Ill-Posedness of a modified Schrödinger equation I am trying to show the ill-posedness of the problem $\partial_{t}u=i\Delta\bar{u}$, with $u(x,0)=u_{0}(x)$. The suggestion of my reference is to differentiate the equation with respect to the variable $t$, and use the conjugate of the equation. Thus, we can conclude that if $u$ solves the problem above, then $u$ is a solution of the problem $\partial_{t}^{2}u=\Delta^{2}u$, with the same initial condition. Under some assumptions, this last problem is ill-posed. However, there is not a equivalence between the two problems, because we can find a plane wave that is a solution for the last problem, but not a solution to the first problem. Thus, if someone can help me with this problem, I will be grateful.
 A: For $\partial_t^2u=\Delta^2u$ with $u(x,0)=u_0(x)$ , i.e. $u_{tt}=u_{xxxx}$ with $u(x,0)=u_0(x)$ , this should be ill-posed. Since there are infinitely many solutions satisfy those.
Similar to Inverse Fourier transform of a hyperbolic cosine:
Let $u(x,t)=X(x)T(t)$ ,
Then $X(x)T''(t)=X''''(x)T(t)$
$\dfrac{T''(t)}{T(t)}=\dfrac{X''''(x)}{X(x)}=s^4$
$\begin{cases}T''(t)-s^4T(t)=0\\X''''(x)-s^4X(x)=0\end{cases}$
$\begin{cases}T(t)=\begin{cases}c_1(s)\sinh ts^2+c_2(s)\cosh ts^2&\text{when}~s\neq0\\c_1t+c_2&\text{when}~s=0\end{cases}\\X(x)=\begin{cases}c_3(s)\sinh xs+c_4(s)\cosh xs+c_5(s)\sin xs+c_6(s)\cos xs&\text{when}~s\neq0\\c_3x^3+c_4x^2+c_5x+c_6&\text{when}~s=0\end{cases}\end{cases}$
$\therefore u(x,t)=\int_0^\infty C_1(s)\sinh ts^2\sinh xs~ds+\int_0^\infty C_2(s)\sinh ts^2\cosh xs~ds+\int_0^\infty C_3(s)\sinh ts^2\sin xs~ds+\int_0^\infty C_4(s)\sinh ts^2\cos xs~ds+\int_0^\infty C_5(s)\cosh ts^2\sinh xs~ds+\int_0^\infty C_6(s)\cosh ts^2\cosh xs~ds+\int_0^\infty C_7(s)\cosh ts^2\sin xs~ds+\int_0^\infty C_8(s)\cosh ts^2\cos xs~ds$
$u(x,0)=u_0(x)$ :
$\int_0^\infty C_5(s)\sinh xs~ds+\int_0^\infty C_6(s)\cosh xs~ds+\int_0^\infty C_7(s)\sin xs~ds+\int_0^\infty C_8(s)\cos xs~ds=u_0(x)$
$\int_0^\infty C_7(s)\sin xs~ds=u_0(x)-\int_0^\infty C_5(s)\sinh xs~ds-\int_0^\infty C_6(s)\cosh xs~ds-\int_0^\infty C_8(s)\cos xs~ds$
$\mathcal{F}_{s,s\to x}\{C_7(s)\}=u_0(x)-\int_0^\infty C_5(s)\sinh xs~ds-\int_0^\infty C_6(s)\cosh xs~ds-\mathcal{F}_{c,s\to x}\{C_8(s)\}$
$C_7(s)=\mathcal{F}^{-1}_{s,x\to s}\{u_0(x)\}-\mathcal{F}^{-1}_{s,x\to s}\{\int_0^\infty C_5(s)\sinh xs~ds\}-\mathcal{F}^{-1}_{s,x\to s}\{\int_0^\infty C_6(s)\cosh xs~ds\}-\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_8(s)\}\}$
$\therefore u(x,t)=\int_0^\infty C_1(s)\sinh ts^2\sinh xs~ds+\int_0^\infty C_2(s)\sinh ts^2\cosh xs~ds+\int_0^\infty C_3(s)\sinh ts^2\sin xs~ds+\int_0^\infty C_4(s)\sinh ts^2\cos xs~ds+\int_0^\infty C_5(s)\cosh ts^2\sinh xs~ds+\int_0^\infty C_6(s)\cosh ts^2\cosh xs~ds+\int_0^\infty C_8(s)\cosh ts^2\cos xs~ds+\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{u_0(x)\}\cosh ts^2\sin xs~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\int_0^\infty C_5(s)\sinh xs~ds\}\cosh ts^2\sin xs~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\int_0^\infty C_6(s)\cosh xs~ds\}\cosh ts^2\sin xs~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_8(s)\}\}\cosh ts^2\sin xs~ds$
It still have infinitely many solutions, so this problem should be ill-posed.
