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To Solve: $\displaystyle 4\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=3u$

My attempt:

Let $\displaystyle u=X(x)Y(y)$. So, $\displaystyle 4X'Y +XY'=3XY$

Separating the variables, $\displaystyle \frac{4X'-3X}{X}=\frac{-Y'}{Y}=k$(say)

$\displaystyle X'-\frac{(3+k)}{4}X=0$

$\displaystyle Y'+kY=0$

Integrating, $\displaystyle X=c_1e^{(k+3)x/4}$

and $\displaystyle Y=c_2e^{-ky}$

So, $\displaystyle u=c_1c_2e^{(k+3)x/4-ky}$

Now I check this against the initial value given as $\displaystyle u=3e^{-y}-e^{5y}$ when $x=0$

The final answer also (obviously) has two components $u=3e^{x-y}-e^{2x-5y}$

How come there are two components here? What did I miss?

square_one
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2 Answers2

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enter image description hereI don't have time to type it out in Mathjax, I have taken an image of the solution and please pardon me if it is not clear. Thanks

Satish Ramanathan
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  • I gather my answer is correct. Have to add the solution to itself to fit into the given solution. This was probably done to test the assimilation of the point mentioned by Vinay above ! Is that right? – square_one Jun 28 '14 at 12:00
  • You are right. Vinay has given the explanation as to why there could be two terms. – Satish Ramanathan Jun 28 '14 at 16:44
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As the PDE is linear and homogeneous, a linear combination of two solutions is another solution. So, even though your assumed solution is of the form $X(x)Y(y)$, you can get a solution of the form $X_1(x)Y_1(y) + X_2(x)Y_2(y)$.

The two terms in the solution are each separate solutions.

Why is the sum of two solutions again a solution? Suppose $u_1$ and $u_2$ two solutions. Then $4\dfrac{\partial (u_1 + u_2)}{\partial x} + \dfrac{\partial (u_1 + u_2)}{\partial y} = 4\dfrac{\partial u_1}{\partial x} + \dfrac{\partial u_1}{\partial y} + 4\dfrac{\partial u_2}{\partial x} + \dfrac{\partial u_2}{\partial y} = 3u_1 + 3u_2 = 3(u_1 + u_2)$, so $u_1 + u_2$ is also a solution.

M. Vinay
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  • I gather my answer is correct. Have to add the solution to itself to fit into the given solution as suggested by Satish. This was probably done to test the assimilation of the point you mentioned ! Is that right? – square_one Jun 28 '14 at 12:01
  • Your answer is already correct. I just explained why it is correct (because your question was how there could be two terms). – M. Vinay Jun 28 '14 at 12:11