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{\bf ASSIGNED EXERCISES OF MATH544: PDE set  5} 
{\it March, 2001}
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\noindent
{\bf  THE METHOD OF RIEMANN}\\
Ronald B Guenther and John W Lee, Partial Differential Equations of
Mathematical Physics and Integral Equations

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\noindent
{\bf 1}.\, Solve the following initial value problem

\begin{eqnarray}
u_{xy}-yu_{x}=1,~~~ y>-x, -\infty \le x \le \infty,\\
u(x,-x)=x,~~ u_{x}(x,-x)=x^2,~~u_{y}(x,-x)=x^2-1,\\
\Gamma=\{{x,y} \in R^2| y=-x \}
\end{eqnarray}

\noindent
{\bf step 1}.\, Define the Riemann function $v(x,y)$ which satisfies
the homogeneous adjoint equation
$$v_{xy}+(yv)_{x}=0$$
Using the equations for $u$ and $v$ we obtain
$$A_{y}+B_{x}=v$$
where
$$A={1 \over 2}(vu_{x}-uv_{x}),~~B={1 \over 2}(vu_{y}-uv_{y})-yuv$$
{\bf step 2}.\, Integrate the above divergence form in a triangular region
with the vertices $M(x_{0},y_{0})$, $P(-y_{0},y_{0})$, and $Q(x_{0}, -x_{0})$
and using Green's theorem we get
$$\int_{C}\,[-A dx+B dy]=\int \int_{T(x_{0},y_{0})}\, v(x,y) dx dy$$
where $C=L_{QM} u L_{MP} u L_{PQ}$. Evaluating each integral
$$\int_{QM}=\int_{QM} B dy={1 \over 2} \int_{QM}\,(vu_{y}-uv_{y}) 
|_{x=x_{0}} dy -\int_{QM}\,yuv|_{x=x_{0}} dy$$
$$={1 \over 2}(uv)|^{M}_{Q}-\int_{QM}\,u(x_{0},y)\,(v_{y}+y v) dy$$
Since the last term contains an unknown value $u(x_{0},y)$ of the function
$u(x,y)$ we then let its coefficient to vanish.
\begin{equation}
v_{y}(x_{0},y)+yv(x_{0},y)=0, ~~ y \le y_{0}
\end{equation}
Hence 
$$\int_{QM}={1 \over 2}u(M) v(M)-{1 \over 2}u(Q)v(Q)$$
Similarly we find that
$$\int_{MP}=-\int_{MP}\,A|_{y=y_{0}} dx={1 \over 2}u(M)v(M)-
{1 \over 2}u(P)v(P)$$
with
\begin{equation}
v_{x}(x,y_{0})=0,~~ x \le x_{0}
\end{equation}

We then find that
\begin{equation}
u(M)v(M)={1 \over 2}u(P)v(P)+{1 \over 2}u(Q)v(Q)-\int_{\Gamma}[-A dx+B dy]+
\int_{T(x_{0},y_{0}}\, v(x,y) dx dy
\end{equation}
Here
$$u(M)=u(x_{0},y_{0}),~~v(M)=v(x_{0},y_{0}),$$
$$u(P)=u(-y_{0},y_{0})=y_{0},~~,v(P)=v(-y_{0},y_{0})$$
$$u(Q)=u(x_{0},-x_{0})=x_{0},~~v(Q)=v(x_{0},-x_{0}),$$
{\bf step 3}. Determination of the Riemann (Green) function: Let us
collect all the conditions on the (Riemann) function $v(x,y)$. We usually
use the notation $v(x,y)=R(x_{0},y_{0},x,y)$
\begin{eqnarray}
v_{xy}(x,y)+y v(x,y)=0,~~y>-x, x \in R,\\
v_{y}(x_{0},y)+y v(x_{0},y)=0,~~ y \le y_{0},\\
v_{x}(x,y_{0})=0, ~~x \le x_{0},\\
v(x_{0},y_{0})=1.
\end{eqnarray}
The complete solution of the above problem (known as the Goursat-Problem)
is easy and given by
$$v(x,y)=e^{{1 \over 2}(-y^2+{y_{0}}^2)}$$
{\bf step 4}. Solution: Inserting the above result in the equation (6)
we obtain that
\begin{eqnarray}
u(x_{0},y_{0})=-{1 \over 2}y_{0}+{1 \over 2}x_{0}e^{({y_{0}}^2-{x_{0}}^2)/2}+
{1 \over 2}\int_{-y_{0}}^{x_{0}}\,x^{2}e^{({y_{0}}^2-x^2)/2} dx\\
-{1 \over 2}\int_{-x_{0}}^{y_{0}}\,e^{({y_{0}}^2-y^2)/2} dy+
\int_{T(x_{0},y_{0})}\,e^{{(y_{0}}^2-y^2)/2} dxdy
\end{eqnarray}

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\noindent
{\bf 2}.\, Solve the above problem without using Riemann's method.

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\noindent
{\bf 3}.\, Solve the initial value problem by the use of Riemann's method:
\begin{equation}
u_{xy}=F(x,y),~~
u(x,x)=f(x),~~u_{x}(x,x)=g(x)
\end{equation}

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\noindent
{\bf 4}.\, Solve the following initial value problem:
\begin{eqnarray}
u_{xy}+xu_{x}=1,~~~ y>x,~~ x \in R,\\
u(x,-x)=u_{x}(x,-x)=0.
\end{eqnarray}

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\noindent
{\bf 5}.\, Solve
\begin{eqnarray}
u_{xy}-u_{x}&=&xy,\\
u(x,x^3)&=&\sin x,~~ u_{x}(x,x^3)=\cos x, ~~u_{y}(x,x^3)=0,
\end{eqnarray}
in the region above the curve $\Gamma:~y=x^3$.
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