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- Thread starter Cbarker1
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Try using the method that Klaas van Aarsen and I discussed with you the other day. Feel free to let me know if there are any issues.

- Jan 30, 2018

- 731

Setting t= 0 in $u(x,t)= F(x+ ct)+ G(x- ct)$ gives $u(x, 0)= F(x)+ G(x)= 0$ so $F(x)= -G(x)$. That is, $u(x,t)= F(x+ ct)- F(x- ct)$. Differentiating that with respect to t, $u_t(x, t)= cF'(x+ ct)+ cF'(x- ct)$. Setting t= 0 in that (I presume you mean "$u_t(x, 0)= \frac{x}{(x^2+1)^2}$) we have $2cF'(x)= \frac{x}{(x^2+1)^2}$ so that $F'(x)= \frac{x}{2c(x^2+ 1)^2}$. Integrate that to find F.Dear Everybody,

I am confused about how to start with the following problem: using the solution from ex. 3:

$u(x,t)=F(x+ct)+G(x-ct)$

"For data u(x,0)=0 and ${u}_{t}=\frac{x}{(x^2+1)^2}$ where x is from neg. infinity to pos. infinity."

Thanks

Cbarker1