Solve this PDE $xu_x+yu_y=u$ with "condition" $x=\cos t, y=\sin t, u=1$ (problem with condition/particular solution)

Question

Find the particular solution for the part differential equation $$xu_x+yu_y=u \\ x=\cos t, y=\sin t, u=1$$

It's the first time I've encountered this PDE with this type "condition" (and I don't know how to solve them)

If I solve it in the "normal" way: $$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{u}$$ $$\ln|C_1 \cdot x|=\ln|y|$$ $$C_1=\frac{y}{x}$$ And: $$\int \frac{dx}{x}=\int \frac{du}{u}$$ $$\ln|C_2 \cdot x|=\ln|u|$$ $$C_2=\frac{u}{x}$$ Because: $$C_2=F(C_1)$$ General solution: $$u(x,y)=xF\left(\frac{y}{x}\right)$$

And what next? How do I set the condition correctly?

$$u(x,y)=xF\left(\frac{y}{x}\right) \Rightarrow 1=\cos t \cdot F\left(\frac{\sin t}{\cos t}\right)$$ So: $$F\left(\frac{\sin t}{\cos t}\right)=\frac{1}{\cos t}$$

What is the particular solution? How to correctly use the conditions given in the task? Or maybe it needs to be done before solving the general equations? -> How?

Please give me hint/help.

Answer 1

You have to write $\frac{1}{\cos t}=\sec t$ as a function of $\frac{\sin t}{\cos t}=\tan t$. That's not difficult: $$ \sec t=\text{sgn}(\sec t)\sqrt{\sec^2 t} =\text{sgn}(\sec t)\sqrt{1+\tan^2 t}. \tag{1} $$ It follows from $(1)$ that $$ F(\tan t)=\text{sgn}(\sec t)\sqrt{1+\tan^2 t} \implies F\!\left(\frac{y}{x}\right) =\text{sgn}(x)\sqrt{1+\frac{y^2}{x^2}}, \tag{2} $$ hence $$ u(x,y)=xF\!\left(\frac{y}{x}\right)=x\,\text{sgn}(x)\sqrt{1+\frac{y^2}{x^2}} =|x|\sqrt{1+\frac{y^2}{x^2}}=\sqrt{x^2+y^2}. \tag{3} $$

Answer 2

Another procedure is to explicitly parametrize the characteristics. This yields the system of ODEs

$$\frac{dx}{ds}=x(s),\quad \frac{dy}{ds}=y(s),\quad \frac{du}{ds}=u(x(s),y(s))$$ as the characteristic equations parametrized by $s$ (since $t$ is taken). If we start with a point on the unit circle, i.e., $$x(0)=\cos t,\quad y(0)=\sin t,\quad u(x(0),y(0))=1$$ then the solutions are immediately $$x(s)=e^s \cos t,\quad y(s)=e^s \sin t,\quad u(x(s),y(s))=e^s.$$ Since $$x(s)^2+y(s)^2=e^{2s}\implies e^s=\sqrt{x(s)^2+y(s)^2},$$ we conclude $$u(x(s),y(s))=\sqrt{x(s)^2+y(s)^2}$$ and thus $u(x,y)=\sqrt{x^2+y^2}$ is the solution.