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Solution satisfying initial conditions for a pde of second order

  • Thread starter maggie56
  • Start date
  • #1
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Homework Statement


I have found the general solution to a second order pde to be
U(x,t) = f(3x + t) + g(-x + t) where f and g are arbitrary functions

I have initial conditions
U(x,0) = sin(x)
Du/dt (x,0) = cos (2x)


The Attempt at a Solution



I have found that
U(x,0) = f(3x) + g(-x) = sin(x)
Du/dt(x,0) = f'(3x) + g'(-x) = cos(2x)

From this point im not sure what to do. I have tried differentiating u(x,0) with respect to x which i think gives me
Du/dx (x,0) = 3f'(3x) - g'(-x) = cos (x)
I thought i would then equate the equations but im not actually sure ths helps.

Could someone please point me in the right direction for this question

Thanks
 

Answers and Replies

  • #2
1,796
53
Ok, didn't get it that other time huh? Then I'll work one through and you do the other one ok.

[tex]f(3x)+g(-x)=\sin(x)[/tex]
[tex]f'(3x)+g'(-x)=\cos(x)[/tex]

I differentiate the top one with respect to x:

[tex]3f'(3x)-g'(-x)=\cos(x)[/tex]
[tex]f'(3x)+g'(-x)=\cos(x)[/tex]

Now add them:

[tex]4f'(3x)=2\cos(x)[/tex]

To get rid of that 3x in there, let u=3x or x=1/3u then:

[tex]\frac{df}{du}=\frac{df}{dx}\frac{dx}{du}=1/3\frac{df}{dx}[/tex]

Then substitute that into that DE with 3x in there:

[tex]12 \frac{df}{du}=2\cos(1/3u)[/tex]

integrate it:

[tex]f(u)=1/2\sin(u/3)[/tex]

so that:

[tex]f(3x+t)=1/2 \sin((3x+t)/3)[/tex]

I believe that's right anyway. Would need to back-substitute it into the PDE to check it.

Ok, now you do the other one.
 
Last edited:

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