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Tsunoyukami

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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!

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##

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.

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).

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##

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})##

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})##

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