These kinds of DE occur in problems involving temperature distribution,
vibrations and potentials. These problems are described by relatively simple
**linear second order PDEs.** The procedures described below are
used to **reduce the problem into two or more linear DE.**

We are going to work with function $u$, which as a function of two variables: either $x(x,y)$ or $u(x,t)$. Then the second order PDE which is discussed can be described as

$$ A\frac{\partial^2 u}{\partial x^2} + B\frac{\partial^2 u}{\partial x \partial y} + C\frac{\partial^2 u}{\partial y^2} + D\frac{\partial u}{\partial x} + E\frac{\partial u}{\partial y} + F(u) = G, $$where the coefficients $A,\ B,\ C,\ \dots ,\ G$ are functions of $x$ and $y$.

We have no ambitions to find the general solution of such DE. That would be too difficult and it is not so useful. We are looking for particular solutions in a form

$$ u(x,y) = X(x)\cdot Y(y) $$Then it is sometimes possible to reduce given linear PDE (in two variables) into two ODEs. Assuming that the solution is $u(x,y) = X(x)Y(y)$ we have to be able to differentiate $u(x,y)$ with respect to $x$, $y$ and so on:

$$ \frac{\partial u}{x} = X'Y, \hskip2em \frac{\partial u}{y} = XY', \hskip2em \frac{\partial^2 u}{x^2} = X''Y, \hskip2em \frac{\partial^2 u}{y^2} = XY''. $$The above statements come from product rule (one function is multiplied by another): let us have for example $u(x,y) = XY = (x^2) (y^3)$. Then $\partial u/\partial x = (2x)(y^3) + (x^2)(0) = 2xy^3 = X'Y$.

If we can separate $X$ to the left hand side of equation and $Y$ to the right
side of equation, then: the left side depends on $x$ and what is on the right
(in terms of $x$) is a constant. The right side depends on $y$ on what is on
the left (in terms of $y$) is constant. So we have two ODE which are solved
against a constant (which is often called *separation constant* $\lambda$). It is easily demonstrated on examples:

Use the method *separation of variables* to find product solution of PDE

We are looking for a solution in the form $y(x,y) = X(x)\cdot Y(y)$. Let us use the solution within given PDE:

$$ \begin{align} X'Y + 3XY' &= 0 \hskip2em / \cdot \frac{1}X \frac 1 Y \nonumber \\ X'\frac 1 X &= -3Y'\frac 1 Y \nonumber \end{align} $$The left side is a constant from the point of view of the right side and conversely. We can solve two ODE

$$ X'\frac 1 X = \lambda, \hskip2em -3Y'\frac 1 Y = \lambda, $$where $\lambda$ is an arbitrary constant, $X$ is $X(x)$, $Y$ is $Y(y)$. Let us solve $X'\frac 1 X = \lambda$ first:

$$ \begin{align} X'\frac 1 X &= \lambda \nonumber \\ X' - \lambda X &= 0 \nonumber \\ m - \lambda &= 0 \implies m = \lambda \nonumber \\ X(x) &= c_1e^{\lambda x} \nonumber \end{align} $$Now let us solve $Y(y)$:

$$ \begin{align} -3Y' \frac 1 Y &= \lambda \nonumber \\ -3Y' - \lambda Y &= 0 \nonumber \\ 3Y' + \lambda Y &= 0 \nonumber \\ 3m + \lambda &= 0 \implies m = -\frac \lambda 3 \nonumber \\ Y(y) &= c_2 e^{-\lambda/3 y} \nonumber \end{align} $$Since we expected solution in the form $u(x,y) = X(x)Y(y)$:

$$ u(x,y) = c_1e^{\lambda x} \cdot c_2 e^{-\lambda/3 y}= c_3 e^{\lambda(x-1/3y)} $$The constant $\lambda$ was laid as an arbitrary constant also. But it would be weird to use $\lambda$, let us rather use $c_4$ instead:

$$ \underline{\underline{u(x,y) = c_3 e^{c_4(3x-y)}}} $$ Use the method *separation of variables* to find product solution of PDE $u_x = u_y + u$

We are looking for a solution in the form $y(x,y) = X(x)\cdot Y(y)$ of PDE

$$ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} + u $$Let us use the solution $y(x,y) = X(x)\cdot Y(y)$ within given PDE:

$$ \begin{align} X'Y &= Y'X + XY \hskip2em / \cdot \frac 1 X \frac 1 Y \nonumber \\ \frac 1 X X' &= \frac 1 Y Y' + 1 \nonumber \end{align} $$As in the previous example, we have to separate variable $x$ from $y$. The consequence is we are solving two ODE $(\ref{ref:PDE_1})$ and $(\ref{ref:PDE_2})$ with arbitrary constant then $\lambda$ instead of PDE:

$$ \begin{align} \frac 1 X X' &= \color{red}{\lambda} = \frac 1 Y Y' + 1 \nonumber \\ \frac 1 X X' &= \lambda \label{ref:PDE_1} \\ \frac 1 Y' + 1 &= \lambda \label{ref:PDE_2} \\ \end{align} $$The first DE $\frac 1 X X' = \lambda$ has been solved in the previous example and its solution is $X(x) = c_1e^{\lambda x}$. The second one has solution

$$ \begin{align} \frac 1 Y Y' + 1 &= \lambda \hskip2em / \cdot Y \nonumber \\ 1Y' + Y(1-\lambda) &= 0 \nonumber \\ m +(1-\lambda) &= 0 \implies m = \lambda - 1 \nonumber \\ Y(y) &= c_2e^{(\lambda - 1)y} \nonumber \end{align} $$Then we collect the solution of two ODE into expected form

$$ u(x,y) = X(x)Y(y) = c_1e^{\lambda x}\cdot c_2e^{(\lambda - 1)y} = c_3e^{\lambda x + \lambda y - y} = c_3e^{\lambda(x+y)-y} $$Finally it is good idea to exchange $\lambda$ to some other symbol:

$$ \underline{\underline{u(x,y) = c_3e^{c_4(x+y)-y}}}. $$We might differentiate $u(x,y)$ to express $\partial u/\partial x$ and $\partial u/\partial y$ in order to substitute back into PDE. That will provide as a confidence that no mistake was made through calculations.

Use the method *separation of variables* to find product solution of PDE

We are looking for a solution $u(x,y) = X(x)Y(y)$. So we have to insert expected solution into DE:

$$ \begin{align} yX'Y + xXY' &= 0 \hskip2em / \cdot \frac 1 {X'} \frac{1}{Y'} \nonumber \\ y \frac 1 {Y'}Y + xX\frac{1}{X'} &= 0 \nonumber \\ y \frac 1 {Y'}Y &= -xX\frac{1}{X'} \nonumber \\ \end{align} $$The variables are separated. On the left is function of $x$ and whatever is now being left on the right side can be considered as constant. And conversely. So we can have two DE which are solved against arbitrary constant $\lambda$:

$$ y\frac 1 {Y'}Y = \lambda $$The above DE is ordinary differential equation and is solved here with changed notation (for convenience) from $Y(y)$ to $y(x)$.

$$ \begin{align} x\frac{1}{y'}y &= \lambda \nonumber \\ xy &= \lambda y' \nonumber \\ y' \color{red}{- \frac 1 {\lambda}x}y &= 0 \nonumber \end{align} $$The DE turned to be solvable as a linear DE with $P(x) = -1/\lambda\ x$. Such DE is solved by means of integrating factor $\mu(x) = e^{\int P(x)\ dx}$

$$ \mu(x) = e^{-1/\lambda \int x\ dx} = e^{-\frac 1 {\lambda}\cdot \frac{x^2}{2}} $$Whole DE is multiplied by the integrating factor and then we anticipate $d/dx\ \mu(x)y$ on the left hand side and such term is trivial to integrate:

$$ \begin{align} \frac{d}{dx}\mu(x) y &= 0\nonumber \\ y'\cdot e^{-\frac{1}{\lambda}\cdot \frac{x^2}{2}} - \frac 1 {\lambda}xy \cdot e^{-\frac{1}{\lambda}\cdot \frac{x^2}{2}} &= 0\nonumber\\ \frac{d}{dx}e^{-\frac{1}{\lambda}\cdot \frac{x^2}{2}} \cdot y &= 0 \nonumber \\ e^{-\frac{1}{\lambda}\cdot \frac{x^2}{2}} \cdot y &= c_1 \nonumber \\ y(x) &= c_1e^{\frac{1}{\lambda}\cdot \frac{x^2}{2}} \nonumber \end{align} $$Note: could be solved easily also by separating variables.

So we have $Y(y) = c_1e^{\frac{1}{\lambda}\cdot \frac{y^2}{2}}$. That implies that the second solution is $X(x) = c_2e^{-\frac{1}{\lambda}\cdot \frac{x^2}{2}}$. Then the solution $u(x,y)$ is

$$ u(x,y) = X(x)\cdot Y(y) = c_1e^{\frac{1}{\lambda}\cdot \frac{y^2}{2}} \cdot c_2e^{-\frac{1}{\lambda}\cdot \frac{x^2}{2}}= c_3e^{\frac{1}{2\lambda}(y^2-x^2)} \nonumber $$After exchanging $c_4$ for $-1/2\lambda$

$$ \underline{\underline{u(x,y)= c_3e^{c_4(x^2-y^2)}}} $$We might have a desire to check the solution for a correctness. In that case, let us compute ${\partial u}/{\partial x},\ {\partial u}/{\partial y}$ and check whether equality holds when substituted into given DE $(\ref{ref:PDE_3})$.

$$ \begin{align} \frac{\partial u}{\partial x} &= c_3\cdot 2x\cdot c_4\cdot e^{c_4(x^2-y^2)} \nonumber \\ \frac{\partial u}{\partial y} &= c_3\cdot (-2y)\cdot c_4\cdot e^{c_4(x^2-y^2)} \nonumber \\ 0 &= y \cdot c_3\cdot 2x\cdot c_4\cdot e^{c_4(x^2-y^2)} + x\cdot c_3\cdot (-2y)\cdot c_4\cdot e^{c_4(x^2-y^2)} \nonumber\\ 0&=0 \nonumber \end{align} $$The other way to check for correctness is to use numerical check by calculator. Let us choose randomly some numbers for arbitrary constants and for $x,\ y$ as well. Let us evaluate numericaly the slope from the solution. When all substituted into DE, the equality have to (roughly) hold.

$$ \begin{align} &c_3 = 2,\ c_4 = 3,\ x= 1,\ y=2 \nonumber \\ &u(x=1, y=2) = 2.468196 \times 10^{-4} = u \nonumber \\ &u(x=1.001, y=2) = 2.483057 \times 10^{-4} = u_{\Delta x} \nonumber \\ &u(x=1, y=2.001) = 2.438747 \times 10^{-4} = u_{\Delta y} \nonumber \\ &\frac{\partial u}{\partial x} = \frac{\Delta u}{\Delta x} = \frac{u_{\Delta x} - u}{0.001} = 1.4861 \times 10^{-3} \nonumber \\ &\frac{\partial u}{\partial y} = \frac{\Delta u}{\Delta y} = \frac{u_{\Delta y} - u}{0.001} = -2.9449 \times 10^{-3} \nonumber \\ \end{align} $$Now substitute prepared values into given DE $(\ref{ref:PDE_3})$:

$$ 2\cdot 1.4861\times 10^{-3} + 1 \cdot (-2.9449\times 10^{-3}) = 0.0273\times 10 ^{-3} $$It can be considered that the equality holds. We did not expect that we are going to get exactly to zero.

Use separation of variables to find solution of

$$ k\frac{\partial^2u}{\partial x^2} - u = \frac{\partial u}{\partial t},\hskip2em k \gt 0. $$We are interested into solution $u(x,y) = X(x)T(t)$:

$$ \begin{align} k\cdot X''T - XT &= XT' \hskip2em /\cdot \frac 1 X \cdot \frac 1 T \nonumber \\ k\cdot X'' \frac 1 X -1 &= T'\frac 1 T \nonumber \\ kX''\frac{1}{X} &= 1 + T'\frac1 T \label{ref:PDE_4} \end{align} $$The variables are separated. Now we can solve two ODE against separation constant $\lambda$.

The left hand side from $(\ref{ref:PDE_4})$ is solved against constant $\lambda$:

$$ \begin{align} kX''\frac 1 X &= \lambda \nonumber \\ kX'' &= \lambda X \nonumber \\ kX''-\lambda X &= 0 \nonumber \end{align} $$Now it depends on the configuration of the problem (on the value of $\lambda$) what the solution would be.

$\boldsymbol{\lambda = 0:}$ $$ \begin{align} kX'' - 0 &= 0 \nonumber \\ X'' &= 0 \nonumber \\ X(x) &= c_1x + c_2 \nonumber \end{align} $$ |
$\boldsymbol{\lambda \gt 0\ (\lambda = \alpha^2k):}$ $$ \begin{align} kX''-\alpha^2k = &0 \nonumber \\ X'' - \alpha^2 = &0 \implies \nonumber \\ \implies m = &\{\alpha, -\alpha\} \nonumber \\ X(x) = &c_3\sinh \alpha x + \nonumber \\ & + c_4 \cosh \alpha x \nonumber \\ \end{align} $$ |
$\boldsymbol{\lambda \lt 0\ (\lambda = -\alpha^2k):}$ $$ \begin{align} kX''+\alpha^2k =& 0 \nonumber \\ X''+\alpha^2 =& 0 \nonumber \implies \nonumber \\ \implies m &= \{i \alpha, -i\alpha\} \nonumber \\ X(x) =& c_5\sin \alpha x + \nonumber \\ & + c_6\cos \alpha x \nonumber \end{align} $$ |

Note: the above solution with $\sinh \alpha x,\ \cosh \alpha x$ is equivalent to $C_3e^{-\alpha x} + C_4e^{\alpha x}$, because $\sinh x = (e^x-e^{-x}) / 2$ and $\cosh x = (e^x+e^{-x}) / 2$.

The right hand side from $(\ref{ref:PDE_4})$ is solved against constant $\lambda$:

$$ \begin{align} 1+ T'\frac 1 T &= \lambda \nonumber \\ T'- (\lambda - 1)T &= 0 \nonumber \\ T'+(1-\lambda)T &= 0 \nonumber \\ m + (1-\lambda) &= 0 \implies m= \lambda -1 \nonumber \\ T(t) &= c_7e^{(\lambda -1 )t} \nonumber \end{align} $$Finally, let us combine $X$ and $T$ into $u(x,y)$:

- When $\lambda=0:$

$\underline{\underline{u(x,y) = (c_1 + c_2x)e^{-t}}}$ - When $\lambda \gt 0\ (\lambda = \alpha^2k):$

$\underline{\underline{u(x,y)}} = c_7e^{(\lambda - 1)t}\cdot (c_3 \sinh \alpha x + c_4 \cosh \alpha x = \underline{\underline{e^{-t}e^{\alpha^2kt}\cdot c_7(c_3\sinh \alpha x + c_4\cosh \alpha x)}}$ - When $\lambda \lt 0\ (\lambda = -\alpha^2k):$

$\underline{\underline{u(x,y)}} = c_7e^{(\lambda -1)t}\cdot (c_5\sin \alpha x + c_6\cos \alpha x) = \underline{\underline{e^{-t}e^{-\alpha^2 kt}\cdot c_7(c_5\sin \alpha x + c_6\cos \alpha x)}}$

Solve the heat equation with given initial and boundary conditions:

$$ \begin{align} & \frac{\partial u}{\partial t} = k\frac{\partial^2 u}{\partial x^2} \hskip2em & u(x,0) = f(x) \nonumber \\ & & u(0,t) = 0 \nonumber \\ & & u(L,t) = 0 \nonumber \\ \end{align} $$We are finding solution $u(x,t) = X(x)T(t)$. Let us observe boundary condition substituted into the solution:

- $u(0,t) = 0 \implies T(t)X(0) = 0$
- $u(L,t) = 0 \implies T(t)X(L) = 0$

That means either $T(t)$ has to be zero or $X(0)$ and $X(L)$ has to be zero. The first one, i.e. $T(t) = 0$ is not reasonable to expect. Because the time is factor of the problem. If $T(t) = 0$, then the solution $u(x,y) = X(x)\cdot T(t) = 0$ can not be useful nor valid. If $T(t)$ is not zero, then we are looking for $X(L) = X(0) = 0$:

$$ \begin{align} XT' &= kX''T \nonumber \\ T'\frac 1 {Tk} &= X''\frac 1 X \label{ref:PDE_6} \end{align} $$We are solving $X''\frac 1 X = \lambda$ from $(\ref{ref:PDE_6})$.

$$ \begin{align} X'' &= \lambda X \nonumber \\ X'' - \lambda X &= 0 \nonumber \end{align} $$Let us replace $\alpha^2$ for $\lambda$, then the expressions and statements are clearer.

- $\lambda = 0:$

$$ X''=0 \implies X(x) = c_1x + c_2 $$ - $\lambda \lt 0\ (\lambda = -\alpha^2):$

$$ \begin{align} X''+\alpha^2 = 0 \implies \color{red}{X(x) = c_1\sin\alpha x + c_2\cos \alpha x} \label{ref:PDE_5} \end{align} $$ - $\lambda \gt 0\ (\lambda = \alpha^2):$

$$ X''-\alpha^2 = 0 \implies X(x) = c_1\sinh\alpha x + c_2\cosh \alpha x $$

All those trivial solutions with $c_1=c_2 = 0$ are really solutions for given initial values but
are not useful and do not solve given problem. Our hope is focused on $\sin \alpha x$ in $(\ref{ref:PDE_5})$. Because if we apply $X(0) = 0$ and
$X(L)=0$ into solution $(\ref{ref:PDE_5})$ we are able to get **nontrivial solution:**

We are solving the left hand side of $(\ref{ref:PDE_6})$:

$$ \begin{align} T'\frac 1 {Tk} &= \lambda \nonumber \\ T' &= \lambda Tk \nonumber \\ T' - \lambda Tk &= 0 \nonumber \\ m- \lambda k &= 0 \implies m=\lambda k \nonumber \\ T &= c_3 e^{\lambda k t} \label{ref:PDE_7}\\ \end{align} $$So we have a **sequence of solutions** for $n=1,2,3, \dots$ combined from $(\ref{ref:PDE_5})$ and $(\ref{ref:PDE_7})$

Now at time $t=0$ inital conditions are given: $u(x, t=0) = f(x)$. So let us investigate the solution there:

$$ u_n(x,t=0) = f(x) = A_n\sin\frac{\pi n}{L}x $$Our solution does not look well. Because if $f(x)$ is *some* function, how
the equality could hold? Is $f(x)$ restricted to be only a sine wave? So we
have to admit that $u_n$ is not a solution. But we know from the superposition
rule, that if $u_1$ and $u_2$ is a solution, then $c_1u_1 + c_2u_2$ is solution
as well. Therefore we can write

Let us set condition at $t=0$ again:

$$ f(x) = \sum_{n=1}^{\infty} A_n \sin\frac{\pi n}{L}x $$That is good enough. That is what we recognize as a Fourier series. Only instead of coefficient $b_n$ we have $A_n$. So the function is now described by Fourier series and we can express its coefficients from known formula $b_n = \frac 1 p \int_{-p}^{p} f(x)\sin nx \frac{\pi}{p}dx$:

$$ b_n = \frac 2 L \int_0^L f(x)\sin nx\frac{\pi}L \ dx $$Then the solution is

$$ \underline{\underline{u(x,t) = \sum_{n=1}^{\infty}\left( \frac 2 L \int_0^L f(x)\sin nx\frac{\pi}L \ dx \right)\cdot e^{-\frac{\pi^2n^2}{L^2}kt} \cdot \sin\frac{\pi n}{L}x}} $$