📂Partial Differential Equations

Solution to the Initial Value Problem for the Heat Equation Given Dirichlet Boundary Conditions

Description

$$\begin{cases} u_{t} = \gamma u_{xx} \\ u(t,0) = u(t,l) = 0 \\ u(0,x) = f(x) \end{cases}$$

The above equation is a Dirichlet boundary condition in a $1$-dimensional space with length of $l$ from Heat Equation

$$\begin{cases} u(t,0) = \alpha (t) \\ u(t,l) = \beta (t) \end{cases}$$

This is given by $\alpha = \beta = 0$ with initial conditions for the heat distribution. Among such problem types, this is the simplest and easiest form. Here, $t$ represents time, $x$ represents position, and $u(t,x)$ represents the distribution of heat at time $t$. $\gamma$ is the thermal diffusivity, the larger it is, the faster the change in distribution. $f$ is the initial condition, specifically representing the distribution of heat when $t=0$.

Solution

Continued from Solution of Heat Equation.

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• Step 4. Check if $\lambda$ is an eigenvalue

  With the candidate solution being $u(t,x) = e^{-\lambda t} v(X)$, due to Dirichlet boundary condition, it follows that

  $$e^{-\lambda t} v(0) = e^{-\lambda t} v(l) = 0$$

  Hence, it’s necessary to verify whether the non-trivial solution $v$ satisfies $v(0) = v(l) = 0$.

  • Case 1. $\lambda < 0$ For some constant $c_{1} , c_{2} \in \mathbb{C}$,

    $$v(x) = c_{1} e^{ \sqrt{ -{{\lambda} \over {\gamma}} } x } + c_{2 } e^{ - \sqrt{ -{{\lambda} \over {\gamma}} } x }$$

    Hence,

    $$v(0) = 0 \implies c_{1} + c_{2} = 0 \\ v(l) = 0 \implies c_{1} e^{ \sqrt{ -{{\lambda} \over {\gamma}} } l } + c_{2 } e^{ - \sqrt{ -{{\lambda} \over {\gamma}} } l } = 0$$

    The only thing satisfying this simultaneously is $c_{1} = c_{2} = 0$. Since $v(x) = 0$, $v$ becomes a trivial solution, making $\lambda$ not an eigenvalue.

  • Case 2. $\lambda = 0$

    For some constant $c_{1} , c_{2} \in \mathbb{C}$, since $\displaystyle v(x) = c_{1} + c_{2} x$,

    $$v(0) = 0 \implies c_{1} = 0 \\ v(l) = 0 \implies c_{1} + c_{2 } l = 0$$

    The only thing satisfying this simultaneously is $c_{1} = c_{2} = 0$. Since $v(x) = 0$, $v$ becomes a trivial solution, making $\lambda$ not an eigenvalue.

  • Case 4. $\lambda \notin \mathbb{R}$

    Let’s express it as $\lambda = r e^{i \theta}$. For some constant $c_{1} , c_{2} \in \mathbb{C}$,

    $$v(x) = c_{1} \exp \left( \sqrt{ {{r} \over {\gamma}} } e^{ i {{ \theta } \over {2}} } x \right) + c_{2 } \exp \left( \sqrt{ {{r} \over {\gamma}} } e^{ i \left( {{ \theta } \over {2}} + \pi\right) } x \right)$$

    Hence, from $v(0) = v(l) = 0$,

    $$\begin{bmatrix} 1 & 1 \\ \exp \left( \sqrt{ {{r} \over {\gamma}} } e^{i {{\theta} \over {2}} } l \right) & \exp \left( \sqrt{ {{r} \over {\gamma}} } e^{i \left( {{\theta} \over {2}} + \pi \right) } l \right) \end{bmatrix} \begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

    Here,

    $$\begin{align*} && \det \begin{bmatrix} 1 & 1 \\ \exp \left( \sqrt{ {{r} \over {\gamma}} } e^{i {{\theta} \over {2}} } l \right) & \exp \left( \sqrt{ {{r} \over {\gamma}} } e^{i \left( {{\theta} \over {2}} + \pi \right) } l \right) \end{bmatrix} \\ =& \exp \left( \sqrt{ {{r} \over {\gamma}} } e^{i \left( {{\theta} \over {2}} + \pi \right) } l \right) - \exp \left( \sqrt{ {{r} \over {\gamma}} } e^{i {{\theta} \over {2}} } l \right) \\ &\ne& 0 \end{align*}$$

    Therefore, $c_{1} = c_{2} = 0$ and since $v(x) = 0$, $v$ becomes a trivial solution, making $\lambda$ not an eigenvalue.

  Hence, in Case 1., Case 2., and Case 4., $u(t,x) = 0$ is found.

  • Case 3. $\lambda > 0$

    For some constant $c_{1} , c_{2} \in \mathbb{C}$,

    $$v(x) = c_{1} e^{ i \sqrt{ -{{\lambda} \over {\gamma}} } x } + c_{2 } e^{ - i \sqrt{ -{{\lambda} \over {\gamma}} } x }$$

    Hence, from $v(0) = v(l) = 0$,

    $$\begin{bmatrix} 1 & 0 \\ \cos \left( \sqrt{ {{ \lambda } \over {\gamma}} } l \right) & \sin \left( \sqrt{ {{ \lambda } \over {\gamma}} } l \right) \end{bmatrix} \begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

    Here,

    $$\det \begin{bmatrix} 1 & 0 \\ \cos \left( \sqrt{ {{ \lambda } \over {\gamma}} } l \right) & \sin \left( \sqrt{ {{ \lambda } \over {\gamma}} } l \right) \end{bmatrix} = \sin \left( \sqrt{ {{ \lambda } \over {\gamma}} } l \right)$$

    i.e., only $\displaystyle \lambda = \lambda_{n} : = \gamma \left( {{n \pi} \over {l}} \right)^2$ that make $\displaystyle \sin \left( \sqrt{ {{ \lambda } \over {\gamma}} } l \right) = 0$ happens are the eigenvalues. In Case 3., we move to Step 5..

• Step 5.

  Let’s assume that $\displaystyle u(t,x) := \sum_{n=1}^{\infty} b_{n} u_{n} (t,X)$ converges for all $x \in [ 0 , l ]$ when

  $$u_{n}(t,x) := \exp \left( - {{\gamma n^2 \pi^2 } \over {l^2}} t \right) \sin \left( {{ n \pi x } \over {l}} \right) \\ b_{n} := {{1 } \over {l}} \int_{-l}^{l} f(x) \sin \left( {{ n \pi x } \over {l}} \right) dx$$

  This solution satisfies the heat equation and meets the Dirichlet boundary conditions. Expanding this:

  $$u(t,x) = \sum_{n=1}^{\infty} \left< f(x) , \sin \left( {{ n \pi x } \over {l}} \right) \right> \exp \left( - {{\gamma n^2 \pi^2 } \over {l^2}} t \right) \sin \left( {{ n \pi x } \over {l}} \right)$$

  When $t=0$,

  $$u(0,x) = \sum_{n=1}^{\infty} \left< f(x) , \sin \left( {{ n \pi x } \over {l}} \right) \right> \sin \left( {{ n \pi x } \over {l}} \right) \sim f(x)$$

  Therefore, if it converges, the initial condition $f(x) = u(0,x)$ is also satisfied.

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From a physical sense, it’s obvious that $\lambda$ only makes sense when positive during the solution process. Since $\lambda$ is a diffusion coefficient, if it could be negative, phenomena violating the Second Law of Thermodynamics would be rampant.

See Also

• Numerical Solution