# Solve PDE using method of characteristics

The following problem I find challeging. Can you help me find a solution? The question is as follows:

Determine the solution (in explicit form), using the method of characteristics, of the following initial-value problem for $u=u(x,y)$: $$\left\{ \begin{array}{l l} u_xu_y=xy \\ u(x,y)=x \quad for \quad y=0 \end{array} \right.$$

Cheers

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I don't know how to use typically with the method in en.wikipedia.org/wiki/…. –  doraemonpaul Jan 20 '13 at 10:18
There's a book called Partial Differential Equations by Bhamra on google books (which you can buy in e-book form from the PHI site) that is full of theory & solved examples. On page 105 there is an example virtually identical to your question (viewable online) so I'd recommend getting this book. Let me know if you do, this is probably the best book I've seen on the subject, even though it did lead to all the questions in my post that you commented on, & it would be good to talk a little about the book & the subject in general. –  sponsoredwalk Jan 23 '13 at 22:47

As doraemonpaul remarked, the method of characteristics is given on this Wikipedia page for fully nonlinear equations. We let $x_1 = x, x_2 = y, p_1 = u_x, p_2 = u_y$, then

$$F(x_1,x_2, u, p_1, p_2) = p_1p_2 - x_1 x_2$$

the Lagrange-Charpit equations simplify to

$$\frac{\dot{x_1}}{p_2} = \frac{\dot{x_2}}{p_1} = \frac{\dot{p_1}}{x_2} = \frac{\dot{p_2}}{x_1} = \frac{\dot{u}}{2p_1 p_2}$$

Cross-multiplying we get

$$\frac{d}{ds}(x^2) = \frac{d}{ds}(u_y^2) \qquad \frac{d}{ds}(y^2) = \frac{d}{ds}(u_x^2)$$

Now let $(x(s),y(s))$ be the integral curve starting at $(x_0,0)$. Note that $u_x(x_0,0) = 1$ and $u_y(x_0,0) = 0$ by the initial condition and the equation. We integrate along the curve to get

$$[u_y(x(s),y(s))]^2 = x(s)^2 - x_0^2 \qquad [u_x(x(s),y(s))]^2 = 1 + y(s)^2$$

So using the equation again we have that $$x^2 y^2 = (x^2 - x_0^2)(1 + y^2) \implies 0 = x^2 - x_0^2(1+ y^2)$$

Solving for $x_0$ and replacing we have that

$$u_y^2 = \frac{x^2 y^2}{1+y^2} \qquad u_x^2 = 1 + y^2$$

and by the equation we have finally that either

$$u_x = \sqrt{1+y^2} \qquad u_y = \frac{xy}{\sqrt{1+y^2}}$$

or

$$u_x = - \sqrt{1+y^2} \qquad u_y = - \frac{xy}{\sqrt{1+y^2}}$$

The latter is inadmissible by the boundary condition. Integrating in $y$ and using the initial value we have that

$$u = x\sqrt{1+y^2}$$

is the final solution.

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I have an extended question math.stackexchange.com/questions/345321, can you help me? –  doraemonpaul Apr 8 '13 at 4:44

I have no idea about how to solve this nonlinear PDE directly by using the method of characteristics. I only know about the following facts about this nonlinear PDE:

Even http://eqworld.ipmnet.ru/en/solutions/fpde/fpdetoc3.htm have not mentioned about this type of PDE.

Let $\begin{cases}p=\dfrac{x^2}{2}\\q=\dfrac{y^2}{2}\end{cases}$ ,

Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=x\dfrac{\partial u}{\partial p}$

$\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial y}=y\dfrac{\partial u}{\partial q}$

$\therefore xu_pyu_q=xy$

$u_pu_q=1$

This belongs to a PDE of the type http://eqworld.ipmnet.ru/en/solutions/fpde/fpde3308.pdf. But even this website is not quite useful as it cannot know its general solution.

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