Hard Partial Differential Equations - Characteristics

$$(x-y)\frac{\partial u}{\partial x}+(x+y)\frac{\partial u}{\partial y}=\alpha u$$ 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, nd have parameterized $$x$$ and $$y$$ but I can't find a way to express $$y$$ in terms of $$x$$! Help please!

• This PDE is linear! – Cesareo May 12 at 18:07

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 $$u(x,y)$$:

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

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

• I haven't heard about this Charpit Euqations in my class, I have edited the question to include the initial conditions, is there any other way to solve this equation without the charpit equations? – MathematicianP May 14 at 16:52
• with condition $u(x,0)=x^2$ the solution is : $$u(x,y)=(x^2+y^2)\exp\left(\alpha \tan^{-1}(\frac{y}{x}) \right)$$ – JJacquelin May 14 at 17:18
• The system of equations $$\frac{dx}{x-y}=\frac{dy}{x+y}=\frac{du}{\alpha u}=ds$$ is an other manner to write $$\begin{cases}\frac{dx}{ds}=x-y \\ \frac{dy}{ds}=x+y\\ \frac{du}{ds}=\alpha u \end{cases}$$ – JJacquelin May 14 at 17:26