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Consider the heat equation $$u_t=u_{xx}, (t,x)\in \mathbb{R}^2$$ with initial condition on $x=0$: \begin{cases} u(0,t)=1+\sin t\\ u_x(0,t)=\sin\left(\dfrac{\pi}{4}+t\right) \end{cases} and with periodic boundary condition on $t: u(x,t)=u(x,t+2\pi)$.

It possibly does not work by simple using the Fourier mode analysis by testing $$u(x,t)=\sum_n e^{int}c_n e^{{1 \over 2}(1-i)nx}$$ since this does not give any feasible solution for given initial condition. By reversing $x,t$ it is quite easy to see that the solution uniquely exists by C-K theorem, but how can I construct such a solution?

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Reversing time in heat equations is a dangerous procedure...you could know the precise temperature profile of almost anything for abitrary times, in the past... –  Avitus Aug 14 '13 at 17:42

2 Answers 2

First, your assumed periodicity in time is physically strange for a solution of a heat equation, since heat diffuses, etc., apart from the initial conditions.

In particular, writing $u(x,t)=\sum_n e^{int}\,f_n(x)$ for a periodic solution, we find $f_n''=in \cdot f_n$, so $f_n$ is a constant multiple of $x\to e^{\pm {\sqrt {in}}\cdot x}$. Yes, those square roots gives complex numbers with arguments odd integer multiples of $\pi/4$, except for $n=0$.

In particular, the only periodic-in-$x$ solution is constant.

Non-constant solutions have exponential blow-up in one direction, decay in the other, in $x$, etc.

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The question does not require the periodic in $x$ . –  doraemonpaul Aug 15 '13 at 3:17

Since the two initial conditions luckily belong to the periodic functions with period $2\pi$ , i.e. which automatically satisfy the periodic boundary condition, the periodic boundary condition is equivalent to nothing.

If you view this question by power series method approach, you will find that it is very easy to solve.

Similar to PDE - solution with power series:

Let $u(x,t)=\sum\limits_{n=0}^\infty\dfrac{x^n}{n!}\dfrac{\partial^nu(0,t)}{\partial x^n}$ ,

Then $u(x,t)=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}\dfrac{\partial^{2n}u(0,t)}{\partial x^{2n}}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(0,t)}{\partial x^{2n+1}}$

$u(x,t)=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}\dfrac{\partial^nu(0,t)}{\partial t^n}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{(2n+1)!}\dfrac{\partial^nu_x(0,t)}{\partial t^n}$

$u(x,t)=1+\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}\sin\left(\dfrac{n\pi}{2}+t\right)+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{(2n+1)!}\sin\biggl(\dfrac{(2n+1)\pi}{4}+t\biggr)$

$u(x,t)=1+\sum\limits_{n=0}^\infty\dfrac{x^{4n}}{(4n)!}\sin(n\pi+t)+\sum\limits_{n=0}^\infty\dfrac{x^{4n+2}}{(4n+2)!}\sin\left(\dfrac{(2n+1)\pi}{2}+t\right)+\sum\limits_{n=0}^\infty\dfrac{x^{4n+1}}{(4n+1)!}\sin\biggl(\dfrac{(4n+1)\pi}{4}+t\biggr)+\sum\limits_{n=0}^\infty\dfrac{x^{4n+3}}{(4n+3)!}\sin\biggl(\dfrac{(4n+3)\pi}{4}+t\biggr)$

$u(x,t)=1+\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{4n}\sin t}{(4n)!}+\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{4n+2}\cos t}{(4n+2)!}+\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{4n+1}(\sin t+\cos t)}{\sqrt2(4n+1)!}+\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{4n+3}(\sin t-\cos t)}{\sqrt2(4n+3)!}$

$u(x,t)=1+\cosh\dfrac{x}{\sqrt2}\cos\dfrac{x}{\sqrt2}\sin t+\sinh\dfrac{x}{\sqrt2}\sin\dfrac{x}{\sqrt2}\cos t+\dfrac{1}{2}\left(\sinh\dfrac{x}{\sqrt2}\cos\dfrac{x}{\sqrt2}+\cosh\dfrac{x}{\sqrt2}\sin\dfrac{x}{\sqrt2}\right)(\sin t+\cos t)+\dfrac{1}{2}\left(\cosh\dfrac{x}{\sqrt2}\sin\dfrac{x}{\sqrt2}-\sinh\dfrac{x}{\sqrt2}\cos\dfrac{x}{\sqrt2}\right)(\sin t-\cos t)$

$u(x,t)=1+\left(\sin\dfrac{x}{\sqrt2}+\cos\dfrac{x}{\sqrt2}\right)\cosh\dfrac{x}{\sqrt2}\sin t+\left(\sin\dfrac{x}{\sqrt2}+\cos\dfrac{x}{\sqrt2}\right)\sinh\dfrac{x}{\sqrt2}\cos t$

$u(x,t)=1+\left(\sin\dfrac{x}{\sqrt2}+\cos\dfrac{x}{\sqrt2}\right)\left(\cosh\dfrac{x}{\sqrt2}\sin t+\sinh\dfrac{x}{\sqrt2}\cos t\right)$

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