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How can one find and prove the general solution to the equation $\dfrac{\partial f(x,t)}{\partial t} =c^2i\dfrac{\partial^2f(x,t)}{\partial x^2}$ ?

I can find the solutions $Ae^{ikx-E_kt}$, so I expect linear sombinations of this to solve the equation, but can, and if so why, every solution be written as $\int a(k)e^{ikx-iE_kt}~dk$?

I think its two questions,
a) (Why) can every 'basis' solution be found by separation of variables?
b) (Why) is the general solution an integral and not a sum of all LI. solutions?

doraemonpaul
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hjg
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    The solution you mentioned can be found by using Fourier transform. It is not always applicable, but in "nice" cases, it is. So,

    (a) Fourier transform is not always applicable. (b) If you think about it, the solution you gave is "like" a sum of all linearly independent solutions. The reason it has to be an integral instead of a sum is because you can have uncountably many linearly independent solutions. If you impose some boundary conditions, you might be able to reduce the set of linearly independent solutions to a countable or even finite set.

     – Tunococ Oct 17 '12 at 11:01

1 Answers1

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Similar to how to solve $ {\partial u \over \partial t} - k {\partial ^2 u \over \partial x^2} =0$:

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

Then $X(x)T'(t)=ic^2X''(x)T(t)$

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

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

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

$\therefore f(x,t)=C_1x+C_2+\int_kC_3(k)e^{-ic^2t(K(k))^2}\sin((x-m)K(k))~dk+\int_kC_4(k)e^{-ic^2t(K(k))^2}\cos((x-m)K(k))~dk$

or $C_1x+C_2+\sum\limits_kC_3(k)e^{-ic^2t(K(k))^2}\sin((x-m)K(k))+\sum\limits_kC_4(k)e^{-ic^2t(K(k))^2}\cos((x-m)K(k))$

The choice of integral or summation depends on number of boundary conditions. If there is only one boundary condition or no boundary conditions, we should choose integral. If there are two boundary conditions, we should choose summation.

Another brilliant method is called the power series method.

Similar to PDE - solution with power series:

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

Then $f(x,t)=\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n}}{(2n)!}\dfrac{\partial^{2n}u(a,t)}{\partial x^{2n}}+\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(a,t)}{\partial x^{2n+1}}=\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n}}{i^nc^{2n}(2n)!}\dfrac{\partial^nu(a,t)}{\partial t^n}+\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n+1}}{i^nc^{2n}(2n+1)!}\dfrac{\partial^{n+1}(a,t)}{\partial t^n\partial x}=\sum\limits_{n=0}^\infty\dfrac{F^{(n)}(t)(x-a)^{2n}}{i^nc^{2n}(2n)!}+\sum\limits_{n=0}^\infty\dfrac{G^{(n)}(t)(x-a)^{2n+1}}{i^nc^{2n}(2n+1)!}=\sum\limits_{n=0}^\infty\dfrac{(-1)^nF^{(2n)}(t)(x-a)^{4n}}{c^{4n}(4n)!}-\sum\limits_{n=0}^\infty\dfrac{i(-1)^nF^{(2n+1)}(t)(x-a)^{4n+2}}{c^{4n+2}(4n+2)!}+\sum\limits_{n=0}^\infty\dfrac{(-1)^nG^{(2n)}(t)(x-a)^{4n+1}}{c^{4n}(4n+1)!}-\sum\limits_{n=0}^\infty\dfrac{i(-1)^nG^{(2n+1)}(t)(x-a)^{4n+3}}{c^{4n+2}(4n+3)!}$

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doraemonpaul
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