[Math] Differential equation with homogeneous Dirichlet-Neumann boundary conditions

heat equationordinary differential equationspartial differential equations

The general solution of the heat equation by separation of the variables is
$$X(x)=a \cos\omega x + b \sin \omega x$$
Dirichlet

with the Dirichlet boundary conditions
$$u_t = \kappa u_{xx}(x,t)\\
u(0,t)=T_0, u(L,t)=T_L\\
u(x,0)=f(x)$$
Since the steady state codition is linear between two Dirichlet boundary conditions, we have
$$u(x,t)=u_s(x)+\nu(x,t)$$
and then we can apply new boundary conditions in the general solution
$$\nu(0,t)=\nu(L,t)=0$$

Neumann

with Nwumann homogenous boundary conditions, we already have
$$u_x(0,t)=u_x(L,t)=0$$
and then the steady state condition is not required, we can apply the boundary conditions to the general solution.

Dirichlet-Neumann

Consider the boundary conditions for a metal bar with an end at a fixed temperature and the end is insulated:
$$u(0,t)=T_1\\
u_x(L,t)=0$$

How can we apply these boundary conditions to the general solution to eliminate one term and obtain the coefficient?

Best Answer

How about defining $u = T_1 + v$ and demanding $v(0,t) = v_x(L,t) = 0$? Notice that the problem for $v$ becomes homogenous (PDE and BCs) with initial condition $v(x,0) = f(x) - T_1$.

Related Solutions

Diffusion Equation – Diffusion Equation with Inhomogeneous Boundary Conditions (The Subtraction Method)

In fact the "subtraction method" you so called is a little trick that pointedly changing some of the conditions from inhomogeneous to become homogeneous by applying the variable transformation on the dependent variable only.

This little trick is briefly introduced in http://maths.swan.ac.uk/staff/vl/pdes-course.pdf#page=32 and http://maths.swan.ac.uk/staff/vl/pdes-course.pdf#page=38.

For example the PDEs with dependent variable $u$ and independent variables $x$ and $t$ , the little trick is as follows:

for pointedly changing $u(0,t)=f(t)$ and $u(a,t)=g(t)$ $(a\neq0)$ : Let $v(x,t)=u(x,t)-f(t)-\dfrac{x}{a}(g(t)-f(t))$

for pointedly changing $u(0,t)=f(t)$ and $u_x(a,t)=g(t)$ : Let $v(x,t)=u(x,t)-f(t)-xg(t)$

for pointedly changing $u_x(0,t)=f(t)$ and $u(a,t)=g(t)$ : Let $v(x,t)=u(x,t)-(x-a)f(t)-g(t)$

for pointedly changing $u_x(0,t)=f(t)$ and $u_x(a,t)=g(t)$ $(a\neq0)$ : Let $v(x,t)=u(x,t)-h(t)-xf(t)-\dfrac{x^2}{2a}(g(t)-f(t))$

The other types of pointedly changing examples leaves yourself to think.

Sometimes this little trick is a must to apply, for example in Boundaries in heat equation and Heat equation with an unknown diffusion coefficient, otherwise you will get the wrong conclusion that has no solution.

For this problem, you might let $v(x,t)=u(x,t)-b(t)-xb(t)$ , but since there are no other conditions exist, so you have not necessarily to apply this trick, just seckilling this problem by one of the these two approaches:

Approach $1$: separation of variables

Case $1$: $\text{Re}(t)\geq0$

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

Then $X(x)T'(t)=X''(x)T(t)$

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

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

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

$\therefore u(x,t)=C_1x+C_2+\int_0^\infty C_3(s^2)e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds$

$u_x(x,t)=C_1+\int_0^\infty sC_3(s^2)e^{-ts^2}\cos xs~ds-\int_0^\infty sC_4(s^2)e^{-ts^2}\sin xs~ds$

$u_x(0,t)=b(t)$ :

$C_1+\int_0^\infty sC_3(s^2)e^{-ts^2}~ds=b(t)$

$\int_0^\infty\dfrac{C_3(s^2)e^{-ts^2}}{2}d(s^2)=b(t)-C_1$

$\int_0^\infty\dfrac{C_3(s)e^{-ts}}{2}ds=b(t)-C_1$

$\mathcal{L}_{s\to t}\biggl\{\dfrac{C_3(s)}{2}\biggr\}=b(t)-C_1$

$C_3(s)=2\mathcal{L}^{-1}_{t\to s}\{b(t)\}-2C_1\delta(s)$

$\therefore u(x,t)=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds-2C_1\int_0^\infty\delta(s^2)e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds-C_1\int_0^\infty\dfrac{\delta(s^2)e^{-ts^2}\sin xs}{s}d(s^2)+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds-C_1\int_0^\infty\dfrac{\delta(s)e^{-ts}\sin x\sqrt{s}}{\sqrt{s}}ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds-C_1\lim\limits_{s\to 0}\dfrac{e^{-ts}\sin x\sqrt{s}}{\sqrt{s}}+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds-C_1\lim\limits_{s\to 0}\dfrac{\dfrac{xe^{-ts}\cos x\sqrt{s}}{2\sqrt{s}}-te^{-ts}\sin x\sqrt{s}}{\dfrac{1}{2\sqrt{s}}}+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds-C_1\lim\limits_{s\to 0}(xe^{-ts}\cos x\sqrt{s}-2t\sqrt{s}e^{-ts}\sin x\sqrt{s})+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds-C_1x+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds$

$u(0,t)=b(t)$ :

$C_2+\int_0^\infty C_4(s^2)e^{-ts^2}~ds=b(t)$

$\int_0^\infty\dfrac{C_4(s^2)e^{-ts^2}}{2s}d(s^2)=b(t)-C_2$

$\int_0^\infty\dfrac{C_4(s)e^{-ts}}{2\sqrt{s}}ds=b(t)-C_2$

$\mathcal{L}_{s\to t}\biggl\{\dfrac{C_4(s)}{2\sqrt{s}}\biggr\}=b(t)-C_2$

$C_4(s)=2\sqrt{s}\mathcal{L}^{-1}_{t\to s}\{b(t)\}-C_2\delta(\sqrt{s})$

$\therefore u(x,t)=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\cos xs~ds-C_2\int_0^\infty\delta(s)e^{-ts^2}\cos xs~ds=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\cos xs~ds-C_2=2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\cos xs~ds$

Case $2$: $\text{Re}(t)\leq0$

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

Then $X(x)T'(t)=X''(x)T(t)$

$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)}{X(x)}=s^2$

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

$\begin{cases}T(t)=c_3(s^2)e^{ts^2}\\X(x)=\begin{cases}c_1(s^2)\sinh xs+c_2(s^2)\cosh xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1x+C_2+\int_0^\infty C_3(s^2)e^{ts^2}\sinh xs~ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds$

$u_x(x,t)=C_1+\int_0^\infty sC_3(s^2)e^{ts^2}\cosh xs~ds+\int_0^\infty sC_4(s^2)e^{ts^2}\sinh xs~ds$

$u_x(0,t)=b(t)$ :

$C_1+\int_0^\infty sC_3(s^2)e^{ts^2}~ds=b(t)$

$\int_0^\infty\dfrac{C_3(s^2)e^{ts^2}}{2}d(s^2)=b(t)-C_1$

$\int_0^\infty\dfrac{C_3(s)e^{ts}}{2}ds=b(t)-C_1$

$\int_0^\infty\dfrac{C_3(s)e^{-ts}}{2}ds=b(-t)-C_1$

$\mathcal{L}_{s\to t}\biggl\{\dfrac{C_3(s)}{2}\biggr\}=b(-t)-C_1$

$C_3(s)=2\mathcal{L}^{-1}_{t\to s}\{b(-t)\}-2C_1\delta(s)$

$\therefore u(x,t)=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds-2C_1\int_0^\infty\delta(s^2)e^{ts^2}\sinh xs~ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds-C_1\int_0^\infty\dfrac{\delta(s^2)e^{ts^2}\sinh xs}{s}d(s^2)+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds-C_1\int_0^\infty\dfrac{\delta(s)e^{ts}\sinh x\sqrt{s}}{\sqrt{s}}ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds-C_1\lim\limits_{s\to 0}\dfrac{e^{ts}\sinh x\sqrt{s}}{\sqrt{s}}+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds-C_1\lim\limits_{s\to 0}\dfrac{\dfrac{xe^{ts}\cosh x\sqrt{s}}{2\sqrt{s}}+te^{ts}\sinh x\sqrt{s}}{\dfrac{1}{2\sqrt{s}}}+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds-C_1\lim\limits_{s\to 0}(xe^{ts}\cosh x\sqrt{s}+2t\sqrt{s}e^{ts}\sinh x\sqrt{s})+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds-C_1x+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds$

$u(0,t)=b(t)$ :

$C_2+\int_0^\infty C_4(s^2)e^{ts^2}~ds=b(t)$

$\int_0^\infty\dfrac{C_4(s^2)e^{ts^2}}{2s}d(s^2)=b(t)-C_2$

$\int_0^\infty\dfrac{C_4(s)e^{ts}}{2\sqrt{s}}ds=b(t)-C_2$

$\int_0^\infty\dfrac{C_4(s)e^{-ts}}{2\sqrt{s}}ds=b(-t)-C_2$

$\mathcal{L}_{s\to t}\biggl\{\dfrac{C_4(s)}{2\sqrt{s}}\biggr\}=b(-t)-C_2$

$C_4(s)=2\sqrt{s}\mathcal{L}^{-1}_{t\to s}\{b(-t)\}-C_2\delta(\sqrt{s})$

$\therefore u(x,t)=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\cosh xs~ds-C_2\int_0^\infty\delta(s)e^{ts^2}\cosh xs~ds=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\cosh xs~ds-C_2=2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\cosh xs~ds$

Hence $u(x,t)=\begin{cases}2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\cos xs~ds&\text{when Re}(t)\geq0\\2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\cosh xs~ds&\text{when Re}(t)\leq0\end{cases}$

Approach $2$: power series method

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

Then $u(x,t)=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}\dfrac{\partial^{2n}u(0,t)}{\partial x^{2n}}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(0,t)}{\partial x^{2n+1}}=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}\dfrac{\partial^nu(0,t)}{\partial t^n}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{(2n+1)!}\dfrac{\partial^nu_x(0,t)}{\partial t^n}=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}\dfrac{\partial^nb(t)}{\partial t^n}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{(2n+1)!}\dfrac{\partial^nb(t)}{\partial t^n}$

[Math] Neumann condition and Dirichlet condition at the same point

applying Dirichlet boundary conditions will override your Neumann boundary conditions in the case of the finite element method (I give this as an example, as you mentioned FEM in your question). Dirichlet and Neumann boundaries should not overlap.

When you impose a prescribed temperature, you are also imposing a heat flux, although implicitly. As you can imagine, that heat flux is directly related to the prescribed temperature and the temperature inside your bar. Given a prescribed temperature, you have a prescribed heat flux (one-to-one relation).

So, to answer your question, in the real world, it does not make sense to impose both, as they are not independent one from another.

Best Answer

Related Solutions

Diffusion Equation – Diffusion Equation with Inhomogeneous Boundary Conditions (The Subtraction Method)

[Math] Neumann condition and Dirichlet condition at the same point

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