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I'm trying to solve this PDE using method of characteristic without coordinate transformation :

$xu_x+yu_y = \sqrt{x^2+y^2}$ for $x^2+y^2>1$, $u(x,y)=x$ on $x^2+y^2 = 1$.

I only know how to use the method of characteristic for this kind of PDE such that initial condition is constant in one variable e.g., $u(0,y) = $ or $u(x,1) = $. I tried to mimick the technique to solve the above pde : Introducing new variable $(\xi,\eta)$ such that $\xi = x,\eta = y$ on $x^2+y^2 = 1$ and using chain rule, $${\partial u\over\partial \eta} = {\partial u\over\partial x}{\partial x\over\partial\eta}+{\partial u\over\partial y}{\partial y\over\partial\eta} = \sqrt{x^2+y^2}$$ so ${\partial x\over\partial\eta}= x,{\partial y\over\partial \eta} = y$. So $x = \phi_1(\xi)e^{\eta}$ and $y = \phi_2(\xi)e^{\eta}$ ..? I'm stuck here and I'm not sure I'm in the right direction. Please help.

one potato two potato
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2 Answers2

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The characteristic curves are $\frac{\mathrm{d}(x,y)}{\mathrm{d}t}=(x,y)$. That is, $$ (x,y)=(x_0,y_0)\,e^t\tag1 $$ Solving for $x_0^2+y_0^2=1$ (where the initial data is given) gives $$ \begin{align} (x_0,y_0)&=\frac{(x,y)}{\sqrt{x^2+y^2}}\tag{2a}\\ t&=\frac12\log\left(x^2+y^2\right)\tag{2b} \end{align} $$ Along those characteristic curves, we have $$ \begin{align} \frac{\mathrm{d}u}{\mathrm{d}t} &=\frac{\partial u}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t} +\frac{\partial u}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}\tag{3a}\\ &=xu_x+yu_y\tag{3b}\\[4pt] &=\textstyle\sqrt{x^2+y^2}\tag{3c}\\[4pt] &=|(x_0,y_0)|\,e^t\tag{3d} \end{align} $$ Explanation:
$\text{(3a)}$: chain rule
$\text{(3b)}$: apply the characteristic equations
$\text{(3c)}$: apply the given differential equation
$\text{(3d)}$: apply $(1)$

Thus, the solution along a characteristic is $$ \begin{align} u(x,y) &=|(x_0,y_0)|\,e^t+c(x_0,y_0)\tag{4a}\\[12pt] &=e^t+x_0-1\tag{4b}\\[9pt] &={\textstyle\sqrt{x^2+y^2}}+\frac{x}{\sqrt{x^2+y^2}}-1\tag{4c}\\ &=\frac{x^2+y^2+x}{\sqrt{x^2+y^2}}-1\tag{4d} \end{align} $$ Explanation:
$\text{(4a)}$: integrate $\text{(3d)}$
$\text{(4b)}$: $|(x_0,y_0)|=1$ and when $t=0$, $u(x_0,y_0)=x_0$
$\text{(4c)}$: solve for $t$ and $x_0$
$\text{(4d)}$: simplify

robjohn
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$$xu_x+yu_y = \sqrt{x^2+y^2}$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{\sqrt{x^2+y^2}}$$ A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{y}$ : $$y=c_1x$$ A second characteristic equation comes from solving $\frac{dx}{x}=\frac{du}{\sqrt{x^2+(c_1x)^2}}$ $$u-x\sqrt{1+c_1^2}=c_2$$ General solution of the PDE on the form of implicit equation $c_2=F(c_1)$ : $$u-x\sqrt{1+\left(\frac{y}{x}\right)^2}=F\left(\frac{y}{x}\right)$$ $F$ is an arbitrary function. $$\boxed{u(x,y)=\sqrt{x^2+y^2}+F\left(\frac{y}{x}\right)}$$ Condition : $u(x,y)=x$ on $x^2+y^2 = 1$ $$\sqrt{x^2+(1-x^2)}+F\left(\frac{\sqrt{1-x^2}}{x}\right)=x$$ $$F\left(\frac{\sqrt{1-x^2}}{x}\right)=x-1$$ Let $X=\frac{\sqrt{1-x^2}}{x}\quad\implies\quad x=\frac{1}{1+X^2}$ $$F(X)=\frac{1}{\sqrt{1+X^2}}-1$$ Now the function $F(X)$ is determined. We put it into the above general solution where $X=\frac{y}{x}$ $$u(x,y)=\sqrt{x^2+y^2}+\frac{1}{\sqrt{1+\left(\frac{y}{x}\right)^2}}-1$$ $$\boxed{u(x,y)=\frac{x^2+y^2+x}{\sqrt{x^2+y^2}}-1}$$

JJacquelin
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