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If you set up the one dimensional heat equation on an infinite line

$$\frac{\partial u}{\partial t} = \frac12 \frac{\partial^2 u}{\partial x^2}$$

knowing the initial conditions $u(x,0) = f(x)$ is enough to solve the equation. But if you have initial conditions along the time axis instead, $u(0,t)=g(t)$ this wouldn't be enough to solve the PDE. E.g. with $g(t)=0$, $u(x,t)= \alpha x$ is a solution for any $\alpha$. It seems like being given the first order space derivative, $u_x(0,t)=h(t)$ as well would be enough, because if we imagine $x$ is now the time variable and $t$ is the space variable, the PDE now depends on the second derivative of the time variable. Though I'm not sure if this is true.

But this is a problem more generally, and with some PDEs there isn't a clear time variable, so even this inuitive approach doesn't work in general for PDEs. So I would like to know whether, more generally, there is a way to tell what initial conditions are enough to specify a unique solution in linear PDEs.

This isn't a precisely stated problem, as I am not trying to solve a specific problem, so I don't know what form an answer to this question would take. But I would like to know any theorems or rules of thumb that can be used to tell when initial conditions are or are not sufficient for linear PDEs. In PDEs with a time dimension, this question usually translates to 'how much information about initial conditions do you need to predict what happens next', but I am also interested in PDEs without a time dimension.

Zoe Allen
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    I'm not quite sure I follow the statement: "Knowing the initial conditions is enough to solve the equation". You always need initial conditions and boundary conditions to solve these equations. Take for example, the infinite line $x > 0$ with initial condition $u(x,0) = f(x)$. How could you solve this without knowing the condition at $x = 0$. – Gregory May 08 '23 at 19:28
  • In general, you need as many conditions as derivatives to fully-specify a problem. – Gregory May 08 '23 at 19:29
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    may be helpful: https://en.wikipedia.org/wiki/Cauchy%E2%80%93Kowalevski_theorem (look at the higher-order part and at the example cited there) – user8268 May 08 '23 at 19:34
  • @user8268 Thank you, that answers how to make the intuition about PDEs with an $n$th time derivative rigorous. – Zoe Allen Jun 01 '23 at 14:02

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