Solutions to Heat Equations
Description
$$ u_{t} = \gamma u_{xx} $$
The above equation is a specialized case of the generalized heat equation
$$ {{\partial} \over {\partial t}} \left( \sigma (x) u \right) = {{\partial} \over {\partial x }} \left( \kappa (x) {{\partial u} \over {\partial x}} \right) $$
where the thermal conductivity $\kappa (x) > 0$ and the heat capacity $\sigma (x) > 0$ are both constants, leading to $\displaystyle \gamma : = {{\kappa} \over {\sigma}}$ being defined as the thermal diffusivity.
Here, $t$ represents time, $x$ represents position, and $u(t,x)$ represents the distribution of heat at time $t$. $\gamma$ represents thermal diffusivity, where a higher value indicates a quicker change in distribution.
Solution
The fundamental idea is derived from the solution of the second-order linear homogeneous differential equation.
Step 1. $\displaystyle L[ u ] := -\gamma {{\partial^2 u} \over { \partial x^2 }} $
Define the linear operator $L$ as follows and assume the solution is represented by $u(t,x) = e^{- \lambda t} v(X)$:
$$ \displaystyle u_{t} = {{ \partial u} \over {\partial t}} = - \lambda e^{-\lambda t} v(x) \\ \displaystyle u_{xx} = {{ \partial ^2 u} \over {\partial x^2}} = e^{-\lambda t} v '' (x) $$
If $u$ is the solution to the given equation, then
$$ u_{t} = \lambda e^{- \lambda t } v = - \gamma e^{- \lambda t } v '' = u_{xx} $$
upon rearrangement, it satisfies $\lambda v = -\gamma v ''$. Displaying this for the linear operator $L$, it can be shown as $\displaystyle L[v] = -\gamma v ''$, hence it could be expressed as $L[v] = \lambda v$. Here, we define $\lambda$ as the eigenvalue for the nontrivial solution $v \ne 0$, and $v$ as the eigenfunction corresponding to $\lambda$.
Step 2. Solving the Ordinary Differential Equation $\displaystyle v '' + {{\lambda } \over {\gamma}} v = 0$
Assuming $v(x) = e^{z x }$ leads to
$$ v '' + {{\lambda } \over {\gamma}} v = 0 \implies z^2 e^{zx} + {{\lambda } \over {\gamma}} e^{zx} = 0 \implies z^2 + {{\lambda } \over {\gamma}} = 0 $$ Essentially, solving $\displaystyle v '' + {{\lambda } \over {\gamma}} v = 0$ means solving the characteristic equation $\displaystyle z^2 + {{\lambda } \over {\gamma}} = 0$.
Step 3.
Having $u(t,x) = e^{- \lambda t} v(X)$ as a candidate for the solution, we just need to find $v(x)$ to conclude the solution.
Case 1. $\lambda < 0$
The solution to the characteristic equation is $\displaystyle z = \pm \sqrt{ - { \lambda } \over { \gamma} }$, so the fundamental solution is
$$ z = e^{ \sqrt{ -{{\lambda} \over {\gamma}} } x }, e^{ - \sqrt{ -{{\lambda} \over {\gamma}} } x } $$
Therefore the solution for some constant $c_{1} , c_{2} \in \mathbb{C}$ is:
$$ u(t,x) = e^{ - \lambda t } \left( c_{1} e^{ \sqrt{ -{{\lambda} \over {\gamma}} } x } + c_{2 } e^{ - \sqrt{ -{{\lambda} \over {\gamma}} } x } \right) $$
Case 2. $\lambda = 0$
The solution to the characteristic equation is $\displaystyle z = 0$, so the fundamental solution is $\displaystyle z = 1, x$; therefore, the solution for some constant $c_{1} , c_{2} \in \mathbb{C}$ is
$$ u(t,x) = c_{1} + c_{2} x $$
Case 3. $\lambda > 0$
The solution to the characteristic equation is $\displaystyle z = \pm i \sqrt{ - { \lambda } \over { \gamma} }$, so the fundamental solution is
$$ z = e^{ i \sqrt{ -{{\lambda} \over {\gamma}} } x }, e^{ - i \sqrt{ -{{\lambda} \over {\gamma}} } x } $$
Therefore, the solution for some constant $c_{1} , c_{2} \in \mathbb{C}$ is
$$ u(t,x) = e^{ - \lambda t } \left( c_{1} e^{ i \sqrt{ -{{\lambda} \over {\gamma}} } x } + c_{2 } e^{ - i \sqrt{ -{{\lambda} \over {\gamma}} } x } \right) $$
Case 4. $\lambda \notin \mathbb{R}$
Let’s represent it as $\lambda = r e^{i \theta}$.
Since $\displaystyle z^2 = - {{\lambda} \over {\gamma}} = {{r } \over {\gamma}} e^{ i \theta}$, the fundamental solution is
$$ z = \sqrt{ {{r} \over {\gamma}} } e^{ i {{ \theta } \over {2}} }, \sqrt{ {{r} \over {\gamma}} } e^{ i \left( {{ \theta } \over {2}} + \pi\right) } $$
Therefore, the solution for some constant $c_{1} , c_{2} \in \mathbb{C}$ is
$$ u(t,x) = \exp \left( r e^{i \theta} t \right) \left[ c_{1} \exp \left( \sqrt{ {{r} \over {\gamma}} } e^{ i {{ \theta } \over {2}} } x \right) + c_{2 } \exp \left( \sqrt{ {{r} \over {\gamma}} } e^{ i \left( {{ \theta } \over {2}} + \pi\right) } x \right) \right] $$
However, a separate check is needed to confirm if $\lambda$ is indeed an eigenvalue.
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