# Trying to solve simple pde $u_t = iu_{xx}+2iu$

I'm trying to solve $$u_t = iu_{xx}+2iu$$ where we know $$u(0,x) = \cos(2\pi x)-i\sin(2\pi x)$$, $$0 \leq x < 1$$, $$0 \leq t$$ with periodic boundary conditions.

This is what I tried:

Assume $$u(t,x) = T(t)X(x)$$, then we have $$T'(t)X(x) = iT(t)X''(x)+2iT(t)X(x)$$, or in other words $$\frac{T'(t)}{T(t)} = i\frac{X''(x)}{X(x)} + 2i = \lambda \in \mathbb C$$

The general solution to the ODE $$\frac{T'(t)}{T(t)} = \lambda$$ is $$T(t) = c_1 e^{\lambda t}$$

The solution to the ODE $$i\frac{X''(x)}{X(x)} + 2i = \lambda$$ is $$X(x) = c_2e^{i\sqrt{2+i\lambda}x} + c_3e^{-i\sqrt{2+i\lambda}x}$$

So overall we have $$u(t,x) = c_1 e^{\lambda t}(c_2e^{i\sqrt{2+i\lambda}x} + c_3e^{-i\sqrt{2+i\lambda}x}) = e^{\lambda t}(\tilde{c_1}e^{i\sqrt{2+i\lambda}x} + \tilde{c_2}e^{-i\sqrt{2+i\lambda}x})$$

Now I have 3 unknowns, $$\tilde{c_1}, \tilde{c_2}, \lambda$$. How can I find them using the initial condition and the periodic boundary conditions? I'm stuck here.

Try it the other way around

$$\begin{cases} \frac{X''}{X} = \lambda \\ \frac{T'}{T} = i(\lambda+2) \end{cases}$$

Solving the $$X$$ equation with periodic B.C.s gives

$$X_n(x) = e^{i2\pi nx}, \quad n = 0, \pm 1, \pm 2, \dots$$

and $$\lambda_n = -(2\pi n)^2$$

The general solution is

$$u(t,x) =\sum_{n=-\infty}^\infty a_ne^{i2\pi nx} e^{i(2-4n^2\pi^2)t}$$

where $$a_n \in \Bbb Z$$.

The initial condition gives

$$u(0,x) = \sum_{n=-\infty}^\infty a_ne^{i2\pi nx} = e^{-i2\pi x}$$

Thus $$a_{-1}=1$$ and every other coefficient is zero.

The final solution is

$$u(t,x) = e^{-i2\pi x}e^{i(2-4\pi^2)t}$$

Edit: The $$X$$ solutions were found by simple observation, since all functions of the form $$e^{i2\pi nx}$$ have a period of $$1$$. You can also consider the general solution $$X(x) = e^{\sqrt{\lambda} x}$$ and apply the B.C.'s

$$X(0) = X(1) \implies 1 = e^{\sqrt{\lambda}} \implies \sqrt{\lambda} = i2\pi n$$

• How did you find the solution for $X$ with periodic B.C? – Oria Gruber Mar 13 at 10:26

Initial condition is $$u(0,x) = \cos(2\pi x)-i\sin(2\pi x)=e^{-2i\pi x}$$ We search solution in form $$u(x,t)=v(t)e^{-2i\pi x}$$ For $$v(t)$$ we get ODE $$v'(t)=-2 i \left( 2 \pi^2-1\right)v(t),\quad v(0)=1.$$ $$\Rightarrow$$ $$v(t)=e^{-2i(2\pi^2-1)t}$$ Final solution is $$u=e^{-2i(\pi x+2\pi^2t-t)}$$

• What makes you think that the solution can be written in the form $u(x,t)=v(t)e^{-2i\pi x}$? – md2perpe Mar 14 at 11:13