# Partial Differential Equations

1. Sep 23, 2012

### Tsunoyukami

I am working on some problems for an assignment in my PDEs class and find myself either not understanding what I am supposed to do or being unsure of my answers or the next step.

I am going to outline my understanding of the problems, provide my attempt at a solution and highlight where I am having difficultly. Please correct me anywhere that I am wrong and offer as much assistance and helpful prods in the right direction as you can. Thank you very much in advance!

Constant Coefficient PDEs
The general solution of a partial differential equation of the form $au_{x} + bu_{y} = 0$ where a and b are constants is of the form $u(x,y) = f(bx -ay)$ where f is any arbitrary function of one variable. You can not find a specific solution to this PDE unless provided with further information, such as an intitial condition (ie. when x or y is equal to 0).

My homework question:

"Consider the equation $u_{x} - 3u_{y} = 0$ in ${x>0, y>0}$ with the initial condition at $x = 0, u = y, y >0$. Where is this solution defined? Is it defined everywhere in ${x>0,y>0}$ or do we need to impose an initial condition at y = 0? If we need to impose such a condition use at $y = 0, u = x (x>0)$ and solve this IVBP."

Given $u_{x} - 3u_{y} = 0$ we find the general solution $u(x,y) = f(-3x -y)$.
Using the initial condition at $x = 0, u = y, y >0$:

$u(x,y) = u(0,y) = f(-y) = y$

Let $w = -y \rightarrow y = -w$; then:

$f(w) = -w$

$u(x,y) = -(-3x-y)$

$u(x,y)= 3x + y$

Now, if I understand the question, I must determine if $u(x,y)= 3x + y$ is a valid solution of $u_{x} - 3u_{y} = 0$ for all x, y such that x>0 and y>0, correct? I'm not sure how to do this exactly. Any help here would be greatly appreciated.

Note: After this question there is a very similar question in which we are asked to do the same thing (given the same initial condition) and we are considering {x<0, y>0} instead.

EDIT:

Variable Coefficient PDEs

My homework question:

"Find the general solution of $xu_{x} + 4yu_{y} = 0$ in ${(x,y)\neq(0,0}$; when is this solution continuous at (0,0)?"

In the one example in my textbook (for $u_{x} + yu_{y} = 0$ they write $\frac{dy}{dx} = \frac{y}{1}$. I do not understand how or why they can write this (they turn it into a separable ODE). Can someone please explain why/how they do this?

If I accept the idea of writing it as a separable equation I can write:

$\frac{dx}{dy} = \frac{x}{4y}$

$\frac{dx}{x} = \frac{dy}{4y}$

Integrating both sides, we find:

$lnx + c = \frac{1}{4} lny + d$

$lnx + c = lny^{\frac{1}{4}} + d$

$lnx + b= lny^{\frac{1}{4}}$, where $b = c-d$

$e^(lnx + b) = e^(lny^{\frac{1}{4}})$

$x e^b = y^{\frac{1}{4}}$, but $e^b$ is a constant, so we say $e^b = C$

$Cx = y^{\frac{1}{4}}$

$(Cx)^4 = y$

So $y = (Cx)^4$ describes the characteristic curves of the ODE, but I'm unsure of what I must do next to solve the PDE.

EDIT:

After browsing around for a little bit I think I've managed to find the general solution for this problem.

$u(x,y) = u(x,[Cx]^4)$
$\frac{du(x, [Cx]^4)}{dx} = \frac{du}{dx} + \frac{du}{dy}\frac{dy}{dx} = \frac{du}{dx} + \frac{du}{dy}\frac{4y}{x} = xu_{x} + 4yu_{y} = 0$; this is the initial question, so we know that this works. Now we know:

$u(x,y) = u(x, [Cx]^4) = f(C)$, but we can solve for C in terms of x and y to find $C = \frac{y^(\frac{1}{4})}{x}$ so we have the general solution:

$u(x,y) = f(\frac{y^(\frac{1}{4})}{x})$

Last edited: Sep 23, 2012
2. Sep 23, 2012

### LCKurtz

It's easy to check. Just plug it into the equation and initial conditions and see if it works.

3. Sep 23, 2012

### Tsunoyukami

Well the solution $u(x,y) = 3x +y$ does satisfy the given PDE, but I'm unsure how to consider the idea of whether or not its valid for all x and y greater than 0.

For example, $u_{x} = 3, u_{y} = 1$ and these satisfy the PDE regardless of the values of x and y because they do not depend on either x or y. Does this mean that it holds true for all x and y greater than 0 or is there more that I have to show?

4. Sep 23, 2012

### LCKurtz

Like you say, it works for all x and y, and that includes x>0 and y>0. If it satisfies the boundary conditions how could anyone argue that it isn't a solution when it is?

5. Sep 23, 2012

### Tsunoyukami

Then by the same reasoning it is also a valid solution provided that x<0, y>0 (for the next part of the problem). I felt like there was more I had to show before I could come to such a conclusion because it seemed so obvious.

Thank you very much for your help.

I've edited my initial post to include a question about Variable Coefficient PDEs; if you have the time would you mind providing me with a hint as to how to proceed? Again, thank you very much!

6. Sep 23, 2012

### LCKurtz

That generally isn't a good idea. You should post new questions in a separate post. With several replies already, many readers may well just skip the thread assuming it is finished. I will answer your first question. Suppose you have $u(x,y) = C$, which defines y implicitly as a function of $x$ and you want to calcuate $dy/dx$. You would differentiate implicitly: $u_x + u_y\frac {dy}{dx} = 0$. This gives$$\frac {dy}{dx} = -\frac{u_x}{u_y}$$So for an equation like $xu_x+4yu_y = 0$ you would have$$\frac {dy}{dx}= -\frac{u_x}{u_y}=\frac{4y}{x}$$ I'm not a PDE expert so I will leave your question about the characteristics to someone else.

Last edited: Sep 23, 2012
7. Sep 23, 2012

### Tsunoyukami

Okay, thank you very much! I think I have the answer to my second question now (just not sure about the continuity aspect). Thanks again for your time!