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Solutions to Heat Equations 📂Partial Differential Equations

Solutions to Heat Equations

Description

ut=γuxx u_{t} = \gamma u_{xx}

The above equation is a specialized case of the generalized heat equation

∂∂t(σ(x)u)=∂∂x(κ(x)∂u∂x) {{\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 κ(x)>0\kappa (x) > 0 and the heat capacity σ(x)>0\sigma (x) > 0 are both constants, leading to γ:=κσ\displaystyle \gamma : = {{\kappa} \over {\sigma}} being defined as the thermal diffusivity.

Here, tt represents time, xx represents position, and u(t,x)u(t,x) represents the distribution of heat at time tt. γ\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. L[u]:=−γ∂2u∂x2\displaystyle L[ u ] := -\gamma {{\partial^2 u} \over { \partial x^2 }}

    Define the linear operator LL as follows and assume the solution is represented by u(t,x)=e−λtv(X)u(t,x) = e^{- \lambda t} v(X):

    ut=∂u∂t=−λe−λtv(x)uxx=∂2u∂x2=e−λtv′′(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 uu is the solution to the given equation, then

    ut=λe−λtv=−γe−λtv′′=uxx u_{t} = \lambda e^{- \lambda t } v = - \gamma e^{- \lambda t } v '' = u_{xx}

    upon rearrangement, it satisfies λv=−γv′′\lambda v = -\gamma v ''. Displaying this for the linear operator LL, it can be shown as L[v]=−γv′′\displaystyle L[v] = -\gamma v '', hence it could be expressed as L[v]=λvL[v] = \lambda v. Here, we define λ\lambda as the eigenvalue for the nontrivial solution v≠0v \ne 0, and vv as the eigenfunction corresponding to λ\lambda.

  • Step 2. Solving the Ordinary Differential Equation v′′+λγv=0\displaystyle v '' + {{\lambda } \over {\gamma}} v = 0

    Assuming v(x)=ezxv(x) = e^{z x } leads to

    v′′+λγv=0  ⟹  z2ezx+λγezx=0  ⟹  z2+λγ=0 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 v′′+λγv=0\displaystyle v '' + {{\lambda } \over {\gamma}} v = 0 means solving the characteristic equation z2+λγ=0\displaystyle z^2 + {{\lambda } \over {\gamma}} = 0.

  • Step 3.

    Having u(t,x)=e−λtv(X)u(t,x) = e^{- \lambda t} v(X) as a candidate for the solution, we just need to find v(x)v(x) to conclude the solution.

    • Case 1. λ<0\lambda < 0

      The solution to the characteristic equation is z=±−λγ\displaystyle z = \pm \sqrt{ - { \lambda } \over { \gamma} }, so the fundamental solution is

      z=e−λγx,e−−λγx z = e^{ \sqrt{ -{{\lambda} \over {\gamma}} } x }, e^{ - \sqrt{ -{{\lambda} \over {\gamma}} } x }

      Therefore the solution for some constant c1,c2∈Cc_{1} , c_{2} \in \mathbb{C} is:

      u(t,x)=e−λt(c1e−λγx+c2e−−λγx) 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. λ=0\lambda = 0

      The solution to the characteristic equation is z=0\displaystyle z = 0, so the fundamental solution is z=1,x\displaystyle z = 1, x; therefore, the solution for some constant c1,c2∈Cc_{1} , c_{2} \in \mathbb{C} is

      u(t,x)=c1+c2x u(t,x) = c_{1} + c_{2} x

    • Case 3. λ>0\lambda > 0

      The solution to the characteristic equation is z=±i−λγ\displaystyle z = \pm i \sqrt{ - { \lambda } \over { \gamma} }, so the fundamental solution is

      z=ei−λγx,e−i−λγx z = e^{ i \sqrt{ -{{\lambda} \over {\gamma}} } x }, e^{ - i \sqrt{ -{{\lambda} \over {\gamma}} } x }

      Therefore, the solution for some constant c1,c2∈Cc_{1} , c_{2} \in \mathbb{C} is

      u(t,x)=e−λt(c1ei−λγx+c2e−i−λγx) 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. λ∉R\lambda \notin \mathbb{R}

      Let’s represent it as λ=reiθ\lambda = r e^{i \theta}.

      Since z2=−λγ=rγeiθ\displaystyle z^2 = - {{\lambda} \over {\gamma}} = {{r } \over {\gamma}} e^{ i \theta}, the fundamental solution is

      z=rγeiθ2,rγei(θ2+π) 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 c1,c2∈Cc_{1} , c_{2} \in \mathbb{C} is

      u(t,x)=exp⁡(reiθt)[c1exp⁡(rγeiθ2x)+c2exp⁡(rγei(θ2+π)x)] 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 λ\lambda is indeed an eigenvalue.

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