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I'm asked to use the method of characteristics to derive the general solution \begin{equation*} u = (x^{2}+y^{2})^{\alpha/2}f\left(\frac{x}{\sqrt{x^{2}+y^{2}}}, \frac{y}{\sqrt{x^{2}+y^{2}}}\right) \end{equation*} from $xu_{x}+yu_{y}=\alpha u$. I begin much like the problem described here: Solving a semilinear partial differential equation with

$\frac{\partial x}{\partial t}=x$, letting $x(0)=x_{0}$, we have $x=x_{0}e^{t}$,

$\frac{\partial y}{\partial t}=y$, letting $y(0)=y_{0}$, we have $y=y_{0}e^{t}$,

and

$\frac{\partial z}{\partial t}=\alpha z$, letting $z(0)=z_{0}$, we have $z=z_{0}e^{\alpha t}$.

From here, I'm not sure how to proceed since $z$ does not contain terms $x$ or $y$.
Also, it is unclear to me why $f$ will be a function of two variables rather than one.

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jpb
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1 Answers1

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Approach $1$:

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$

$\dfrac{dy}{dt}=y$ , letting $y(0)=y_0$ , we have $y=y_0e^t=y_0x$

$\dfrac{du}{dt}=\alpha u$ , letting $u(0)=f(y_0)$ , we have $u(x,y)=f(y_0)e^{\alpha t}=f\left(\dfrac{y}{x}\right)x^{\alpha}$

Approach $2$:

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dy}{dt}=y$ , letting $y(0)=1$ , we have $y=e^t$

$\dfrac{dx}{dt}=x$ , letting $x(0)=x_0$ , we have $x=x_0e^t=x_0y$

$\dfrac{du}{dt}=\alpha u$ , letting $u(0)=f(x_0)$ , we have $u(x,y)=f(x_0)e^{\alpha t}=f\left(\dfrac{x}{y}\right)y^{\alpha}$

doraemonpaul
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