11 Direct solving of some simple PDE’s

Some simple PDE’s can be solved by direct integration, here are few examples.

Example 1

\[ \frac {\partial z\left ( x,y\right ) }{\partial x}=0 \]

Integrating w.r.t. \(x\)., and remembering that now constant of integration will be function of \(y\), hence

\[ z\left ( x,y\right ) =f\left ( y\right ) \]

Example 2

\[ \frac {\partial ^{2}z\left ( x,y\right ) }{\partial x^{2}}=x \]

Integrating once w.r.t. \(x\) gives

\[ \frac {\partial z\left ( x,y\right ) }{\partial x}=\frac {x^{2}}{2}+f\left ( y\right ) \]

Integrating again gives

\[ z\left ( x,y\right ) =\frac {x^{3}}{6}+xf\left ( y\right ) +g\left ( y\right ) \]

Example 3

\[ \frac {\partial ^{2}z\left ( x,y\right ) }{\partial y^{2}}=y \]

Integrating once w.r.t. \(y\) gives

\[ \frac {\partial z\left ( x,y\right ) }{\partial y}=\frac {y^{2}}{2}+f\left ( x\right ) \]

Integrating again gives

\[ z\left ( x,y\right ) =\frac {y^{3}}{6}+yf\left ( x\right ) +g\left ( x\right ) \]

Example 4

\[ \frac {\partial ^{2}z\left ( x,y\right ) }{\partial x\partial y}=0 \]

Integrating once w.r.t \(x\) gives

\[ \frac {\partial z\left ( x,y\right ) }{\partial y}=f\left ( y\right ) \]

Integrating again w.r.t. \(y\) gives

\[ z\left ( x,y\right ) =\int f\left ( y\right ) dy+g\left ( x\right ) \]

Example 5

Solve \(u_{t}+u_{x}=0\) with \(u\left ( x,1\right ) =\frac {x}{1+x^{2}}\). Let \(u\equiv u\left ( x\left ( t\right ) ,t\right ) \), therefore

\[ \frac {du}{dt}=\frac {\partial u}{\partial t}+\frac {\partial u}{\partial x}\frac {dx}{dt}\]

Comparing the above with the given PDE, we see that if \(\frac {dx}{dt}=1\) then \(\frac {du}{dt}=0\) or \(u\left ( x\left ( t\right ) ,t\right ) \) is constant. At \(t=1\) we are given that

\begin{equation} u=\frac {x\left ( 1\right ) }{1+x\left ( 1\right ) ^{2}} \tag {1}\end{equation}

To find \(x\left ( 1\right ) \), from \(\frac {dx}{dt}=1\) we obtain that \(x\left ( t\right ) =t+c\). At \(t=1\), \(c=x\left ( 1\right ) -1\). Hence \(x\left ( t\right ) =t+x\left ( 1\right ) -1\) or

\[ x\left ( 1\right ) =x\left ( t\right ) +1-t \]

Hence solution from (1) becomes

\[ u=\frac {x-t+1}{1+\left ( x-t+1\right ) ^{2}}\]

Example 6

Solve \(u_{t}+u_{x}+u^{2}=0\).

Let \(u\equiv u\left ( x\left ( t\right ) ,t\right ) \), therefore

\[ \frac {du}{dt}=\frac {\partial u}{\partial t}+\frac {\partial u}{\partial x}\frac {dx}{dt}\]

Comparing the above with the given PDE, we see that if \(\frac {dx}{dt}=1\) then \(\frac {du}{dt}=-u^{2}\) or \(\frac {-1}{u}=-t+c.\) Hence

\[ u=\frac {1}{t+c}\]

At \(t=0\), \(c=\frac {1}{u\left ( x\left ( 0\right ) ,0\right ) }\). Let \(u\left ( x\left ( 0\right ) ,0\right ) =f\left ( x\left ( 0\right ) \right ) \). Therefore

\[ u=\frac {1}{t+\frac {1}{f\left ( x\left ( 0\right ) \right ) }}\]

Now we need to find \(x\left ( 0\right ) \). From \(\frac {dx}{dt}=1\), then \(x=t+c\) or \(c=x\left ( 0\right ) \), hence \(x\left ( 0\right ) =x-t\) and the above becomes

\[ u\left ( x,t\right ) =\frac {1}{t+\frac {1}{f\left ( x-t\right ) }}=\frac {f\left ( x-t\right ) }{tf\left ( x-t\right ) +1}\]