Hard Partial Differential Equations - Characteristics

Question

$$ (x-y)\frac{\partial u}{\partial x}+(x+y)\frac{\partial u}{\partial y}=\alpha u $$ where α is a constant, with initial condition $u(x, 0) = x^2$ , $x > 0$. How do I solve this partial differential equation? I have tried to use the method of characteristics and parameterize $x$ and $y$ but I can't find a way to express $y$ in terms of $x$! Help please!

Answer 1

As @Cesareo already noted, the PDE is indeed linear. And you don't need necessarily to express $y$ as a function of $x$ to solve your PDE.

First of all, observe that your PDE is defined in the whole plane $\mathbb{R}^2$, that the coefficients $x-y$ and $x+y$ of $u_x$ and $u_y$ are smooth and vanish simultaneously in $(0,0)$ (i.e., $(0,0)$ is a singular point for your equation). Therefore your PDE has smooth solutions in $\mathbb{R}^2 \setminus \{(0,0)\}$.

In order to find the solution $u$, you need to put your PDE in implicit form:

$$\underbrace{(x-y) u_x + (x+y)u_y - \alpha u}_{=:F(x,y,u,u_x,u_y)} = 0\; ,$$

or $F(x,y,u,p,q)=0$, in which the socalled Monge notation $p=u_x,\ q=u_y$ is used.

Since $F$ is linear, the characteristic system you need to solve reduces to the following linear system of ODEs:

$$\begin{cases} x^\prime (s) = F_p(x(s),y(s),u(s),p(s),q(s)) \\ y^\prime (s) = F_q(x(s),y(s),u(s),p(s),q(s)) \\ u^\prime (s) = p(s)\ F_p(x(s),y(s),u(s),p(s),q(s)) +q(s)\ F_q(x(s),y(s),u(s),p(s),q(s)) \end{cases}$$

which for $F(x,y,u,p,q)=(x-y)p + (x+y)q - \alpha u$ rewrites:

$$\tag{CS} \begin{cases} x^\prime (s) = x(s) - y(s) \\ y^\prime (s) = x(s) + y(s) \\ u^\prime (s) = \alpha u(s) \end{cases}$$

(in the last equation I used the PDE $(x-y)p + (x+y)q=\alpha u$); moreover, you can couple to (CS) initial conditions:

$$\begin{cases} x(0) = x_0\\ y(0) = y_0 \\ u(0) = u_0\end{cases}\; ,$$

with $(x_0,y_0) \in \Gamma$ ($\Gamma \subset \mathbb{R}^2 \setminus \{(0,0)\}$ being a suitable smooth boundary curve) and $u_0=u(x_0,y_0) \in \mathbb{R}$, in order to obtain a Cauchy problem.

It's easily seen that $u(s) = u_0 e^{\alpha s}$ solves the subproblem:

$$\tag{CS-u}\begin{cases} u^\prime (s) = \alpha u(s) \\ u(0) = u_0\end{cases}\; ,$$

hence you only have to solve:

$$\tag{CS-(x,y)}\begin{cases} x^\prime (s) = x(s) - y(s) \\ y^\prime (s) = x(s) + y(s) \\ x(0) = x_0 \\ y(0) = y_0 \end{cases}\; .$$

Multiplying the first equation by $x$, the second by $y$ and subtracting you got:

$$\begin{cases}\frac{\text{d}}{\text{d} s} [x^2 (s) + y^2 (s)] = 2 (x^2(s) + y^2(s)) \\ x^2(0) + y^2(0) = x_0^2 + y_0^2 \end{cases}$$

hence:

$$x^2(s) + y^2(s) = (x_0^2 + y_0^2) e^{2s}\; ,$$

which simplifies into:

$$\tag{1} x^2(s) + y^2 (s) = e^{2s}$$

if you choose the unitary circle $x^2 + y^2 = 1$ as boundary curve $\Gamma$.

From (1) you get:

$$s= \frac{1}{2}\ \log (x^2 + y^2)$$

which you can plug into $u(s) = u_0 e^{\alpha s}$ in order to obtain the solution $u(x,y)$:

$$u(x,y) = u_0\ (x^2 + y^2)^{\alpha/2}\; .$$

Answer 2

$$ (x-y)\frac{\partial u}{\partial x}+(x+y)\frac{\partial u}{\partial y}=\alpha u $$ Note that it should be easier to solve it in polar coordinates since the PDE would be $\rho\frac{\partial u}{\partial \rho}+\frac{\partial u}{\partial \theta}=\alpha u$. Nevertheless the solving below is carried out in Cartesian coordinates.

The Charpit-Lagrange system of equations is : $$\frac{dx}{x-y}=\frac{dy}{x+y}=\frac{du}{\alpha u}$$ A first characteristic equation comes from $\frac{dx}{x-y}=\frac{dy}{x+y}$ which solving leads to : $$\ln(x^2+y^2)-2 \tan^{-1}(\frac{y}{x})=c_1$$ A second characteristic equation comes from

$\frac{dx}{x-y}=\frac{dy}{x+y}=\frac{xdx+ydy}{x(x-y)+y(x+y)}=\frac{du}{\alpha u}\quad\implies\quad\frac{d(x^2+y^2)}{x^2+y^2}=2\frac{du}{\alpha u}$ $$u(x^2+y^2)^{-\alpha/2}=c_2$$ The solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ is : $\quad u(x^2+y^2)^{-\alpha/2}=F\left(\ln(x^2+y^2)-2 \tan^{-1}(\frac{y}{x})\right)$ .

$F$ is an arbitrary function, to be determined according to some boundary condition (not specified in the wording of the question). $$\boxed{u(x,y)=(x^2+y^2)^{\alpha/2}F\left(\ln(x^2+y^2)-2 \tan^{-1}(\frac{y}{x})\right)}$$

Note that I agree with the particular solution proposed by Pacciu : $u(x,y)=u_0(x^2+y^2)^{\alpha/2}$ belongs to the above general family of solutions. In this particular case $F(X)=u_0$ . But they are an infinity of other solutions. For example, with $F(X)=e^{-\alpha X/2}$ the solution $u(x,y)=\exp\left(\alpha \tan^{-1}(\frac{y}{x}) \right)$ is a solution which satisfies the PDE as well.

Other example : Suppose that the OP specifies the condition $u(x,0)=x^2$ obviously $u(x,y)=u_0(x^2+y^2)^{\alpha/2}$ cannot be a solution of the problem, what ever $u_0$ is. In this particular case the solution is : $u(x,y)=(x^2+y^2)\exp\left(\alpha \tan^{-1}(\frac{y}{x}) \right)$ .

IN ADDITION, after that the condition $u(x,0)=x^2$ was specified by MathematicianP.

$u(x,0)=x^2=(x^2+0^2)^{\alpha/2}F\left(\ln(x^2+0^2)-2 \tan^{-1}(\frac{0}{x})\right) = x^{\alpha}F\left(2\ln(x)\right)$

Let : $\quad X=2\ln(x)\quad;\quad x=e^{X/2}$ .

$x^2=x^{\alpha}F\left(2\ln(x)\right)=e^X=e^{\alpha X/2}F(X) \quad\implies\quad F(X)=\exp\left((1-\frac{\alpha}{2})X\right)$

So, the function $F(X)$ is determined. We put it into the above general solution where $X=\ln(x^2+y^2)-2 \tan^{-1}(\frac{y}{x})$ .

$u(x,y)=(x^2+y^2)^{\alpha/2}\exp\left(\left(1-\frac{\alpha}{2}\right)\left(\ln(x^2+y^2)-2 \tan^{-1}(\frac{y}{x})\right)\right)$

After simplification : $$u(x,y)=(x^2+y^2)\exp\left(\alpha \tan^{-1}(\frac{y}{x}) \right)$$