Two-dimensional heat conduction boundary value problem

Let us consider the two-dimensional nonsymmetric heat conduction problem in a disk of radius $R$. The heat conduction equation (1.8$\displaystyle \frac{\partial T(M, t)}{\partial t}$ $\displaystyle = a^2 \cdot \Delta T(M, t)$ $\displaystyle + f(M, t),$) in polar coordinates takes the form

The Dirichlet boundary conditions (1.13$\displaystyle T(M, t)$ $\displaystyle = \Phi(M, t), \quad M$ $\displaystyle \in S.$) in the two-dimensional case have the form

where $T_0(r, \theta)$ is the initial condition, $\Phi(\theta, t)$ is the boundary condition, which is, of course, periodic in $\theta$.

To get rid of the inhomogeneity, we introduce $T(r, \theta, t)$ $= \widehat{T}(r, \theta, t)$ $+ U(\theta, t)$, let us choose the function $U(\theta, t)$ $= \Phi(\theta, t)$, in accordance with (2.2$\displaystyle \widehat{T}(M, t)$ $\displaystyle + U(M, t)$ $\displaystyle = \Phi(M, t), \quad \widehat{T}(M, t)$ $\displaystyle = 0 \;$ $\displaystyle \Rightarrow\; U(M, t)$ $\displaystyle = \Phi(M, t), \quad M$ $\displaystyle \in S.$). Then equation (2.47$\displaystyle \frac{\partial T(r, \theta, t)}{\partial t}$ $\displaystyle = a^2 \cdot \frac{\partial^2 T(r, \theta, t)}{\partial r^2}$ $\displaystyle + a^2 \cdot \frac{1}{r} \cdot \frac{\partial T(r, \theta, t)}{\partial r}$ $\displaystyle + a^2 \cdot \frac{1}{r^2} \cdot \frac{\partial^2 T(r, \theta, t)}{\partial \theta^2}$ $\displaystyle + f(r, \theta, t).$) and the boundary conditions (2.48$\displaystyle T(r, \theta, 0)$ $\displaystyle = T_0(r, \theta),$$\displaystyle T(R, \theta, t)$ $\displaystyle = \Phi(\theta, t),$$\displaystyle T(r, \theta, t)$ $\displaystyle = T(r, \theta + 2\pi, t),$) take the form

where $\widehat{f}(r, \theta, t)$ $= f(r, \theta, t)$ $- \frac{\displaystyle \partial U(\theta, t)}{\displaystyle \partial t}$ $+ a^2 \cdot \frac{\displaystyle 1}{\displaystyle r^2} \cdot \frac{\displaystyle \partial^2 U(\theta, t)}{\displaystyle \partial \theta^2}$.

The Neumann boundary conditions (1.14$\displaystyle \lambda \cdot \frac{\partial T(M, t)}{\partial \vec{n}}$ $\displaystyle = \Phi(M, t), \quad M$ $\displaystyle \in S,$) in the two-dimensional case have the form

where $\frac{\displaystyle \partial T(r, \theta, t)}{\displaystyle \partial \mathbf{n}} \bigg|_{r=R}$ is the normal derivative of the temperature on the circle of radius $R$; the direction of the normal coincides with the direction of the coordinate $r$.

To get rid of the inhomogeneity, we introduce $T(r, \theta, t)$ $= \widehat{T}(r, \theta, t)$ $+ U(r, \theta, t)$, let us choose the function $U(r, \theta, t)$ $= \Phi(\theta, t) \cdot r$ — it satisfies condition (2.3$\displaystyle \frac{\partial \widehat{T}(M, t)}{\partial \vec{n}}$ $\displaystyle + \frac{\partial U(M, t)}{\partial \vec{n}}$ $\displaystyle = \Phi(M, t), \quad \frac{\partial \widehat{T}(M, t)}{\partial \vec{n}}$ $\displaystyle = 0 \;$ $\displaystyle \Rightarrow\; \frac{\partial U(M, t)}{\partial \vec{n}}$ $\displaystyle = \Phi(M, t), \quad M$ $\displaystyle \in S.$): $\frac{\displaystyle \partial U(r, \theta, t)}{\displaystyle \partial \mathbf{n}} \bigg|_{r=R}$ $= \Phi(\theta, t)$. Then equation (2.47$\displaystyle \frac{\partial T(r, \theta, t)}{\partial t}$ $\displaystyle = a^2 \cdot \frac{\partial^2 T(r, \theta, t)}{\partial r^2}$ $\displaystyle + a^2 \cdot \frac{1}{r} \cdot \frac{\partial T(r, \theta, t)}{\partial r}$ $\displaystyle + a^2 \cdot \frac{1}{r^2} \cdot \frac{\partial^2 T(r, \theta, t)}{\partial \theta^2}$ $\displaystyle + f(r, \theta, t).$) and the boundary conditions (2.50$\displaystyle T(r, \theta, 0)$ $\displaystyle = T_0(r, \theta),$$\displaystyle \frac{\displaystyle \partial T(r, \theta, t)}{\displaystyle \partial \mathbf{n}} \bigg|_{r=R}$ $\displaystyle = \Phi(\theta, t),$$\displaystyle T(r, \theta, t)$ $\displaystyle = T(r, \theta + 2\pi, t),$) take the form

where $\widehat{f}(r, \theta, t)$ $= f(r, \theta, t)$ $- \frac{\displaystyle \partial U(r, \theta, t)}{\displaystyle \partial t}$ $+ a^2 \cdot \frac{\displaystyle 1}{\displaystyle r} \cdot \frac{\displaystyle \partial U(r, \theta, t)}{\displaystyle \partial r}$ $+ a^2 \cdot \frac{\displaystyle 1}{\displaystyle r^2} \cdot \frac{\displaystyle \partial^2 U(r, \theta, t)}{\displaystyle \partial \theta^2}$.

Equation (2.8$\displaystyle \Delta \Psi(M)$ $\displaystyle + \gamma^2 \cdot \Psi(M)$ $\displaystyle = 0.$) takes the form

Let us represent the function $\Psi(r, \theta)$ as a product of two functions $\Psi(r, \theta)$ $= \Lambda(r) \cdot \Upsilon(\theta)$, substitute it into equation (2.52$\displaystyle \frac{\displaystyle \partial^2 \Psi(r, \theta)}{\displaystyle \partial r^2}$ $\displaystyle + \frac{\displaystyle 1}{\displaystyle r} \cdot \frac{\displaystyle \partial \Psi(r, \theta)}{\displaystyle \partial r}$ $\displaystyle + \frac{\displaystyle 1}{\displaystyle r^2} \cdot \frac{\displaystyle \partial^2 \Psi(r, \theta)}{\displaystyle \partial \theta^2}$ $\displaystyle + \gamma^2 \cdot \Psi(r, \theta)$ $\displaystyle = 0.$) and carry out the transformations

We introduced the constant $m^2$ by analogy with the separation of variables for time and geometry. Now let us rewrite it as a system of two equations: one in the angle $\theta$, the other in the radius $r$.

The solution of the equation for the angle $\theta$ can be represented by a sine or a cosine, since both of these functions are periodic and satisfy the boundary condition $\Upsilon(\theta)$ $= \Upsilon(\theta + 2\pi)$, which imposes a restriction on $m$ — it must be an integer. We get two families of solutions; later we will take their superposition.

Let us take the coefficients $C_{1m}$ and $C_{2m}$ in front of the sine and cosine equal to one — these factors are absorbed into the expansion coefficients anyway. The solution of the equation for the radius $r$ has the form

where $J_m(\gamma \cdot r)$ and $Y_m(\gamma \cdot r)$ are the Bessel functions of the first and second kind respectively, $A_{1m}$ and $A_{2m}$ are coefficients determined from the boundary conditions, and the weight ensuring the orthogonality of the eigenfunctions is $\rho(r, \theta)$ $= r$. Moreover, for the function $\Lambda_m(r)$ to be bounded as $r$ $\to 0$, it is necessary that $A_{2m}$ $= 0$: the Bessel function of the second kind is infinite at zero, as seen in figure (B.2).

Thus, the solutions of equation (2.52$\displaystyle \frac{\displaystyle \partial^2 \Psi(r, \theta)}{\displaystyle \partial r^2}$ $\displaystyle + \frac{\displaystyle 1}{\displaystyle r} \cdot \frac{\displaystyle \partial \Psi(r, \theta)}{\displaystyle \partial r}$ $\displaystyle + \frac{\displaystyle 1}{\displaystyle r^2} \cdot \frac{\displaystyle \partial^2 \Psi(r, \theta)}{\displaystyle \partial \theta^2}$ $\displaystyle + \gamma^2 \cdot \Psi(r, \theta)$ $\displaystyle = 0.$) have the form

Let us substitute the solutions (2.56$\displaystyle \Psi_{1m}(r, \theta)$ $\displaystyle = \Lambda_m(r) \cdot \Upsilon_{1m}(\theta)$ $\displaystyle = A_{1m} \cdot J_m(\gamma \cdot r) \cdot \sin(m \cdot \theta), m$ $\displaystyle \in (1..\infty),$$\displaystyle \Psi_{2m}(r, \theta)$ $\displaystyle = \Lambda_m(r) \cdot \Upsilon_{2m}(\theta)$ $\displaystyle = A_{1m} \cdot J_m(\gamma \cdot r) \cdot \cos(m \cdot \theta), m$ $\displaystyle \in (0..\infty).$) into the boundary conditions (2.49$\displaystyle \frac{\displaystyle \partial \widehat{T}(r, \theta, t)}{\displaystyle \partial t}$ $\displaystyle = a^2 \cdot \frac{\displaystyle \partial^2 \widehat{T}(r, \theta, t)}{\displaystyle \partial r^2}$ $\displaystyle + a^2 \cdot \frac{\displaystyle 1}{\displaystyle r} \cdot \frac{\displaystyle \partial \widehat{T}(r, \theta, t)}{\displaystyle \partial r}$ $\displaystyle + a^2 \cdot \frac{\displaystyle 1}{\displaystyle r^2} \cdot \frac{\displaystyle \partial^2 \widehat{T}(r, \theta, t)}{\displaystyle \partial \theta^2}$ $\displaystyle + \widehat{f}(r, \theta, t),$$\displaystyle \widehat{T}(r, \theta, 0)$ $\displaystyle = T_0(r, \theta)$ $\displaystyle - U(\theta, 0),$$\displaystyle \widehat{T}(R, \theta, t)$ $\displaystyle = 0,$$\displaystyle \widehat{T}(r, \theta, t)$ $\displaystyle = \widehat{T}(r, \theta + 2\pi, t),$) and obtain

Clearly, the solution makes sense only when $A_{1m}$ $\neq 0$, which is possible only when $J_m(\gamma \cdot R)$ $= 0$. Let us denote the roots of the equation $J_m(\mu)$ $= 0$ by $\mu_{mk}$ — then the eigenvalues take the form

Zero and negative roots are no good: for $\mu$ $= 0$ we have $\gamma$ $= 0$, and for $m$ $\ge 1$ the eigenfunction $J_m(0 \cdot r)$ $\equiv 0$ — identically zero, which is not an eigenfunction, while for $m$ $= 0$ zero is not a root at all: $J_0(0)$ $= 1$. Negative roots give no new solutions: $J_m(-\mu)$ $= (-1)^m \cdot J_m(\mu)$ — the eigenfunction is the same up to sign. Therefore we take only the positive roots: $\mu_{mk}, k$ $\in (1..\infty)$.

Thus, the eigenvalues and eigenfunctions have the form

where $A_{1mk}$ is an arbitrary constant factor: the eigenfunction is defined up to it. Let us set $A_{1mk}$ $= 1$ — this factor is absorbed into the expansion coefficients anyway, as in the general solution.

To expand functions into a Fourier series in $\Psi_{1mk}(r, \theta)$ and $\Psi_{2mk}(r, \theta)$ we need to compute the norms $\|\Psi_{1mk}(r, \theta)\|^2$ and $\|\Psi_{2mk}(r, \theta)\|^2$; the weight for polar coordinates is $\rho(r, \theta)$ $= r$.

Let us make the substitution $x$ $= \gamma_{mk} \cdot r$ $= \frac{\displaystyle \mu_{mk}}{\displaystyle R} \cdot r$.

We use formula (C.6$\displaystyle \int_0^{\mu} r \cdot J_m^2(r) \,dr$ $\displaystyle = \frac{\displaystyle \mu^2}{\displaystyle 2} \cdot J_{m+1}^2(\mu).$) to compute the norm

Finally we obtain

The norms $\|\Psi_{1mk}(r, \theta)\|^2$ and $\|\Psi_{2mk}(r, \theta)\|^2$ have the form

Now let us substitute the solution (2.56$\displaystyle \Psi_{1m}(r, \theta)$ $\displaystyle = \Lambda_m(r) \cdot \Upsilon_{1m}(\theta)$ $\displaystyle = A_{1m} \cdot J_m(\gamma \cdot r) \cdot \sin(m \cdot \theta), m$ $\displaystyle \in (1..\infty),$$\displaystyle \Psi_{2m}(r, \theta)$ $\displaystyle = \Lambda_m(r) \cdot \Upsilon_{2m}(\theta)$ $\displaystyle = A_{1m} \cdot J_m(\gamma \cdot r) \cdot \cos(m \cdot \theta), m$ $\displaystyle \in (0..\infty).$) into the boundary conditions (2.51$\displaystyle \frac{\displaystyle \partial \widehat{T}(r, \theta, t)}{\displaystyle \partial t}$ $\displaystyle = a^2 \cdot \frac{\displaystyle \partial^2 \widehat{T}(r, \theta, t)}{\displaystyle \partial r^2}$ $\displaystyle + a^2 \cdot \frac{\displaystyle 1}{\displaystyle r} \cdot \frac{\displaystyle \partial \widehat{T}(r, \theta, t)}{\displaystyle \partial r}$ $\displaystyle + a^2 \cdot \frac{\displaystyle 1}{\displaystyle r^2} \cdot \frac{\displaystyle \partial^2 \widehat{T}(r, \theta, t)}{\displaystyle \partial \theta^2}$ $\displaystyle + \widehat{f}(r, \theta, t),$$\displaystyle \widehat{T}(r, \theta, 0)$ $\displaystyle = T_0(r, \theta)$ $\displaystyle - U(r, \theta, 0),$$\displaystyle \frac{\displaystyle \partial \widehat{T}(r, \theta, t)}{\displaystyle \partial \mathbf{n}} \bigg|_{r=R}$ $\displaystyle = 0,$$\displaystyle \widehat{T}(r, \theta, t)$ $\displaystyle = \widehat{T}(r, \theta + 2\pi, t),$) and obtain

Clearly, the solution makes sense only when $A_{1m}$ $\neq 0$, which is possible only when $J_m'(\gamma \cdot R)$ $= 0$. Let us denote the roots of the equation $J_m'(\mu)$ $= 0$ by $\mu_{mk}$ — then the eigenvalues take the form

Here, unlike the Dirichlet problem, the zero mode survives: for $m$ $= 0$ zero is a root of the equation $J_0'(\mu)$ $= 0$, and it corresponds to $\gamma$ $= 0$ and the eigenfunction $\Psi(r, \theta)$ $= J_0(0)$ $= 1$ — a nonzero constant, that is, a fully fledged eigenfunction (the constant mode). We will count it as the first root: $\mu_{01}$ $= 0$. For $m$ $\ge 1$ the zero root gives no eigenfunction ($J_m(0 \cdot r)$ $\equiv 0$), so there we still take only the positive roots.

Thus, the eigenvalues and eigenfunctions have the form

where $A_{1mk}$ is an arbitrary constant factor; as before, we set it equal to one.

We compute the norms of the eigenfunctions in the same way as in the Dirichlet problem: the integral over the angle remains the same (2.59$\displaystyle \int_0^{2\pi} \sin^2(m \cdot \theta) \,d\theta$ $\displaystyle = \pi, m$ $\displaystyle \in (1..\infty),$$\displaystyle \int_0^{2\pi} \cos^2(m \cdot \theta) \,d\theta$ $\displaystyle = \begin{cases} 2 \cdot \pi, & m = 0, \\ \pi, & m \geq 1 \end{cases}, m$ $\displaystyle \in (0..\infty).$), while the integral over the radius takes a different value — we use formula (D.2$\displaystyle \int_0^{\mu} r \cdot J_m^2(r) \,dr$ $\displaystyle = \left( \mu^2 - m^2 \right) \cdot \frac{\displaystyle J_m^2(\mu)}{\displaystyle 2}.$) to compute the norm

Finally we obtain

The special case $\mu_{01}$ $= 0$ (for $\gamma_{01}$ $= 0$ the eigenfunction $\Psi_{201}(r, \theta)$ $= 1$) we compute separately

The norms $\|\Psi_{1mk}(r, \theta)\|^2$ and $\|\Psi_{2mk}(r, \theta)\|^2$ have the form

The final solution of the two-dimensional Dirichlet boundary value problem with inhomogeneous boundary conditions takes the form