{1}}\right)} \end{align*} $$

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Quasi-Linear

The following describes cases where the given differential equation is linear with respect to the highest order of differentiation. As we are dealing with first-order differential equations, it refers to cases where they are linear with respect to first-order derivatives.

$$ F(Du,\ u,\ x)=\mathbf{b}(x,\ u(x))\cdot Du(x)+c(x,\ u(x))=0 $$

Here, if we set the variables of $F$ as $p, z, x$, it would be as follows.

$$ \begin{equation} F(p, z, x)=\mathbf{b}(x, z)\cdot p + c(x, z)=b_{1}p_{1} + \cdots + b_{n} p_{n} +c=0 \label{eq3} \end{equation} $$

If we calculate $D_{p}F$, it would be as follows.

$$ D_{p}F=(F_{p_{1}},\ \cdots,\ F_{p_{n}})=(b_{1},\ \cdots,\ b_{n})=\mathbf{b}(x,\ z) $$

Then, the characteristic equation is as follows.

$$ \begin{align*} \dot{\mathbf{x}}(s) &= \mathbf{b}(\mathbf{x}(s),\ z(s)) \\ \dot{z}(s) &= \mathbf{b}(\mathbf{x}(s),\ z(s))\mathbf{p}(s)=-c(\mathbf{x}(s),\ z(s)) \end{align*} $$

The second equality for $\dot{z}$ is valid due to $(3)$. Even in this case, the condition for $\mathbf{p}(s)$ is not necessary to solve the problem.

Example

Suppose the following differential equation is given.

$$ \left\{ \begin{align*} u_{x_{1}} + u_{x_{2}} &= u^{2} && \text{in } \Omega \\ u&=g && \text{on } \Gamma \end{align*} \right. $$

• $\Omega=\left\{ x_{2}>0 \right\}$
• $\Gamma=\left\{x_{2}=0 \right\}$

Then in $(3)$, $\mathbf{b}=(1,\ 1)$, $c=-z^{2}$. Thus, the characteristic equation is as follows.

$$ \left\{ \begin{align*} \dot{x}^{1} &=1, \dot{x}^{2}=1 \\ \dot{z} &= z^{2} \end{align*} \right. $$

Since these are simple ordinary differential equations, they can be solved as follows.

$$ \left\{ \begin{align*} x^{1}(s) &= x^{0}+s, x^{2}(s)=s \\ z(s)&=\frac{z^{0}}{1-sz^{0}}=\frac{g(x^{0})}{1-sg(x^{0})} \end{align*} \right. $$ $x^{0}$ is a constant set to pass through the $x_{2}-$axis $(\Gamma)$ when $s=0$. Now, fix the point $(x_{1},\ x_{2}) \in \Omega$.

$$ (x_{1}, x_{2})=(x^{1}(s), x^{2}(s))=(x^{0}+s, s) $$

Then, for $s>0, x^{0} \in \mathbb{R}$, we obtain the following. $$ s=x_{2}, \quad x^{0}=x_{1}-x_{2} $$

Therefore, the solution of the equation is as follows.

$$ \begin{align*} u(x) &= u(x^{1}(s),\ x^{2}(s)) \\ &= z(s) \\ &= \frac{g(x^{0})}{1-sg(x^{0})} \\ &= \frac{g(x_{1}-x_{2})}{1-x_{2}g(x_{1}-x_{2})} \end{align*} $$

Of course, this is only valid when $1-x^{2}g(x_{1}-x_{2})\ne 0$.

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Fully Nonlinear

Suppose the following differential equation is given.

$$ \begin{align*} u_{x_{1}}u_{x_{2}} &= u && \text{in } \Omega \\ u &= x_{2}^{2} && \text{on } \Gamma \end{align*} $$

• $\Omega=\left\{ x_{1}>0 \right\}$
• $\Gamma=\left\{x_{1}=0 \right\}$

If we set the variables of $F$ as $P, z, x$, it would be as follows.

$$ F(p, z, x)=p_{1}p_{2}-z $$

Thus, the characteristic equation is as follows.

$$ \begin{align*} \dot{p}^{1} &= p^{1},\quad \dot{p}^{2}=p^{2} \\ \dot{z} &= 2p^{1}p^{2} \\ \dot{x}^{1} &= p^{2},\quad \dot{x}^{2}=p^{1} \end{align*} $$

First, if we solve the differential equation for $p$, it would be as follows.

$$ p^{1}(s)=p_{1}^{0}e^s,\ \ p^{2}(s)=p_{2}^{0}e^s $$

Here, $p_{1}^{0}=p(0)$, $p_{2}^{0}=p(0)$. Then, since $\dot{z}(s)=2p_{1}^{0}p_{2}^{0}e^{2s}$, $z$ is as follows.

$$ z(s)=p_{1}^{0}p_{2}^{0}e^{2s}+C $$

Since $z(0)=z^{0}=p_{1}^{0}p_{2}^{0}+C$, $C=z^{0}-p_{1}^{0}p_{2}^{0}$. Therefore, it is as follows.

$$ z(s)=z^{0}+p_{1}^{0}p_{2}^{0}(e^{2s}-1) $$

Similarly, if we calculate $x^{1}$ and $x^{2}$, it would be as follows.

$$ \begin{equation} \left\{ \begin{aligned} p^{1}(s) &= p_{1}^{0}e^s \\ p^{2}(s) &= p_{2}^{0}e^s \\ z(s) &= z^{0}+p_{1}^{0}p_{2}^{0}(e^{2s}-1) \\ x^{1}(s) &= p_{2}^{0}(e^s-1) \\ x^{2}(s) &= x^{0}+p_{1}^{0}(e^s-1) \end{aligned} \right. \label{eq4} \end{equation} $$

Here, $x^{0}$ is a constant set to pass through the $x_{1}-$axis $(\Gamma)$ when $s=0$. Since $u_{x_{2}}=p^{2}$ and by the boundary condition $u=x_{2}^{2}$ when $x_{1}=0$ ($s=0$), $p_{2}^{0}=u(0,\ x^{0})=2x^{0}$. Also, as the given differential equation is $u_{x_{1}}u_{x_{2}}=u$, $p_{1}^{0}p_{2}^{0}=z^{0}=(x^{0})^{2}$ and $p_{1}^{0}=\frac{x^{0}}{2}$. If we substitute all these into $(4)$, we obtain the following.

$$ \left\{ \begin{align*} p^{1}(s) &= \frac{x^{0}}{2}e^s \\ p^{2}(s) &= 2x^{0}e^s \\ z(s) &= (x^{0})^{2}e^{2s} \\ x^{1}(s) &= 2x^{0}(e^s-1) \\ x^{2}(s) &= x^{0}+\frac{x^{0}}{2}(e^s-1) \end{align*} \right. $$

Now, fix the point $(x_{1}, x_{2})\in \Omega$.

$$ (x_{1}, x_{2})=(x^{1}(s), x^{2}(s))=\left( 2x^{0}(e^s -1), \frac{x^{0}}{2}(e^s+1) \right) $$

Then, for $s,\ x^{0}$, we obtain the following.

$$ x^{0}=\frac{4x_{2}-x_{1}}{4},\ \ e^s=\frac{x_{1}+4x_{2}}{4x_{2}-x_{1}} $$

Therefore, the solution of the equation is as follows.

$$ u(x)=u(x^{1}(s),\ x^{2}(s))=z(s)=(x^{0})^{2}e^{2s}=\dfrac{(x_{1}+4x_{2})^{2}}{16} $$

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