How is the Method of Characteristics used to solve a differential equation?

[BrainHurts]

Homework Statement

Use the method of characteristics to solve the problem:

$-xu_x + yu_y = 2xyu$ and $u(x,x)=x$

Homework Equations

The Attempt at a Solution

let x=x(t), y=y(t), u=u(x(t),y(t))

so

$\frac{du}{dt} = \frac{∂u}{∂x} \frac{dx}{dt} + \frac{∂u}{∂y} \frac{dy}{dt}$

and

$\frac{dx}{dt} = -x$, $\frac{dy}{dt}=y$, $\frac{du}{dt}=2xyu$

so solving these differential equations I get

$x=e^-t$ , $y=e^t$, and $u= Ae^{2t}$

so do are the solutions just the following?

x(t) = e^-t
y(t) = e^t

and not sure what to do with u.

 

[christoff]

To start, you haven't used the initial condition. Remember when you solve the characteristic equations (eg. the $dx/dt=-x$), these need initial conditions. In fact, the way you've written out your solutions for x and y, you seem to be assuming $x(0)=1=y(0)$, which is not what you have.

Do you know how to convert the initial condition $u(x,x)=x$ into a set of initial conditions for your characteristics? If you don't, let me know and I'll explain. If you do, then do it and solve the characteristic equations again. Let me know how it goes.

 

[BrainHurts]

Oh could you please explain that part on the initial condition? I don't know what you mean by converting the initial condition to a set of initial conditions for my characteristics.

 

[christoff]

Without going into a full explanation on why the method works, what you typically do is: if your initial condition is written in the form $u(x,x)=g(x)$, then your initial conditions for your characteristics become $$x(0)=x_0\\<br /> y(0)=x_0\\<br /> u(0)=g(x_0).$$ Sometimes the initial condition might look different. For example, if you had the initial condition $u(5,y)=g(y)$, then your initial condition would be$$x(0)=5\\<br /> y(0)=y_0\\<br /> u(0)=g(y_0).$$ Essentially, you parameterize the initial curve. In the case where you have $n$ independent variables, so $u=u(x_1,\cdots,x_n)$, you would parameterize the initial curve using at most $n-1$ parameters. Here, we had n=2, and I denoted the parameter by $x_0$. We had the initial curve $u(x,x)=x$, so the "x" coordinate is represented by $x_0$, the "y" coordinate is also represented by $x_0$, as is the "u" coordinate. This curve is actually just a line in $\mathbb{R}^3$; we can write it as $\{(x,y,u)\in\mathbb{R}^3 : x=y=u\}$.

For example, we could look at an initial curve in a system with three independent variables, but now the curve would (in general) be represented with up to two parameters. eg: if we had $u(x,x,z)=3z-\log(x^2)$, the we could parameterize this by $x_0,z_0$ and we would get$$x(0)=x_0\\<br /> y(0)=x_0\\<br /> z(0)=z_0\\<br /> u(0)=3z_0-\log(x_0^2).$$ Do you understand the procedure?

If you want to understand how/why this works, then any decent PDE textbook with an emphasis on explicitly solving PDEs should cover this. There are also numerous resources online.

 

[BrainHurts]

Ok first I want to thank you Christoff for helping me out

But just to make sure if I understand the procedure let's see

$-xu_x + yu_y = 2xyu$ and $u(x,x)=x$

so x=x(t), y=y(t), u=u(x(t), y(t))

so $\frac{du}{dt} = \frac{∂u}{∂x} \frac{dx}{dt} + \frac{∂u}{∂y} \frac{dy}{dt}$

and we have the given 3 diff eqs.

$\frac{dx}{dt}=-x$ (1)

$\frac{dy}{dt}=y$ (2)

$\frac{du}{dt}=2xyu$ (3)

solving (1) and applying initial conditions I get

$x(t) = x_oe^{-t}$

Similarly $y(t) = x_oe^t$

and lastly solving 3

$\frac{du(t)}{dt} = 2xyu$ plugging in solutions from (1) and (2) we get

$\frac{du}{u} = 2xydt \Rightarrow \frac{du}{u} = 2(x_oe^{-t})(x_oe^t)dt = 2x^{2}_{o}dt$

and applying initial conditions we have that $u(t) = x_oe^{2x^{2}_{o}t}$

 

[christoff]

Yes, you're on the right track. Now, you have obtained $u(t)=x_0e^{2x_0^2t}$. This solution is not explicitly a function of $x$ and $y$; it is a function of the parameter $x_0$ and the variable $t$. In the end, we want to get rid of the $t$'s and $x_0$'s and to somehow re-introduce the variables $x,y$.

We know that $x=x_0e^{-t}$, and so $x_0=xe^t$. Also, $y=x_0e^t$, so that $x_0=ye^{-t}$. Using these two together, we can obtain a very nice expression for $x_0^2$ that doesn't introduce any more $t$'s into our solution; namely$$x_0^2=x_0 x_0=xe^t ye^{-t} = xy.$$ In this way, we can get rid of the parameter $x_0$ from our expression for $u$, and we arrive at$$u(t;x,y)=x_0e^{2x_0^2t}=xe^te^{2xyt}=xe^t(e^t)^{2xy}.$$ Now, we need to get rid of the $t$, but without bringing the parameter $x_0$ back in. Can you manipulate the equations $x=x_0e^{-t}$ and $y=x_0e^t$ to obtain $t=t(x,y)$ which is independent of the parameter $x_0$? Equivalently, write $e^t=G(x,y)$ for some function $G$ which doesn't depend on $x_0$.

 

[BrainHurts]

ok after all is said and done, I hope the algebra is right, but going off what you said, I have that

$u(t; x,y) = x\sqrt{\frac{y}{x}}(\frac{y}{x})^{xy}$

just to double check I took $\frac{y}{x}=e^{2t}$ and it follows that $e^{t}=\sqrt{\frac{y}{x}}$

so plugging that all in I got u, and I think that's the answer.

 

[christoff]

That's exactly what I got in the end. I plugged that function into Maple and checked the partial derivatives; it is definitely the solution. One final thing of course: since your solution doesn't depend on $t$, you can drop the dependence on it and safely write $$u(x,y)=x\sqrt{\frac{y}{x}}\left(\frac{y}{x}\right)^{xy}.$$

 

[BrainHurts]

Thanks so much! They need to add like a "buy this person a beer" tab somewhere on this website! I would have bought quite a few :).