Partial Differential Equations with MATLAB

178 An Introduction to Partial Differential Equations with MATLAB©R

with solution
y=cex or ye−x=c.

Then, our transformation is

ξ=x,

η=ye−x,

or

x=ξ,

y=ηeξ

and our PDE becomes
uξ=ξ

with solution

u=

ξ^2

2

+g(η),

=

x^2

2

+g(ye−x).

Finally,

u(1,y)=cosy=

1

2

+g(ye−^1 ),

and, lettingz=ye,y=ez,weget

g(z)=cosez−

1

2

.

Therefore, our unique solution is

u=

x^2

2

+cos(ye^1 −x)−

1

2

.

Exercises 5.2

In Exercises 1–6, first find all possible solutions of the PDE, and check your
answer. Then, solve the initial-value problem, and check your answer. Also,
sketch the initial curve and some of the characteristics.

1.yux−xuy=0,u(x, 2 x)=x^4

2. (1 +x^2 )ux+uy=0,u(1,y)=cosy

3.ux+3x^2 uy=0,u(0,y)=sin3y

4. (1 +y^2 )ux+uy=0,u(x,0) =e−x
   2

5.ux+3x^2 uy−u=0,u(2,y)=3y+1

6.ux+xuy=x^2 y,u

(

x, 2 x+x

2

2

)

=5x

7.MATLAB:Plot the solution of the given problem inx-y-uspace. On

a separate graph, plot some of the characteristic curves.