Partial Diff Qn: Can u(x,y) be a Product of fx and fy?

[fredrick08]

Homework Statement

Does the following differential equation for u(x,y) have solutions which have the form of a product of functions of each independent variable:

$$\partial$$²u/$$\partial$$x$$\partial$$y=u3. The Attempt at a Solution [/]

=>$$\int$$$$\int$$u dx dy =uxy=f(x)f(y) ?

 

[fredrick08]

im not sure if that even makes sense, but if u(x,y)=f(x)f(y) and d*f(x)/dx=u/constant and d*f(y)/dy=constant... i think, i don't know I am confused..

 

[Dick]

Set u(x,y)=f(x)*g(y). What is your partial derivative in terms of f(x), f'(x), g(y), and g'(y)?

 

[fredrick08]

so $$\partial$$²u/$$\partial$$x$$\partial$$y=f(x)g(y)??
so in terms of f'(x) and g'(y) I am not sure..

 

[Cyosis]

You have an equation that equates the derivative of u to u itself. You can already guess what u will be, but let's not guess.

Setting u(x,y)=f(x)g(y), then the equation becomes $$\frac{\partial^2 f(x)g(y)}{\partial x \partial y}=f(x)g(y)$$. Now what Dick asks you to do is to calculate $$\frac{\partial^2 f(x)g(y)}{\partial x \partial y}$$.

Hint:

$$\frac{d}{dx}f(x)= ?$$

In terms of f'(x)

 

[Dick]

I actually meant u(x,y)=f(x)g(y). There's no reason why the function of x would have to be the same as the function of y. And yes, evaluate the derivative.

 

[Cyosis]

Sorry, that was sloppy of me I actually meant f(x)g(x). I will fix it.

 

[fredrick08]

hmm d/dx of f(x).. isn't that just f'(x)??... is it something like the integral of f'(x)g'(y)??

 

[fredrick08]

so... $$\frac{\partial^2 f(x)g(y)}{\partial x \partial y}$$ =$$\int$$$$\int$$f'(x)g'(y) dx dy??

 

[Cyosis]

What you've written down now is the equivalent to $\frac{d}{dx}f(x)=\int f'(x) dx$. Which is wrong. Let's test it, f(x)=x then $\frac{d}{dx}x=1 \neq \int 1 dx=x+c$.

hmm d/dx of f(x).. isn't that just f'(x)??

This is right, so if you evaluate $$\frac{\partial^2 f(x)g(y)}{\partial x \partial y}$$ you get...?

 

[fredrick08]

u get f'(x)g'(y)?? I am confused becasue when u take partial of f(x), u get d/dy(d/dx*f(x))=0??

 

[Cyosis]

Yes that's correct, but you're not taking the partial of f(x), you're talking the partial derivative of f(x)g(y).

$$\frac{\partial^2 f(x)g(y)} {\partial x \partial y}=\frac{\partial}{\partial x} f(x) \frac{\partial}{\partial y}g(y)=f'(x)g'(y)$$

 

[fredrick08]

ok so since f(x)g(y)=f'(x)g(y)=u(x,y)=??

 

[Cyosis]

You forgot a prime it should be $f'(x)g'(y)=f(x)g(y)=u(x,y)$. You can instantly see which function satisfies the equation, but you can also separate the variables.

 

[fredrick08]

e^x+y?? so f(x)=e^x and g(y)=e^y same as the derivativeS?

 

[Cyosis]

That would be a solution indeed, you're on the right track. it is not the general solution however. The general solution will involve integration constants and a constant in the argument of the exponential. Can you see how to integrate the differential equation?

 

[fredrick08]

e^xy?? so f(x)=e^x and g(y)=e^y same as the derrivativeS?

srry this was an accident

 

[Cyosis]

Don't randomly guess, $f'(x)g'(y)=y e^{xy}xe^{xy} \neq e^{xy}$. Separate the variables first, that is all the x dependent stuff on one side and all the y dependent stuff on the other side.

 

[fredrick08]

na I am confused now... like f(x)=e^c1 g(y)=e^c2 and u(x,y)=e^c1+c2??

 

[Cyosis]

If c are constants then f'(x)=g'(y)=0, which doesn't fit either. Can you please separate the variables first.

 

[fredrick08]

im confused because when u take the derivative of a constant its zero?

 

[Cyosis]

Yes when you take the derivative of a constant it's zero. You may want to elaborate on your confusion.

 

[fredrick08]

oh ok i will try we only just learned this... U(x,y)=F(x)G(y)

F'(x)G'(y)=F(x)G(y) divide by U(x,y)=F'(x)G'(y)/F(x)G(y)=1
im not sure the ones i have done arent as confusing as this one

 

[fredrick08]

no that's not rite... umm F'/F=G/G'=gamma?

 

[Cyosis]

No you have the y and x terms on the same side now. Separation of variables means that you put all the functions depending on x on one side and all functions depending on y on the other side. Can you do this?

 

[fredrick08]

please help.. I am sure this is not rite, and if so, where am i going with this?

F'=gamma*F and G'=G/gamma??

 

[Cyosis]

Don't worry I won't run away. Where does the gamma come from and no that's not right.

The equation we have is $f'(x)g'(y)=f(x)g(y)$ dividing by f(x) and g'(y) ensures that x and y are both on separate sides.

$$\frac{f'(x)}{f(x)}=\frac{g(y)}{g'(y)}$$

Now if this equality holds for every x and y, then what do both sides need to be equal to?

 

[fredrick08]

ummm ok sorry this gamma thing, is what we do in our notes... both side equal gamma, then gamma has different conditions... ok well if

F'/F=G/G'=a constant?

 

[Cyosis]

That is correct, they are equal to a constant. So now you have to solve two separate differential equations.

Can you solve $f'(x)/f(x)=a$?

 

[fredrick08]

na, how can i solve for that if the answer is a constant? what am i solving for? do i make f(x)= like 3x^2 or something?

 

[fredrick08]

ohh i see. so a=constant/x? or maybe not, F'=F*constant and G'=G/constant

 

[Cyosis]

No it is a differential equation, you're solving for f(x) and no a does not depend on x. Solving this differential equation for x means that you need to integrate.

$$\int \frac{f'(x)}{f(x)}dx=\int a dx$$

 

[fredrick08]

or am i solving for F=F'/constant and G=G'*constant? then the general solution would be?oh ok then well then integral(a)dx=ax?

 

[Cyosis]

Yes, asides from the integration constant. But you also need to solve the right hand side, this is the interesting side. So evaluate the integral.

 

[fredrick08]

$$\int \frac{fsingle-quote(x)}{f(x)}dx=\int a dx$$ = f(x)/F(x)=ax=>f(x)=a*x*F(x).. something like that?