MHB How Do I Solve the Wave Equation with Given Initial Data?

cbarker1
Gold Member
MHB
Messages
345
Reaction score
23
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
 
Physics news on Phys.org
Hi Cbarker1,

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.
 
Cbarker1 said:
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

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.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Back
Top