solution of standard types of partial differential equations

First-Order Linear Partial Differential Equations

1.1. Equations of the Form f(x, y)wx + g(x, y)wy = 0

  1. wx + [f(x)y + g(x)]wy = 0.
  2. wx + [f(x)y + g(x)yk]wy = 0.
  3. wx + [f(x)eλy + g(x)]wy = 0.
  4. f(x)wx + g(y)wy = 0.
  5. [f(y) + amxnym−1]wx − [g(x) + anxn−1ym]wy = 0.
  6. [eαxf(y) + cβ]wx − [eβyg(x) + cα]wy = 0.
  7. wx + f(ax + by + c)wy = 0.
  8. wx + f(y/x)wy = 0.
  9. xwx + yf(xnym)wy = 0.
  10. wx + yf(eαxym)wy = 0.
  11. xwx + f(xneαy)wy = 0.

1.2. Equations of the Form f(x, y)wx + g(x, y)wy = h(x, y)

  1. awx + bwy = f(x).
  2. wx + awy = f(x)yk.
  3. wx + awy = f(x)eλy.
  4. awx + bwy = f(x) + g(y).
  5. wx + awy = f(x)g(y).
  6. wx + awy = f(x, y).
  7. wx + [ay + f(x)]wy = g(x).
  8. wx + [ay + f(x)]wy = g(x)h(y).
  9. wx + [f(x)y + g(x)yk]wy = h(x).
  10. wx + [f(x) + g(x)eλy]wy = h(x).
  11. axwx + bywy = f(x, y).
  12. f(x)wx + g(y)wy = h1(x) + h2(y).
  13. f(x)wx + g(y)wy = h(x, y).
  14. f(y)wx + g(x)wy = h(x, y).

1.3. Equations of the Form f(x, y)wx + g(x, y)wy = h(x, y)w + r(x, y)

  1. awx + bwy = f(x)w.
  2. awx + bwy = f(x)w + g(x).
  3. awx + bwy = [f(x) + g(y)]w.
  4. wx + awy = f(x, y)w.
  5. wx + awy = f(x, y)w + g(x, y).
  6. axwx + bywy = f(x)w + g(x).
  7. axwx + bywy = f(x, y)w.
  8. xwx + aywy = f(x, y)w + g(x, y).
  9. f(x)wx + g(y)wy = [h1(x) + h2(y)]w.
  10. f1(x)wx + f2(y)wy = aw + g1(x) + g2(y).
  11. f(x)wx + g(y)wy = h(x, y)w + r(x, y).
  12. f(y)wx + g(x)wy = h(x, y)w + r(x, y).  courtesy:eqworld.ipmnet.ru/en/solutions/fpde/fpdetoc1.htm

    Definitions of Partial Differential Equations

    A partial differential equation is an equation that involves an unknown function and its partial derivatives. The order of the highest derivative defines the order of the equation. The equation is called linear if the unknown function only appears in a linear form. Finally, the equation is homogeneous if every term involves the unknown function or its partial derivatives and inhomogeneous if it does not.

    The course will only consider linear partial differential equations of second order. Examples of important linear partial differential equations are:
     

    \begin{displaymath}<br />
                {\partial^{2}u\over \partial t^{2}} = c^{2}{\partial^{2}u\over \partial<br />
                x^{2}},<br />
                \end{displaymath} (2.1)



     

     

    \begin{displaymath}<br />
                {\partial u\over \partial t} = k^{2}{\partial^{2}u\over \partial<br />
                x^{2}},<br />
                \end{displaymath} (2.2)



     

     

    \begin{displaymath}<br />
                {\partial^{2}u\over \partial x^{2}} + {\partial^{2}u\over \partial<br />
                y^{2}} = 0.<br />
                \end{displaymath} (2.3)



     

     

    Equation (2.1) is the one dimensional wave equation, equation (2.2) is the one dimensional heat (or diffusion) equation and equation (2.3) is the two dimensional Laplace equation. They are all examples of homogeneous, linear partial differential equations.

    An example of an inhomogeneous equation is the two dimensional Poisson equation, namely
     

    \begin{displaymath}<br />
                {\partial^{2}u\over \partial x^{2}} + {\partial^{2}u\over \partial<br />
                y^{2}} = f(x,y).<br />
                \end{displaymath} (2.4)



     

     

    The solutions to the above equations are numerous. For example, if one considers equation (2.3) then it is easily verified that all of the functions
     

     

    \begin{displaymath}<br />
    u = x^{2} - y^{2}, \qquad u = e^{x}\cos y, \qquad \ln (x^{2} +<br />
    y^{2}),<br />
    \end{displaymath}


     

    are solutions. So how do we determine the actual solution we are looking for? The answer, of course, lies in the application of boundary conditions. Once the initial conditions (conditions at $t=0$) and the boundary conditions (conditions at specific values of $x$), where appropriate, are specified, there will be a unique solution to the linear partial differential equation.

    Using our knowledge of linear, ordinary differential equations, we expect that it should be possible to linearly superimpose solutions to the equations to obtain the most general solution. This is indeed the case. Thus, if $u_{1}(x,t)$ and $u_{2}(x,t)$ are both solutions to (2.1), then
     

     

    \begin{displaymath}<br />
    u = c_{1}u_{1}(x,t) + c_{2}u_{2}(x,t),<br />
    \end{displaymath}


     

    is also a solution if $c_{1}$ and $c_{2}$ are constants. However, unlike second order ordinary differential equations where there are two linearly independent solutions and two arbitrary constants, linear partial differential equations may well have an infinte number of linearly independent solutions and we may have to add together solutions involving an infinite number of constants. We will return to this later on.

     

    Example 2. .12Obtain a solution, $u(x,t)$, to the equation
     

     

    \begin{displaymath}<br />
    {\partial^{2}u\over \partial t^{2}} + u = 0.<br />
    \end{displaymath}


     

    Note that there are only derivatives with respect to $t$ and none with respect to $x$. Thus, we can treat the equation like an ordinary differential equation and use first year work to write the solution.

     

    \begin{displaymath}<br />
    u(x,t) = A \cos t + B \sin t .<br />
    \end{displaymath}


     

    However, unlike the ordinary differential equation case $A$ and $B$ are not constants but are, in fact, functions of the other independent variable. Hence, the actual solution is

     

    \begin{displaymath}<br />
    u(x,t) = A(x) \cos t + B(x) \sin t ,<br />
    \end{displaymath}


     

    as can be verified by differenitating and substituting into the differential equation.

     

    Example 2. .13Consider the equation
     

     

    \begin{displaymath}<br />
    {\partial ^{2} u\over \partial t\partial x} \equiv {\partia...<br />
    ...\over \partial<br />
    x}\right ) = - {\partial u \over \partial x}.<br />
    \end{displaymath}


     

    To solve this we set

    $p = \partial u /\partial x$ so that the equation becomes

     

    \begin{displaymath}<br />
    {\partial p \over \partial t} = - p.<br />
    \end{displaymath}


     

    Thus, the solution is

     

    \begin{displaymath}<br />
    p = c(x) e^{-t}.<br />
    \end{displaymath}


     

    Now we must integrate with respect to $x$ to get the solution $u(x,t)$. Therefore, we get

     

    \begin{displaymath}<br />
    u(x,t) = f(x)e^{-t} + g(t),<br />
    \end{displaymath}


     

    where

    $f(x) = \int c(x) dx$ and $g(t)$ is an arbitrary function of integration when we integrate with respect to $x$.

    Modelling: Derivation of the wave equation

    The wave equation has many physical applications from sound waves in air to magnetic waves in the Sun's atmosphere. However, one of the simplest systems to visualise and describe are waves on a stretched elastic string.

    Initially the string is horizontal. Then we distort it by displacing it in the vertical direction and at some time, say $t=0$, we release it and the string starts to oscillate. The aim is to try and determine the vertical displacement of the string, $u(x,t)$, as a function of space, $x$, and time $t$. The typical situation is illustrated in Figure 2.1 at a typical time $t$.

    To derive the wave equation we need to make some simplifying assumptions
    (i) The density of the string, $\rho $, is constant so that the mass of the string between $P$ and $Q$ is simply $\rho $ times the length of the string between $P$ and $Q$, namely $\Delta s$. Thus,
     

     

    \begin{displaymath}<br />
    \Delta s=\sqrt{(\Delta x)^{2} + (\Delta<br />
    u)^{2}} = \Delta x \...<br />
    ...\approx \Delta x<br />
    \sqrt{1+({\partial u \over \partial x})^{2}}.<br />
    \end{displaymath}


     

    (ii) The displacement, $u(x,t)$, and its derivatives are assumed small so that

     

    \begin{displaymath}<br />
    \Delta s \approx \Delta x<br />
    \end{displaymath}


     

    and the mass of the portion of the string is

     

    \begin{displaymath}<br />
    \rho \Delta x.<br />
    \end{displaymath}


     

    (iii) The only forces acting on this portion of the string are the tensions, $T_{1}$ at $P$ and $T_2$ at $Q$. The gravitational force is neglected.
    (iv) The motions are purely vertical. There is no horizontal motion of any portion of the string.

    Now we consider the forces acting on the typical string portion shown in Figure 2.1. Tension acts tangential to the string and the gradient of the tangent is, of course, the slope of $u(x,t)$ or simply

    ${\partial u \over \partial x}$. Now we resolve the forces into their horizontal and vertical components.

    Horizontal: At $P$ the tension force is

    $T_{1} \cos \Theta_{1}$ and it acts to the left, whereas at $Q$ the force is

    $T_{2} \cos \Theta_{2}$, acting to the right. Since there is no horizontal motion, these forces must balance and so
     

    \begin{displaymath}<br />
                T_{1} \cos \Theta_{1} = T_{2} \cos \Theta_{2} = T = {\rm const}.<br />
                \end{displaymath} (2.5)



     

     

    Vertical: From Figure 2.1 it is clear that the force at $Q$ is

    $T_{2} \sin<br />
    \Theta_{2}$ and at $P$ is

    $-T_{1} \sin \Theta_{1}$. Then Newton's law of motion gives
     

    $\displaystyle {\rm mass \times acceleration = applied \, force}$      


     

    so that

    $\displaystyle \rho \Delta x {\partial^{2} u \over \partial t^{2}} = T_{2} \sin \Theta_{2}-<br />
                T_{1} \sin \Theta_{1}$      


     

     

    Figure 2.1: The string at a typical time $t$.

    \includegraphics [scale=0.7]{fig1.ps}

     

    Divide by $T$ and use (2.5) to give

    $\displaystyle {\rho \Delta x \over T} {\partial^{2} u\over \partial t^{2}} = {T...<br />
                ...sin \Theta_{1} \over T_{1}<br />
                \cos \Theta_{1}} = \tan \Theta_{2} - \tan \Theta_{1}$      


     

    $\tan \Theta_{2}$ is the gradient of the tangent to $u(x,t)$ at $x=x+\Delta x$ and this is just

    $({\partial u \over \partial x})_{x+ \Delta x}$. In a similar manner,

    $\displaystyle \tan \Theta_{1} = ({\partial u \over \partial x})_{x} .$      


     

    Thus, we get

    $\displaystyle {\rho \over T} {\partial^{2} u \over \partial t^{2}} = {1 \over \...<br />
                ...ght )_{x + \Delta x} -<br />
                \left ({\partial u \over<br />
                \partial x}\right )_{x} \right]$      


     

    Using the definition of the derivative

    $\displaystyle \stackrel{\lim}{\Delta x} \rightarrow 0 \left\{ {f(x+\Delta x) - f(x) \over<br />
                \Delta x} \right\} ,$      


     

    we let $\Delta x$ tend to zero to obtain

    \begin{displaymath}<br />
                {1 \over c^{2}} {\partial^{2} u \over \partial t^{2}} = {\partial^{2} u \over<br />
                \partial x^{2}} .<br />
                \end{displaymath} (2.6)



     

     

    where

    $c^{2} = T/ \rho$, the tension divided by the density, is the wave speed squared.

    D'Alembert's solution of the Wave Equation

    A particularly neat solution to the wave equation, that is valid when the string is so long that it may be approximated by one of infinite length, was obtained by d'Alembert. The idea is to change coordinates from $x$ and $t$ to $\xi$ and $\eta$ in order to simplify the equation. Anticipating the final result, we choose the following linear transformation
     

    $\displaystyle \xi = x-ct \qquad {\rm and} \qquad \eta =x +ct.$      


     

    Thus,

    $u(x,t)=u(\xi, \eta) = u(\xi(x,t), \eta (x,t))$ and we must use the chain rule to express derivatives in terms of $x$ and $t$ as derivatives in terms of $\xi$ and $\eta$. Hence,

    $\displaystyle {\partial u \over \partial x}$ $\textstyle =$ $\displaystyle {\partial \xi \over \partial x} {\partial u<br />
                \over \partial \xi} +...<br />
                ...tial \eta} = {\partial u \over \partial \xi} + {\partial u \over \partial<br />
                \eta}$  
    $\displaystyle {\partial u \over \partial t}$ $\textstyle =$ $\displaystyle {\partial \xi \over \partial t} {\partial u<br />
                \over \partial \xi} +...<br />
                ...\eta} = - c{\partial u \over \partial \xi} + c{\partial u \over \partial<br />
                \eta}.$  


     

    The second derivatives require a bit of care.
     

    $\displaystyle {\partial^{2} u \over \partial x} = {\partial \over \partial x} (...<br />
                ...tial \eta} ({\partial u \over \partial \xi} +<br />
                {\partial u \over \partial \eta})$      


     

    Thus,

    $\displaystyle {\partial^{2} u \over \partial x^{2}} = {\partial^{2} u \over \pa...<br />
                ...2}u \over \partial \xi \partial \eta} + {\partial^{2}u \over<br />
                \partial \eta^{2}}$      


     

    and in a similar manner

    $\displaystyle {\partial^{2} u \over \partial t^{2}} = c^{2} {\partial^{2} u \ov...<br />
                ... \partial \xi \partial \eta} + c^{2}<br />
                {\partial^{2} u \over \partial \eta^{2}} .$      


     

    Thus, (2.6) becomes

      $\textstyle \,$ $\displaystyle {\partial^{2} u \over \partial \xi^{2}} - 2 {\partial^{2}u \over ...<br />
                ...{2}u \over \partial \xi \partial \eta} +<br />
                {\partial^{2}u \over \partial \xi^{2}}$  
      $\textstyle \,$ $\displaystyle \Rightarrow {\partial^{2} u \over \partial \xi \partial \eta} = 0 .$  


     

    This equation is much simpler and can be solved by direct integration. First of all integrate with respect to $\xi$ to give
     

    $\displaystyle {\partial u \over \partial \eta} = g(\eta)$      


     

    where $g(\eta)$ is an arbitrary function of $\eta$. Then integrate with respect to $\eta$ to obtain

    $\displaystyle u (\xi, \eta) = F(\xi) + G(\eta)$      


     

    where $F(\xi)$ is an arbitrary function of $\xi$ and

    $G(\eta) = \int g(\eta)d<br />
    \eta$.

    Finally, we replace $\xi$ and $\eta$ by their expressions in terms of $x$ and $t$ to obtain
     

    $\displaystyle u(x,t) = F(x-ct)+G(x+ct).$     (2.7)


     

     

    Example 2. .14Verify that

    $\sin (x-ct) + \sin (x+ct)$ is a solution to the wave equation. Here

    $F(\xi) = \sin \xi$ and

    $G(\eta) = \sin \eta$.

    Thus

     

    \begin{eqnarray*}<br />
    {\partial u \over \partial x} &=& \cos (x-ct) + \cos (x+ct), \...<br />
    ...l t^{2}} &=& - c^{2} \sin (x-ct) - c^{2} \sin (x+ct).<br />
    \nonumber<br />
    \end{eqnarray*}

     

     


     


     

    Therefore,

     

    \begin{displaymath}<br />
    {\partial^{2} u \over \partial x^{2}} = {1 \over c^{2}}<br />
    {\partial^{2} u \over \partial t^{2}}.<br />
    \end{displaymath}


     

    Note that we may use the trigonometric identities for

    $\sin (A \pm B)$ to obtain
     

    $\displaystyle \sin (x-ct) + \sin (x+ct) = 2 \sin x \cos ct$      


     

    and we remark that this is the product of a function of $x$ with a function of $t$. This result will be used later.

    D'Alembert's solution involves two arbitrary functions that are determined (normally) by two initial conditions. If the initial conditions are
     

    $\displaystyle u(x,o)$ $\textstyle =$ $\displaystyle f(x)$  
    $\displaystyle {\partial u \over \partial t} (x,o)$ $\textstyle =$ $\displaystyle h(x).$ (2.8)


     

    Then, from (2.7), we have

    \begin{displaymath}<br />
                F(x) +G(x) = f(x).<br />
                \end{displaymath} (2.9)



     

     

    Note that $F$ is a function of $\xi =x-ct$ but $\xi = x$ at $t=0$. The tricky part is calculating

    ${\partial u \over \partial t}$

    $\displaystyle {\partial u \over \partial t}$ $\textstyle =$ $\displaystyle {\partial \xi \over \partial t}{dF(\xi)<br />
                \over d\xi} + {\partial \eta \over \partial t}{dG \over d \eta}<br />
                (\eta)$  
    $\displaystyle {\partial u \over \partial t}$ $\textstyle =$ $\displaystyle -c {dF(\xi) \over d \xi} + c {dG \over d \eta}<br />
                (\eta)$  


     

    but at $t=0, \xi = x$ and $\eta = x$. Thus, the functional form of $F$ and $G$ are obtained by replacing $\xi$ by $x$ and $\eta = x$.

     

    Example 2. .15If, for example,

    $ u(x,t) = \sin (\xi) = \sin (x-ct)$, then
     

    $\displaystyle \begin{array}{ll}<br />
                u(x,t)=F(\xi)= \sin \xi & u(x,0)=F(x)= \sin x\\...<br />
                ...& {\partial u<br />
                \over \partial t} (x,0) =-cF^{\prime} (x) = -c \cos x<br />
                \end{array}$      


     

    The initial speed, that is

    $\partial u / \partial t$ evaluated at $t=0$, is then
     

    $\displaystyle -cF^{\prime}(x)+cG^{\prime}(x)=h(x).$      


     

    Integrating with respect to $x$, as both $F$ and $G$ are functions of $x$ when $t=0$, we get

    \begin{displaymath}<br />
                -F(x) +G(x) = {1 \over c} H(x) = {1 \over c} \int h(x) dx<br />
                \end{displaymath} (2.10)



     

     

    Subtracting (2.10) from (2.9) gives

    $\displaystyle F(x)$ $\textstyle =$ $\displaystyle {1 \over 2} \{ f(x) - {1 \over c} H(x) \}$  
    $\displaystyle F(\xi)$ $\textstyle =$ $\displaystyle {1 \over 2} \{ f(\xi) - {1 \over c} H {\xi} \} = {1 \over 2} \{ f<br />
                (x-ct)-{1 \over c} H(x-ct) \}$  


     

    Adding (2.9) and (2.10) gives
     

    $\displaystyle G(x)$ $\textstyle =$ $\displaystyle {1 \over 2} \{ f(x) + {1 \over c} H(x) \}$  
    $\displaystyle \Rightarrow G(\eta)$ $\textstyle =$ $\displaystyle {1 \over 2} \{ f(\eta) + {1 \over c} H(\eta) \} = {1<br />
                \over 2} \{ f(x+ct) + {1 \over c} H(x+ct) \}$  


     

    Hence, d'Alembert's solution that satisfies the initial conditions (2.8) is
     

    $\displaystyle u(x,t)$ $\textstyle =$ $\displaystyle {1 \over 2} \left \{ f(x-ct)+f(x+ct) +{1 \over c} \left (H(x+ct) -<br />
                H(x-ct)\right ) \right \}$  
      $\textstyle =$ $\displaystyle {1 \over 2} \left \{ f(x-ct) + f(x+ct) + {1 \over c}<br />
                \int^{x+ct}_{x-ct} h(s) ds \right \}.$  


     

     

    Example 2. .16$f(x-ct)$ can be thought of as a ``shape", defined by the function $f(\xi)$, moving to the right. We can see this if we consider the value $\xi = 0$. Say
     

    $\displaystyle f(0) =1.$      


     

     

    
     Then $f=1$ at 		$x=0$ when 		$ct=0$ 
    
    $x=1$ when $ct=1$
    $x=2$ when $ct=2$

    Thus, we are moving to the right to larger values of $x$. Similarily $G(x+ct)$ corresponds to a wave propagating to the left.

     

    Example 2. .17

    $F(x) = a \cos kx \quad G(x) = 0$ Thus, the solution to the wave equation is
     

    $\displaystyle u= a \cos k(x-ct)$      


     

    corresponding to a wave travelling to the right (increasing $x$).

     

    
    Here 		 $a$ is the amplitude, 
    
    $k$ is the wavenumber - $k= {2 \pi \over \lambda}$,
    $\lambda$ is the wavelength - distance from one crest to the next,
    $kc$ is the angular frequency $kc = 2\pi / \tau$,
    $\tau$ is the period of the wave - time from one crest to the next to pass afixed point.

     

    Figure 2.2:

    \includegraphics [scale=0.7]{fig2.ps}

     

     

    Example 2. .18

    $F(x)= a \sin kx \qquad G(x) = a \sin kx$ This gives the solution
     

    $\displaystyle u$ $\textstyle =$ $\displaystyle a \{ \sin k (x-ct)+ \sin k (x+ct) \},$  
      $\textstyle =$ $\displaystyle a \{ \sin kx \cos kct + \cos kx \sin kct + \sin kx \cos kct - \cos kx \sin<br />
                kct \},$  
      $\textstyle =$ $\displaystyle 2a \sin kx \cos kct.$  


     

    The solution is a standing wave as shown in Figure 2.2. The maximum and minimum displacements are at

    $x=0, \pm {\pi \over k}, \ldots,<br />
    \pm {N \pi \over k}$ and the nodes (where $u=0$) are at

    $x= \pm {\pi \over 2k},<br />
    \ldots , \pm (N+ {\textstyle{1 \over 2}}) {\pi \over K}.$

     

    Example 2. .19

     

    Figure 2.3:

    \includegraphics [scale=0.7]{fig3.ps}

     

    Consider the case of the string initially at rest with the above initial displacement. Thus,

    $\displaystyle u(x,0)=f(x) = \left\{ \begin{array}{cc}<br />
                0 & \vert x\vert \geq L, ...<br />
                ... -L \leq x \leq 0, \\<br />
                \delta (1-x/L) & 0 \leq x \leq L,\\<br />
                \end{array} \right.$      


     

    where $\delta$ is a constant. In Figure 2.3, $\delta = 2$.

    $\displaystyle {\partial u \over \partial t} (x,0) = 0 .$      


     

    From d'Alembert's solution we have

    $\displaystyle F(x) + G(x)$ $\textstyle =$ $\displaystyle f(x)$  
    $\displaystyle -c F^{\prime} (x) + cG^{\prime} (x)$ $\textstyle =$ $\displaystyle 0 \Rightarrow F(x) = G(x) + C,$  


     

    where $C$ is a constant. Adding the two solutions we have

    $\displaystyle F(x)$ $\textstyle =$ $\displaystyle {\textstyle{1 \over 2}} f(x) + {1 \over 2}C,$  
    $\displaystyle G(x)$ $\textstyle =$ $\displaystyle {1\over 2} f(x) - {1\over 2} C$  


     

    Hence, the solution is

    $\displaystyle u(x,t) = {\textstyle{1 \over 2}} f(x-ct) + {\textstyle{1 \over 2}} f(x+ct),$      


     

    and the arbitrary constant, $C$, does not appear in the solution. Thus, we may, in fact, take $C=0$ without any loss of generality. So we have 2 equal triangles of height $\delta /2$ going to the right and the left. This is shown in Figure 2.4.

     

    Figure 2.4: The solution obtained in Example 19 is shown for $\delta = 2$ and $ c = 1$ at various times.

    \includegraphics [scale=0.7]{fig4.ps}

    courtesy:solar.mcs.st-and.ac.uk/~alan/MT2003/PDE/node12.html

 


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Tue, 05/26/2009 - 10:21

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