Solve the PDE $-u_{tx} + cu_{xx}=0$ given initial/boundary conditions $u(0,x)=sin(x)$, $u_t(t,0)=0$.
I substituted $u(x,t) = X(x)T(t)$ into the PDE and did 1 extra step to get $$\frac{T'(t)}{cT(t)} = \frac{X''(x)}{X'(x)}=\lambda$$ and then I got the two equations:
$$T'(t)-\lambda cT(t)=0$$ $$X''(x)-\lambda X'(x)=0$$.
Now I'm stuck. No idea what to do with the initial/boundary conditions.
On
Recall for the separation of variables, we want a solution of the form $u(x,t)=X(x)T(t)$
The first ODE is $(1)$ $T'-\lambda cT=0$, which has solution $T(t)=exp(\lambda c t)$ (it's a 1st order linear ODE).
The second ODE is $(2)$ $X''-\lambda X'=0$. The characteristic equation is $r^2-r=0$ which factors to $r(r-1)=0$, so our solution becomes $X(x)=\frac{k_1exp(c\lambda x)}{c\lambda}+k_2$.
$X'(x)=\frac{k_1c\lambda}{c\lambda}exp(c\lambda x)+k_2=k_1exp(c\lambda x)+k_2$,
With the boundary condition, we obtain $X'(0)=k_1+k_2=0$, so $k_1=-k_2$.
Putting this together, we get $u(x,t)=\sum_{n=1}^{\infty}A_n(k_1exp(c\lambda x)+k_2)exp(c\lambda t)$. Now our initial condition is $u(0,x)=sin(x)$ so we obtain:
$u(0,x)=\sum_{n=1}^{\infty}A_n(k_1exp(c\lambda x)+k_2)$ as the solution.
Note that unlike the case where we have the wave equation or heat equation, the value of $\lambda$ is not restricted, so all values of $\lambda$ solve $(1)$ and $(2)$.
I'll present two ways of looking at this problem, both of which lead to same conclusion: we can't solve the PDE with the given boundary/initial conditions.
Round 1: Separation of Variables
We assume that our solution can be written as a product of single variable functions: $$u(t,x)=T(t)X(x)$$
We can check that $$u_{tx}=T'(t)X'(x)\text{, and }u_{xx}=T(t)X''(x)$$
Thus, we can rewrite our PDE in the following form: $$-T'(t)X'(x)+cT(t)X''(x)=0$$
We can divide through by $T(t)X'(x)$ to obtain $$-\dfrac{T'(t)}{T(t)}+c\dfrac{X''(x)}{X'(x)}=0$$
If we subtract $c\dfrac{X''(x)}{X'(x)}$ from both sides, we obtain $$\dfrac{T'(t)}{T(t)}=c\dfrac{X''(x)}{X'(x)}$$
Since the LHS only depends on $t$, and the RHS only depends on $x$, and they are always equal, it must be the case that both are constant. Hence, $$\dfrac{T'(t)}{T(t)}=c\dfrac{X''(x)}{X'(x)}=\lambda$$
This brings us to the two ODEs: $$T'(t)=\lambda T(t)$$$$X''(x)=\dfrac{\lambda}{c}X'(x)$$
For convenience sake, we let $\dfrac{\lambda}{c}=r$. The second equation becomes $$X''(x)=rX'(x)$$
So really, we have two versions of the same equation. To solve the $T$ equations, we actually go a step back to separate variables: $$\dfrac{T'(t)}{T(t)}=\lambda$$ Integrating both sides with respect to $t$, we obtain $$\ln(T(t))=\lambda t+c$$
We exponentiate, and obtain $$T(t)=e^{\lambda t}e^c=Ae^{\lambda t}$$ Note that we replaced the arbitrary constant $e^c$ with the arbitrary constant $A$.
Similarly, in the $X$ equation, we obtain $$X'(x)=B'e^{rx}$$ We integrate to find $$X(x)=\frac{B'}{k}e^{rx}+C=Be^{rx}+C$$.
Notice that the constant $\lambda$ is arbitrary - it could be any number for all we know. Also, since our PDE is linear, any linear combination of solutions will also be a solutions. Therefore, our solution will take a form $$u(x,t)=\sum\left[\left(B_ke^{r_kx}+C_k\right)\left(A_ke^{\lambda_kt}\right)\right]$$ where $r_k=\dfrac{\lambda_k}{c}$
If we plug in $t=0$, we get $$u(x,0)=\sum\left[\left(B_ke^{r_kx}+C_k\right)A_k\right]$$ It may not be immediately obvious how we can make this look like $\sin(x)$. At this point, we have to rely on complex numbers if we are to find our solutions. The formula we will need is that $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ This suggests that $$u(x,0)=\frac{1}{2i}e^{ix}-\frac{1}{2i}e^{-ix}$$ We can accomplish this by taking $C_1,C_{-1}=0$ and $r_1=i$, $B_1A_1=\dfrac{1}{2i}$ and $r_{-1}=-i$, $A_{-1}B_{-1}=\dfrac{-1}{2i}$ with no other terms to the sum.
Thus, we obtain $$u(x,t)=\frac{1}{2i}e^{ix}e^{cit}-\frac{1}{2i}e^{-ix}e^{-cit}$$ Using our formula from earlier, we can rewrite this as $$u(x,t)=\sin(x+ct)$$ This still satisfies $$u(x,0)=\sin(x)$$ We also have $$u_t(0, t)=c\cos(ct)$$ which is unfortunately not the stated condition, unless $c=0$.
Round 2: Transport Equation
We could also recognize that this PDE is equivalent to taking an $x$ derivative of the transport equation $$-v_t+cv_x=0$$
The solution of this, given $v(x,0)=f(x)$ is well known to be $$v(x,t)=f(x+ct)$$ Thus, given $v_x=u$, we should take $$v(x, 0)=-\cos(x)$$ and obtain $$v(x,t)=-\cos(x+ct)$$ and $$u(x,t)=v_x(x,t)=\sin(x+ct)$$ as we found earlier. The fact remains, the initial conditions given are not compatible, so this PDE does not have a solutions.