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I have problems solving this simple pde. I don't know what I'm doing wrong:

$$yu_x - u_y = y, \ \ \ u(x,0 ) = \frac{1}{x}$$

Here is how I do it. The equation is equivalent to this one:

$$\frac{dx}{y} = \frac{dy}{-1} = \frac{du}{y}$$

So $\frac{dx}{y} = - dy, \ \ dx=du$

This means that $dx = -ydy$ and $u = x+ C$

So $x + C' = - \frac{y^2}{2} $ and $u = x+C$

This above gives us an integral curve $F(x,y) = - x - \frac{y^2}{2}$ of $yu_x - u_y=0$, but not of the main pde.

If we plug $F(x,0) = -x$, we get $u(x,0) = - \frac{1}{F(x,0)}$, so $u(x,y) = - \frac{1}{F(x,y)} = \frac{1}{x+ \frac{y^2}{2}}$, which is not the solution.

Could you help me fix it?

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Use method of characteristics.

First, reparameterise your curve letting

$$x \to x(s) \\ y \to y(s)$$

Hence you have

$$u = u(x(s), y(s))$$

Taking the total derivative wrt to $s$

$$\begin{align} \implies \frac{d}{ds} u &= \frac{\partial u}{\partial x} \frac{dx}{ds} + \frac{\partial u}{\partial y} \frac{dy}{ds} \\ &= \frac{\partial u}{\partial x} \cdot y + \frac{\partial u}{\partial y} \cdot (-1) \\ &= y \\ \end{align}$$

Equating, we find

$$\begin{align} \frac{dy}{ds} &= -1 \\ \implies dy &= -ds \ \ \ \ \ (1) \\ \frac{dx}{ds} &= y \\ \implies \frac{dx}{-dy} &= y \ \ \ \ \ \ \ \ \ \ (2) \\ \frac{du}{ds} &= y \\ \implies \frac{du}{-dy} &= y \ \ \ \ \ \ \ \ \ \ (3) \\ \end{align}$$

Solving

$$\begin{align} (2) \implies x(s) &= \frac{-y^{2}}{2} + x_{0} \\ (3) \implies u(x, y) &= \frac{-y^{2}}{2} + f(x_0) \\ &= \frac{-y^{2}}{2} + f \bigg(x + \frac{y^{2}}{2} \bigg) \ \ \ \ \ \ (4) \\ \end{align}$$

Using our initial condition

$$\begin{align} u(x,0) &= f(x) \\ &= \frac{1}{x} \\ \implies f \bigg(x + \frac{y^{2}}{2} \bigg) &= \frac{1}{x + \frac{y^{2}}{2}} \\ \end{align}$$

Substituting into $(4)$, we find

$$u(x, y) = \frac{-y^{2}}{2} + \frac{1}{x + \frac{y^{2}}{2}}$$

You can check by differentiation that this satisfies the PDE and initial condition.

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  • $\begingroup$ If you need anymore help, just post below this comment. Also, if you could accept and upvote this answer, that would be great. $\endgroup$ – mattos Feb 10 '15 at 13:30
  • $\begingroup$ No, everything is clear. Thank you :) $\endgroup$ – Bilbo Feb 10 '15 at 13:38

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