Finding the constants for a PDE

Question

Here we have the straight forward PDE

$U_{xx}=\frac{1}{4}U_{tt}$

Solving it by separation of variables, we get the two ODEs:

\begin{equation} \begin{array} f\frac{F_{xx}}{F}=k\\ \frac{G_{tt}}{G}=k \end{array} \end{equation}

which are easily solved individually by finding their characteristic equation without the first order derivative term, giving the solutions for F: $\lambda=\frac{\pm\sqrt{-4\cdot(-k)\cdot 4}}{2}=\pm\frac{1}{2}\sqrt{k}$; for G: $\lambda=\frac{\pm\sqrt{-4\cdot(-k)}}{2}=\pm\sqrt{k}$.

Now we have the two solutions:

\begin{equation} \begin{array} rF(x)=c_1e^{1/2\sqrt{k}x}+c_2e^{-1/2\sqrt{k}x}\\ G(t)=c_3e^{\sqrt{k}t}+c_4e^{-\sqrt{k}t} \end{array} \end{equation} With initial conditions, $U(0,t)=U(x,0)=0$ and $U(\frac{\pi}{2},t)=\sin2t$ we should have for F(x):

$0=c_1e+c_2e \longrightarrow c_1=c_2=A$:

$F(x)=A(e^{1/2\sqrt{k}x}+e^{-1/2\sqrt{k}x})$.

The same can be found for G, yielding:

$G(t)=B(e^{\sqrt{k}t}+e^{-\sqrt{k}t})$

But at this stage, we still have 3 unknown coefficients including k, because the next initial condition would act on the product of these two equations, since u(x,t)=f(x)y(t):

\begin{equation} u(x,t)= A(e^{1/2\sqrt{k}x}+e^{-1/2\sqrt{k}x})B(e^{\sqrt{k}t}+e^{-\sqrt{k}t}) \end{equation}

So with the third condition, and renaming $AB=C$ we get:

\begin{equation} \sin2t= C(e^{1/2\sqrt{k}\frac{\pi}{2}}+e^{-1/2\sqrt{k}\frac{\pi}{2}})(e^{\sqrt{k}t}+e^{-\sqrt{k}t}) \end{equation}

which gives that the coefficient A: \begin{equation} C= \frac{\sin2t}{(e^{1/2\sqrt{k}\frac{\pi}{2}}+e^{-1/2\sqrt{k}\frac{\pi}{2}})(e^{\sqrt{k}t}+e^{-\sqrt{k}t})} \end{equation}

which would give:

\begin{equation} u(x,t)= \frac{\sin2t(e^{1/2\sqrt{k}x}+e^{-1/2\sqrt{k}x})(e^{\sqrt{k}t}+e^{-\sqrt{k}t})}{(e^{1/2\sqrt{k}\frac{\pi}{2}}+e^{-1/2\sqrt{k}\frac{\pi}{2}})(e^{\sqrt{k}t}+e^{-\sqrt{k}t})} \end{equation}

then, cancelling the common terms we get

\begin{equation} u(x,t)= \frac{\sin2t(e^{1/2\sqrt{k}\frac{\pi}{2}}+e^{-1/2\sqrt{k}\frac{\pi}{2}})}{(e^{1/2\sqrt{k}x}+e^{-1/2\sqrt{k}x})} \end{equation}

But this is not a normal answer for a PDE problem as such, usually the k-constant is solve for too?

When I put in an arbitrary value for k, I get the following plot:

Though this looks fine, it is strange that the k-coefficient is left unknown.

Thanks

Answer 1

Supposing an additional condition, for instance $U_t(0,x)=0$ we can solve the problem by using the Laplace Transform. After transforming, the problem to solve for $x$ now is

$$ s^2U(x,s)=4U_{xx}(x,s),\ \ U(0,s)=0, U\left(\frac{\pi}{2},s\right)= \frac{2}{s^2+4} $$

thus we obtain the transformed solution

$$ U(x,s) = \frac{2 e^{\frac{\pi s}{4}-\frac{s x}{2}} \left(e^{s x}-1\right)}{\left(e^{\frac{\pi s}{2}}-1\right) \left(s^2+4\right)} $$

now to invert $U(x,s)$ we use the residues at the solutions for $\left(e^{\frac{\pi s}{2}}-1\right) \left(s^2+4\right)=0$ which are $s=\{\pm 2i\}\cup\{ \pm 4i k\}, k\in \mathbb{Z}$

and the residues at $\{\pm 2i\}$ are

$$ \lim_{s\to\pm 2i}\frac{2 e^{\frac{\pi s}{4}-\frac{s x}{2}} \left(e^{s x}-1\right)}{\left(e^{\frac{\pi s}{2}}-1\right) \left(s\pm 2i\right)} $$

giving $\frac{2 \sin (x)}{s^2+4}$ and the residues at $\pm 4ik,\ \ k\in \mathbb{Z}$

$$ \lim_{s\to\pm 4ik}\frac{2 e^{\frac{\pi s}{4}-\frac{s x}{2}} \left(e^{s x}-1\right)}{\left(s^2+4\right)} $$

giving

$$ \frac{2 \sin (2 k x) (s \sin (\pi k)+4 k \cos (\pi k))}{\left(4 k^2-1\right) \left(16 k^2+s^2\right)},\ \ k\in \mathbb{Z}_{\ge 0} $$

and after inversion we have

$$ U(x,t) = \sin (2 t) \sin (x)+\sum_{k=0}^n\frac{2 \sin (2 k x) (\sin (\pi k) \cos (4 k t)+\cos (\pi k) \sin (4 k t))}{4 k^2-1} $$

Follows a plot with $n=10$