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I have this equation

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

Is it possible for me to use both the wave and heat equations to solve this equation? I understand both, I just wanted to see if it was possible.

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    $\begingroup$ I edited your question to $\LaTeX$ify the equations and added the "pde" tag; you can learn a little $\LaTeX$ by checking out my edits. Cheers! $\endgroup$ – Robert Lewis May 19 '15 at 20:25
  • $\begingroup$ I think you could use the usual techniques we solve the wave and heat equations to solve this problem. Can you just take a solution off the shelf to solve it? I don't think it fits the standard templates I know. $\endgroup$ – James S. Cook May 19 '15 at 20:28
  • $\begingroup$ I would suggest just to use Fourier transform. $\endgroup$ – demitau May 19 '15 at 20:37
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    $\begingroup$ I don't think it would work. When you derive a solution for the wave equation, we don't consider the first order differential. Nor the second order differential for the heat equation. I would just start from $U=X(x)T(t)$ which gives you $\frac{X''}{X} = \frac{T''}{T} + 2\frac{T'}{T} = \lambda$ $\endgroup$ – user211337 May 19 '15 at 20:58
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No, one cannot obtain a solution of this PDE by adding a solution of heat equation to a solution of the wave equation. (With linear PDE, we can combine solutions of the same equation to make new ones; but your situation is different).

Your PDE is known as damped wave equation and is solved here. It does inherit some features from the heat and wave equation.

  • oscillation is possible (as in wave equation)
  • with homogeneous boundary conditions, solution dissipates to zero (as in heat equation)

It is instructive to generalize to $u_{tt}+cu_t = u_{xx}$ and consider varying $c$. As $c$ increases, the diffusion aspect begins to dominate in that some low-frequency harmonics get overdamped (they don't oscillate at all, simply return to zero). At the same time, higher frequencies are still able to oscillate.

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    $\begingroup$ Yes: No, not really. $\endgroup$ – Soham Chowdhury May 20 '15 at 2:51
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Let $u=e^{-t}v$ ,

Then $\dfrac{\partial u}{\partial t}=e^{-t}\dfrac{\partial v}{\partial t}-e^{-t}v$

$\dfrac{\partial^2u}{\partial t^2}=e^{-t}\dfrac{\partial^2v}{\partial t^2}-e^{-t}\dfrac{\partial v}{\partial t}-e^{-t}\dfrac{\partial v}{\partial t}+e^{-t}v=e^{-t}\dfrac{\partial^2v}{\partial t^2}-2e^{-t}\dfrac{\partial v}{\partial t}+e^{-t}v$

$\dfrac{\partial u}{\partial x}=e^{-t}\dfrac{\partial v}{\partial x}$

$\dfrac{\partial^2u}{\partial x^2}=e^{-t}\dfrac{\partial^2v}{\partial x^2}$

$\therefore e^{-t}\dfrac{\partial^2v}{\partial x^2}=2e^{-t}\dfrac{\partial v}{\partial t}-2e^{-t}v+e^{-t}\dfrac{\partial^2v}{\partial t^2}-2e^{-t}\dfrac{\partial v}{\partial t}+e^{-t}v$

$\dfrac{\partial^2v}{\partial t^2}=v+\dfrac{\partial^2v}{\partial x^2}$

Similar to Solve initial value problem for $u_{tt} - u_{xx} - u = 0$ using characteristics

Consider $v(x,a)=f(x)$ and $v_t(x,a)=g(x)$ ,

Let $v(x,t)=\sum\limits_{n=0}^\infty\dfrac{(t-a)^n}{n!}\dfrac{\partial^nv(x,a)}{\partial t^n}$ ,

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

$\dfrac{\partial^4v}{\partial t^4}=\dfrac{\partial^2v}{\partial t^2}+\dfrac{\partial^4v}{\partial x^2\partial t^2}=v+\dfrac{\partial^2v}{\partial x^2}+\dfrac{\partial^2v}{\partial x^2}+\dfrac{\partial^4v}{\partial x^4}=v+2\dfrac{\partial^2v}{\partial x^2}+\dfrac{\partial^4v}{\partial x^4}$

Similarly, $\dfrac{\partial^{2n}v}{\partial t^{2n}}=\sum\limits_{k=0}^nC_k^n\dfrac{\partial^{2k}v}{\partial x^{2k}}$

$\dfrac{\partial^3v}{\partial t^3}=\dfrac{\partial v}{\partial t}+\dfrac{\partial^3v}{\partial x^2\partial t}$

$\dfrac{\partial^5v}{\partial t^5}=\dfrac{\partial^3v}{\partial t^3}+\dfrac{\partial^5v}{\partial x^2\partial t^3}=\dfrac{\partial v}{\partial t}+\dfrac{\partial^3v}{\partial x^2\partial t}+\dfrac{\partial^3v}{\partial x^2\partial t}+\dfrac{\partial^5v}{\partial x^4\partial t}=\dfrac{\partial v}{\partial t}+2\dfrac{\partial^3v}{\partial x^2\partial t}+\dfrac{\partial^5v}{\partial x^4\partial t}$

Similarly, $\dfrac{\partial^{2n+1}v}{\partial t^{2n+1}}=\sum\limits_{k=0}^nC_k^n\dfrac{\partial^{2k+1}v}{\partial x^{2k}\partial t}$

$\therefore v(x,t)=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nf^{(2k)}(x)(t-a)^{2n}}{(2n)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^ng^{(2k)}(x)(t-a)^{2n+1}}{(2n+1)!}$

Hence $u(x,t)=e^{-t}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nf^{(2k)}(x)(t-a)^{2n}}{(2n)!}+e^{-t}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^ng^{(2k)}(x)(t-a)^{2n+1}}{(2n+1)!}$

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