3

I have a PDE:

$$\frac{1}{x}u_x+\frac{1}{y}u_y=0$$ with the boudaries

$$a)u(0,y)=y, b)u(1,1)=1$$

Using Lagrange Chapite we get the characteristics: $$ln|y|=ln|x|+C \rightarrow C=\frac{y}{x}$$

For the boundary b) it seems like i can just use the characteristics $$u(x,y)=\frac{y}{x}$$

for the boundary a) that's where i seem to hav e a problem because the solution is in the form:

$$u(x,y)=f(\frac{y}{x}) \rightarrow u(0,y)=f(\frac{y}{0})$$

The only way i can avoid it is put $x$ in front of it, i.e. $$u(x,y)=y$$

But that seems like i'm cheating. Any advice?

Jerry Holmes
  • 159

1 Answers1

3

Remark: Notice that if $u(x,y)=y/x$, then $\frac{1}{x}u_{x}+\frac{1}{y}u_{y}\not\equiv 0$ and if $u(x,y)=y$, then $\frac{1}{x}u_{x}+\frac{1}{y}u_{y}\not\equiv 0$.

Since \begin{align*}\frac{1}{x}u_{x}+\frac{1}{y}u_{y}&=0,\\u(0,y)=y,\quad u(1,1)=1 \end{align*}

General solution:

The characteristics equations are given by $$\underbrace{\frac{dx}{dt}=\frac{1}{x}}_{(1)},\quad \underbrace{\frac{dy}{dt}=\frac{1}{y}}_{(2)},\quad \underbrace{\frac{du}{dt}=0}_{(3)}$$

  • Using $(1)$ and $(2)$, we have $$\frac{dx}{1/x}=\frac{dy}{1/y}\implies \boxed{y^{2}-x^{2}=c_{1}}$$
  • Using $(3)$ and $(1)$ we have $$0\cdot x\, dx=\, du \implies du=0\implies \boxed{u=c_{2}}$$

Thus, the solution of the equation satisfies $$F(u,y^{2}-x^{2})=0\iff u(x,y)=G(y^{2}-x^{2}) $$ for some arbitrary function $F$ or $G$.

Particular solution:

  • If $u=y$ when $x=0$, then $y=G(y^{2})$. Setting $t=y^{2}\iff y=\pm \sqrt{t}$. Then $G(t)=\pm \sqrt{t}$. Therefore since $u(x,y)=G(y^{2}-x^{2})=\pm \sqrt{y^{2}-x^{2}}$. Therefore, $u(x,y)=\pm (y^{2}-x^{2})^{1/2}$.

  • If $u=1$ when $(x,y)=(1,1)$, then $1=G(0)$, then in particular $u(x,y)=y^{2}-x^{2}+1$ works.

A. P.
  • 5,978