2

Related Question:

I am doing a PDE problem from a course that I will take next semester. I watched this video in advance. It appears that I am doing something wrong with the initial data.

Consider the following first order PDE

\begin{cases} uu_x+yu_y=x, \\ u(x,1)=2x. \end{cases}

  1. State the condition which guarantees that the initial surface $\Gamma$ is not characteristic.
  2. Use the method of characteristics to find a solution of the PDE and discuss for which $(x,y)\in\mathbb R^2$ the solution exists.

For the second part, the general form of the method of characteristics is $au_x+bu_y=f$. Therefore, we have that

$$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{f}$$ or $$\frac{dx}{u}=\frac{dy}{y}=\frac{du}{x}$$

In order to solve these equations, we need to find the value of two constants $C_1$ and $C_2$. First, let

$$\frac{dx}{u}=\frac{du}{x}$$

Then,

$$ \frac{du}{dx}=\frac{x}{u}~~\Rightarrow~~~ udu = xdx ~~\Rightarrow~~~ \frac{u^2}{2}=\frac{x^2}{2}+C ~~\Rightarrow~~~ C_1=x^2-u^2$$

Next, let

$$\frac{dx}{u}=\frac{dy}{y}$$

Then,

$$\frac{dy}{dx}=\frac{y}{u} ~~\Rightarrow~~~ \frac{1}{y}dy = \frac{1}{u}dx ~~\Rightarrow~~~ \ln(y)=\frac{x}{u}+C ~~\Rightarrow~~~ C_2=y+e^{\frac{x}{u}} $$

So, we have that $C_1=x^2-u^2$ and $C_2=y+e^{\frac{x}{u}}$. We can combine the two constants such that $C_2=F(C_1)$ where $F$ is an arbitrary differentiable function. Then,

$$y+e^{\frac{x}{u}}=F(x^2-u^2)$$

We now need to apply our initial data. We are given that $u(x,1)=2x$. Therefore,

$$1+e^{\frac{1}{2}}=F(-3x^2)$$

This doesn't appear to help us find the solution. If instead we set $C_1=F(C_2)$ then

$$x^2-u^2=F(y+e^{\frac{x}{u}})$$

Applying the initial data produces

$$-3x^2=F(1+e^{\frac{1}{2}})$$

That also appears to be incorrect. I must have made a mistake in applying the initial data. How can we apply the initial data to solve for $u$?

PinkyWay
  • 4,565
Axion004
  • 10,056

1 Answers1

1

Similar to Method of characteristics inhomogeneous nonlinear wave equation:

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dy}{dt}=y$ , letting $y(0)=1$ , we have $y=e^t$

$\begin{cases}\dfrac{dx}{dt}=u\\\dfrac{du}{dt}=x\end{cases}$

$\therefore\dfrac{d^2x}{dt^2}=x$

$x=C_1\sinh t+C_2\cosh t$

$\therefore u=C_1\cosh t+C_2\sinh t$

Hence $\begin{cases}x=C_1\sinh t+C_2\cosh t\\u=C_1\cosh t+C_2\sinh t\end{cases}$

$x(0)=x_0$ , $u(0)=F(x_0)$ :

$\begin{cases}C_1=F(x_0)\\C_2=x_0\end{cases}$

$\therefore\begin{cases}x=F(x_0)\sinh t+x_0\cosh t\\u=F(x_0)\cosh t+x_0\sinh t\end{cases}$

$\therefore\begin{cases}x_0=x\cosh t-u\sinh t=x\cosh\ln y-u\sinh\ln y\\F(x_0)=u\cosh t-x\sinh t=u\cosh\ln y-x\sinh\ln y\end{cases}$

Hence $u\cosh\ln y-x\sinh\ln y=F(x\cosh\ln y-u\sinh\ln y)$

$u(x,1)=2x$ :

$F(x)=2x$

$\therefore u\cosh\ln y-x\sinh\ln y=2x\cosh\ln y-2u\sinh\ln y$

$u(x,y)=\dfrac{x(\sinh\ln y+2\cosh\ln y)}{2\sinh\ln y+\cosh\ln y}$

doraemonpaul
  • 16,178
  • 3
  • 31
  • 75