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:

\partial²u/\partialx\partialy=u3. The Attempt at a Solution [/]

=>\int\intu 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 \partial²u/\partialx\partialy=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:

<br /> \frac{d}{dx}f(x)= ?<br />

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... <br /> \frac{\partial^2 f(x)g(y)}{\partial x \partial y}<br /> =\int\intf'(x)g'(y) dx dy??

 

[Cyosis]

What you've written down now is the equivalent to \frac{d}{dx}f(x)=\int f&#039;(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).

<br /> \frac{\partial^2 f(x)g(y)} {\partial x \partial y}=\frac{\partial}{\partial x} f(x) \frac{\partial}{\partial y}g(y)=f&#039;(x)g&#039;(y)<br />

 

[fredrick08]

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

 

[Cyosis]

You forgot a prime it should be f&#039;(x)g&#039;(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&#039;(x)g&#039;(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&#039;(x)g&#039;(y)=f(x)g(y) dividing by f(x) and g'(y) ensures that x and y are both on separate sides.

<br /> \frac{f&#039;(x)}{f(x)}=\frac{g(y)}{g&#039;(y)}<br />

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&#039;(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.

<br /> \int \frac{f&#039;(x)}{f(x)}dx=\int a dx<br />

 

[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]

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

 

[fredrick08]

and for g(y)/g'(y)=a Integral of all of it is G(y)/g(y)=ay=>g(y)=G(y)/(a*y)??

 

[Cyosis]

Are you just randomly guessing or? We want to SOLVE f(x). We want to find all expressions for f(x) that satisfy this differential equation. Have you never solved a single variable differential equation before?

You're now saying that the integral of a fraction is the integral of the numerator divided by the integral of the denominator. This is wrong.

Example:
<br /> \int \frac{1}{x}dx=\log x \neq \frac{\int 1 dx}{\int x dx}=\frac{2}{x}<br />

Hint: make a substitution u=f(x)

 

[fredrick08]

ok sorry... i didnt realize how stupid that was... integral of all it is constant*ln(x)=ax...

 

[Cyosis]

That is almost correct. It should be \ln (f(x))=ax+cst. Do you see why? If not use the hint in my previous post, which I edited in just before your reply.

 

[fredrick08]

oh yes ln(f(x))=ax+c... so then for g... (g(y)/2)y^2=ay+c?

 

[Cyosis]

The integral we want to calculate is:

<br /> \int \frac{f&#039;(x)}{f(x)}dx<br />

Use the substitution u=f(x). You do know how to integrate by using substitutions right?

 

[fredrick08]

its something liek that... because no matter what number you put infront of g... eg 3y^2, then g'=6y integral of g/g'= ((3*2)/2)y^2

ok ignore this

 

[Cyosis]

Can you please just evaluate the integral and stop guessing? And if you guess don't guess things that are so obviously wrong, g/g'=constant. Yet in your example it's suddenly a function of y?

Evaluate the integral with the substitution I have given you!

 

[fredrick08]

ok then so int(f'/f)=ln(f(x))=ax+c... yes i can intergrate by substituion... but what am i intergating? ln(u)=ax+c

 

[Cyosis]

You're asking me what we are integrating yet you give the answer of the integration. This means to me that you did not evaluate the integration. Show me the steps leading up to your answer.

 

[fredrick08]

anrt i meant to solve for u? therefore u=f(x)=e^(ax+c)?

 

[fredrick08]

ok we are integrating f'/f=a so the answer would be ln(f(x))=ax+c (as u said), then you said something about substitution...

 

[Cyosis]

Fredrick, if you just ignore what I ask of you we are going nowhere. You do not need to solve for u. I want you to evaluate the integral, in other words simply solve the differential equation f'(x)/f(x)=a.

Can you do this or are we going to continue to beat around the bush?

ok we are integrating f'/f=a so the answer would be ln(f(x))=ax+c (as u said), then you said something about substitution...

How did you get that answer, that is what I want to know.

Understanding how you arrive at that answer will allow you to understand how to solve the entire system.

 

[fredrick08]

i don't know that is wat u told me... but if use f(x)=3x^2 then int of f'(x)/f(x) dx would equal 2 ln(x).. this is what i am confused about as don't understand what you mean by solve the differential equation... as i am unsure how to do so, without a function...

 

[Cyosis]

To be able to solve partial differential equations you must first be able to solve single variable differential equations. You must also be able to integrate by substitutions. Have you followed these courses?

I have given you a substitution that allows you to solve the integral, u=f(x). What is du?