Solving boundary value problem (Wave Equation)

[mizzcriss]

Homework Statement

Show that the boundary-value problem $$u_{tt}=u_{xx}\qquad u(x,0)=2f(x)\qquad u_t(x,0)=2g(x)$$ has the solution $$u(x,t)=f(x+t)+f(x-t)+G(x+t)-G(x-t)$$ where $G$ is an antiderivative/indefinite integral of $g$. Here, we assume that $-\infty<x<\infty$ and $t\geq 0$

Homework Equations

I know that the solution for a wave equation of the form $u_{tt}=a^2u_{xx}$ is $u(x,t)=f(x+at)+g(x-at)$

The Attempt at a Solution

Using the "knowledge" above, I tried to take the first partial derivative of $u(x,t)=f(x+t)+g(x-t)$ with respect to $t$ since $a=1$. I got $u_t(x,t)=f'(x+t)-g'(x-t)$ Then using the conditions $u(x,0)=2f(x)$ and $u_t(x,0)=2g(x)$, I substituted 0 for $t$ and set them equal to $2f(x)$ and $2g(x)$.

I came up with $f(x)=g(x)=\frac{1}{2}(f'(x)-g'(x))$, which I thought could also be $\frac{1}{2}(f'(x)-G(x))$. But then I got stuck because I have no idea how to get to $$u(x,t)=f(x+t)+f(x-t)+G(x+t)-G(x-t)$$

Part of my confusion is that I only have initial conditions and no boundary conditions, and all the examples I can find have boundary conditions as well, not just initial.

Thanks so much if you can help I might just be missing something because I'm in freak out mode about my test tomorrow on this stuff! This is a numerical methods for differential equations class by the way.

 

[LCKurtz]

Homework Statement

Show that the boundary-value problem $$u_{tt}=u_{xx}\qquad u(x,0)=2f(x)\qquad u_t(x,0)=2g(x)$$ has the solution $$u(x,t)=f(x+t)+f(x-t)+G(x+t)-G(x-t)$$ where $G$ is an antiderivative/indefinite integral of $g$. Here, we assume that $-\infty<x<\infty$ and $t\geq 0$

Homework Equations

I know that the solution for a wave equation of the form $u_{tt}=a^2u_{xx}$ is $u(x,t)=f(x+at)+g(x-at)$

Let's say $u(x,t) = F(x+t) + G(x-t)$ since we don't want to confuse the general solution with the $f$ and $g$ given in the problem. And we have $a=1$.

The Attempt at a Solution

Using the "knowledge" above, I tried to take the first partial derivative of $u(x,t)=f(x+t)+g(x-t)$ with respect to $t$ since $a=1$. I got $u_t(x,t)=f'(x+t)-g'(x-t)$

Good so far, but let's use the capital letters.

$u(x,t)=F(x+t)+G(x-t)$
$u_t(x,t)=F'(x+t)-G'(x-t)$

Then using the conditions $u(x,0)=2f(x)$ and $u_t(x,0)=2g(x)$, I substituted 0 for $t$ and set them equal to $2f(x)$ and $2g(x)$.

This should give you

$F(x)+G(x) = 2f(x)$
$F'(x)-G'(x) = 2g(x)$

I came up with $f(x)=g(x)=\frac{1}{2}(f'(x)-g'(x))$, which I thought could also be $\frac{1}{2}(f'(x)-G(x))$. But then I got stuck because I have no idea how to get to $$u(x,t)=f(x+t)+f(x-t)+G(x+t)-G(x-t)$$

Part of your trouble is confusing the capital F and G with the little f and g. What you want to do next is integrate the equation $F'(x)-G'(x) = 2g(x)$ giving$$
F(x) - G(x) = 2\int g(x)\, dx =2 H(x)$$
where I am just using $H$ for the antiderivative of $g$.

Now work with these two equations:

$F(x)+G(x) = 2f(x)$
$F(x) - G(x) = 2H(x)$

to solve for $F$ and $G$.

 

[mizzcriss]

Wow thank you so so much!

From the 2 equations I came up with $$F(x)=f(x)+H(x)$$ and $$G(x)=f(x)-H(x)$$ Then adding them gives me essentially exactly what I need but in the problem the solution $u(x,t)=f(x+t)+f(x−t)+G(x+t)−G(x−t)$ uses lower case f and capital G, those aren't like typos or something. You were exactly right that I was thinking that the $f$ and $g$ from the general solution and the conditions were the same thing, when apparently they aren't So could I just use completely different letters/symbols for the general solution to keep the variables separate?

 

[LCKurtz]

Wow thank you so so much!

You're welcome

From the 2 equations I came up with $$F(x)=f(x)+H(x)$$ and $$G(x)=f(x)-H(x)$$ Then adding them gives me essentially exactly what I need but in the problem the solution $u(x,t)=f(x+t)+f(x−t)+G(x+t)−G(x−t)$ uses lower case f and capital G, those aren't like typos or something. You were exactly right that I was thinking that the $f$ and $g$ from the general solution and the conditions were the same thing, when apparently they aren't So could I just use completely different letters/symbols for the general solution to keep the variables separate?

Yes, that is what you should do. When I wrote up my original reply I hadn't noticed you had already used G or I wouldn't have used it either.