# Ill-conditioned Heat equation Cauchy problem or just typo problem

I have been sailing in circles with this problem for one week. In fact I have asked for help from my instructor but it's really annoying since I know that something is wrong.

The problem is \begin{align*} \begin{cases} u_{t}-u_{xx}=0 &x>0,\,t>0\\ \begin{cases} u(0,x)=f(x) &x\geq 0\\ u_{t}(0,x)=0 & \end{cases} \end{cases} \end{align*} or, other version my instructor has put as the "right version" \begin{align*} \begin{cases} u_{t}-u_{xx}=0 &x\in\mathbb{R},\,t>0\\ \begin{cases} u(0,x)=f(x) &x\geq 0\\ u_{t}(0,x)=0 &x\leq 0 \end{cases} \end{cases} \end{align*}

I think both versions are flawed since constraing time derivative in a equation that only has time derivatives is overdetermining problem. Other flaw in the former is that you have two initial conditions for a equation that only has first order time derivatives. What do you think?

I think the first could be $u_{x}(0,x)=0$ for you need a boundary condition in semi-infinite problems in order to know how extend function for using the fundamental solution. But the latter version I don't know how to solve it at all. Please help.

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Maybe the latter can be solved by superposition, but the flaw with the time derivative for left quarter plane problem is not good. What do you think? – elessartelkontar Nov 19 '12 at 22:54
For the second one note that $u_t=0$ can just as well be written as $u_{xx}=0$... – fedja Nov 21 '12 at 4:01
Someone post a more exciting problem math.stackexchange.com/questions/239732, I am still investigating whether the problem has solution or not. – doraemonpaul Dec 3 '12 at 3:27

You should note that some of the current analyses such as "2nd-order PDE problem with only one condition but get unique solution" for example provided in http://en.wikipedia.org/wiki/Heat_equation#Fundamental_solutions should be in fact misleading, or say nastily, brainwashing! Because they bring us the wrong concept that "2nd-order PDE can get unique solution by just only one condition" . In fact the analysis even get the wrong conclusion starting in the "fundamental solutions" part, as it get the "unique" fundamental solutions, therefore it eventually brings the wrong conclusion to the general solution with the condition of the general case.

You can easily to see how the analysis mention in above has problems. Whatever you solve the heat equation without any conditions by separation of variables, kernel method, or even kernel transform, you will always find that the general solution contains two distinct arbitrary functions. This should be impossible to determine the two arbitrary functions at the same time with just only one condition.

In fact the current PDE analyses still contains many misleading ones, for example "results by separation of variables can only allow to take discrete form of superposition" , "two same type conditions automatically imply the range" , "solutions involve unkown type of some inverse transformations should be a wrong question" , etc. These in fact make me disappointing. As even in current Mathematics there are still many misleading or even brainwashing analyses flooding.

It should note that the ranges stated in the questions are only provide the minimum requirements of the domain of the solutions required, you are always welcomed if you smart enough to find the solutions which the domain larger than the ranges stated in the questions.

So $u_t-u_{xx}=0$ with $u(0,x)=f(x)$ and $u_t(0,x)=0$ , even it is relatively strange comparing to the current, is still in fact a just-determining problem.

Let $u(t,x)=T(t)X(x)$ ,

Then $T'(t)X(x)-T(t)X''(x)=0$

$T'(t)X(x)=T(t)X''(x)$

$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-s^2\\X''(x)+s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(t,x)=C_1x+C_2+\int_0^\infty C_3(s)e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds$

$u_t(t,x)=-\int_0^\infty s^2C_3(s)e^{-ts^2}\sin xs~ds-\int_0^\infty s^2C_4(s)e^{-ts^2}\cos xs~ds$

Substituting $u(0,x)=f(x)$ and $u_t(0,x)=0$ respectively , you will get this system of equations $\begin{cases}C_1x+C_2+\int_0^\infty C_3(s)\sin xs~ds+\int_0^\infty C_4(s)\cos xs~ds=f(x)\\\int_0^\infty s^2C_3(s)\sin xs~ds+\int_0^\infty s^2C_4(s)\cos xs~ds=0\end{cases}$ .

The only thing is that $C_3(s)$ and $C_4(s)$ are in fact really difficult to find.

Since we can find the solutions that already suitable for $t>0$ and $x\in\mathbb{R}$ , that means the two questions are still identical even after the modifications of some ranges from your instructor.

Note that even when $u_t(0,x)=0$ is changed to $u_x(0,x)=0$ , it is still in fact a just-determining problem, but you should handle this system of equations instead $\begin{cases}C_1x+C_2+\int_0^\infty C_3(s)\sin xs~ds+\int_0^\infty C_4(s)\cos xs~ds=f(x)\\C_1+\int_0^\infty sC_3(s)\cos xs~ds-\int_0^\infty sC_4(s)\sin xs~ds=0\end{cases}$ .

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