partial differential equations
Solve $$u_{tt}=7u_{xx},-\infty< x <\infty$$ $$u(x,0)=x^2, u_t(x,0)=\cos(3x),-\infty< x <\infty$$
If you use d'Alembert's solution for this problem after doing change of variables and everything to get the solution $u(x,t)=F(x-ct)+G(x+ct)$ do you use the conditions $u(x,0)=x^2, u_t(x,0)=\cos(3x)$ for F and G? Example would the solution be $u(x,t)=(x-\sqrt{7}t)^2+\cos(3(x+\sqrt{7}t))$? If this is not the case where do we use the conditions?
a) Yes, even extension is appropriate for $u_x=0$ boundary condition.
b) The first step is to express the extension of $\hat \psi$ in terms of Heaviside function $H$. This does not do much mathematically, other than introduce convenient notation. In your case, the extended function $$\Psi(x ) = \begin{cases} c ,\quad & -1<x<1 \\ 0,\quad & \text{otherwise}\end{cases}$$ can be written as $$\Psi(x ) = cH(x+1) -cH(x-1)$$ because it jumps up by $c$ at $x=-1$ (as $cH(x+1)$ does), and then down by $c$ at $x=1$ (as $-cH(x-1)$ does).
In general, a piecewise constant function that jumps by amount $\delta_j$ at point $x_j$ would be written as $\sum_j \delta_j H(x-x_j)$.
Then you need antiderivative of $\Psi$, for the purpose of integration in d'Alembert's formula. Note that $xH(x)$ is an antiderivative of $H(x)$, and more generally $(x-a)H(x-a)$ is an antiderivative of $H(x-a)$. So, $$F(x) = c(x+1)H(x+1) -c(x-1)H(x-1)$$ as an antiderivative of $\Psi$. Finally, plug into d'Alembert's formula: $$ u(x,t) = \frac{1}{2c}\int_{x - ct}^{x + ct}\psi(s)\,ds =\frac{1}{2c}(F(x+ct)-F(x-ct)) $$ and you'll get the answer as in your post.
You should only ask questions about a single problem. This is not the place for a homework dump.
With that being said, I'm going to solve the first problem. First, define the primitive of $u_t(x,0)=g(x)$ as
$$ G(s) = \int g(s)\ ds = \begin{cases} \dfrac{1}{\pi}\sin\big(\pi(s-1)\big), & 1 < s < 2 \\ 0, & \text{otherwise} \end{cases} $$
Then the solution given by d'Alembert's formula is
$$ u(x,t) = \frac{G(x+t) - G(x-t)}{2} $$
where you have to check case by case for both $x+t$ and $x-t$. Below is a graph ($x$ vs. $t$) of the 4 characteristic lines going through $x=1$ and $x=2$.
The numbered regions are as follows:
\begin{array}{rrr} \text{I}: && x - t < 1, && x + t < 1 \\ \text{II}: && x - t < 1, && 1 < x + t < 2 \\ \text{III}: && x - 1 < 1, && x + t > 2 \\ \text{IV}: && 1 < x - t < 2, && 1 < x + t < 2 \\ \text{V}: && 1 < x - t < 2, && x + t > 2 \\ \text{VI}: && x - t > 2, && x + t > 2 \\ \end{array} Then, you can simplify the solution
$$ u(x,t) = \begin{cases} 0, && (x,t) \in \text{I, III, VI} \\ \dfrac{1}{2\pi}\sin\big(\pi(x+t-1)\big), && (x,t) \in \text{II} \\ \dfrac{1}{2\pi}\Big[\sin\big(\pi(x+t-1)\big) - \sin\big(\pi(x-t-1)\big) \Big], && (x,t) \in \text{IV} \\ -\dfrac{1}{2\pi}\sin\big(\pi(x-t-1)\big), && (x,t) \in \text{V} \end{cases} $$
Best Answer
You memorized wrongly about d'Alembert's formula.
d'Alembert's formula should be $\dfrac{f(x+ct)+f(x-ct)}{2}+\dfrac{1}{2c}\int_{x-ct}^{x+ct}g(s)~ds$ , not $f(x-ct)+g(x+ct)$ !
The solution of this question should be $u(x,t)=\dfrac{(x+\sqrt{7}t)^2+(x-\sqrt{7}t)^2}{2}+\dfrac{\sin(3(x+\sqrt{7}t))-\sin(3(x-\sqrt{7}t))}{6\sqrt{7}}$ , not $u(x,t)=(x-\sqrt{7}t)^2+\cos(3(x+\sqrt{7}t))$ .