2
$\begingroup$

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 $u$?

$\endgroup$
1
$\begingroup$

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

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

$F(x)=2x$

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

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

$\endgroup$
0
$\begingroup$

For the first question, on $\Gamma$ it is specified that $u(x,y)=2x$. Therefore, $u_x=2$ and $u_y=0$. Putting them into the PDE leads to $uu_x+yu_y=4x$ which is contradictory with $uu_x+yu_y=x$. So, the boundary condition is not on a characteristic.

For the second question, we should follow the method of characteristics. Lets first write the general form as $au_x+bu_y=f$. 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,

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

Next, observe that $\frac{a}{b}=\frac{c}{d} \iff \frac{a+c}{b+d}=\frac{c}{d}$. So,

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

can be written as

$$\frac{dx+du}{u+x}=\frac{dy}{y} ~~\Rightarrow~~~ \ln(x+u) = \ln(y)+C ~~\Rightarrow~~~ \ln\Big(\frac{x+u}{y}\Big)=C ~~\Rightarrow~~~ C_2=\frac{x+u}{y} $$

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

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

or

$$u=-x+yF(u^2-x^2)$$

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

$$2x=-x+f((2x)^2-x^2) ~~\Rightarrow~~~ 3x=F(3x^2)$$

Letting $w=3x^2$, we see that

$$ x^2=\frac{w}{3} ~~\Rightarrow~~~ x=\pm\sqrt{\frac{w}{3}}$$

So, $F(w)=\pm3\sqrt{\frac{w}{3}}$. Therefore,

$$u=-x+yF(u^2-x^2) = -x+y\pm3\sqrt{\frac{u^2-x^2}{3}} = -x+y\pm\sqrt{3(u^2-x^2)}$$

Hence,

$$u(x,y)=-x+y\pm\sqrt{3(u^2-x^2)}$$ $$u_x=-1\pm\frac{\sqrt{3}{x}}{\sqrt{u^2-x^2}}\quad\text{and}\quad u_y=1$$

Thus,

$$uu_x+yu_y=u\Big(-1\pm\frac{\sqrt{3}{x}}{\sqrt{u^2-x^2}})+y=...=x$$

No, it doesn't. I must have made a mistake somewhere.

$\endgroup$
  • $\begingroup$ It appears that a mistake is made after setting $3x=F(3x^2)$. This produces $u=-x+yF(u^2-x^2) = -x+y\pm3\sqrt{\frac{u^2-x^2}{3}} = -x+y\pm\sqrt{3(u^2-x^2)}$, which is not a function of x and y. $\endgroup$ – Axion004 Jan 12 '19 at 19:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.