# Method of Characteristics: Prove the u and v is a solution

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• October 23rd 2010, 09:41 AM
bugatti79
Method of Characteristics: Prove the u and v is a solution
Dear Folks,

Can anyone get me started on how to approach the following question:

Consider the following 1st order linear PDE $a(x,y) \frac{\partial u}{\partial x} + b(x,y) \frac{\partial u}{\partial y} + c(x,y)u=0$
If u and v are 2 solutions shown that a) u + v and b) ku for any constant k are also solutions?
Just a tip would be good, I want to tackle it myself.
Thanks
• October 23rd 2010, 10:35 AM
Ackbeet
Just plug u + v into the pde and use the linearity of everything. Do the same for ku.
• October 23rd 2010, 10:51 AM
bugatti79
but we dont know what the unknown functions u and v are so that we can substitute back into PDE.... Thats my understanding of things...
• October 23rd 2010, 11:00 AM
Ackbeet
That's true, we don't know what u and v are. At least, we don't have a closed-form expression for them. However, we do know they are solutions of the pde, and that's all we need. Try this:

$\displaystyle a(x,y)\frac{\partial}{\partial x}(u+v)+b(x,y)\frac{\partial}{\partial y}(u+v)+c(x,y)(u+v)=?$
• October 23rd 2010, 12:03 PM
bugatti79
Quote:

Originally Posted by Ackbeet
That's true, we don't know what u and v are. At least, we don't have a closed-form expression for them. However, we do know they are solutions of the pde, and that's all we need. Try this:

$\displaystyle a(x,y)\frac{\partial}{\partial x}(u+v)+b(x,y)\frac{\partial}{\partial y}(u+v)+c(x,y)(u+v)=?$

Well I could multiply out the brackets like

$a(x,y)\frac{\partial u(x,y)}{\partial x} +a (x,y)\frac{\partial v(x,y)}{\partial x}.....$

Put all v terms on one side and u terms the other side....
I have difficulty seeing the 'link'..... (Shake)
• October 23rd 2010, 12:07 PM
Ackbeet
No need to put them on different sides of the equation. Just group them differently. You can separate out all the u terms, and shove them to the left. Group the v terms, and shove them to the right, but all terms are still on the LHS of the equation. Write this out. Then what can you say?
• October 23rd 2010, 12:23 PM
bugatti79
Quote:

Originally Posted by Ackbeet
No need to put them on different sides of the equation. Just group them differently. You can separate out all the u terms, and shove them to the left. Group the v terms, and shove them to the right, but all terms are still on the LHS of the equation. Write this out. Then what can you say?

$a(x,y)\frac{\partial u(x,y)}{\partial x} +b (x,y)\frac{\partial u(x,y)}{\partial x} +c(x,y)u+a(x,y)\frac{\partial v(x,y)}{\partial x} +b (x,y)\frac{\partial v(x,y)}{\partial x}+c(x,y)v=0$

Ok, just multiplying out the brackets and grouping like terms etc and all this is equal to 0 gives me the impression that u + v is also a solution.....but its still not a 'proof'.
I merely substituted and rearranged.... (Doh)
• October 23rd 2010, 12:26 PM
Ackbeet
But $u$ is a solution, and hence

$a(x,y)\dfrac{\partial u(x,y)}{\partial x} +b (x,y)\dfrac{\partial u(x,y)}{\partial x} +c(x,y)u=0.$

Also, $v$ is a solution, and hence

$a(x,y)\dfrac{\partial v(x,y)}{\partial x} +b (x,y)\dfrac{\partial v(x,y)}{\partial x}+c(x,y)v=0.$

So you see why you're done now?
• October 23rd 2010, 12:35 PM
bugatti79
Quote:

Originally Posted by Ackbeet
But $u$ is a solution, and hence

$a(x,y)\dfrac{\partial u(x,y)}{\partial x} +b (x,y)\dfrac{\partial u(x,y)}{\partial x} +c(x,y)u=0.$

Also, $v$ is a solution, and hence

$a(x,y)\dfrac{\partial v(x,y)}{\partial x} +b (x,y)\dfrac{\partial v(x,y)}{\partial x}+c(x,y)v=0.$

So you see why you're done now?

Yes sure!!! From above 0+0 = 0. So obvious. I was just fixated on some elaborate proof. Its a pleasure learning from you, thanks (Happy)
• October 23rd 2010, 12:53 PM
Ackbeet
The other proof works in a very similar vein. Yeah, it can sometimes be tough knowing when something simple is perfectly adequate, as opposed to the latest complicated proof technique. Incidentally, can you see how this proof would fail if, say, you had u^2 multiplying the c(x,y)?
• October 23rd 2010, 12:56 PM
bugatti79
Quote:

Originally Posted by Ackbeet
The other proof works in a very similar vein. Yeah, it can sometimes be tough knowing when something simple is perfectly adequate, as opposed to the latest complicated proof technique. Incidentally, can you see how this proof would fail if, say, you had u^2 multiplying the c(x,y)?

I will have a look at it tomorrow...Im going for a pint of guinneas now :-)

Thanks
• October 23rd 2010, 01:05 PM
Ackbeet
Sounds good! I always take Sundays off from the MHF, but I'll be back tomorrow.
• October 24th 2010, 06:10 AM
bugatti79
Quote:

Originally Posted by Ackbeet
The other proof works in a very similar vein. Yeah, it can sometimes be tough knowing when something simple is perfectly adequate, as opposed to the latest complicated proof technique. Incidentally, can you see how this proof would fail if, say, you had u^2 multiplying the c(x,y)?

Well, because of the unknown function u is squared, the 1st order PDE becomes nonlinear. But not sure how to prove it mathematically...
I could multiply out the terms and get something like the 'partial terms' + c(x,y)u^2+c(x,y)v^2=0

I dont see how it would fail....

Also, I have attempted the ku problem...see attached jpg (part 2).

Thanks
• October 25th 2010, 02:10 AM
Ackbeet
No, no. Your solution for ku is not correct. You DON'T know that all those partial derivatives are zero. In fact, they are probably nonzero. However, you do know that k, a constant, behaves in a certain fashion under derivative signs, right? I mean, how could you simplify the expression

$\dfrac{\partial(ku)}{\partial x}?$
• October 25th 2010, 03:13 AM
bugatti79
Quote:

Originally Posted by Ackbeet
No, no. Your solution for ku is not correct. You DON'T know that all those partial derivatives are zero. In fact, they are probably nonzero. However, you do know that k, a constant, behaves in a certain fashion under derivative signs, right? I mean, how could you simplify the expression

$\dfrac{\partial(ku)}{\partial x}?$

Your right, they cant be 0, poor judgement on my part. I guess you can pull out the k's and divide across by k. And you could do this for any constant k. See attached... Hope im right!

I also replied to your query about what if u was squared?... Thanks

Attachment 19464
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